contestId int64 0 1.01k | index stringclasses 57 values | name stringlengths 2 58 | type stringclasses 2 values | rating int64 0 3.5k | tags listlengths 0 11 | title stringclasses 522 values | time-limit stringclasses 8 values | memory-limit stringclasses 8 values | problem-description stringlengths 0 7.15k | input-specification stringlengths 0 2.05k | output-specification stringlengths 0 1.5k | demo-input listlengths 0 7 | demo-output listlengths 0 7 | note stringlengths 0 5.24k | points float64 0 425k | test_cases listlengths 0 402 | creationTimeSeconds int64 1.37B 1.7B | relativeTimeSeconds int64 8 2.15B | programmingLanguage stringclasses 3 values | verdict stringclasses 14 values | testset stringclasses 12 values | passedTestCount int64 0 1k | timeConsumedMillis int64 0 15k | memoryConsumedBytes int64 0 805M | code stringlengths 3 65.5k | prompt stringlengths 262 8.2k | response stringlengths 17 65.5k | score float64 -1 3.99 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
755 | A | PolandBall and Hypothesis | PROGRAMMING | 800 | [
"brute force",
"graphs",
"math",
"number theory"
] | null | null | PolandBall is a young, clever Ball. He is interested in prime numbers. He has stated a following hypothesis: "There exists such a positive integer *n* that for each positive integer *m* number *n*·*m*<=+<=1 is a prime number".
Unfortunately, PolandBall is not experienced yet and doesn't know that his hypothesis is incorrect. Could you prove it wrong? Write a program that finds a counterexample for any *n*. | The only number in the input is *n* (1<=≤<=*n*<=≤<=1000) — number from the PolandBall's hypothesis. | Output such *m* that *n*·*m*<=+<=1 is not a prime number. Your answer will be considered correct if you output any suitable *m* such that 1<=≤<=*m*<=≤<=103. It is guaranteed the the answer exists. | [
"3\n",
"4\n"
] | [
"1",
"2"
] | A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.
For the first sample testcase, 3·1 + 1 = 4. We can output 1.
In the second sample testcase, 4·1 + 1 = 5. We cannot output 1 because 5 is prime. However, *m* = 2 is okay since 4·2 + 1 = 9, which is not a prime number. | 500 | [
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "2"
},
{
"input": "10",
"output": "2"
},
{
"input": "153",
"output": "1"
},
{
"input": "1000",
"output": "1"
},
{
"input": "1",
"output": "3"
},
{
"input": "2",
"output": "4"
},
{
"input": "5",
"output": "1"
},
{
"input": "6",
"output": "4"
},
{
"input": "7",
"output": "1"
},
{
"input": "8",
"output": "1"
},
{
"input": "9",
"output": "1"
},
{
"input": "11",
"output": "1"
},
{
"input": "998",
"output": "1"
},
{
"input": "996",
"output": "3"
},
{
"input": "36",
"output": "4"
},
{
"input": "210",
"output": "4"
},
{
"input": "270",
"output": "4"
},
{
"input": "306",
"output": "4"
},
{
"input": "330",
"output": "5"
},
{
"input": "336",
"output": "4"
},
{
"input": "600",
"output": "4"
},
{
"input": "726",
"output": "4"
},
{
"input": "988",
"output": "1"
},
{
"input": "12",
"output": "2"
},
{
"input": "987",
"output": "1"
},
{
"input": "13",
"output": "1"
},
{
"input": "986",
"output": "1"
},
{
"input": "14",
"output": "1"
},
{
"input": "985",
"output": "1"
},
{
"input": "15",
"output": "1"
},
{
"input": "984",
"output": "1"
},
{
"input": "16",
"output": "2"
},
{
"input": "983",
"output": "1"
},
{
"input": "17",
"output": "1"
},
{
"input": "982",
"output": "2"
},
{
"input": "18",
"output": "3"
},
{
"input": "981",
"output": "1"
},
{
"input": "19",
"output": "1"
},
{
"input": "980",
"output": "1"
},
{
"input": "20",
"output": "1"
},
{
"input": "979",
"output": "1"
},
{
"input": "21",
"output": "1"
},
{
"input": "978",
"output": "1"
},
{
"input": "22",
"output": "2"
},
{
"input": "977",
"output": "1"
},
{
"input": "23",
"output": "1"
},
{
"input": "976",
"output": "2"
},
{
"input": "24",
"output": "1"
},
{
"input": "975",
"output": "1"
},
{
"input": "25",
"output": "1"
},
{
"input": "2",
"output": "4"
},
{
"input": "6",
"output": "4"
},
{
"input": "996",
"output": "3"
},
{
"input": "8",
"output": "1"
},
{
"input": "998",
"output": "1"
},
{
"input": "330",
"output": "5"
},
{
"input": "18",
"output": "3"
},
{
"input": "12",
"output": "2"
},
{
"input": "24",
"output": "1"
},
{
"input": "999",
"output": "1"
},
{
"input": "726",
"output": "4"
},
{
"input": "876",
"output": "3"
}
] | 1,691,174,083 | 2,147,483,647 | Python 3 | OK | TESTS | 63 | 46 | 0 | encontrado = False
n = int(input())
i = 1
m = 1
"""
Well known algorithm got from chat gpt
"""
def isPrime(n):
if n <= 1:
return False
if n <= 3:
return True
if n % 2 == 0 or n % 3 == 0:
return False
i = 5
while i * i <= n:
if n % i == 0 or n % (i + 2) == 0:
return False
i += 6
return True
while not encontrado:
if not isPrime(n * i + 1):
encontrado = True
m = i
i += 1
print(m)
| Title: PolandBall and Hypothesis
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
PolandBall is a young, clever Ball. He is interested in prime numbers. He has stated a following hypothesis: "There exists such a positive integer *n* that for each positive integer *m* number *n*·*m*<=+<=1 is a prime number".
Unfortunately, PolandBall is not experienced yet and doesn't know that his hypothesis is incorrect. Could you prove it wrong? Write a program that finds a counterexample for any *n*.
Input Specification:
The only number in the input is *n* (1<=≤<=*n*<=≤<=1000) — number from the PolandBall's hypothesis.
Output Specification:
Output such *m* that *n*·*m*<=+<=1 is not a prime number. Your answer will be considered correct if you output any suitable *m* such that 1<=≤<=*m*<=≤<=103. It is guaranteed the the answer exists.
Demo Input:
['3\n', '4\n']
Demo Output:
['1', '2']
Note:
A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.
For the first sample testcase, 3·1 + 1 = 4. We can output 1.
In the second sample testcase, 4·1 + 1 = 5. We cannot output 1 because 5 is prime. However, *m* = 2 is okay since 4·2 + 1 = 9, which is not a prime number. | ```python
encontrado = False
n = int(input())
i = 1
m = 1
"""
Well known algorithm got from chat gpt
"""
def isPrime(n):
if n <= 1:
return False
if n <= 3:
return True
if n % 2 == 0 or n % 3 == 0:
return False
i = 5
while i * i <= n:
if n % i == 0 or n % (i + 2) == 0:
return False
i += 6
return True
while not encontrado:
if not isPrime(n * i + 1):
encontrado = True
m = i
i += 1
print(m)
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | You are an experienced Codeforces user. Today you found out that during your activity on Codeforces you have made *y* submissions, out of which *x* have been successful. Thus, your current success rate on Codeforces is equal to *x*<=/<=*y*.
Your favorite rational number in the [0;1] range is *p*<=/<=*q*. Now you wonder: what is the smallest number of submissions you have to make if you want your success rate to be *p*<=/<=*q*? | The first line contains a single integer *t* (1<=≤<=*t*<=≤<=1000) — the number of test cases.
Each of the next *t* lines contains four integers *x*, *y*, *p* and *q* (0<=≤<=*x*<=≤<=*y*<=≤<=109; 0<=≤<=*p*<=≤<=*q*<=≤<=109; *y*<=><=0; *q*<=><=0).
It is guaranteed that *p*<=/<=*q* is an irreducible fraction.
Hacks. For hacks, an additional constraint of *t*<=≤<=5 must be met. | For each test case, output a single integer equal to the smallest number of submissions you have to make if you want your success rate to be equal to your favorite rational number, or -1 if this is impossible to achieve. | [
"4\n3 10 1 2\n7 14 3 8\n20 70 2 7\n5 6 1 1\n"
] | [
"4\n10\n0\n-1\n"
] | In the first example, you have to make 4 successful submissions. Your success rate will be equal to 7 / 14, or 1 / 2.
In the second example, you have to make 2 successful and 8 unsuccessful submissions. Your success rate will be equal to 9 / 24, or 3 / 8.
In the third example, there is no need to make any new submissions. Your success rate is already equal to 20 / 70, or 2 / 7.
In the fourth example, the only unsuccessful submission breaks your hopes of having the success rate equal to 1. | 0 | [
{
"input": "4\n3 10 1 2\n7 14 3 8\n20 70 2 7\n5 6 1 1",
"output": "4\n10\n0\n-1"
},
{
"input": "8\n0 1 0 1\n0 2 1 2\n0 3 1 1\n1 2 0 1\n1 2 1 1\n2 2 0 1\n3 3 1 2\n4 4 1 1",
"output": "0\n2\n-1\n-1\n-1\n-1\n3\n0"
},
{
"input": "5\n1 1000000000 1 2\n1 1000000000 1 2\n1 1000000000 1 2\n1 1000000000 1 2\n1 1000000000 1 2",
"output": "999999998\n999999998\n999999998\n999999998\n999999998"
},
{
"input": "5\n999999999 1000000000 1 1000000000\n999999999 1000000000 1 1000000000\n999999999 1000000000 1 1000000000\n999999999 1000000000 1 1000000000\n999999999 1000000000 1 1000000000",
"output": "999999998000000000\n999999998000000000\n999999998000000000\n999999998000000000\n999999998000000000"
},
{
"input": "5\n0 1000000000 999999999 1000000000\n0 1000000000 999999999 1000000000\n0 1000000000 999999999 1000000000\n0 1000000000 999999999 1000000000\n0 1000000000 999999999 1000000000",
"output": "999999999000000000\n999999999000000000\n999999999000000000\n999999999000000000\n999999999000000000"
},
{
"input": "1\n999999999 1000000000 1 2",
"output": "999999998"
},
{
"input": "1\n50 50 1 1",
"output": "0"
},
{
"input": "1\n100000000 100000000 1 2",
"output": "100000000"
},
{
"input": "1\n3 999999990 1 1000000000",
"output": "2000000010"
},
{
"input": "5\n3 10 1 2\n7 14 3 8\n20 70 2 7\n5 6 1 1\n1 1 1 1",
"output": "4\n10\n0\n-1\n0"
},
{
"input": "5\n9999999 10000000 1 1000000000\n9999999 10000000 1 1000000000\n9999999 10000000 1 1000000000\n9999999 10000000 1 1000000000\n9999999 10000000 1 1000000000",
"output": "9999998990000000\n9999998990000000\n9999998990000000\n9999998990000000\n9999998990000000"
},
{
"input": "1\n0 1000000000 999999999 1000000000",
"output": "999999999000000000"
},
{
"input": "5\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000",
"output": "999999998000000000\n999999998000000000\n999999998000000000\n999999998000000000\n999999998000000000"
},
{
"input": "5\n1 1000000000 999999999 1000000000\n2 1000000000 999999999 1000000000\n3 1000000000 999999999 1000000000\n4 1000000000 999999999 1000000000\n5 1000000000 999999999 1000000000",
"output": "999999998000000000\n999999997000000000\n999999996000000000\n999999995000000000\n999999994000000000"
},
{
"input": "1\n1 1 1 1",
"output": "0"
},
{
"input": "5\n999999997 999999998 2 999999999\n999999997 999999998 2 999999999\n999999997 999999998 2 999999999\n999999997 999999998 2 999999999\n999999997 999999998 2 999999999",
"output": "499999997500000003\n499999997500000003\n499999997500000003\n499999997500000003\n499999997500000003"
},
{
"input": "5\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000",
"output": "999999999000000000\n999999999000000000\n999999999000000000\n999999999000000000\n999999999000000000"
},
{
"input": "5\n99999997 999999998 999999998 999999999\n99999996 999999997 999999997 999999999\n99999997 999999998 999999998 999999999\n99999996 999999997 999999997 999999999\n99999997 999999998 999999998 999999999",
"output": "899999999100000001\n449999999550000002\n899999999100000001\n449999999550000002\n899999999100000001"
},
{
"input": "1\n1000000000 1000000000 1 1000000000",
"output": "999999999000000000"
},
{
"input": "1\n7 7 1 1",
"output": "0"
},
{
"input": "5\n1000000000 1000000000 1 2\n1000000000 1000000000 1 2\n1000000000 1000000000 1 2\n1000000000 1000000000 1 2\n1000000000 1000000000 1 2",
"output": "1000000000\n1000000000\n1000000000\n1000000000\n1000000000"
},
{
"input": "1\n1000000000 1000000000 1 1",
"output": "0"
},
{
"input": "1\n1 1000000000 999999999 1000000000",
"output": "999999998000000000"
},
{
"input": "4\n1 1000000000 999999999 1000000000\n999999999 1000000000 1 1000000000\n1 2 0 1\n0 1 0 1",
"output": "999999998000000000\n999999998000000000\n-1\n0"
},
{
"input": "1\n1 1000000000 1 2",
"output": "999999998"
},
{
"input": "5\n1 982449707 1 2\n1 982449707 1 2\n1 982449707 1 2\n1 982449707 1 2\n1 982449707 1 2",
"output": "982449705\n982449705\n982449705\n982449705\n982449705"
},
{
"input": "5\n13 900000007 900000007 900000009\n13 900000007 900000007 900000009\n13 900000007 900000007 900000009\n13 900000007 900000007 900000009\n13 900000007 900000007 900000009",
"output": "405000000449999966\n405000000449999966\n405000000449999966\n405000000449999966\n405000000449999966"
},
{
"input": "1\n5 10 0 1",
"output": "-1"
},
{
"input": "1\n2 2 1 1",
"output": "0"
},
{
"input": "5\n0 999999999 999999999 1000000000\n0 999999999 999999999 1000000000\n0 999999999 999999999 1000000000\n0 999999999 999999999 1000000000\n0 999999999 999999999 1000000000",
"output": "999999998000000001\n999999998000000001\n999999998000000001\n999999998000000001\n999999998000000001"
},
{
"input": "1\n0 5 0 1",
"output": "0"
},
{
"input": "5\n999999999 1000000000 1 9999\n999999999 1000000000 1 9999\n999999999 1000000000 1 9999\n999999999 1000000000 1 9999\n999999999 1000000000 1 9999",
"output": "9997999990001\n9997999990001\n9997999990001\n9997999990001\n9997999990001"
},
{
"input": "5\n999999997 1000000000 3 1000000000\n999999997 1000000000 3 1000000000\n999999997 1000000000 3 1000000000\n999999997 1000000000 3 1000000000\n999999997 1000000000 3 1000000000",
"output": "333333332000000000\n333333332000000000\n333333332000000000\n333333332000000000\n333333332000000000"
},
{
"input": "5\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000",
"output": "999999999000000000\n999999999000000000\n999999998000000000\n999999998000000000\n999999998000000000"
},
{
"input": "5\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000\n1000000000 1000000000 1 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000",
"output": "999999999000000000\n999999999000000000\n999999999000000000\n999999998000000000\n999999998000000000"
},
{
"input": "1\n999999998 999999999 1 10",
"output": "8999999981"
},
{
"input": "5\n1 1 1 1\n2 2 1 1\n100 100 1 1\n1000000000 1000000000 1 1\n1000000000 1000000000 1 1",
"output": "0\n0\n0\n0\n0"
},
{
"input": "4\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000\n1 1000000000 999999999 1000000000",
"output": "999999998000000000\n999999998000000000\n999999998000000000\n999999998000000000"
}
] | 1,689,437,471 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | print("_RANDOM_GUESS_1689437471.7355886")# 1689437471.7356088 | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are an experienced Codeforces user. Today you found out that during your activity on Codeforces you have made *y* submissions, out of which *x* have been successful. Thus, your current success rate on Codeforces is equal to *x*<=/<=*y*.
Your favorite rational number in the [0;1] range is *p*<=/<=*q*. Now you wonder: what is the smallest number of submissions you have to make if you want your success rate to be *p*<=/<=*q*?
Input Specification:
The first line contains a single integer *t* (1<=≤<=*t*<=≤<=1000) — the number of test cases.
Each of the next *t* lines contains four integers *x*, *y*, *p* and *q* (0<=≤<=*x*<=≤<=*y*<=≤<=109; 0<=≤<=*p*<=≤<=*q*<=≤<=109; *y*<=><=0; *q*<=><=0).
It is guaranteed that *p*<=/<=*q* is an irreducible fraction.
Hacks. For hacks, an additional constraint of *t*<=≤<=5 must be met.
Output Specification:
For each test case, output a single integer equal to the smallest number of submissions you have to make if you want your success rate to be equal to your favorite rational number, or -1 if this is impossible to achieve.
Demo Input:
['4\n3 10 1 2\n7 14 3 8\n20 70 2 7\n5 6 1 1\n']
Demo Output:
['4\n10\n0\n-1\n']
Note:
In the first example, you have to make 4 successful submissions. Your success rate will be equal to 7 / 14, or 1 / 2.
In the second example, you have to make 2 successful and 8 unsuccessful submissions. Your success rate will be equal to 9 / 24, or 3 / 8.
In the third example, there is no need to make any new submissions. Your success rate is already equal to 20 / 70, or 2 / 7.
In the fourth example, the only unsuccessful submission breaks your hopes of having the success rate equal to 1. | ```python
print("_RANDOM_GUESS_1689437471.7355886")# 1689437471.7356088
``` | 0 | |
262 | B | Roma and Changing Signs | PROGRAMMING | 1,200 | [
"greedy"
] | null | null | Roma works in a company that sells TVs. Now he has to prepare a report for the last year.
Roma has got a list of the company's incomes. The list is a sequence that consists of *n* integers. The total income of the company is the sum of all integers in sequence. Roma decided to perform exactly *k* changes of signs of several numbers in the sequence. He can also change the sign of a number one, two or more times.
The operation of changing a number's sign is the operation of multiplying this number by -1.
Help Roma perform the changes so as to make the total income of the company (the sum of numbers in the resulting sequence) maximum. Note that Roma should perform exactly *k* changes. | The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=105), showing, how many numbers are in the sequence and how many swaps are to be made.
The second line contains a non-decreasing sequence, consisting of *n* integers *a**i* (|*a**i*|<=≤<=104).
The numbers in the lines are separated by single spaces. Please note that the given sequence is sorted in non-decreasing order. | In the single line print the answer to the problem — the maximum total income that we can obtain after exactly *k* changes. | [
"3 2\n-1 -1 1\n",
"3 1\n-1 -1 1\n"
] | [
"3\n",
"1\n"
] | In the first sample we can get sequence [1, 1, 1], thus the total income equals 3.
In the second test, the optimal strategy is to get sequence [-1, 1, 1], thus the total income equals 1. | 1,000 | [
{
"input": "3 2\n-1 -1 1",
"output": "3"
},
{
"input": "3 1\n-1 -1 1",
"output": "1"
},
{
"input": "17 27\n257 320 676 1136 2068 2505 2639 4225 4951 5786 7677 7697 7851 8337 8429 8469 9343",
"output": "81852"
},
{
"input": "69 28\n-9822 -9264 -9253 -9221 -9139 -9126 -9096 -8981 -8521 -8313 -8257 -8253 -7591 -7587 -7301 -7161 -7001 -6847 -6441 -6241 -5949 -5896 -5713 -5692 -5644 -5601 -5545 -5525 -5331 -5253 -5041 -5000 -4951 -4855 -4384 -4293 -4251 -4001 -3991 -3762 -3544 -3481 -3261 -2983 -2882 -2857 -2713 -2691 -2681 -2653 -2221 -2043 -2011 -1997 -1601 -1471 -1448 -1363 -1217 -1217 -1129 -961 -926 -801 -376 -327 -305 -174 -91",
"output": "102443"
},
{
"input": "12 28\n-6652 -6621 -6471 -5559 -5326 -4551 -4401 -4326 -3294 -1175 -1069 -43",
"output": "49488"
},
{
"input": "78 13\n-9961 -9922 -9817 -9813 -9521 -9368 -9361 -9207 -9153 -9124 -9008 -8981 -8951 -8911 -8551 -8479 -8245 -8216 -7988 -7841 -7748 -7741 -7734 -7101 -6846 -6804 -6651 -6526 -6519 -6463 -6297 -6148 -6090 -5845 -5209 -5201 -5161 -5061 -4537 -4529 -4433 -4370 -4266 -4189 -4125 -3945 -3843 -3777 -3751 -3476 -3461 -3279 -3205 -3001 -2889 -2761 -2661 -2521 -2481 -2305 -2278 -2269 -2225 -1648 -1524 -1476 -1353 -1097 -867 -785 -741 -711 -692 -440 -401 -225 -65 -41",
"output": "-147832"
},
{
"input": "4 1\n218 3441 4901 7601",
"output": "15725"
},
{
"input": "73 26\n-8497 -8363 -7603 -7388 -6830 -6827 -6685 -6389 -6237 -6099 -6013 -5565 -5465 -4965 -4947 -4201 -3851 -3793 -3421 -3410 -3201 -3169 -3156 -2976 -2701 -2623 -2321 -2169 -1469 -1221 -950 -926 -9 47 236 457 773 1321 1485 1545 1671 1736 2014 2137 2174 2301 2625 3181 3536 3851 4041 4685 4981 4987 5145 5163 5209 5249 6011 6337 6790 7254 7361 7407 7969 7982 8083 8251 8407 8735 9660 9855 9957",
"output": "315919"
},
{
"input": "53 5\n-9821 -9429 -9146 -8973 -8807 -8801 -8321 -7361 -7222 -7161 -6913 -5961 -4877 -4756 -4753 -4661 -3375 -3031 -2950 -2661 -2161 -2041 -1111 -1071 -905 -697 -397 323 772 1617 1752 2736 2737 3201 3465 4029 4121 4463 4561 4637 4814 6119 6610 6641 6961 7217 7523 8045 8610 8915 9004 9265 9576",
"output": "92703"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "1 1\n10000",
"output": "-10000"
},
{
"input": "1 2\n-1",
"output": "-1"
},
{
"input": "2 1\n0 1",
"output": "1"
},
{
"input": "2 2\n-1 0",
"output": "1"
},
{
"input": "5 6\n-3 -2 -1 5 6",
"output": "15"
},
{
"input": "3 3\n-50 -10 30",
"output": "70"
},
{
"input": "4 4\n-100 -90 -80 1",
"output": "269"
},
{
"input": "4 3\n-3 -2 1 2",
"output": "6"
},
{
"input": "6 4\n-6 -3 -2 1 2 3",
"output": "15"
},
{
"input": "5 6\n-10 -9 -8 1 2",
"output": "28"
},
{
"input": "1 2\n1",
"output": "1"
},
{
"input": "2 2\n-1 3",
"output": "2"
},
{
"input": "4 7\n-3 -2 1 6",
"output": "10"
},
{
"input": "4 3\n-7 -6 1 5",
"output": "17"
},
{
"input": "4 4\n-5 -1 1 2",
"output": "9"
}
] | 1,639,468,609 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 4 | 92 | 0 | n,k=list(map(int,input().split(" ")))
profits = list(map(int,input().split(" ")))
i=0
while i<k:
if profits[i]<0:
profits[i]= profits[i]*-1
else:
if (k-i)%2!=0:
profits[i]=profits[i]*-1
break
i+=1
print(sum(profits)) | Title: Roma and Changing Signs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Roma works in a company that sells TVs. Now he has to prepare a report for the last year.
Roma has got a list of the company's incomes. The list is a sequence that consists of *n* integers. The total income of the company is the sum of all integers in sequence. Roma decided to perform exactly *k* changes of signs of several numbers in the sequence. He can also change the sign of a number one, two or more times.
The operation of changing a number's sign is the operation of multiplying this number by -1.
Help Roma perform the changes so as to make the total income of the company (the sum of numbers in the resulting sequence) maximum. Note that Roma should perform exactly *k* changes.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=105), showing, how many numbers are in the sequence and how many swaps are to be made.
The second line contains a non-decreasing sequence, consisting of *n* integers *a**i* (|*a**i*|<=≤<=104).
The numbers in the lines are separated by single spaces. Please note that the given sequence is sorted in non-decreasing order.
Output Specification:
In the single line print the answer to the problem — the maximum total income that we can obtain after exactly *k* changes.
Demo Input:
['3 2\n-1 -1 1\n', '3 1\n-1 -1 1\n']
Demo Output:
['3\n', '1\n']
Note:
In the first sample we can get sequence [1, 1, 1], thus the total income equals 3.
In the second test, the optimal strategy is to get sequence [-1, 1, 1], thus the total income equals 1. | ```python
n,k=list(map(int,input().split(" ")))
profits = list(map(int,input().split(" ")))
i=0
while i<k:
if profits[i]<0:
profits[i]= profits[i]*-1
else:
if (k-i)%2!=0:
profits[i]=profits[i]*-1
break
i+=1
print(sum(profits))
``` | -1 | |
112 | A | Petya and Strings | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Petya and Strings | 2 | 256 | Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison. | Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters. | If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared. | [
"aaaa\naaaA\n",
"abs\nAbz\n",
"abcdefg\nAbCdEfF\n"
] | [
"0\n",
"-1\n",
"1\n"
] | If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
- http://en.wikipedia.org/wiki/Lexicographical_order | 500 | [
{
"input": "aaaa\naaaA",
"output": "0"
},
{
"input": "abs\nAbz",
"output": "-1"
},
{
"input": "abcdefg\nAbCdEfF",
"output": "1"
},
{
"input": "asadasdasd\nasdwasdawd",
"output": "-1"
},
{
"input": "aslkjlkasdd\nasdlkjdajwi",
"output": "1"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "0"
},
{
"input": "aAaaaAAaAaaAzZsssSsdDfeEaeqZlpP\nAaaaAaaAaaAaZzSSSSsDdFeeAeQZLpp",
"output": "0"
},
{
"input": "bwuEhEveouaTECagLZiqmUdxEmhRSOzMauJRWLQMppZOumxhAmwuGeDIkvkBLvMXwUoFmpAfDprBcFtEwOULcZWRQhcTbTbX\nHhoDWbcxwiMnCNexOsKsujLiSGcLllXOkRSbnOzThAjnnliLYFFmsYkOfpTxRNEfBsoUHfoLTiqAINRPxWRqrTJhgfkKcDOH",
"output": "-1"
},
{
"input": "kGWUuguKzcvxqKTNpxeDWXpXkrXDvGMFGoXKDfPBZvWSDUyIYBynbKOUonHvmZaKeirUhfmVRKtGhAdBfKMWXDUoqvbfpfHYcg\ncvOULleuIIiYVVxcLZmHVpNGXuEpzcWZZWyMOwIwbpkKPwCfkVbKkUuosvxYCKjqfVmHfJKbdrsAcatPYgrCABaFcoBuOmMfFt",
"output": "1"
},
{
"input": "nCeNVIzHqPceNhjHeHvJvgBsNFiXBATRrjSTXJzhLMDMxiJztphxBRlDlqwDFImWeEPkggZCXSRwelOdpNrYnTepiOqpvkr\nHJbjJFtlvNxIbkKlxQUwmZHJFVNMwPAPDRslIoXISBYHHfymyIaQHLgECPxAmqnOCizwXnIUBRmpYUBVPenoUKhCobKdOjL",
"output": "1"
},
{
"input": "ttXjenUAlfixytHEOrPkgXmkKTSGYuyVXGIHYmWWYGlBYpHkujueqBSgjLguSgiMGJWATIGEUjjAjKXdMiVbHozZUmqQtFrT\nJziDBFBDmDJCcGqFsQwDFBYdOidLxxhBCtScznnDgnsiStlWFnEXQrJxqTXKPxZyIGfLIToETKWZBPUIBmLeImrlSBWCkTNo",
"output": "1"
},
{
"input": "AjQhPqSVhwQQjcgCycjKorWBgFCRuQBwgdVuAPSMJAvTyxGVuFHjfJzkKfsmfhFbKqFrFIohSZBbpjgEHebezmVlGLTPSCTMf\nXhxWuSnMmKFrCUOwkTUmvKAfbTbHWzzOTzxJatLLCdlGnHVaBUnxDlsqpvjLHMThOPAFBggVKDyKBrZAmjnjrhHlrnSkyzBja",
"output": "-1"
},
{
"input": "HCIgYtnqcMyjVngziNflxKHtdTmcRJhzMAjFAsNdWXFJYEhiTzsQUtFNkAbdrFBRmvLirkuirqTDvIpEfyiIqkrwsjvpPWTEdI\nErqiiWKsmIjyZuzgTlTqxYZwlrpvRyaVhRTOYUqtPMVGGtWOkDCOOQRKrkkRzPftyQCkYkzKkzTPqqXmeZhvvEEiEhkdOmoMvy",
"output": "1"
},
{
"input": "mtBeJYILXcECGyEVSyzLFdQJbiVnnfkbsYYsdUJSIRmyzLfTTtFwIBmRLVnwcewIqcuydkcLpflHAFyDaToLiFMgeHvQorTVbI\nClLvyejznjbRfCDcrCzkLvqQaGzTjwmWONBdCctJAPJBcQrcYvHaSLQgPIJbmkFBhFzuQLBiRzAdNHulCjIAkBvZxxlkdzUWLR",
"output": "1"
},
{
"input": "tjucSbGESVmVridTBjTmpVBCwwdWKBPeBvmgdxgIVLwQxveETnSdxkTVJpXoperWSgdpPMKNmwDiGeHfxnuqaDissgXPlMuNZIr\nHfjOOJhomqNIKHvqSgfySjlsWJQBuWYwhLQhlZYlpZwboMpoLoluGsBmhhlYgeIouwdkPfiaAIrkYRlxtiFazOPOllPsNZHcIZd",
"output": "1"
},
{
"input": "AanbDfbZNlUodtBQlvPMyomStKNhgvSGhSbTdabxGFGGXCdpsJDimsAykKjfBDPMulkhBMsqLmVKLDoesHZsRAEEdEzqigueXInY\ncwfyjoppiJNrjrOLNZkqcGimrpTsiyFBVgMWEPXsMrxLJDDbtYzerXiFGuLBcQYitLdqhGHBpdjRnkUegmnwhGHAKXGyFtscWDSI",
"output": "-1"
},
{
"input": "HRfxniwuJCaHOcaOVgjOGHXKrwxrDQxJpppeGDXnTAowyKbCsCQPbchCKeTWOcKbySSYnoaTJDnmRcyGPbfXJyZoPcARHBu\nxkLXvwkvGIWSQaFTznLOctUXNuzzBBOlqvzmVfTSejekTAlwidRrsxkbZTsGGeEWxCXHzqWVuLGoCyrGjKkQoHqduXwYQKC",
"output": "-1"
},
{
"input": "OjYwwNuPESIazoyLFREpObIaMKhCaKAMWMfRGgucEuyNYRantwdwQkmflzfqbcFRaXBnZoIUGsFqXZHGKwlaBUXABBcQEWWPvkjW\nRxLqGcTTpBwHrHltCOllnTpRKLDofBUqqHxnOtVWPgvGaeHIevgUSOeeDOJubfqonFpVNGVbHFcAhjnyFvrrqnRgKhkYqQZmRfUl",
"output": "-1"
},
{
"input": "tatuhQPIzjptlzzJpCAPXSRTKZRlwgfoCIsFjJquRoIDyZZYRSPdFUTjjUPhLBBfeEIfLQpygKXRcyQFiQsEtRtLnZErBqW\ntkHUjllbafLUWhVCnvblKjgYIEoHhsjVmrDBmAWbvtkHxDbRFvsXAjHIrujaDbYwOZmacknhZPeCcorbRgHjjgAgoJdjvLo",
"output": "-1"
},
{
"input": "cymCPGqdXKUdADEWDdUaLEEMHiXHsdAZuDnJDMUvxvrLRBrPSDpXPAgMRoGplLtniFRTomDTAHXWAdgUveTxaqKVSvnOyhOwiRN\nuhmyEWzapiRNPFDisvHTbenXMfeZaHqOFlKjrfQjUBwdFktNpeiRoDWuBftZLcCZZAVfioOihZVNqiNCNDIsUdIhvbcaxpTRWoV",
"output": "-1"
},
{
"input": "sSvpcITJAwghVfJaLKBmyjOkhltTGjYJVLWCYMFUomiJaKQYhXTajvZVHIMHbyckYROGQZzjWyWCcnmDmrkvTKfHSSzCIhsXgEZa\nvhCXkCwAmErGVBPBAnkSYEYvseFKbWSktoqaHYXUmYkHfOkRwuEyBRoGoBrOXBKVxXycjZGStuvDarnXMbZLWrbjrisDoJBdSvWJ",
"output": "-1"
},
{
"input": "hJDANKUNBisOOINDsTixJmYgHNogtpwswwcvVMptfGwIjvqgwTYFcqTdyAqaqlnhOCMtsnWXQqtjFwQlEcBtMFAtSqnqthVb\nrNquIcjNWESjpPVWmzUJFrelpUZeGDmSvCurCqVmKHKVAAPkaHksniOlzjiKYIJtvbuQWZRufMebpTFPqyxIWWjfPaWYiNlK",
"output": "-1"
},
{
"input": "ycLoapxsfsDTHMSfAAPIUpiEhQKUIXUcXEiopMBuuZLHtfPpLmCHwNMNQUwsEXxCEmKHTBSnKhtQhGWUvppUFZUgSpbeChX\ndCZhgVXofkGousCzObxZSJwXcHIaqUDSCPKzXntcVmPxtNcXmVcjsetZYxedmgQzXTZHMvzjoaXCMKsncGciSDqQWIIRlys",
"output": "1"
},
{
"input": "nvUbnrywIePXcoukIhwTfUVcHUEgXcsMyNQhmMlTltZiCooyZiIKRIGVHMCnTKgzXXIuvoNDEZswKoACOBGSyVNqTNQqMhAG\nplxuGSsyyJjdvpddrSebOARSAYcZKEaKjqbCwvjhNykuaECoQVHTVFMKXwvrQXRaqXsHsBaGVhCxGRxNyGUbMlxOarMZNXxy",
"output": "-1"
},
{
"input": "EncmXtAblQzcVRzMQqdDqXfAhXbtJKQwZVWyHoWUckohnZqfoCmNJDzexFgFJYrwNHGgzCJTzQQFnxGlhmvQTpicTkEeVICKac\nNIUNZoMLFMyAjVgQLITELJSodIXcGSDWfhFypRoGYuogJpnqGTotWxVqpvBHjFOWcDRDtARsaHarHaOkeNWEHGTaGOFCOFEwvK",
"output": "-1"
},
{
"input": "UG\nak",
"output": "1"
},
{
"input": "JZR\nVae",
"output": "-1"
},
{
"input": "a\nZ",
"output": "-1"
},
{
"input": "rk\nkv",
"output": "1"
},
{
"input": "RvuT\nbJzE",
"output": "1"
},
{
"input": "PPS\nydq",
"output": "-1"
},
{
"input": "q\nq",
"output": "0"
},
{
"input": "peOw\nIgSJ",
"output": "1"
},
{
"input": "PyK\noKN",
"output": "1"
},
{
"input": "O\ni",
"output": "1"
},
{
"input": "NmGY\npDlP",
"output": "-1"
},
{
"input": "nG\nZf",
"output": "-1"
},
{
"input": "m\na",
"output": "1"
},
{
"input": "MWyB\nWZEV",
"output": "-1"
},
{
"input": "Gre\nfxc",
"output": "1"
},
{
"input": "Ooq\nwap",
"output": "-1"
},
{
"input": "XId\nlbB",
"output": "1"
},
{
"input": "lfFpECEqUMEOJhipvkZjDPcpDNJedOVXiSMgBvBZbtfzIKekcvpWPCazKAhJyHircRtgcBIJwwstpHaLAgxFOngAWUZRgCef\nLfFPEcequmeojHIpVkzjDPcpdNJEDOVXiSmGBVBZBtfZikEKcvPwpCAzKAHJyHIrCRTgCbIJWwSTphALagXfOnGAwUzRGcEF",
"output": "0"
},
{
"input": "DQBdtSEDtFGiNRUeJNbOIfDZnsryUlzJHGTXGFXnwsVyxNtLgmklmFvRCzYETBVdmkpJJIvIOkMDgCFHZOTODiYrkwXd\nDQbDtsEdTFginRUEJNBOIfdZnsryulZJHGtxGFxnwSvYxnTLgmKlmFVRCzyEtBVdmKpJjiVioKMDgCFhzoTODiYrKwXD",
"output": "0"
},
{
"input": "tYWRijFQSzHBpCjUzqBtNvBKyzZRnIdWEuyqnORBQTLyOQglIGfYJIRjuxnbLvkqZakNqPiGDvgpWYkfxYNXsdoKXZtRkSasfa\nTYwRiJfqsZHBPcJuZQBTnVbkyZZRnidwEuYQnorbQTLYOqGligFyjirJUxnblVKqZaknQpigDVGPwyKfxyNXSDoKxztRKSaSFA",
"output": "0"
},
{
"input": "KhScXYiErQIUtmVhNTCXSLAviefIeHIIdiGhsYnPkSBaDTvMkyanfMLBOvDWgRybLtDqvXVdVjccNunDyijhhZEAKBrdz\nkHsCXyiErqIuTMVHNTCxSLaViEFIEhIIDiGHsYNpKsBAdTvMKyANFMLBovdwGRYbLtdQVxvDVJCcNUndYiJHhzeakBrdZ",
"output": "0"
},
{
"input": "cpPQMpjRQJKQVXjWDYECXbagSmNcVfOuBWNZxihdERraVuiOpSVDCPgTGuSQALNoVjySceHcKXwOEpSzXrEqWwwrYeppNiWhDVg\nCPPqmPjRqJkQvxJwdyECXBAGsMNcVfOuBWNzxIhderRavUiOpSvDCpGTgusqAlNovjyScEhCKXwoePSZxrEQwWwryEPPniWHDvG",
"output": "0"
},
{
"input": "SajcCGMepaLjZIWLRBGFcrZRCRvvoCsIyKsQerbrwsIamxxpRmQSZSalasJLVFbCHCuXJlubciQAvLxXYBazLsMKLHLdDQ\nsaJcCgmEpaLJziWlrBgFcRzrCrVVOcSIykSQerBrwSIamxxPrMqSzSalASjLVFbChCUxjLUbCIQAVlxxybAZLsmkLhLDdQ",
"output": "0"
},
{
"input": "kigPrWNTOUNDBskAfefjhHYZNYdnfZWuXWzHiBxFQryBbAkPtenFwWvCSTYGpzOntUNzNUhxRWjKmicTwLwJAnbAxj\nkigpRWntOUNdBsKaFEFjhhYZnYDNfzWuXwZhibxFQRybbakPteNfwwvcStyGPzoNTunznuHXrWjKMIctWLWJANBAxJ",
"output": "0"
},
{
"input": "nTomZZuTTRTAAPoUsySVFGElrpQRNLjqvFmcYytiheQnjUhPLnqNBiYtQkljbcvmjuNAVKbvQOWpqqFlQhAhULIhquoCnjUI\nntOmzZuttrtAAPOUSySVFgeLRPQrNLjQvfmCyYTiHEQnjuHPlNQNbIYtqKLJBCVmjunavkbvQOWPQQFlqHaHULIHQuOcnJUi",
"output": "0"
},
{
"input": "abac\nadaa",
"output": "-1"
},
{
"input": "Bbc\nabc",
"output": "1"
},
{
"input": "aaaba\naaaab",
"output": "1"
}
] | 1,695,699,876 | 2,147,483,647 | Python 3 | OK | TESTS | 53 | 92 | 0 | #2300012105 刘乐天，生命科学学院
a=input().lower()
b=input().lower()
if a==b:
print(0)
elif a>b:
print(1)
else:
print(-1)
| Title: Petya and Strings
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.
Input Specification:
Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.
Output Specification:
If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.
Demo Input:
['aaaa\naaaA\n', 'abs\nAbz\n', 'abcdefg\nAbCdEfF\n']
Demo Output:
['0\n', '-1\n', '1\n']
Note:
If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
- http://en.wikipedia.org/wiki/Lexicographical_order | ```python
#2300012105 刘乐天，生命科学学院
a=input().lower()
b=input().lower()
if a==b:
print(0)
elif a>b:
print(1)
else:
print(-1)
``` | 3.977 |
59 | A | Word | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Word | 2 | 256 | Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. | The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. | Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. | [
"HoUse\n",
"ViP\n",
"maTRIx\n"
] | [
"house\n",
"VIP\n",
"matrix\n"
] | none | 500 | [
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,652,670,210 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 92 | 4,300,800 | word = input()
countu = 0
countl = 0
for i in word:
if i.isupper():
countu = countu + 1
else:
countl = countl + 1
if countu > countl :
print(word.upper())
else:
print(word.lower()) | Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none | ```python
word = input()
countu = 0
countl = 0
for i in word:
if i.isupper():
countu = countu + 1
else:
countl = countl + 1
if countu > countl :
print(word.upper())
else:
print(word.lower())
``` | 3.968989 |
25 | A | IQ test | PROGRAMMING | 1,300 | [
"brute force"
] | A. IQ test | 2 | 256 | Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness. | The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness. | Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order. | [
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] | [
"3\n",
"2\n"
] | none | 0 | [
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,636,163,112 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 62 | 4,505,600 | import sys
sys.stdin = open("iq_test.in", "r")
n = int(input())
numbers = [int(x) for x in input().split()]
num_even = 0
evens = []
num_odd = 0
odds = []
counter = 1
for num in numbers:
if num % 2 == 0:
num_even += 1
evens.append(counter)
else:
num_odd += 1
odds.append(counter)
counter += 1
if num_even > num_odd:
print(odds[0])
else:
print(evens[0])
| Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none | ```python
import sys
sys.stdin = open("iq_test.in", "r")
n = int(input())
numbers = [int(x) for x in input().split()]
num_even = 0
evens = []
num_odd = 0
odds = []
counter = 1
for num in numbers:
if num % 2 == 0:
num_even += 1
evens.append(counter)
else:
num_odd += 1
odds.append(counter)
counter += 1
if num_even > num_odd:
print(odds[0])
else:
print(evens[0])
``` | -1 |
485 | B | Valuable Resources | PROGRAMMING | 1,300 | [
"brute force",
"greedy"
] | null | null | Many computer strategy games require building cities, recruiting army, conquering tribes, collecting resources. Sometimes it leads to interesting problems.
Let's suppose that your task is to build a square city. The world map uses the Cartesian coordinates. The sides of the city should be parallel to coordinate axes. The map contains mines with valuable resources, located at some points with integer coordinates. The sizes of mines are relatively small, i.e. they can be treated as points. The city should be built in such a way that all the mines are inside or on the border of the city square.
Building a city takes large amount of money depending on the size of the city, so you have to build the city with the minimum area. Given the positions of the mines find the minimum possible area of the city. | The first line of the input contains number *n* — the number of mines on the map (2<=≤<=*n*<=≤<=1000). Each of the next *n* lines contains a pair of integers *x**i* and *y**i* — the coordinates of the corresponding mine (<=-<=109<=≤<=*x**i*,<=*y**i*<=≤<=109). All points are pairwise distinct. | Print the minimum area of the city that can cover all the mines with valuable resources. | [
"2\n0 0\n2 2\n",
"2\n0 0\n0 3\n"
] | [
"4\n",
"9\n"
] | none | 500 | [
{
"input": "2\n0 0\n2 2",
"output": "4"
},
{
"input": "2\n0 0\n0 3",
"output": "9"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "3\n2 2\n1 1\n3 3",
"output": "4"
},
{
"input": "3\n3 1\n1 3\n2 2",
"output": "4"
},
{
"input": "3\n0 1\n1 0\n2 2",
"output": "4"
},
{
"input": "2\n-1000000000 -1000000000\n1000000000 1000000000",
"output": "4000000000000000000"
},
{
"input": "2\n1000000000 -1000000000\n-1000000000 1000000000",
"output": "4000000000000000000"
},
{
"input": "5\n-851545463 -208880322\n-154983867 -781305244\n293363100 785256340\n833468900 -593065920\n-920692803 -637662144",
"output": "3077083280271860209"
},
{
"input": "10\n-260530833 169589238\n-681955770 -35391010\n223450511 24504262\n479795061 -26191863\n-291344265 21153856\n714700263 -328447419\n-858655942 161086142\n-270884153 462537328\n-501424901 977460517\n115284904 -151626824",
"output": "2475449747812002025"
},
{
"input": "10\n917139470 819990899\n-69828590 691215072\n-846815289 112372447\n560780737 -890423729\n243241705 284240970\n-47397355 -263709479\n759162072 709456353\n-330469400 -597545533\n436509256 728506920\n133368867 668789238",
"output": "3111536391798748081"
},
{
"input": "10\n-200157522 -824574736\n299208799 -287211553\n-160170880 148363130\n103709327 245344406\n482860382 547328085\n895537733 -545816336\n671947380 910981768\n-43209851 585461399\n-573679087 427675821\n151452830 27262384",
"output": "3012156378576702016"
},
{
"input": "2\n-2 -2\n-3 -3",
"output": "1"
},
{
"input": "2\n-1000 -1000\n-1100 -1100",
"output": "10000"
},
{
"input": "2\n-5 -5\n-4 -4",
"output": "1"
},
{
"input": "2\n-10 0\n-9 0",
"output": "1"
},
{
"input": "2\n-10 -10\n-20 -20",
"output": "100"
},
{
"input": "2\n-1000000 -1000000\n-100 -100",
"output": "999800010000"
},
{
"input": "2\n100000000 100000000\n200000000 200000000",
"output": "10000000000000000"
},
{
"input": "2\n-10 10\n-2 3",
"output": "64"
},
{
"input": "2\n-999999999 -999999999\n-999999991 -999999991",
"output": "64"
},
{
"input": "2\n-1000 -1000\n-999 -999",
"output": "1"
},
{
"input": "2\n-3 0\n-5 0",
"output": "4"
},
{
"input": "2\n999999999 999999999\n999999991 999999991",
"output": "64"
},
{
"input": "2\n100000012 100000012\n100000012 100000013",
"output": "1"
}
] | 1,501,990,380 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
#define pb push_back
#define pi 3.1415926535897932384626433
#define INF 1e18
#define MOD 1000000007
#define all(s) s.begin(),s.end()
#define mp make_pair
#define vc vector<int>
#define vp vector<pair<ll,ll> >
#define sec second
#define sz(x) (int)((x).size())
#define timesaver ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL)
const int N = 1e6 + 5;
ll gcd(ll a, ll b){ if(a==0) return b; else return gcd(b%a,a);}
ll power(ll base,ll exp,ll mod=MOD) {ll ans=1;while(exp){if(exp&1) ans=(ans*base)%mod;exp>>=1,base=(base*base)%mod;}return ans;}
int main()
{
timesaver;
string s, u;
int maxi = 0;
cin >> s >> u;
for (int i = 0; i < sz(s); i += 1)
{
int ct = 0;
for (int j = 0; j < min(sz(u),sz(s) - i); j += 1)
{
if(s[j + i] == u[j])
ct++;
}
maxi = max(ct, maxi);
}
for (int j = 0; j < sz(u); j += 1)
{
int ct = 0;
for (int i = 0; i < min(sz(s), sz(u) - j); i += 1)
{
if(u[j + i] == s[i])
ct++;
}
maxi = max(ct, maxi);
}
cout << sz(u) - maxi;
}
| Title: Valuable Resources
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Many computer strategy games require building cities, recruiting army, conquering tribes, collecting resources. Sometimes it leads to interesting problems.
Let's suppose that your task is to build a square city. The world map uses the Cartesian coordinates. The sides of the city should be parallel to coordinate axes. The map contains mines with valuable resources, located at some points with integer coordinates. The sizes of mines are relatively small, i.e. they can be treated as points. The city should be built in such a way that all the mines are inside or on the border of the city square.
Building a city takes large amount of money depending on the size of the city, so you have to build the city with the minimum area. Given the positions of the mines find the minimum possible area of the city.
Input Specification:
The first line of the input contains number *n* — the number of mines on the map (2<=≤<=*n*<=≤<=1000). Each of the next *n* lines contains a pair of integers *x**i* and *y**i* — the coordinates of the corresponding mine (<=-<=109<=≤<=*x**i*,<=*y**i*<=≤<=109). All points are pairwise distinct.
Output Specification:
Print the minimum area of the city that can cover all the mines with valuable resources.
Demo Input:
['2\n0 0\n2 2\n', '2\n0 0\n0 3\n']
Demo Output:
['4\n', '9\n']
Note:
none | ```python
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
#define pb push_back
#define pi 3.1415926535897932384626433
#define INF 1e18
#define MOD 1000000007
#define all(s) s.begin(),s.end()
#define mp make_pair
#define vc vector<int>
#define vp vector<pair<ll,ll> >
#define sec second
#define sz(x) (int)((x).size())
#define timesaver ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL)
const int N = 1e6 + 5;
ll gcd(ll a, ll b){ if(a==0) return b; else return gcd(b%a,a);}
ll power(ll base,ll exp,ll mod=MOD) {ll ans=1;while(exp){if(exp&1) ans=(ans*base)%mod;exp>>=1,base=(base*base)%mod;}return ans;}
int main()
{
timesaver;
string s, u;
int maxi = 0;
cin >> s >> u;
for (int i = 0; i < sz(s); i += 1)
{
int ct = 0;
for (int j = 0; j < min(sz(u),sz(s) - i); j += 1)
{
if(s[j + i] == u[j])
ct++;
}
maxi = max(ct, maxi);
}
for (int j = 0; j < sz(u); j += 1)
{
int ct = 0;
for (int i = 0; i < min(sz(s), sz(u) - j); i += 1)
{
if(u[j + i] == s[i])
ct++;
}
maxi = max(ct, maxi);
}
cout << sz(u) - maxi;
}
``` | -1 | |
227 | B | Effective Approach | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat. | Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier. | [
"2\n1 2\n1\n1\n",
"2\n2 1\n1\n1\n",
"3\n3 1 2\n3\n1 2 3\n"
] | [
"1 2\n",
"2 1\n",
"6 6\n"
] | In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | 1,000 | [
{
"input": "2\n1 2\n1\n1",
"output": "1 2"
},
{
"input": "2\n2 1\n1\n1",
"output": "2 1"
},
{
"input": "3\n3 1 2\n3\n1 2 3",
"output": "6 6"
},
{
"input": "9\n2 9 3 1 6 4 7 8 5\n9\n5 1 5 2 8 4 4 4 5",
"output": "58 32"
},
{
"input": "10\n3 10 9 2 7 6 5 8 4 1\n1\n4",
"output": "9 2"
},
{
"input": "10\n5 2 10 8 3 1 9 7 6 4\n9\n2 5 9 2 3 2 5 9 8",
"output": "31 68"
},
{
"input": "9\n3 8 4 7 1 2 5 6 9\n3\n2 7 1",
"output": "15 15"
},
{
"input": "9\n5 3 8 4 2 6 1 7 9\n4\n6 1 9 2",
"output": "27 13"
},
{
"input": "4\n1 3 2 4\n4\n3 1 2 3",
"output": "8 12"
},
{
"input": "3\n1 2 3\n8\n3 2 1 1 2 3 1 2",
"output": "15 17"
}
] | 1,655,881,254 | 2,147,483,647 | PyPy 3 | OK | TESTS | 34 | 560 | 14,336,000 | from sys import stdin
input = stdin.readline
n = int(input())
a = list(map(int,input().split()))
q = int(input())
qa = list(map(int,input().split()))
d = {}
for i in range(n):
d[a[i]] = [i + 1, n - i]
v ,p =0,0
for i in range(q):
v += d[qa[i]][0]
p += d[qa[i]][1]
print(f"{v} {p}") | Title: Effective Approach
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
Output Specification:
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Demo Input:
['2\n1 2\n1\n1\n', '2\n2 1\n1\n1\n', '3\n3 1 2\n3\n1 2 3\n']
Demo Output:
['1 2\n', '2 1\n', '6 6\n']
Note:
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | ```python
from sys import stdin
input = stdin.readline
n = int(input())
a = list(map(int,input().split()))
q = int(input())
qa = list(map(int,input().split()))
d = {}
for i in range(n):
d[a[i]] = [i + 1, n - i]
v ,p =0,0
for i in range(q):
v += d[qa[i]][0]
p += d[qa[i]][1]
print(f"{v} {p}")
``` | 3 | |
344 | A | Magnets | PROGRAMMING | 800 | [
"implementation"
] | null | null | Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed. | The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position. | On the single line of the output print the number of groups of magnets. | [
"6\n10\n10\n10\n01\n10\n10\n",
"4\n01\n01\n10\n10\n"
] | [
"3\n",
"2\n"
] | The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets. | 500 | [
{
"input": "6\n10\n10\n10\n01\n10\n10",
"output": "3"
},
{
"input": "4\n01\n01\n10\n10",
"output": "2"
},
{
"input": "1\n10",
"output": "1"
},
{
"input": "2\n01\n10",
"output": "2"
},
{
"input": "2\n10\n10",
"output": "1"
},
{
"input": "3\n10\n01\n10",
"output": "3"
},
{
"input": "1\n01",
"output": "1"
},
{
"input": "2\n01\n01",
"output": "1"
},
{
"input": "2\n10\n01",
"output": "2"
},
{
"input": "3\n01\n01\n01",
"output": "1"
},
{
"input": "3\n10\n10\n01",
"output": "2"
},
{
"input": "3\n01\n10\n10",
"output": "2"
},
{
"input": "115\n10\n10\n10\n10\n01\n01\n10\n10\n10\n01\n01\n10\n01\n01\n10\n10\n10\n01\n10\n01\n10\n10\n01\n01\n10\n10\n10\n10\n01\n10\n01\n01\n10\n10\n10\n10\n01\n10\n10\n10\n01\n10\n01\n10\n10\n10\n10\n01\n01\n01\n10\n10\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n01\n10\n01\n10\n10\n01\n01\n01\n10\n01\n01\n10\n10\n01\n01\n01\n01\n01\n10\n01\n10\n01\n10\n01\n01\n01\n10\n01\n10\n10\n01\n10\n10\n01\n01\n01\n10\n10\n10\n10\n10\n10\n10\n10",
"output": "55"
}
] | 1,699,076,675 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 62 | 0 | n = int(input())
stroka = []
count = 0
for i in range(n):
a = input()
stroka.append(a)
for i in range(len(stroka)-1):
if stroka[i] == stroka[i+1]:
count+=1
print(count) | Title: Magnets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.
Output Specification:
On the single line of the output print the number of groups of magnets.
Demo Input:
['6\n10\n10\n10\n01\n10\n10\n', '4\n01\n01\n10\n10\n']
Demo Output:
['3\n', '2\n']
Note:
The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets. | ```python
n = int(input())
stroka = []
count = 0
for i in range(n):
a = input()
stroka.append(a)
for i in range(len(stroka)-1):
if stroka[i] == stroka[i+1]:
count+=1
print(count)
``` | 0 | |
312 | B | Archer | PROGRAMMING | 1,300 | [
"math",
"probabilities"
] | null | null | SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is for SmallR while for Zanoes. The one who shoots in the target first should be the winner.
Output the probability that SmallR will win the match. | A single line contains four integers . | Print a single real number, the probability that SmallR will win the match.
The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6. | [
"1 2 1 2\n"
] | [
"0.666666666667"
] | none | 1,000 | [
{
"input": "1 2 1 2",
"output": "0.666666666667"
},
{
"input": "1 3 1 3",
"output": "0.600000000000"
},
{
"input": "1 3 2 3",
"output": "0.428571428571"
},
{
"input": "3 4 3 4",
"output": "0.800000000000"
},
{
"input": "1 2 10 11",
"output": "0.523809523810"
},
{
"input": "4 5 4 5",
"output": "0.833333333333"
},
{
"input": "466 701 95 721",
"output": "0.937693791148"
},
{
"input": "268 470 444 885",
"output": "0.725614009325"
},
{
"input": "632 916 713 821",
"output": "0.719292895126"
},
{
"input": "269 656 918 992",
"output": "0.428937461623"
},
{
"input": "71 657 187 695",
"output": "0.310488463257"
},
{
"input": "435 852 973 978",
"output": "0.511844133157"
},
{
"input": "518 816 243 359",
"output": "0.719734031025"
},
{
"input": "882 962 311 811",
"output": "0.966386645447"
},
{
"input": "684 774 580 736",
"output": "0.906051574446"
},
{
"input": "486 868 929 999",
"output": "0.577723252958"
},
{
"input": "132 359 996 998",
"output": "0.368154532345"
},
{
"input": "933 977 266 450",
"output": "0.972879407907"
},
{
"input": "298 833 615 872",
"output": "0.441270817024"
},
{
"input": "34 554 14 958",
"output": "0.817324099167"
},
{
"input": "836 934 800 905",
"output": "0.906105535462"
},
{
"input": "482 815 69 509",
"output": "0.914365577772"
},
{
"input": "284 423 137 521",
"output": "0.885974839378"
},
{
"input": "648 881 486 703",
"output": "0.800911421248"
},
{
"input": "450 885 755 836",
"output": "0.533901011176"
},
{
"input": "533 773 823 998",
"output": "0.729222130525"
},
{
"input": "897 957 92 898",
"output": "0.993193806364"
},
{
"input": "699 925 441 928",
"output": "0.866816866175"
},
{
"input": "64 704 148 603",
"output": "0.289486317811"
},
{
"input": "719 735 626 990",
"output": "0.986124079764"
},
{
"input": "1 1000 1 1000",
"output": "0.500250125063"
}
] | 1,640,740,477 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 31 | 124 | 1,228,800 | a, b, c, d = [int(i) for i in input().split()]
p1 = a / b
p2 = 1 - c / d
def v(c, p1, p2):
return (1 - p1) ** (c - 1) * p2 ** (c - 1) * p1
s = 0
for i in range(10000):
s += v(i + 1, p1, p2)
print(round(s, 6)) | Title: Archer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is for SmallR while for Zanoes. The one who shoots in the target first should be the winner.
Output the probability that SmallR will win the match.
Input Specification:
A single line contains four integers .
Output Specification:
Print a single real number, the probability that SmallR will win the match.
The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6.
Demo Input:
['1 2 1 2\n']
Demo Output:
['0.666666666667']
Note:
none | ```python
a, b, c, d = [int(i) for i in input().split()]
p1 = a / b
p2 = 1 - c / d
def v(c, p1, p2):
return (1 - p1) ** (c - 1) * p2 ** (c - 1) * p1
s = 0
for i in range(10000):
s += v(i + 1, p1, p2)
print(round(s, 6))
``` | 3 | |
69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,591,521,090 | 2,147,483,647 | Python 3 | OK | TESTS | 81 | 218 | 0 | n=int(input())
ti=0
tj=0
tk=0
for i in range(n):
i,j,k=map(int,input().split())
ti+=i
tj+=j
tk+=k
if(ti==0 and tj==0 and tk==0):
print("YES")
else:
print("NO")
| Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
n=int(input())
ti=0
tj=0
tk=0
for i in range(n):
i,j,k=map(int,input().split())
ti+=i
tj+=j
tk+=k
if(ti==0 and tj==0 and tk==0):
print("YES")
else:
print("NO")
``` | 3.9455 |
501 | A | Contest | PROGRAMMING | 900 | [
"implementation"
] | null | null | Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points.
Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth. | The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180).
It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round). | Output on a single line:
"Misha" (without the quotes), if Misha got more points than Vasya.
"Vasya" (without the quotes), if Vasya got more points than Misha.
"Tie" (without the quotes), if both of them got the same number of points. | [
"500 1000 20 30\n",
"1000 1000 1 1\n",
"1500 1000 176 177\n"
] | [
"Vasya\n",
"Tie\n",
"Misha\n"
] | none | 500 | [
{
"input": "500 1000 20 30",
"output": "Vasya"
},
{
"input": "1000 1000 1 1",
"output": "Tie"
},
{
"input": "1500 1000 176 177",
"output": "Misha"
},
{
"input": "1500 1000 74 177",
"output": "Misha"
},
{
"input": "750 2500 175 178",
"output": "Vasya"
},
{
"input": "750 1000 54 103",
"output": "Tie"
},
{
"input": "2000 1250 176 130",
"output": "Tie"
},
{
"input": "1250 1750 145 179",
"output": "Tie"
},
{
"input": "2000 2000 176 179",
"output": "Tie"
},
{
"input": "1500 1500 148 148",
"output": "Tie"
},
{
"input": "2750 1750 134 147",
"output": "Misha"
},
{
"input": "3250 250 175 173",
"output": "Misha"
},
{
"input": "500 500 170 176",
"output": "Misha"
},
{
"input": "250 1000 179 178",
"output": "Vasya"
},
{
"input": "3250 1000 160 138",
"output": "Misha"
},
{
"input": "3000 2000 162 118",
"output": "Tie"
},
{
"input": "1500 1250 180 160",
"output": "Tie"
},
{
"input": "1250 2500 100 176",
"output": "Tie"
},
{
"input": "3500 3500 177 178",
"output": "Tie"
},
{
"input": "3000 3250 16 34",
"output": "Tie"
},
{
"input": "1750 3000 137 49",
"output": "Vasya"
},
{
"input": "500 1500 179 71",
"output": "Vasya"
},
{
"input": "1250 2000 101 180",
"output": "Misha"
},
{
"input": "250 750 180 176",
"output": "Vasya"
},
{
"input": "2250 2250 163 145",
"output": "Vasya"
},
{
"input": "3000 3000 176 78",
"output": "Vasya"
},
{
"input": "250 3500 8 178",
"output": "Vasya"
},
{
"input": "1750 1250 179 180",
"output": "Misha"
},
{
"input": "2750 1750 13 164",
"output": "Misha"
},
{
"input": "1750 2250 178 53",
"output": "Vasya"
},
{
"input": "2500 2750 73 179",
"output": "Misha"
},
{
"input": "1000 3500 178 175",
"output": "Vasya"
},
{
"input": "1000 500 7 162",
"output": "Misha"
},
{
"input": "1000 250 175 48",
"output": "Misha"
},
{
"input": "1750 500 166 177",
"output": "Misha"
},
{
"input": "250 250 0 0",
"output": "Tie"
},
{
"input": "250 3500 0 0",
"output": "Vasya"
},
{
"input": "250 3500 0 180",
"output": "Vasya"
},
{
"input": "3500 3500 180 180",
"output": "Tie"
},
{
"input": "3500 250 0 180",
"output": "Misha"
}
] | 1,590,456,426 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 108 | 0 | a,b,c,d=map(int,input().split())
p1=max(3*a/10,a-((a*c)/250))
p2=max(3*b/10,b-((b*d)/250))
if p1==p2:
print('Tie')
exit(0)
print(['Misha','Vasya'][p1<p2]) | Title: Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points.
Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth.
Input Specification:
The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180).
It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round).
Output Specification:
Output on a single line:
"Misha" (without the quotes), if Misha got more points than Vasya.
"Vasya" (without the quotes), if Vasya got more points than Misha.
"Tie" (without the quotes), if both of them got the same number of points.
Demo Input:
['500 1000 20 30\n', '1000 1000 1 1\n', '1500 1000 176 177\n']
Demo Output:
['Vasya\n', 'Tie\n', 'Misha\n']
Note:
none | ```python
a,b,c,d=map(int,input().split())
p1=max(3*a/10,a-((a*c)/250))
p2=max(3*b/10,b-((b*d)/250))
if p1==p2:
print('Tie')
exit(0)
print(['Misha','Vasya'][p1<p2])
``` | 3 | |
765 | B | Code obfuscation | PROGRAMMING | 1,100 | [
"greedy",
"implementation",
"strings"
] | null | null | Kostya likes Codeforces contests very much. However, he is very disappointed that his solutions are frequently hacked. That's why he decided to obfuscate (intentionally make less readable) his code before upcoming contest.
To obfuscate the code, Kostya first looks at the first variable name used in his program and replaces all its occurrences with a single symbol *a*, then he looks at the second variable name that has not been replaced yet, and replaces all its occurrences with *b*, and so on. Kostya is well-mannered, so he doesn't use any one-letter names before obfuscation. Moreover, there are at most 26 unique identifiers in his programs.
You are given a list of identifiers of some program with removed spaces and line breaks. Check if this program can be a result of Kostya's obfuscation. | In the only line of input there is a string *S* of lowercase English letters (1<=≤<=|*S*|<=≤<=500) — the identifiers of a program with removed whitespace characters. | If this program can be a result of Kostya's obfuscation, print "YES" (without quotes), otherwise print "NO". | [
"abacaba\n",
"jinotega\n"
] | [
"YES\n",
"NO\n"
] | In the first sample case, one possible list of identifiers would be "number string number character number string number". Here how Kostya would obfuscate the program:
- replace all occurences of number with a, the result would be "a string a character a string a",- replace all occurences of string with b, the result would be "a b a character a b a",- replace all occurences of character with c, the result would be "a b a c a b a",- all identifiers have been replaced, thus the obfuscation is finished. | 1,000 | [
{
"input": "abacaba",
"output": "YES"
},
{
"input": "jinotega",
"output": "NO"
},
{
"input": "aaaaaaaaaaa",
"output": "YES"
},
{
"input": "aba",
"output": "YES"
},
{
"input": "bab",
"output": "NO"
},
{
"input": "a",
"output": "YES"
},
{
"input": "abcdefghijklmnopqrstuvwxyz",
"output": "YES"
},
{
"input": "fihyxmbnzq",
"output": "NO"
},
{
"input": "aamlaswqzotaanasdhcvjoaiwdhctezzawagkdgfffeqkyrvbcrfqgkdsvximsnvmkmjyofswmtjdoxgwamsaatngenqvsvrvwlbzuoeaolfcnmdacrmdleafbsmerwmxzyylfhemnkoayuhtpbikm",
"output": "NO"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "YES"
},
{
"input": "darbbbcwynbbbbaacbkvbakavabbbabzajlbajryaabbbccxraakgniagbtsswcfbkubdmcasccepybkaefcfsbzdddxgcjadybcfjtmqbspflqrdghgfwnccfveogdmifkociqscahdejctacwzbkhihajfilrgcjiofwfklifobozikcmvcfeqlidrgsgdfxffaaebzjxngsjxiclyolhjokqpdbfffooticxsezpgqkhhzmbmqgskkqvefzyijrwhpftcmbedmaflapmeljaudllojfpgfkpvgylaglrhrslxlprbhgknrctilngqccbddvpamhifsbmyowohczizjcbleehfrecjbqtxertnpfmalejmbxkhkkbyopuwlhkxuqellsybgcndvniyyxfoufalstdsdfjoxlnmigkqwmgojsppaannfstxytelluvvkdcezlqfsperwyjsdsmkvgjdbksswamhmoukcawiigkggztr",
"output": "NO"
},
{
"input": "bbbbbb",
"output": "NO"
},
{
"input": "aabbbd",
"output": "NO"
},
{
"input": "abdefghijklmnopqrstuvwxyz",
"output": "NO"
},
{
"input": "abcdeghijklmnopqrstuvwxyz",
"output": "NO"
},
{
"input": "abcdefghijklmnopqrsuvwxyz",
"output": "NO"
},
{
"input": "abcdefghijklmnopqrstuvwxy",
"output": "YES"
},
{
"input": "abcdefghijklmnopqrsutvwxyz",
"output": "NO"
},
{
"input": "acdef",
"output": "NO"
},
{
"input": "z",
"output": "NO"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaababaaababaabababccbabdbcbadccacdbdedabbeecbcabbdcaecdabbedddafeffaccgeacefbcahabfiiegecdbebabhhbdgfeghhbfahgagefbgghdbhadeicbdfgdchhefhigfcgdhcihecacfhadfgfejccibcjkfhbigbealjjkfldiecfdcafbamgfkbjlbifldghmiifkkglaflmjfmkfdjlbliijkgfdelklfnadbifgbmklfbqkhirhcadoadhmjrghlmelmjfpakqkdfcgqdkaeqpbcdoeqglqrarkipncckpfmajrqsfffldegbmahsfcqdfdqtrgrouqajgsojmmukptgerpanpcbejmergqtavwsvtveufdseuemwrhfmjqinxjodddnpcgqullrhmogflsxgsbapoghortiwcovejtinncozk",
"output": "NO"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "YES"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbabbbabbaaabbaaaaabaabbaa",
"output": "YES"
},
{
"input": "aababbabbaabbbbbaabababaabbbaaaaabbabbabbaabbbbabaabbaaababbaaacbbabbbbbbcbcababbccaaacbaccaccaababbccaacccaabaaccaaabacacbaabacbaacbaaabcbbbcbbaacaabcbcbccbacabbcbabcaccaaaaaabcbacabcbabbbbbabccbbcacbaaabbccbbaaaaaaaaaaaadbbbabdacabdaddddbaabbddbdabbdacbacbacaaaabbacadbcddddadaddabbdccaddbaaacbceebbceadbeaadecddbbbcaaecbdeaebaddbbdebbcbaabcacbdcdc",
"output": "YES"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbaabaabaababbbabbacacbbbacbbaaaabbccacbaabaaccbbbbbcbbbacabbccaaabbaaacabcbacbcabbbbecbecadcbacbaadeeadabeacdebccdbbcaecdbeeebbebcaaaeacdcbdeccdbbdcdebdcbdacebcecbacddeeaebcedffedfggbeedceacaecagdfedfabcfchffceachgcbicbcffeeebgcgiefcafhibhceiedgbfebbccegbehhibhhfedbaeedbghggffehggaeaidifhdhaggdjcfjhiaieaichjacedchejg",
"output": "NO"
},
{
"input": "b",
"output": "NO"
},
{
"input": "ac",
"output": "NO"
},
{
"input": "cde",
"output": "NO"
},
{
"input": "abd",
"output": "NO"
},
{
"input": "zx",
"output": "NO"
},
{
"input": "bcd",
"output": "NO"
},
{
"input": "aaac",
"output": "NO"
},
{
"input": "aacb",
"output": "NO"
},
{
"input": "acd",
"output": "NO"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaz",
"output": "NO"
},
{
"input": "abcdefghijklmnopqrstuvwxyzz",
"output": "YES"
},
{
"input": "bc",
"output": "NO"
},
{
"input": "aaaaaaaaad",
"output": "NO"
},
{
"input": "abb",
"output": "YES"
},
{
"input": "abcb",
"output": "YES"
},
{
"input": "aac",
"output": "NO"
},
{
"input": "abcbcb",
"output": "YES"
},
{
"input": "bb",
"output": "NO"
},
{
"input": "abbb",
"output": "YES"
},
{
"input": "bbb",
"output": "NO"
},
{
"input": "x",
"output": "NO"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaazz",
"output": "NO"
},
{
"input": "acbccccccccccc",
"output": "NO"
},
{
"input": "za",
"output": "NO"
},
{
"input": "ade",
"output": "NO"
},
{
"input": "bbbbbbbbbb",
"output": "NO"
},
{
"input": "bac",
"output": "NO"
},
{
"input": "bcddcb",
"output": "NO"
},
{
"input": "aaacb",
"output": "NO"
},
{
"input": "aaaaac",
"output": "NO"
},
{
"input": "aaaaaaaaaaad",
"output": "NO"
},
{
"input": "c",
"output": "NO"
},
{
"input": "abcccccccc",
"output": "YES"
},
{
"input": "aaaaaaac",
"output": "NO"
}
] | 1,502,966,757 | 2,147,483,647 | Python 3 | OK | TESTS | 59 | 62 | 0 | a=input()
t=97
d=1
while(a!=""):
b=""
if(ord(a[0])==t):
for i in a:
if(ord(i)!=t):
b=b+(i)
else:
d=0
t+=1
a=b
if(d==1):print("YES")
else:print("NO") | Title: Code obfuscation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kostya likes Codeforces contests very much. However, he is very disappointed that his solutions are frequently hacked. That's why he decided to obfuscate (intentionally make less readable) his code before upcoming contest.
To obfuscate the code, Kostya first looks at the first variable name used in his program and replaces all its occurrences with a single symbol *a*, then he looks at the second variable name that has not been replaced yet, and replaces all its occurrences with *b*, and so on. Kostya is well-mannered, so he doesn't use any one-letter names before obfuscation. Moreover, there are at most 26 unique identifiers in his programs.
You are given a list of identifiers of some program with removed spaces and line breaks. Check if this program can be a result of Kostya's obfuscation.
Input Specification:
In the only line of input there is a string *S* of lowercase English letters (1<=≤<=|*S*|<=≤<=500) — the identifiers of a program with removed whitespace characters.
Output Specification:
If this program can be a result of Kostya's obfuscation, print "YES" (without quotes), otherwise print "NO".
Demo Input:
['abacaba\n', 'jinotega\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample case, one possible list of identifiers would be "number string number character number string number". Here how Kostya would obfuscate the program:
- replace all occurences of number with a, the result would be "a string a character a string a",- replace all occurences of string with b, the result would be "a b a character a b a",- replace all occurences of character with c, the result would be "a b a c a b a",- all identifiers have been replaced, thus the obfuscation is finished. | ```python
a=input()
t=97
d=1
while(a!=""):
b=""
if(ord(a[0])==t):
for i in a:
if(ord(i)!=t):
b=b+(i)
else:
d=0
t+=1
a=b
if(d==1):print("YES")
else:print("NO")
``` | 3 | |
1,000 | B | Light It Up | PROGRAMMING | 1,500 | [
"greedy"
] | null | null | Recently, you bought a brand new smart lamp with programming features. At first, you set up a schedule to the lamp. Every day it will turn power on at moment $0$ and turn power off at moment $M$. Moreover, the lamp allows you to set a program of switching its state (states are "lights on" and "lights off"). Unfortunately, some program is already installed into the lamp.
The lamp allows only good programs. Good program can be represented as a non-empty array $a$, where $0 < a_1 < a_2 < \dots < a_{|a|} < M$. All $a_i$ must be integers. Of course, preinstalled program is a good program.
The lamp follows program $a$ in next manner: at moment $0$ turns power and light on. Then at moment $a_i$ the lamp flips its state to opposite (if it was lit, it turns off, and vice versa). The state of the lamp flips instantly: for example, if you turn the light off at moment $1$ and then do nothing, the total time when the lamp is lit will be $1$. Finally, at moment $M$ the lamp is turning its power off regardless of its state.
Since you are not among those people who read instructions, and you don't understand the language it's written in, you realize (after some testing) the only possible way to alter the preinstalled program. You can insert at most one element into the program $a$, so it still should be a good program after alteration. Insertion can be done between any pair of consecutive elements of $a$, or even at the begining or at the end of $a$.
Find such a way to alter the program that the total time when the lamp is lit is maximum possible. Maybe you should leave program untouched. If the lamp is lit from $x$ till moment $y$, then its lit for $y - x$ units of time. Segments of time when the lamp is lit are summed up. | First line contains two space separated integers $n$ and $M$ ($1 \le n \le 10^5$, $2 \le M \le 10^9$) — the length of program $a$ and the moment when power turns off.
Second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($0 < a_1 < a_2 < \dots < a_n < M$) — initially installed program $a$. | Print the only integer — maximum possible total time when the lamp is lit. | [
"3 10\n4 6 7\n",
"2 12\n1 10\n",
"2 7\n3 4\n"
] | [
"8\n",
"9\n",
"6\n"
] | In the first example, one of possible optimal solutions is to insert value $x = 3$ before $a_1$, so program will be $[3, 4, 6, 7]$ and time of lamp being lit equals $(3 - 0) + (6 - 4) + (10 - 7) = 8$. Other possible solution is to insert $x = 5$ in appropriate place.
In the second example, there is only one optimal solution: to insert $x = 2$ between $a_1$ and $a_2$. Program will become $[1, 2, 10]$, and answer will be $(1 - 0) + (10 - 2) = 9$.
In the third example, optimal answer is to leave program untouched, so answer will be $(3 - 0) + (7 - 4) = 6$. | 0 | [
{
"input": "3 10\n4 6 7",
"output": "8"
},
{
"input": "2 12\n1 10",
"output": "9"
},
{
"input": "2 7\n3 4",
"output": "6"
},
{
"input": "1 2\n1",
"output": "1"
},
{
"input": "5 10\n1 3 5 6 8",
"output": "6"
},
{
"input": "7 1000000000\n1 10001 10011 20011 20021 40021 40031",
"output": "999999969"
},
{
"input": "7 1000000000\n3 10001 10011 20011 20021 40021 40031",
"output": "999999969"
},
{
"input": "1 10\n1",
"output": "9"
},
{
"input": "1 10000000\n1",
"output": "9999999"
},
{
"input": "1 8\n1",
"output": "7"
},
{
"input": "7 17\n1 5 9 10 11 14 16",
"output": "9"
},
{
"input": "4 17\n1 5 9 10",
"output": "12"
},
{
"input": "5 12\n1 2 3 4 5",
"output": "9"
},
{
"input": "2 1000000000\n100 111",
"output": "999999989"
},
{
"input": "3 90591\n90579 90580 90581",
"output": "90589"
},
{
"input": "5 16\n1 2 3 4 5",
"output": "13"
},
{
"input": "6 981231233\n1 6 15 30 130 1875",
"output": "981229468"
},
{
"input": "1 3\n1",
"output": "2"
},
{
"input": "3 12\n4 7 11",
"output": "8"
},
{
"input": "1 100000\n1",
"output": "99999"
},
{
"input": "2 5\n1 3",
"output": "3"
},
{
"input": "1 1000000000\n1",
"output": "999999999"
},
{
"input": "1 1000000000\n999999999",
"output": "999999999"
},
{
"input": "3 4\n1 2 3",
"output": "2"
},
{
"input": "3 5\n1 2 3",
"output": "3"
},
{
"input": "1 4\n2",
"output": "3"
},
{
"input": "3 5\n1 3 4",
"output": "3"
},
{
"input": "9 20\n5 9 11 12 14 15 16 17 19",
"output": "12"
},
{
"input": "2 4\n1 2",
"output": "3"
},
{
"input": "1 10\n2",
"output": "9"
},
{
"input": "2 6\n2 4",
"output": "4"
},
{
"input": "2 4\n1 3",
"output": "2"
},
{
"input": "3 6\n1 2 4",
"output": "4"
},
{
"input": "7 9863\n65 96 97 98 101 112 1115",
"output": "9819"
}
] | 1,666,165,964 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 19 | 1,000 | 17,305,600 | n,M=map(int,input().split())
p=list(map(int,input().split()))
p=[0]+p+[M]
k=[]
g=[]
for i in range(0,len(p),2):
if i+1<len(p):
k.append(p[i+1]-p[i])
for i in range(1,len(p),2):
if i+1<len(p):
g.append(p[i+1]-p[i])
t0=sum(k)
T=[t0]
for i in range(2,n+2,2):
if p[i-1]+1!=p[i]:
t=p[i]-p[i-1]-1
t+=sum(g[i//2:])+sum(k[0:i//2])
T.append(t)
print(max(T)) | Title: Light It Up
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, you bought a brand new smart lamp with programming features. At first, you set up a schedule to the lamp. Every day it will turn power on at moment $0$ and turn power off at moment $M$. Moreover, the lamp allows you to set a program of switching its state (states are "lights on" and "lights off"). Unfortunately, some program is already installed into the lamp.
The lamp allows only good programs. Good program can be represented as a non-empty array $a$, where $0 < a_1 < a_2 < \dots < a_{|a|} < M$. All $a_i$ must be integers. Of course, preinstalled program is a good program.
The lamp follows program $a$ in next manner: at moment $0$ turns power and light on. Then at moment $a_i$ the lamp flips its state to opposite (if it was lit, it turns off, and vice versa). The state of the lamp flips instantly: for example, if you turn the light off at moment $1$ and then do nothing, the total time when the lamp is lit will be $1$. Finally, at moment $M$ the lamp is turning its power off regardless of its state.
Since you are not among those people who read instructions, and you don't understand the language it's written in, you realize (after some testing) the only possible way to alter the preinstalled program. You can insert at most one element into the program $a$, so it still should be a good program after alteration. Insertion can be done between any pair of consecutive elements of $a$, or even at the begining or at the end of $a$.
Find such a way to alter the program that the total time when the lamp is lit is maximum possible. Maybe you should leave program untouched. If the lamp is lit from $x$ till moment $y$, then its lit for $y - x$ units of time. Segments of time when the lamp is lit are summed up.
Input Specification:
First line contains two space separated integers $n$ and $M$ ($1 \le n \le 10^5$, $2 \le M \le 10^9$) — the length of program $a$ and the moment when power turns off.
Second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($0 < a_1 < a_2 < \dots < a_n < M$) — initially installed program $a$.
Output Specification:
Print the only integer — maximum possible total time when the lamp is lit.
Demo Input:
['3 10\n4 6 7\n', '2 12\n1 10\n', '2 7\n3 4\n']
Demo Output:
['8\n', '9\n', '6\n']
Note:
In the first example, one of possible optimal solutions is to insert value $x = 3$ before $a_1$, so program will be $[3, 4, 6, 7]$ and time of lamp being lit equals $(3 - 0) + (6 - 4) + (10 - 7) = 8$. Other possible solution is to insert $x = 5$ in appropriate place.
In the second example, there is only one optimal solution: to insert $x = 2$ between $a_1$ and $a_2$. Program will become $[1, 2, 10]$, and answer will be $(1 - 0) + (10 - 2) = 9$.
In the third example, optimal answer is to leave program untouched, so answer will be $(3 - 0) + (7 - 4) = 6$. | ```python
n,M=map(int,input().split())
p=list(map(int,input().split()))
p=[0]+p+[M]
k=[]
g=[]
for i in range(0,len(p),2):
if i+1<len(p):
k.append(p[i+1]-p[i])
for i in range(1,len(p),2):
if i+1<len(p):
g.append(p[i+1]-p[i])
t0=sum(k)
T=[t0]
for i in range(2,n+2,2):
if p[i-1]+1!=p[i]:
t=p[i]-p[i-1]-1
t+=sum(g[i//2:])+sum(k[0:i//2])
T.append(t)
print(max(T))
``` | 0 | |
11 | A | Increasing Sequence | PROGRAMMING | 900 | [
"constructive algorithms",
"implementation",
"math"
] | A. Increasing Sequence | 1 | 64 | A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*.
You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing? | The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106). | Output the minimal number of moves needed to make the sequence increasing. | [
"4 2\n1 3 3 2\n"
] | [
"3\n"
] | none | 0 | [
{
"input": "4 2\n1 3 3 2",
"output": "3"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "2 1\n2 5",
"output": "0"
},
{
"input": "2 1\n1 2",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "2 7\n10 20",
"output": "0"
},
{
"input": "2 7\n1 1",
"output": "1"
},
{
"input": "3 3\n18 1 9",
"output": "10"
},
{
"input": "3 3\n15 17 9",
"output": "3"
},
{
"input": "3 3\n10 9 12",
"output": "2"
},
{
"input": "10 3\n2 1 17 10 5 16 8 4 15 17",
"output": "31"
},
{
"input": "10 3\n6 11 4 12 22 15 23 26 24 26",
"output": "13"
},
{
"input": "10 3\n10 24 13 15 18 14 15 26 33 35",
"output": "29"
},
{
"input": "100 3\n529 178 280 403 326 531 671 427 188 866 669 646 421 804 494 609 53 1012 211 243 887 833 900 543 226 42 859 718 454 372 971 692 846 770 511 395 499 479 641 756 115 269 206 45 1039 727 400 779 859 614 146 214 196 919 702 959 380 830 535 878 859 784 316 305 782 924 536 243 236 978 564 150 291 877 808 983 537 839 490 120 168 838 267 650 900 170 211 504 326 771 895 984 994 483 776 100 471 1078 317 580",
"output": "15717"
},
{
"input": "100 3\n329 226 331 909 962 112 837 1005 194 818 506 416 125 648 367 459 400 582 989 547 329 438 234 121 272 226 821 376 834 427 718 164 834 113 654 177 737 212 169 696 744 180 89 944 233 147 667 990 809 1072 1085 1093 814 265 1067 312 833 572 303 901 1032 504 185 817 389 158 613 723 239 269 911 352 769 404 225 822 897 606 947 323 913 804 923 1084 552 901 486 249 209 898 847 610 728 1122 986 669 1116 1076 367 327",
"output": "16133"
}
] | 1,643,427,621 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 15 | 1,000 | 0 | info = input().split()
listlen = int(info[0])
addnum = int(info[1])
addtimes = 0
list = [int(x) for x in input().split()]
for i in range(1, listlen):
while list[i] <= list[i-1]:
list[i] += addnum
addtimes += 1
print(addtimes)
| Title: Increasing Sequence
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
A sequence *a*0,<=*a*1,<=...,<=*a**t*<=-<=1 is called increasing if *a**i*<=-<=1<=<<=*a**i* for each *i*:<=0<=<<=*i*<=<<=*t*.
You are given a sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 and a positive integer *d*. In each move you may choose one element of the given sequence and add *d* to it. What is the least number of moves required to make the given sequence increasing?
Input Specification:
The first line of the input contains two integer numbers *n* and *d* (2<=≤<=*n*<=≤<=2000,<=1<=≤<=*d*<=≤<=106). The second line contains space separated sequence *b*0,<=*b*1,<=...,<=*b**n*<=-<=1 (1<=≤<=*b**i*<=≤<=106).
Output Specification:
Output the minimal number of moves needed to make the sequence increasing.
Demo Input:
['4 2\n1 3 3 2\n']
Demo Output:
['3\n']
Note:
none | ```python
info = input().split()
listlen = int(info[0])
addnum = int(info[1])
addtimes = 0
list = [int(x) for x in input().split()]
for i in range(1, listlen):
while list[i] <= list[i-1]:
list[i] += addnum
addtimes += 1
print(addtimes)
``` | 0 |
0 | none | none | none | 0 | [
"none"
] | null | null | Little Nastya has a hobby, she likes to remove some letters from word, to obtain another word. But it turns out to be pretty hard for her, because she is too young. Therefore, her brother Sergey always helps her.
Sergey gives Nastya the word *t* and wants to get the word *p* out of it. Nastya removes letters in a certain order (one after another, in this order strictly), which is specified by permutation of letters' indices of the word *t*: *a*1... *a*|*t*|. We denote the length of word *x* as |*x*|. Note that after removing one letter, the indices of other letters don't change. For example, if *t*<==<="nastya" and *a*<==<=[4,<=1,<=5,<=3,<=2,<=6] then removals make the following sequence of words "nastya" "nastya" "nastya" "nastya" "nastya" "nastya" "nastya".
Sergey knows this permutation. His goal is to stop his sister at some point and continue removing by himself to get the word *p*. Since Nastya likes this activity, Sergey wants to stop her as late as possible. Your task is to determine, how many letters Nastya can remove before she will be stopped by Sergey.
It is guaranteed that the word *p* can be obtained by removing the letters from word *t*. | The first and second lines of the input contain the words *t* and *p*, respectively. Words are composed of lowercase letters of the Latin alphabet (1<=≤<=|*p*|<=<<=|*t*|<=≤<=200<=000). It is guaranteed that the word *p* can be obtained by removing the letters from word *t*.
Next line contains a permutation *a*1,<=*a*2,<=...,<=*a*|*t*| of letter indices that specifies the order in which Nastya removes letters of *t* (1<=≤<=*a**i*<=≤<=|*t*|, all *a**i* are distinct). | Print a single integer number, the maximum number of letters that Nastya can remove. | [
"ababcba\nabb\n5 3 4 1 7 6 2\n",
"bbbabb\nbb\n1 6 3 4 2 5\n"
] | [
"3",
"4"
] | In the first sample test sequence of removing made by Nastya looks like this:
"ababcba" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "ababcba" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "ababcba" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "ababcba"
Nastya can not continue, because it is impossible to get word "abb" from word "ababcba".
So, Nastya will remove only three letters. | 0 | [
{
"input": "ababcba\nabb\n5 3 4 1 7 6 2",
"output": "3"
},
{
"input": "bbbabb\nbb\n1 6 3 4 2 5",
"output": "4"
},
{
"input": "cacaccccccacccc\ncacc\n10 9 14 5 1 7 15 3 6 12 4 8 11 13 2",
"output": "9"
},
{
"input": "aaaabaaabaabaaaaaaaa\naaaa\n18 5 4 6 13 9 1 3 7 8 16 10 12 19 17 15 14 11 20 2",
"output": "16"
},
{
"input": "aaaaaaaadbaaabbbbbddaaabdadbbbbbdbbabbbabaabdbbdababbbddddbdaabbddbbbbabbbbbabadaadabaaaadbbabbbaddb\naaaaaaaaaaaaaa\n61 52 5 43 53 81 7 96 6 9 34 78 79 12 8 63 22 76 18 46 41 56 3 20 57 21 75 73 100 94 35 69 32 4 70 95 88 44 68 10 71 98 23 89 36 62 28 51 24 30 74 55 27 80 38 48 93 1 19 84 13 11 86 60 87 33 39 29 83 91 67 72 54 2 17 85 82 14 15 90 64 50 99 26 66 65 31 49 40 45 77 37 25 42 97 47 58 92 59 16",
"output": "57"
}
] | 1,488,100,320 | 4,020 | Python 3 | RUNTIME_ERROR | PRETESTS | 0 | 218 | 45,670,400 | #Fahim Ahmed :: Dhaka Residential Model College
import sys
sys.setrecursionlimit(10000001); #Maximizing Recursion limit
import math
import os
import filecmp
import fileinput
import cmath
import cmd
import operator
import collections
# freopen("in.txt", "r", sys.stdin)
# freopen("out.txt","w",sys.stdout)
def freopen(f,option,stream):
oldf = open(f,option)
oldfd = oldf.fileno()
newfd = stream.fileno()
os.close(newfd)
os.dup2(oldfd, newfd)
#BITWISE operations
def check(i,k):
return bool(i & (1<<k));
def On(i,k):
return (i | (1<<k));
def Off(i,k):
if(check(i,k) == True):
v=1
else:
v=0
return (i-(v << k));
#gcd,lcm
def gcd(a,b):
while (b > 0):
a=a%b;
a,b=b,a;
return a;
def lcm(a,b):
return a* (b//gcd(a,b));
#Section SIV()
#notprime[i] = True means i is not a prime
notprime = [False]*(10000001);
primes=[];
def siv(n):
sq = int(math.sqrt(n));
notprime[0] = True;
notprime[1] = True;
for i in range(4,n+1,+2):
notprime[i] = True;
for i in range(3,sq+1, +2):
if (notprime[i]==False):
j=i*i;
while(j<=n):
notprime[j]=True;
j+=2*i;
primes.append(2);
for i in range(3, n+1, +2):
if(notprime[i] == False):
primes.append(i);
n=int();
t=int();
res=int();
squares=[];
m=0;
def prime_square_generator(n):
sq = int(math.sqrt(n));
siv(sq);
ln = len(primes);
for i in range(0, ln, +1):
squares.append(primes[i]*primes[i]);
def sq_free(current_index, product_b4, counter):
global n,m;
global res;
# print(current_index, product_b4,m, bool(product_b4>n));
if(product_b4 > n):
return;
if(current_index >= m):
if(counter % 2 == 0):
res+=(n//product_b4);
else:
res-=(n//product_b4);
return;
v= lcm(product_b4,squares[current_index]);
if(v < n):
sq_free(current_index+1, v, counter+1);
sq_free(current_index+1, product_b4, counter); #skipping the current prime_square
def main():
mp[_]=0;
t=input();
p=input();
pos=list(map(int,input()));
n = len(t);
for i in range(n):
mp[t[i]]+=1;
needed[_]=0;
lp = len(p);
for i in range(lp):
needed[p[i]]+=1;
for i in range(n):
if(mp[t[pos[i]-1]]-1 < needed[t[pos[i]-1]]):
print(i);
return;
if __name__=="__main__":
main(); | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Nastya has a hobby, she likes to remove some letters from word, to obtain another word. But it turns out to be pretty hard for her, because she is too young. Therefore, her brother Sergey always helps her.
Sergey gives Nastya the word *t* and wants to get the word *p* out of it. Nastya removes letters in a certain order (one after another, in this order strictly), which is specified by permutation of letters' indices of the word *t*: *a*1... *a*|*t*|. We denote the length of word *x* as |*x*|. Note that after removing one letter, the indices of other letters don't change. For example, if *t*<==<="nastya" and *a*<==<=[4,<=1,<=5,<=3,<=2,<=6] then removals make the following sequence of words "nastya" "nastya" "nastya" "nastya" "nastya" "nastya" "nastya".
Sergey knows this permutation. His goal is to stop his sister at some point and continue removing by himself to get the word *p*. Since Nastya likes this activity, Sergey wants to stop her as late as possible. Your task is to determine, how many letters Nastya can remove before she will be stopped by Sergey.
It is guaranteed that the word *p* can be obtained by removing the letters from word *t*.
Input Specification:
The first and second lines of the input contain the words *t* and *p*, respectively. Words are composed of lowercase letters of the Latin alphabet (1<=≤<=|*p*|<=<<=|*t*|<=≤<=200<=000). It is guaranteed that the word *p* can be obtained by removing the letters from word *t*.
Next line contains a permutation *a*1,<=*a*2,<=...,<=*a*|*t*| of letter indices that specifies the order in which Nastya removes letters of *t* (1<=≤<=*a**i*<=≤<=|*t*|, all *a**i* are distinct).
Output Specification:
Print a single integer number, the maximum number of letters that Nastya can remove.
Demo Input:
['ababcba\nabb\n5 3 4 1 7 6 2\n', 'bbbabb\nbb\n1 6 3 4 2 5\n']
Demo Output:
['3', '4']
Note:
In the first sample test sequence of removing made by Nastya looks like this:
"ababcba" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "ababcba" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "ababcba" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "ababcba"
Nastya can not continue, because it is impossible to get word "abb" from word "ababcba".
So, Nastya will remove only three letters. | ```python
#Fahim Ahmed :: Dhaka Residential Model College
import sys
sys.setrecursionlimit(10000001); #Maximizing Recursion limit
import math
import os
import filecmp
import fileinput
import cmath
import cmd
import operator
import collections
# freopen("in.txt", "r", sys.stdin)
# freopen("out.txt","w",sys.stdout)
def freopen(f,option,stream):
oldf = open(f,option)
oldfd = oldf.fileno()
newfd = stream.fileno()
os.close(newfd)
os.dup2(oldfd, newfd)
#BITWISE operations
def check(i,k):
return bool(i & (1<<k));
def On(i,k):
return (i | (1<<k));
def Off(i,k):
if(check(i,k) == True):
v=1
else:
v=0
return (i-(v << k));
#gcd,lcm
def gcd(a,b):
while (b > 0):
a=a%b;
a,b=b,a;
return a;
def lcm(a,b):
return a* (b//gcd(a,b));
#Section SIV()
#notprime[i] = True means i is not a prime
notprime = [False]*(10000001);
primes=[];
def siv(n):
sq = int(math.sqrt(n));
notprime[0] = True;
notprime[1] = True;
for i in range(4,n+1,+2):
notprime[i] = True;
for i in range(3,sq+1, +2):
if (notprime[i]==False):
j=i*i;
while(j<=n):
notprime[j]=True;
j+=2*i;
primes.append(2);
for i in range(3, n+1, +2):
if(notprime[i] == False):
primes.append(i);
n=int();
t=int();
res=int();
squares=[];
m=0;
def prime_square_generator(n):
sq = int(math.sqrt(n));
siv(sq);
ln = len(primes);
for i in range(0, ln, +1):
squares.append(primes[i]*primes[i]);
def sq_free(current_index, product_b4, counter):
global n,m;
global res;
# print(current_index, product_b4,m, bool(product_b4>n));
if(product_b4 > n):
return;
if(current_index >= m):
if(counter % 2 == 0):
res+=(n//product_b4);
else:
res-=(n//product_b4);
return;
v= lcm(product_b4,squares[current_index]);
if(v < n):
sq_free(current_index+1, v, counter+1);
sq_free(current_index+1, product_b4, counter); #skipping the current prime_square
def main():
mp[_]=0;
t=input();
p=input();
pos=list(map(int,input()));
n = len(t);
for i in range(n):
mp[t[i]]+=1;
needed[_]=0;
lp = len(p);
for i in range(lp):
needed[p[i]]+=1;
for i in range(n):
if(mp[t[pos[i]-1]]-1 < needed[t[pos[i]-1]]):
print(i);
return;
if __name__=="__main__":
main();
``` | -1 | |
805 | B | 3-palindrome | PROGRAMMING | 1,000 | [
"constructive algorithms"
] | null | null | In the beginning of the new year Keivan decided to reverse his name. He doesn't like palindromes, so he changed Naviek to Navick.
He is too selfish, so for a given *n* he wants to obtain a string of *n* characters, each of which is either 'a', 'b' or 'c', with no palindromes of length 3 appearing in the string as a substring. For example, the strings "abc" and "abca" suit him, while the string "aba" doesn't. He also want the number of letters 'c' in his string to be as little as possible. | The first line contains single integer *n* (1<=≤<=*n*<=≤<=2·105) — the length of the string. | Print the string that satisfies all the constraints.
If there are multiple answers, print any of them. | [
"2\n",
"3\n"
] | [
"aa\n",
"bba\n"
] | A palindrome is a sequence of characters which reads the same backward and forward. | 1,000 | [
{
"input": "2",
"output": "aa"
},
{
"input": "3",
"output": "aab"
},
{
"input": "38",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaa"
},
{
"input": "47",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab"
},
{
"input": "59",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab"
},
{
"input": "67",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab"
},
{
"input": "77",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabba"
},
{
"input": "89",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabba"
},
{
"input": "98",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaa"
},
{
"input": "109",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabba"
},
{
"input": "117",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabba"
},
{
"input": "1",
"output": "a"
},
{
"input": "200000",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "100000",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "143670",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "104217",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "17879",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "131809",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "140873",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "77859",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "153022",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "179227",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "182801",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "5188",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "86539",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "12802",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "120289",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "132866",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "133377",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "31775",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "160397",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "161415",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "163623",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "22942",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "134767",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "52257",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "64905",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "183758",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "60131",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "146883",
"output": "aabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaabbaab..."
},
{
"input": "6",
"output": "aabbaa"
},
{
"input": "4",
"output": "aabb"
}
] | 1,677,775,723 | 2,147,483,647 | Python 3 | OK | TESTS | 42 | 61 | 819,200 | n = int(input())
ans=['a']*n
for i in range(2,n,4):
ans[i] = 'b'
if i+1<n:
ans[i+1] = 'b'
print(''.join(ans)) | Title: 3-palindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In the beginning of the new year Keivan decided to reverse his name. He doesn't like palindromes, so he changed Naviek to Navick.
He is too selfish, so for a given *n* he wants to obtain a string of *n* characters, each of which is either 'a', 'b' or 'c', with no palindromes of length 3 appearing in the string as a substring. For example, the strings "abc" and "abca" suit him, while the string "aba" doesn't. He also want the number of letters 'c' in his string to be as little as possible.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=2·105) — the length of the string.
Output Specification:
Print the string that satisfies all the constraints.
If there are multiple answers, print any of them.
Demo Input:
['2\n', '3\n']
Demo Output:
['aa\n', 'bba\n']
Note:
A palindrome is a sequence of characters which reads the same backward and forward. | ```python
n = int(input())
ans=['a']*n
for i in range(2,n,4):
ans[i] = 'b'
if i+1<n:
ans[i+1] = 'b'
print(''.join(ans))
``` | 3 | |
513 | A | Game | PROGRAMMING | 800 | [
"constructive algorithms",
"math"
] | null | null | Two players play a simple game. Each player is provided with a box with balls. First player's box contains exactly *n*1 balls and second player's box contains exactly *n*2 balls. In one move first player can take from 1 to *k*1 balls from his box and throw them away. Similarly, the second player can take from 1 to *k*2 balls from his box in his move. Players alternate turns and the first player starts the game. The one who can't make a move loses. Your task is to determine who wins if both players play optimally. | The first line contains four integers *n*1,<=*n*2,<=*k*1,<=*k*2. All numbers in the input are from 1 to 50.
This problem doesn't have subproblems. You will get 3 points for the correct submission. | Output "First" if the first player wins and "Second" otherwise. | [
"2 2 1 2\n",
"2 1 1 1\n"
] | [
"Second\n",
"First\n"
] | Consider the first sample test. Each player has a box with 2 balls. The first player draws a single ball from his box in one move and the second player can either take 1 or 2 balls from his box in one move. No matter how the first player acts, the second player can always win if he plays wisely. | 3 | [
{
"input": "2 2 1 2",
"output": "Second"
},
{
"input": "2 1 1 1",
"output": "First"
},
{
"input": "5 7 4 1",
"output": "Second"
},
{
"input": "5 7 1 4",
"output": "Second"
},
{
"input": "5 7 10 10",
"output": "Second"
},
{
"input": "5 7 1 10",
"output": "Second"
},
{
"input": "1 1 1 1",
"output": "Second"
},
{
"input": "50 50 50 50",
"output": "Second"
},
{
"input": "50 49 1 2",
"output": "First"
},
{
"input": "50 48 3 1",
"output": "First"
},
{
"input": "48 50 12 11",
"output": "Second"
},
{
"input": "49 50 11 12",
"output": "Second"
},
{
"input": "49 49 4 1",
"output": "Second"
},
{
"input": "49 49 3 3",
"output": "Second"
},
{
"input": "1 50 1 50",
"output": "Second"
},
{
"input": "1 50 50 50",
"output": "Second"
},
{
"input": "50 1 1 1",
"output": "First"
},
{
"input": "50 1 1 50",
"output": "First"
},
{
"input": "32 31 10 9",
"output": "First"
},
{
"input": "32 4 17 3",
"output": "First"
}
] | 1,480,774,271 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 62 | 0 | n1, n2, k1, k2 = [int(i) for i in input().split()]
print('Second' if n1 <= n2 else 'First') | Title: Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players play a simple game. Each player is provided with a box with balls. First player's box contains exactly *n*1 balls and second player's box contains exactly *n*2 balls. In one move first player can take from 1 to *k*1 balls from his box and throw them away. Similarly, the second player can take from 1 to *k*2 balls from his box in his move. Players alternate turns and the first player starts the game. The one who can't make a move loses. Your task is to determine who wins if both players play optimally.
Input Specification:
The first line contains four integers *n*1,<=*n*2,<=*k*1,<=*k*2. All numbers in the input are from 1 to 50.
This problem doesn't have subproblems. You will get 3 points for the correct submission.
Output Specification:
Output "First" if the first player wins and "Second" otherwise.
Demo Input:
['2 2 1 2\n', '2 1 1 1\n']
Demo Output:
['Second\n', 'First\n']
Note:
Consider the first sample test. Each player has a box with 2 balls. The first player draws a single ball from his box in one move and the second player can either take 1 or 2 balls from his box in one move. No matter how the first player acts, the second player can always win if he plays wisely. | ```python
n1, n2, k1, k2 = [int(i) for i in input().split()]
print('Second' if n1 <= n2 else 'First')
``` | 3 | |
488 | A | Giga Tower | PROGRAMMING | 1,100 | [
"brute force"
] | null | null | Giga Tower is the tallest and deepest building in Cyberland. There are 17<=777<=777<=777 floors, numbered from <=-<=8<=888<=888<=888 to 8<=888<=888<=888. In particular, there is floor 0 between floor <=-<=1 and floor 1. Every day, thousands of tourists come to this place to enjoy the wonderful view.
In Cyberland, it is believed that the number "8" is a lucky number (that's why Giga Tower has 8<=888<=888<=888 floors above the ground), and, an integer is lucky, if and only if its decimal notation contains at least one digit "8". For example, 8,<=<=-<=180,<=808 are all lucky while 42,<=<=-<=10 are not. In the Giga Tower, if you write code at a floor with lucky floor number, good luck will always be with you (Well, this round is #278, also lucky, huh?).
Tourist Henry goes to the tower to seek good luck. Now he is at the floor numbered *a*. He wants to find the minimum positive integer *b*, such that, if he walks *b* floors higher, he will arrive at a floor with a lucky number. | The only line of input contains an integer *a* (<=-<=109<=≤<=*a*<=≤<=109). | Print the minimum *b* in a line. | [
"179\n",
"-1\n",
"18\n"
] | [
"1\n",
"9\n",
"10\n"
] | For the first sample, he has to arrive at the floor numbered 180.
For the second sample, he will arrive at 8.
Note that *b* should be positive, so the answer for the third sample is 10, not 0. | 500 | [
{
"input": "179",
"output": "1"
},
{
"input": "-1",
"output": "9"
},
{
"input": "18",
"output": "10"
},
{
"input": "-410058385",
"output": "1"
},
{
"input": "-586825624",
"output": "1"
},
{
"input": "852318890",
"output": "1"
},
{
"input": "919067153",
"output": "5"
},
{
"input": "690422411",
"output": "7"
},
{
"input": "-408490162",
"output": "1"
},
{
"input": "-8",
"output": "16"
},
{
"input": "-6",
"output": "14"
},
{
"input": "-4",
"output": "12"
},
{
"input": "-2",
"output": "10"
},
{
"input": "0",
"output": "8"
},
{
"input": "2",
"output": "6"
},
{
"input": "4",
"output": "4"
},
{
"input": "6",
"output": "2"
},
{
"input": "8",
"output": "10"
},
{
"input": "1000000000",
"output": "8"
},
{
"input": "-1000000000",
"output": "2"
},
{
"input": "88888",
"output": "1"
},
{
"input": "89",
"output": "9"
},
{
"input": "-80000000",
"output": "2"
},
{
"input": "-8888",
"output": "1"
},
{
"input": "-17",
"output": "9"
},
{
"input": "78",
"output": "2"
},
{
"input": "-19",
"output": "1"
},
{
"input": "-999999998",
"output": "9"
},
{
"input": "-999999997",
"output": "8"
},
{
"input": "999999997",
"output": "1"
},
{
"input": "811111111",
"output": "1"
},
{
"input": "-8",
"output": "16"
},
{
"input": "-5",
"output": "13"
},
{
"input": "-7",
"output": "15"
},
{
"input": "1000000000",
"output": "8"
}
] | 1,664,377,404 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 46 | 0 | n=int(input())
c=0
while True:
n+=1
c+=1
if '8' in str(n):print(c);break
| Title: Giga Tower
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Giga Tower is the tallest and deepest building in Cyberland. There are 17<=777<=777<=777 floors, numbered from <=-<=8<=888<=888<=888 to 8<=888<=888<=888. In particular, there is floor 0 between floor <=-<=1 and floor 1. Every day, thousands of tourists come to this place to enjoy the wonderful view.
In Cyberland, it is believed that the number "8" is a lucky number (that's why Giga Tower has 8<=888<=888<=888 floors above the ground), and, an integer is lucky, if and only if its decimal notation contains at least one digit "8". For example, 8,<=<=-<=180,<=808 are all lucky while 42,<=<=-<=10 are not. In the Giga Tower, if you write code at a floor with lucky floor number, good luck will always be with you (Well, this round is #278, also lucky, huh?).
Tourist Henry goes to the tower to seek good luck. Now he is at the floor numbered *a*. He wants to find the minimum positive integer *b*, such that, if he walks *b* floors higher, he will arrive at a floor with a lucky number.
Input Specification:
The only line of input contains an integer *a* (<=-<=109<=≤<=*a*<=≤<=109).
Output Specification:
Print the minimum *b* in a line.
Demo Input:
['179\n', '-1\n', '18\n']
Demo Output:
['1\n', '9\n', '10\n']
Note:
For the first sample, he has to arrive at the floor numbered 180.
For the second sample, he will arrive at 8.
Note that *b* should be positive, so the answer for the third sample is 10, not 0. | ```python
n=int(input())
c=0
while True:
n+=1
c+=1
if '8' in str(n):print(c);break
``` | 3 | |
200 | B | Drinks | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink. | The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space. | Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4. | [
"3\n50 50 100\n",
"4\n0 25 50 75\n"
] | [
"66.666666666667\n",
"37.500000000000\n"
] | Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent. | 500 | [
{
"input": "3\n50 50 100",
"output": "66.666666666667"
},
{
"input": "4\n0 25 50 75",
"output": "37.500000000000"
},
{
"input": "3\n0 1 8",
"output": "3.000000000000"
},
{
"input": "5\n96 89 93 95 70",
"output": "88.600000000000"
},
{
"input": "7\n62 41 78 4 38 39 75",
"output": "48.142857142857"
},
{
"input": "13\n2 22 7 0 1 17 3 17 11 2 21 26 22",
"output": "11.615384615385"
},
{
"input": "21\n5 4 11 7 0 5 45 21 0 14 51 6 0 16 10 19 8 9 7 12 18",
"output": "12.761904761905"
},
{
"input": "26\n95 70 93 74 94 70 91 70 39 79 80 57 87 75 37 93 48 67 51 90 85 26 23 64 66 84",
"output": "69.538461538462"
},
{
"input": "29\n84 99 72 96 83 92 95 98 97 93 76 84 99 93 81 76 93 99 99 100 95 100 96 95 97 100 71 98 94",
"output": "91.551724137931"
},
{
"input": "33\n100 99 100 100 99 99 99 100 100 100 99 99 99 100 100 100 100 99 100 99 100 100 97 100 100 100 100 100 100 100 98 98 100",
"output": "99.515151515152"
},
{
"input": "34\n14 9 10 5 4 26 18 23 0 1 0 20 18 15 2 2 3 5 14 1 9 4 2 15 7 1 7 19 10 0 0 11 0 2",
"output": "8.147058823529"
},
{
"input": "38\n99 98 100 100 99 92 99 99 98 84 88 94 86 99 93 100 98 99 65 98 85 84 64 97 96 89 79 96 91 84 99 93 72 96 94 97 96 93",
"output": "91.921052631579"
},
{
"input": "52\n100 94 99 98 99 99 99 95 97 97 98 100 100 98 97 100 98 90 100 99 97 94 90 98 100 100 90 99 100 95 98 95 94 85 97 94 96 94 99 99 99 98 100 100 94 99 99 100 98 87 100 100",
"output": "97.019230769231"
},
{
"input": "58\n10 70 12 89 1 82 100 53 40 100 21 69 92 91 67 66 99 77 25 48 8 63 93 39 46 79 82 14 44 42 1 79 0 69 56 73 67 17 59 4 65 80 20 60 77 52 3 61 16 76 33 18 46 100 28 59 9 6",
"output": "50.965517241379"
},
{
"input": "85\n7 8 1 16 0 15 1 7 0 11 15 6 2 12 2 8 9 8 2 0 3 7 15 7 1 8 5 7 2 26 0 3 11 1 8 10 31 0 7 6 1 8 1 0 9 14 4 8 7 16 9 1 0 16 10 9 6 1 1 4 2 7 4 5 4 1 20 6 16 16 1 1 10 17 8 12 14 19 3 8 1 7 10 23 10",
"output": "7.505882352941"
},
{
"input": "74\n5 3 0 7 13 10 12 10 18 5 0 18 2 13 7 17 2 7 5 2 40 19 0 2 2 3 0 45 4 20 0 4 2 8 1 19 3 9 17 1 15 0 16 1 9 4 0 9 32 2 6 18 11 18 1 15 16 12 7 19 5 3 9 28 26 8 3 10 33 29 4 13 28 6",
"output": "10.418918918919"
},
{
"input": "98\n42 9 21 11 9 11 22 12 52 20 10 6 56 9 26 27 1 29 29 14 38 17 41 21 7 45 15 5 29 4 51 20 6 8 34 17 13 53 30 45 0 10 16 41 4 5 6 4 14 2 31 6 0 11 13 3 3 43 13 36 51 0 7 16 28 23 8 36 30 22 8 54 21 45 39 4 50 15 1 30 17 8 18 10 2 20 16 50 6 68 15 6 38 7 28 8 29 41",
"output": "20.928571428571"
},
{
"input": "99\n60 65 40 63 57 44 30 84 3 10 39 53 40 45 72 20 76 11 61 32 4 26 97 55 14 57 86 96 34 69 52 22 26 79 31 4 21 35 82 47 81 28 72 70 93 84 40 4 69 39 83 58 30 7 32 73 74 12 92 23 61 88 9 58 70 32 75 40 63 71 46 55 39 36 14 97 32 16 95 41 28 20 85 40 5 50 50 50 75 6 10 64 38 19 77 91 50 72 96",
"output": "49.191919191919"
},
{
"input": "99\n100 88 40 30 81 80 91 98 69 73 88 96 79 58 14 100 87 84 52 91 83 88 72 83 99 35 54 80 46 79 52 72 85 32 99 39 79 79 45 83 88 50 75 75 50 59 65 75 97 63 92 58 89 46 93 80 89 33 69 86 99 99 66 85 72 74 79 98 85 95 46 63 77 97 49 81 89 39 70 76 68 91 90 56 31 93 51 87 73 95 74 69 87 95 57 68 49 95 92",
"output": "73.484848484848"
},
{
"input": "100\n18 15 17 0 3 3 0 4 1 8 2 22 7 21 5 0 0 8 3 16 1 0 2 9 9 3 10 8 17 20 5 4 8 12 2 3 1 1 3 2 23 0 1 0 5 7 4 0 1 3 3 4 25 2 2 14 8 4 9 3 0 11 0 3 12 3 14 16 7 7 14 1 17 9 0 35 42 12 3 1 25 9 3 8 5 3 2 8 22 14 11 6 3 9 6 8 7 7 4 6",
"output": "7.640000000000"
},
{
"input": "100\n88 77 65 87 100 63 91 96 92 89 77 95 76 80 84 83 100 71 85 98 26 54 74 78 69 59 96 86 88 91 95 26 52 88 64 70 84 81 76 84 94 82 100 66 97 98 43 94 59 94 100 80 98 73 69 83 94 70 74 79 91 31 62 88 69 55 62 97 40 64 62 83 87 85 50 90 69 72 67 49 100 51 69 96 81 90 83 91 86 34 79 69 100 66 97 98 47 97 74 100",
"output": "77.660000000000"
},
{
"input": "100\n91 92 90 91 98 84 85 96 83 98 99 87 94 70 87 75 86 90 89 88 82 83 91 94 88 86 90 99 100 98 97 75 95 99 95 100 91 92 76 93 95 97 88 93 95 81 96 89 88 100 98 87 90 96 100 99 58 90 96 77 92 82 100 100 93 93 98 99 79 88 97 95 98 66 96 83 96 100 99 92 98 98 92 93 100 97 98 100 98 97 100 100 94 90 99 100 98 79 80 81",
"output": "91.480000000000"
},
{
"input": "1\n0",
"output": "0.000000000000"
},
{
"input": "1\n100",
"output": "100.000000000000"
},
{
"input": "1\n78",
"output": "78.000000000000"
},
{
"input": "2\n0 100",
"output": "50.000000000000"
},
{
"input": "2\n100 100",
"output": "100.000000000000"
},
{
"input": "5\n0 0 0 0 1",
"output": "0.200000000000"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99",
"output": "99.990000000000"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1",
"output": "0.010000000000"
},
{
"input": "5\n100 100 100 100 100",
"output": "100.000000000000"
}
] | 1,691,163,881 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 92 | 0 | def calculate_orange_cocktail_fraction(n, fractions):
total_fraction = sum(fractions) / n
return total_fraction
n = int(input())
fractions = list(map(int, input().split()))
result = calculate_orange_cocktail_fraction(n, fractions)
print("{:.10f}".format(result))
| Title: Drinks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space.
Output Specification:
Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4.
Demo Input:
['3\n50 50 100\n', '4\n0 25 50 75\n']
Demo Output:
['66.666666666667\n', '37.500000000000\n']
Note:
Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent. | ```python
def calculate_orange_cocktail_fraction(n, fractions):
total_fraction = sum(fractions) / n
return total_fraction
n = int(input())
fractions = list(map(int, input().split()))
result = calculate_orange_cocktail_fraction(n, fractions)
print("{:.10f}".format(result))
``` | 3 | |
761 | B | Dasha and friends | PROGRAMMING | 1,300 | [
"brute force",
"implementation",
"math"
] | null | null | Running with barriers on the circle track is very popular in the country where Dasha lives, so no wonder that on her way to classes she saw the following situation:
The track is the circle with length *L*, in distinct points of which there are *n* barriers. Athlete always run the track in counterclockwise direction if you look on him from above. All barriers are located at integer distance from each other along the track.
Her friends the parrot Kefa and the leopard Sasha participated in competitions and each of them ran one lap. Each of the friends started from some integral point on the track. Both friends wrote the distance from their start along the track to each of the *n* barriers. Thus, each of them wrote *n* integers in the ascending order, each of them was between 0 and *L*<=-<=1, inclusively.
There are several tracks in the country, all of them have same length and same number of barriers, but the positions of the barriers can differ among different tracks. Now Dasha is interested if it is possible that Kefa and Sasha ran the same track or they participated on different tracks.
Write the program which will check that Kefa's and Sasha's tracks coincide (it means that one can be obtained from the other by changing the start position). Note that they always run the track in one direction — counterclockwise, if you look on a track from above. | The first line contains two integers *n* and *L* (1<=≤<=*n*<=≤<=50, *n*<=≤<=*L*<=≤<=100) — the number of barriers on a track and its length.
The second line contains *n* distinct integers in the ascending order — the distance from Kefa's start to each barrier in the order of its appearance. All integers are in the range from 0 to *L*<=-<=1 inclusively.
The second line contains *n* distinct integers in the ascending order — the distance from Sasha's start to each barrier in the order of its overcoming. All integers are in the range from 0 to *L*<=-<=1 inclusively. | Print "YES" (without quotes), if Kefa and Sasha ran the coinciding tracks (it means that the position of all barriers coincides, if they start running from the same points on the track). Otherwise print "NO" (without quotes). | [
"3 8\n2 4 6\n1 5 7\n",
"4 9\n2 3 5 8\n0 1 3 6\n",
"2 4\n1 3\n1 2\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | The first test is analyzed in the statement. | 1,000 | [
{
"input": "3 8\n2 4 6\n1 5 7",
"output": "YES"
},
{
"input": "4 9\n2 3 5 8\n0 1 3 6",
"output": "YES"
},
{
"input": "2 4\n1 3\n1 2",
"output": "NO"
},
{
"input": "5 9\n0 2 5 6 7\n1 3 6 7 8",
"output": "YES"
},
{
"input": "5 60\n7 26 27 40 59\n14 22 41 42 55",
"output": "YES"
},
{
"input": "20 29\n0 1 2 4 5 8 9 12 14 15 17 19 20 21 22 23 25 26 27 28\n0 2 4 5 6 7 8 10 11 12 13 14 15 16 18 19 22 23 26 28",
"output": "YES"
},
{
"input": "35 41\n0 1 2 3 4 5 6 7 9 10 11 12 13 14 18 19 20 21 22 23 24 25 26 28 30 31 32 33 34 35 36 37 38 39 40\n0 1 2 3 4 5 7 8 9 10 11 12 16 17 18 19 20 21 22 23 24 26 28 29 30 31 32 33 34 35 36 37 38 39 40",
"output": "YES"
},
{
"input": "40 63\n0 2 3 4 5 6 9 10 12 15 17 19 23 25 26 27 28 29 30 31 33 34 36 37 38 39 40 43 45 49 50 52 53 54 55 57 58 60 61 62\n1 2 3 4 5 8 10 14 15 17 18 19 20 22 23 25 26 27 28 30 31 32 33 34 37 38 40 43 46 47 51 53 54 55 56 57 58 59 61 62",
"output": "NO"
},
{
"input": "50 97\n1 2 3 4 6 9 10 11 12 13 14 21 22 23 24 25 28 29 30 31 32 33 34 36 37 40 41 45 53 56 59 64 65 69 70 71 72 73 74 77 81 84 85 86 87 89 91 92 95 96\n0 1 2 3 6 10 13 14 15 16 18 20 21 24 25 27 28 29 30 33 35 36 37 38 39 40 47 48 49 50 51 54 55 56 57 58 59 60 62 63 66 67 71 79 82 85 90 91 95 96",
"output": "NO"
},
{
"input": "50 100\n0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98\n1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99",
"output": "YES"
},
{
"input": "1 2\n0\n0",
"output": "YES"
},
{
"input": "1 2\n0\n1",
"output": "YES"
},
{
"input": "1 2\n1\n0",
"output": "YES"
},
{
"input": "1 2\n1\n1",
"output": "YES"
},
{
"input": "1 1\n0\n0",
"output": "YES"
},
{
"input": "5 12\n2 3 4 8 10\n2 3 4 8 10",
"output": "YES"
},
{
"input": "1 18\n3\n10",
"output": "YES"
},
{
"input": "1 75\n65\n8",
"output": "YES"
},
{
"input": "2 16\n4 13\n2 11",
"output": "YES"
},
{
"input": "2 95\n45 59\n3 84",
"output": "YES"
},
{
"input": "3 53\n29 43 50\n29 43 50",
"output": "YES"
},
{
"input": "3 60\n39 46 51\n43 50 55",
"output": "YES"
},
{
"input": "4 4\n0 1 2 3\n0 1 2 3",
"output": "YES"
},
{
"input": "4 93\n45 48 50 90\n20 68 71 73",
"output": "YES"
},
{
"input": "6 18\n0 3 8 11 15 16\n2 7 10 14 15 17",
"output": "YES"
},
{
"input": "6 87\n0 1 21 31 34 66\n11 12 32 42 45 77",
"output": "YES"
},
{
"input": "7 26\n0 3 9 13 14 19 20\n4 7 13 17 18 23 24",
"output": "YES"
},
{
"input": "7 81\n0 12 19 24 25 35 59\n1 8 13 14 24 48 70",
"output": "YES"
},
{
"input": "8 20\n0 1 2 3 5 6 14 15\n1 2 10 11 16 17 18 19",
"output": "YES"
},
{
"input": "8 94\n0 8 11 27 38 54 57 89\n1 33 38 46 49 65 76 92",
"output": "YES"
},
{
"input": "9 18\n1 3 6 8 11 12 13 16 17\n0 2 5 6 7 10 11 13 15",
"output": "YES"
},
{
"input": "9 90\n10 11 27 33 34 55 63 84 87\n9 12 25 26 42 48 49 70 78",
"output": "YES"
},
{
"input": "10 42\n4 9 10 14 15 16 19 33 36 40\n0 14 17 21 27 32 33 37 38 39",
"output": "YES"
},
{
"input": "10 73\n4 5 15 19 20 25 28 42 57 58\n3 4 9 12 26 41 42 61 62 72",
"output": "YES"
},
{
"input": "11 11\n0 1 2 3 4 5 6 7 8 9 10\n0 1 2 3 4 5 6 7 8 9 10",
"output": "YES"
},
{
"input": "11 57\n1 4 27 30 31 35 37 41 50 52 56\n22 25 26 30 32 36 45 47 51 53 56",
"output": "YES"
},
{
"input": "12 73\n5 9 11 20 25 36 40 41 44 48 56 60\n12 16 18 27 32 43 47 48 51 55 63 67",
"output": "YES"
},
{
"input": "12 95\n1 37 42 46 56 58 59 62 64 71 76 80\n2 18 54 59 63 73 75 76 79 81 88 93",
"output": "YES"
},
{
"input": "13 29\n2 5 6 9 12 17 18 19 20 21 22 24 27\n0 3 6 11 12 13 14 15 16 18 21 25 28",
"output": "YES"
},
{
"input": "13 90\n9 18 23 30 31 36 39 44 58 59 74 82 87\n1 6 18 27 32 39 40 45 48 53 67 68 83",
"output": "YES"
},
{
"input": "14 29\n1 2 3 4 5 7 9 12 13 20 21 22 23 24\n0 3 4 11 12 13 14 15 21 22 23 24 25 27",
"output": "YES"
},
{
"input": "14 94\n7 8 9 21 34 35 36 37 38 43 46 52 84 93\n2 3 4 16 29 30 31 32 33 38 41 47 79 88",
"output": "YES"
},
{
"input": "15 19\n1 2 3 4 5 6 7 8 9 10 11 13 14 16 17\n0 1 2 3 4 5 6 7 8 9 10 12 13 15 16",
"output": "YES"
},
{
"input": "15 27\n2 3 4 5 6 7 8 9 10 11 12 14 17 24 26\n2 3 4 5 6 7 8 9 10 11 12 14 17 24 26",
"output": "YES"
},
{
"input": "16 28\n3 5 6 7 9 10 11 12 13 14 17 19 20 25 26 27\n0 5 6 7 11 13 14 15 17 18 19 20 21 22 25 27",
"output": "YES"
},
{
"input": "16 93\n5 6 10 11 13 14 41 43 46 61 63 70 74 79 83 92\n0 9 15 16 20 21 23 24 51 53 56 71 73 80 84 89",
"output": "YES"
},
{
"input": "17 49\n2 5 11 12 16 18 19 21 22 24 36 37 38 39 40 44 47\n1 7 8 12 14 15 17 18 20 32 33 34 35 36 40 43 47",
"output": "YES"
},
{
"input": "17 86\n16 17 25 33 39 41 50 51 54 56 66 70 72 73 77 80 85\n3 9 11 20 21 24 26 36 40 42 43 47 50 55 72 73 81",
"output": "YES"
},
{
"input": "18 20\n0 1 2 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19\n0 1 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19",
"output": "YES"
},
{
"input": "18 82\n0 5 10 13 14 16 21 28 29 30 44 46 61 64 69 71 77 78\n0 5 8 9 11 16 23 24 25 39 41 56 59 64 66 72 73 77",
"output": "YES"
},
{
"input": "19 25\n0 1 2 3 5 7 9 10 12 13 16 17 18 19 20 21 22 23 24\n0 3 4 5 6 7 8 9 10 11 12 13 14 15 17 19 21 22 24",
"output": "YES"
},
{
"input": "19 91\n5 17 18 20 22 25 26 31 32 33 43 47 54 61 62 64 77 80 87\n4 5 6 16 20 27 34 35 37 50 53 60 69 81 82 84 86 89 90",
"output": "YES"
},
{
"input": "20 53\n2 6 8 9 16 17 20 21 22 23 25 26 35 36 38 39 44 46 47 50\n4 5 8 9 10 11 13 14 23 24 26 27 32 34 35 38 43 47 49 50",
"output": "YES"
},
{
"input": "21 44\n0 1 3 4 6 7 8 9 10 11 12 15 17 18 21 22 27 29 34 36 42\n1 7 9 10 12 13 15 16 17 18 19 20 21 24 26 27 30 31 36 38 43",
"output": "YES"
},
{
"input": "21 94\n3 5 6 8 9 15 16 20 28 31 35 39 49 50 53 61 71 82 85 89 90\n6 17 20 24 25 32 34 35 37 38 44 45 49 57 60 64 68 78 79 82 90",
"output": "YES"
},
{
"input": "22 24\n0 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 22 23\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 21 22 23",
"output": "YES"
},
{
"input": "22 85\n3 5 7 14 18 21 25 32 38 41 53 58 61 62 66 70 71 73 75 76 79 83\n3 6 18 23 26 27 31 35 36 38 40 41 44 48 53 55 57 64 68 71 75 82",
"output": "YES"
},
{
"input": "23 38\n0 2 4 5 7 8 12 13 14 16 17 18 21 22 24 27 28 30 31 32 35 36 37\n0 1 2 3 5 7 8 10 11 15 16 17 19 20 21 24 25 27 30 31 33 34 35",
"output": "YES"
},
{
"input": "23 93\n1 3 5 10 19 22 26 27 30 35 39 53 55 60 66 67 75 76 77 80 82 89 90\n9 11 16 22 23 31 32 33 36 38 45 46 50 52 54 59 68 71 75 76 79 84 88",
"output": "YES"
},
{
"input": "24 37\n1 4 5 6 8 11 12 13 15 16 17 19 20 21 23 26 27 28 30 31 33 34 35 36\n0 3 4 5 7 8 10 11 12 13 15 18 19 20 22 25 26 27 29 30 31 33 34 35",
"output": "YES"
},
{
"input": "24 94\n9 10 13 14 16 18 19 22 24 29 32 35 48 55 57 63 64 69 72 77 78 85 90 92\n1 7 8 13 16 21 22 29 34 36 47 48 51 52 54 56 57 60 62 67 70 73 86 93",
"output": "YES"
},
{
"input": "25 45\n0 1 2 4 6 7 8 9 13 14 17 19 21 22 23 25 28 29 30 31 34 36 38 39 42\n1 3 4 5 7 10 11 12 13 16 18 20 21 24 27 28 29 31 33 34 35 36 40 41 44",
"output": "YES"
},
{
"input": "25 72\n1 2 6 8 9 11 15 18 19 20 26 29 31 33 34 40 41 43 45 48 58 60 68 69 71\n0 6 9 11 13 14 20 21 23 25 28 38 40 48 49 51 53 54 58 60 61 63 67 70 71",
"output": "YES"
},
{
"input": "26 47\n0 2 5 7 8 9 10 12 13 14 20 22 23 25 27 29 31 32 33 35 36 37 38 42 44 45\n0 2 4 6 8 9 10 12 13 14 15 19 21 22 24 26 29 31 32 33 34 36 37 38 44 46",
"output": "YES"
},
{
"input": "26 99\n0 1 13 20 21 22 25 26 27 28 32 39 44 47 56 58 60 62 71 81 83 87 89 93 94 98\n6 8 12 14 18 19 23 24 25 37 44 45 46 49 50 51 52 56 63 68 71 80 82 84 86 95",
"output": "YES"
},
{
"input": "27 35\n0 2 3 4 5 6 7 8 10 11 12 13 14 15 16 17 19 20 21 23 26 27 29 30 31 32 33\n0 1 2 3 5 7 8 9 10 11 12 13 15 16 17 18 19 20 21 22 24 25 26 28 31 32 34",
"output": "YES"
},
{
"input": "27 51\n1 2 4 7 8 11 13 17 20 21 23 24 25 28 29 30 34 35 37 38 40 43 45 46 47 48 50\n0 1 2 4 6 7 9 12 13 16 18 22 25 26 28 29 30 33 34 35 39 40 42 43 45 48 50",
"output": "YES"
},
{
"input": "28 38\n1 4 5 7 8 9 10 11 12 14 15 16 18 19 20 21 22 23 24 25 28 29 30 32 33 35 36 37\n0 1 2 3 4 5 6 9 10 11 13 14 16 17 18 20 23 24 26 27 28 29 30 31 33 34 35 37",
"output": "YES"
},
{
"input": "28 67\n0 1 2 3 6 9 10 15 18 22 24 25 30 35 36 38 39 47 48 49 51 53 55 56 58 62 63 64\n4 7 11 13 14 19 24 25 27 28 36 37 38 40 42 44 45 47 51 52 53 56 57 58 59 62 65 66",
"output": "YES"
},
{
"input": "29 29\n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28\n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28",
"output": "YES"
},
{
"input": "29 93\n1 2 11 13 18 21 27 28 30 38 41 42 46 54 55 56 60 61 63 64 66 69 71 72 77 81 83 89 90\n2 10 11 12 16 17 19 20 22 25 27 28 33 37 39 45 46 50 51 60 62 67 70 76 77 79 87 90 91",
"output": "YES"
},
{
"input": "30 63\n0 2 3 5 6 7 8 10 13 18 19 21 22 23 26 32 35 37 38 39 40 41 43 44 49 51 53 54 58 61\n0 2 3 5 6 7 8 10 13 18 19 21 22 23 26 32 35 37 38 39 40 41 43 44 49 51 53 54 58 61",
"output": "YES"
},
{
"input": "30 91\n1 2 3 7 8 9 13 16 17 19 27 29 38 45 47 52 53 55 61 62 66 77 78 79 80 81 82 84 88 89\n3 4 5 9 12 13 15 23 25 34 41 43 48 49 51 57 58 62 73 74 75 76 77 78 80 84 85 88 89 90",
"output": "YES"
},
{
"input": "31 39\n0 1 2 3 4 5 6 7 8 10 11 13 14 17 18 20 21 23 24 25 27 28 29 30 31 33 34 35 36 37 38\n0 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 18 19 21 22 25 26 28 29 31 32 33 35 36 37 38",
"output": "YES"
},
{
"input": "31 95\n9 12 14 15 21 23 26 28 30 36 37 42 47 51 54 56 59 62 64 65 66 70 72 74 75 79 82 85 87 91 93\n0 2 3 7 10 13 15 19 21 32 35 37 38 44 46 49 51 53 59 60 65 70 74 77 79 82 85 87 88 89 93",
"output": "YES"
},
{
"input": "32 61\n0 2 3 5 7 10 13 14 15 18 19 20 21 22 23 24 26 32 33 34 36 38 43 46 47 51 54 55 56 57 58 59\n1 2 4 6 9 12 13 14 17 18 19 20 21 22 23 25 31 32 33 35 37 42 45 46 50 53 54 55 56 57 58 60",
"output": "YES"
},
{
"input": "32 86\n5 7 9 10 13 17 18 19 25 26 28 32 33 37 38 43 45 47 50 53 57 58 60 69 73 74 75 77 80 82 83 85\n7 11 12 13 15 18 20 21 23 29 31 33 34 37 41 42 43 49 50 52 56 57 61 62 67 69 71 74 77 81 82 84",
"output": "YES"
},
{
"input": "33 44\n0 1 2 3 5 9 10 11 12 13 14 15 17 18 20 21 22 23 24 25 26 27 28 30 31 32 35 36 38 39 41 42 43\n0 2 3 4 7 8 10 11 13 14 15 16 17 18 19 21 25 26 27 28 29 30 31 33 34 36 37 38 39 40 41 42 43",
"output": "YES"
},
{
"input": "33 73\n3 6 7 8 9 10 11 13 14 15 17 19 22 23 26 27 28 31 33 34 35 37 42 44 48 52 54 57 62 63 64 67 68\n2 3 4 7 8 16 19 20 21 22 23 24 26 27 28 30 32 35 36 39 40 41 44 46 47 48 50 55 57 61 65 67 70",
"output": "YES"
},
{
"input": "34 52\n1 2 3 4 5 6 8 9 10 12 13 14 15 16 17 19 21 24 26 27 28 29 31 33 35 36 37 39 40 45 46 49 50 51\n0 1 2 3 4 6 7 8 10 11 12 13 14 15 17 19 22 24 25 26 27 29 31 33 34 35 37 38 43 44 47 48 49 51",
"output": "YES"
},
{
"input": "34 68\n0 7 9 10 11 14 15 16 20 21 22 24 26 32 34 35 37 38 40 41 42 43 44 45 47 50 53 55 57 58 59 62 64 65\n0 1 2 3 5 8 11 13 15 16 17 20 22 23 26 33 35 36 37 40 41 42 46 47 48 50 52 58 60 61 63 64 66 67",
"output": "YES"
},
{
"input": "35 90\n4 5 7 8 10 11 12 13 14 22 27 29 31 33 34 38 46 49 52 53 54 55 56 57 60 61 64 69 77 81 83 86 87 88 89\n4 7 10 11 12 13 14 15 18 19 22 27 35 39 41 44 45 46 47 52 53 55 56 58 59 60 61 62 70 75 77 79 81 82 86",
"output": "YES"
},
{
"input": "36 43\n1 2 3 4 6 7 8 9 10 11 14 16 17 18 19 20 21 22 23 24 25 26 27 29 30 31 32 33 34 35 36 37 38 39 40 42\n0 1 2 3 4 5 6 8 9 10 11 12 13 14 15 16 17 18 19 21 23 24 25 26 28 29 30 31 32 33 36 38 39 40 41 42",
"output": "YES"
},
{
"input": "36 84\n1 3 6 13 15 16 17 18 19 21 23 26 29 33 38 40 42 45 49 50 53 54 57 58 60 61 64 65 67 70 73 76 78 79 81 83\n0 2 5 8 12 17 19 21 24 28 29 32 33 36 37 39 40 43 44 46 49 52 55 57 58 60 62 64 66 69 76 78 79 80 81 82",
"output": "YES"
},
{
"input": "37 46\n0 1 3 6 7 8 9 10 12 13 14 16 17 19 20 21 22 23 24 25 26 27 28 29 30 31 33 34 35 36 37 39 40 41 42 43 44\n0 3 4 5 6 7 9 10 11 13 14 16 17 18 19 20 21 22 23 24 25 26 27 28 30 31 32 33 34 36 37 38 39 40 41 43 44",
"output": "YES"
},
{
"input": "37 97\n0 5 10 11 12 15 16 18 19 25 28 29 34 35 36 37 38 40 46 47 48 49 55 58 60 61 62 64 65 70 76 77 80 82 88 94 96\n1 7 13 15 16 21 26 27 28 31 32 34 35 41 44 45 50 51 52 53 54 56 62 63 64 65 71 74 76 77 78 80 81 86 92 93 96",
"output": "YES"
},
{
"input": "38 58\n1 2 3 4 5 8 9 11 12 13 15 16 17 22 23 24 25 26 27 29 30 31 32 33 34 36 37 40 41 43 46 47 48 52 53 55 56 57\n1 2 3 5 6 7 8 9 12 13 15 16 17 19 20 21 26 27 28 29 30 31 33 34 35 36 37 38 40 41 44 45 47 50 51 52 56 57",
"output": "YES"
},
{
"input": "38 92\n1 2 3 5 6 7 12 14 15 16 17 18 20 22 29 31 33 34 38 41 43 49 54 55 57 58 61 63 66 67 69 73 75 76 82 85 88 90\n1 3 4 10 13 16 18 21 22 23 25 26 27 32 34 35 36 37 38 40 42 49 51 53 54 58 61 63 69 74 75 77 78 81 83 86 87 89",
"output": "YES"
},
{
"input": "39 59\n0 1 2 3 5 6 7 8 9 10 11 12 13 15 16 17 19 24 25 28 29 31 32 33 35 37 38 40 41 42 43 45 46 47 49 50 53 55 56\n0 1 3 4 5 6 8 9 10 12 13 16 18 19 22 23 24 25 27 28 29 30 31 32 33 34 35 37 38 39 41 46 47 50 51 53 54 55 57",
"output": "YES"
},
{
"input": "39 67\n1 3 5 7 8 16 18 20 21 23 24 25 27 28 29 31 32 34 36 38 40 43 44 46 47 48 49 50 52 53 54 55 58 59 61 62 63 64 66\n0 1 2 4 6 8 10 12 13 21 23 25 26 28 29 30 32 33 34 36 37 39 41 43 45 48 49 51 52 53 54 55 57 58 59 60 63 64 66",
"output": "YES"
},
{
"input": "40 63\n0 2 3 4 5 6 9 10 12 15 18 19 23 25 26 27 28 29 30 31 33 34 36 37 38 39 40 43 45 49 50 52 53 54 55 57 58 60 61 62\n1 2 3 4 5 8 10 14 15 17 18 19 20 22 23 25 26 27 28 30 31 32 33 34 37 38 40 43 46 47 51 53 54 55 56 57 58 59 61 62",
"output": "YES"
},
{
"input": "40 96\n5 11 12 13 14 16 17 18 19 24 30 31 32 33 37 42 46 50 53 54 55 58 60 61 64 67 68 69 70 72 75 76 77 81 84 85 89 91 92 93\n2 7 11 15 18 19 20 23 25 26 29 32 33 34 35 37 40 41 42 46 49 50 54 56 57 58 66 72 73 74 75 77 78 79 80 85 91 92 93 94",
"output": "YES"
},
{
"input": "41 67\n0 2 3 5 8 10 11 12 13 14 15 19 20 21 22 26 29 30 31 32 34 35 37 38 40 41 44 45 46 47 49 51 52 53 54 56 57 58 59 63 66\n2 3 4 5 9 12 13 14 15 17 18 20 21 23 24 27 28 29 30 32 34 35 36 37 39 40 41 42 46 49 50 52 53 55 58 60 61 62 63 64 65",
"output": "YES"
},
{
"input": "41 72\n0 3 4 6 7 8 9 12 13 14 16 21 23 24 25 26 27 29 31 32 33 34 35 38 40 41 45 47 49 50 51 52 56 57 58 59 61 62 65 66 69\n0 1 4 5 6 8 13 15 16 17 18 19 21 23 24 25 26 27 30 32 33 37 39 41 42 43 44 48 49 50 51 53 54 57 58 61 64 67 68 70 71",
"output": "YES"
},
{
"input": "42 48\n0 1 2 3 4 7 8 9 10 11 12 13 15 16 17 18 19 20 21 22 23 24 26 27 28 29 30 32 33 34 35 36 37 38 40 41 42 43 44 45 46 47\n0 1 2 3 4 5 6 8 9 10 11 12 14 15 16 17 18 19 20 22 23 24 25 26 27 28 29 30 31 32 33 34 37 38 39 40 41 42 43 45 46 47",
"output": "YES"
},
{
"input": "42 81\n0 1 3 6 7 8 11 13 17 18 19 21 22 24 29 30 31 32 34 35 38 44 46 48 49 50 51 52 53 55 59 61 62 63 65 66 67 69 70 72 77 80\n0 1 3 4 6 11 12 13 14 16 17 20 26 28 30 31 32 33 34 35 37 41 43 44 45 47 48 49 51 52 54 59 62 63 64 66 69 70 71 74 76 80",
"output": "YES"
},
{
"input": "43 55\n0 1 2 3 4 5 6 7 8 12 14 15 17 18 19 20 21 22 23 26 27 28 29 31 32 33 35 36 37 38 40 42 43 44 45 46 47 48 49 50 51 53 54\n1 2 4 5 6 7 8 9 10 13 14 15 16 18 19 20 22 23 24 25 27 29 30 31 32 33 34 35 36 37 38 40 41 42 43 44 45 46 47 48 49 50 54",
"output": "YES"
},
{
"input": "43 81\n2 3 4 5 6 7 9 10 12 13 18 19 20 21 23 26 27 29 30 32 34 38 39 43 46 47 48 50 51 52 54 55 58 62 64 67 69 70 71 72 73 75 80\n0 3 5 6 7 8 9 11 16 19 20 21 22 23 24 26 27 29 30 35 36 37 38 40 43 44 46 47 49 51 55 56 60 63 64 65 67 68 69 71 72 75 79",
"output": "YES"
},
{
"input": "44 54\n0 1 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 21 22 23 24 25 26 27 28 29 31 33 34 35 36 37 39 40 41 43 44 47 49 50 52 53\n0 1 2 3 4 5 6 7 8 10 12 13 14 15 16 18 19 20 22 23 26 28 29 31 32 33 34 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52",
"output": "YES"
},
{
"input": "44 93\n1 5 6 7 8 10 14 17 19 21 25 26 27 30 33 34 35 36 38 41 45 48 49 51 53 55 57 60 66 67 69 70 73 76 78 79 80 81 82 83 85 87 88 90\n0 2 4 8 9 10 13 16 17 18 19 21 24 28 31 32 34 36 38 40 43 49 50 52 53 56 59 61 62 63 64 65 66 68 70 71 73 77 81 82 83 84 86 90",
"output": "YES"
},
{
"input": "45 47\n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 43 44 45 46\n0 1 2 3 4 5 6 7 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 33 34 35 36 37 38 39 40 41 42 43 44 45 46",
"output": "YES"
},
{
"input": "45 71\n0 2 3 7 8 11 12 13 14 15 16 17 20 21 22 23 24 26 28 30 32 37 39 41 42 43 44 45 47 48 50 52 54 55 56 57 58 59 60 61 62 64 66 68 70\n0 1 2 3 4 7 8 9 10 11 13 15 17 19 24 26 28 29 30 31 32 34 35 37 39 41 42 43 44 45 46 47 48 49 51 53 55 57 58 60 61 65 66 69 70",
"output": "YES"
},
{
"input": "46 46\n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45\n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45",
"output": "YES"
},
{
"input": "46 93\n0 1 2 6 13 16 17 18 19 21 27 29 32 34 37 38 39 40 41 44 45 49 50 52 54 56 57 61 64 65 66 67 69 71 73 75 77 78 79 83 85 87 88 90 91 92\n0 2 4 5 7 8 9 10 11 12 16 23 26 27 28 29 31 37 39 42 44 47 48 49 50 51 54 55 59 60 62 64 66 67 71 74 75 76 77 79 81 83 85 87 88 89",
"output": "YES"
},
{
"input": "47 49\n0 1 2 3 4 5 6 7 9 10 11 12 13 14 15 16 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48\n0 1 2 3 4 5 6 7 8 9 10 11 13 14 15 16 17 18 19 20 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48",
"output": "YES"
},
{
"input": "47 94\n0 1 3 4 5 7 8 9 14 18 19 26 30 33 34 35 37 40 42 45 46 49 50 51 52 53 55 56 60 61 62 63 64 65 66 69 71 73 75 79 84 86 87 88 90 92 93\n1 2 3 4 6 7 8 10 11 12 17 21 22 29 33 36 37 38 40 43 45 48 49 52 53 54 55 56 58 59 63 64 65 66 67 68 69 72 74 76 78 82 87 89 90 91 93",
"output": "YES"
},
{
"input": "48 65\n0 1 2 4 5 6 7 8 9 10 11 12 15 16 17 20 22 24 25 26 27 28 30 32 33 34 35 37 38 39 44 45 46 47 48 50 51 52 53 54 55 56 57 58 59 61 62 63\n0 1 4 6 8 9 10 11 12 14 16 17 18 19 21 22 23 28 29 30 31 32 34 35 36 37 38 39 40 41 42 43 45 46 47 49 50 51 53 54 55 56 57 58 59 60 61 64",
"output": "YES"
},
{
"input": "48 90\n1 3 4 5 8 9 11 13 14 15 16 18 20 21 24 26 29 30 31 33 34 36 37 38 39 40 42 43 44 46 47 48 51 52 55 58 59 61 62 63 65 66 68 78 79 81 82 89\n0 3 4 6 8 9 10 11 13 15 16 19 21 24 25 26 28 29 31 32 33 34 35 37 38 39 41 42 43 46 47 50 53 54 56 57 58 60 61 63 73 74 76 77 84 86 88 89",
"output": "YES"
},
{
"input": "49 60\n0 1 2 5 7 8 9 10 11 12 13 14 15 16 17 19 20 21 23 25 26 27 28 29 30 31 32 33 34 36 38 39 40 41 42 43 44 46 47 48 49 50 51 52 53 54 55 58 59\n0 1 2 3 4 5 6 7 8 10 11 12 14 16 17 18 19 20 21 22 23 24 25 27 29 30 31 32 33 34 35 37 38 39 40 41 42 43 44 45 46 49 50 51 52 53 56 58 59",
"output": "YES"
},
{
"input": "49 97\n0 1 2 3 6 8 11 14 19 23 26 29 32 34 35 37 39 41 43 44 45 46 51 53 63 64 65 66 67 70 71 72 73 76 77 78 79 81 83 84 86 87 90 91 92 93 94 95 96\n0 3 4 5 6 7 8 9 10 11 12 13 16 18 21 24 29 33 36 39 42 44 45 47 49 51 53 54 55 56 61 63 73 74 75 76 77 80 81 82 83 86 87 88 89 91 93 94 96",
"output": "YES"
},
{
"input": "50 58\n0 1 2 3 5 6 7 8 10 11 12 13 14 15 16 17 18 19 21 22 23 24 25 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 49 50 54 55 56 57\n0 1 3 4 5 6 7 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 31 32 36 37 38 39 40 41 42 43 45 46 47 48 50 51 52 53 54 55 56 57",
"output": "YES"
},
{
"input": "50 97\n1 2 3 4 7 9 10 11 12 13 14 21 22 23 24 25 28 29 30 31 32 33 34 36 37 40 41 45 53 56 59 64 65 69 70 71 72 73 74 77 81 84 85 86 87 89 91 92 95 96\n0 1 2 3 6 10 13 14 15 16 18 20 21 24 25 27 28 29 30 33 35 36 37 38 39 40 47 48 49 50 51 54 55 56 57 58 59 60 62 63 66 67 71 79 82 85 90 91 95 96",
"output": "YES"
},
{
"input": "40 96\n5 11 12 13 14 16 17 18 19 24 30 31 32 33 37 42 46 50 53 54 55 58 60 61 64 67 68 69 70 72 75 76 77 81 84 85 88 91 92 93\n2 7 11 15 18 19 20 23 25 26 29 32 33 34 35 37 40 41 42 46 49 50 54 56 57 58 66 72 73 74 75 77 78 79 80 85 91 92 93 94",
"output": "NO"
},
{
"input": "41 67\n0 2 3 5 8 10 11 12 13 14 15 19 20 21 22 25 29 30 31 32 34 35 37 38 40 41 44 45 46 47 49 51 52 53 54 56 57 58 59 63 66\n2 3 4 5 9 12 13 14 15 17 18 20 21 23 24 27 28 29 30 32 34 35 36 37 39 40 41 42 46 49 50 52 53 55 58 60 61 62 63 64 65",
"output": "NO"
},
{
"input": "41 72\n0 3 4 6 7 8 9 12 13 14 16 21 23 24 25 26 27 28 31 32 33 34 35 38 40 41 45 47 49 50 51 52 56 57 58 59 61 62 65 66 69\n0 1 4 5 6 8 13 15 16 17 18 19 21 23 24 25 26 27 30 32 33 37 39 41 42 43 44 48 49 50 51 53 54 57 58 61 64 67 68 70 71",
"output": "NO"
},
{
"input": "42 48\n0 1 2 3 4 7 8 9 10 11 12 13 15 16 17 18 19 20 21 22 23 24 25 27 28 29 30 32 33 34 35 36 37 38 40 41 42 43 44 45 46 47\n0 1 2 3 4 5 6 8 9 10 11 12 14 15 16 17 18 19 20 22 23 24 25 26 27 28 29 30 31 32 33 34 37 38 39 40 41 42 43 45 46 47",
"output": "NO"
},
{
"input": "42 81\n0 1 3 6 7 8 11 13 17 18 19 20 22 24 29 30 31 32 34 35 38 44 46 48 49 50 51 52 53 55 59 61 62 63 65 66 67 69 70 72 77 80\n0 1 3 4 6 11 12 13 14 16 17 20 26 28 30 31 32 33 34 35 37 41 43 44 45 47 48 49 51 52 54 59 62 63 64 66 69 70 71 74 76 80",
"output": "NO"
},
{
"input": "43 55\n0 1 2 3 4 5 6 7 8 12 14 15 17 18 19 20 21 22 23 26 27 28 29 31 32 33 34 36 37 38 40 42 43 44 45 46 47 48 49 50 51 53 54\n1 2 4 5 6 7 8 9 10 13 14 15 16 18 19 20 22 23 24 25 27 29 30 31 32 33 34 35 36 37 38 40 41 42 43 44 45 46 47 48 49 50 54",
"output": "NO"
},
{
"input": "43 81\n2 3 4 5 6 7 9 10 12 13 17 19 20 21 23 26 27 29 30 32 34 38 39 43 46 47 48 50 51 52 54 55 58 62 64 67 69 70 71 72 73 75 80\n0 3 5 6 7 8 9 11 16 19 20 21 22 23 24 26 27 29 30 35 36 37 38 40 43 44 46 47 49 51 55 56 60 63 64 65 67 68 69 71 72 75 79",
"output": "NO"
},
{
"input": "44 54\n0 1 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 21 22 23 24 25 26 27 28 29 31 33 34 35 36 37 38 40 41 43 44 47 49 50 52 53\n0 1 2 3 4 5 6 7 8 10 12 13 14 15 16 18 19 20 22 23 26 28 29 31 32 33 34 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52",
"output": "NO"
},
{
"input": "44 93\n1 5 6 7 8 10 14 17 19 21 25 26 27 30 33 34 35 36 38 41 45 48 49 51 53 55 57 60 66 67 69 70 73 76 78 79 80 81 82 83 84 87 88 90\n0 2 4 8 9 10 13 16 17 18 19 21 24 28 31 32 34 36 38 40 43 49 50 52 53 56 59 61 62 63 64 65 66 68 70 71 73 77 81 82 83 84 86 90",
"output": "NO"
},
{
"input": "45 47\n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 44 45 46\n0 1 2 3 4 5 6 7 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 33 34 35 36 37 38 39 40 41 42 43 44 45 46",
"output": "YES"
},
{
"input": "45 71\n0 2 3 7 8 11 12 13 14 15 16 17 20 21 22 23 24 26 28 30 32 37 39 40 42 43 44 45 47 48 50 52 54 55 56 57 58 59 60 61 62 64 66 68 70\n0 1 2 3 4 7 8 9 10 11 13 15 17 19 24 26 28 29 30 31 32 34 35 37 39 41 42 43 44 45 46 47 48 49 51 53 55 57 58 60 61 65 66 69 70",
"output": "NO"
},
{
"input": "46 46\n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45\n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45",
"output": "YES"
},
{
"input": "46 93\n0 1 2 6 13 16 17 18 19 21 27 29 32 34 37 38 39 40 41 44 45 49 50 52 54 56 57 61 64 65 66 67 69 71 73 75 77 78 79 83 85 86 88 90 91 92\n0 2 4 5 7 8 9 10 11 12 16 23 26 27 28 29 31 37 39 42 44 47 48 49 50 51 54 55 59 60 62 64 66 67 71 74 75 76 77 79 81 83 85 87 88 89",
"output": "NO"
},
{
"input": "47 49\n0 1 2 3 4 5 6 7 9 10 11 12 13 14 15 16 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48\n0 1 2 3 4 5 6 7 8 9 10 11 13 14 15 16 17 18 19 20 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48",
"output": "YES"
},
{
"input": "47 94\n0 1 3 4 5 7 8 9 14 18 19 26 30 33 34 35 37 40 42 44 46 49 50 51 52 53 55 56 60 61 62 63 64 65 66 69 71 73 75 79 84 86 87 88 90 92 93\n1 2 3 4 6 7 8 10 11 12 17 21 22 29 33 36 37 38 40 43 45 48 49 52 53 54 55 56 58 59 63 64 65 66 67 68 69 72 74 76 78 82 87 89 90 91 93",
"output": "NO"
},
{
"input": "48 65\n0 1 2 4 5 6 7 8 9 10 11 12 15 16 17 20 21 24 25 26 27 28 30 32 33 34 35 37 38 39 44 45 46 47 48 50 51 52 53 54 55 56 57 58 59 61 62 63\n0 1 4 6 8 9 10 11 12 14 16 17 18 19 21 22 23 28 29 30 31 32 34 35 36 37 38 39 40 41 42 43 45 46 47 49 50 51 53 54 55 56 57 58 59 60 61 64",
"output": "NO"
},
{
"input": "48 90\n1 3 4 5 8 9 11 13 14 15 16 17 20 21 24 26 29 30 31 33 34 36 37 38 39 40 42 43 44 46 47 48 51 52 55 58 59 61 62 63 65 66 68 78 79 81 82 89\n0 3 4 6 8 9 10 11 13 15 16 19 21 24 25 26 28 29 31 32 33 34 35 37 38 39 41 42 43 46 47 50 53 54 56 57 58 60 61 63 73 74 76 77 84 86 88 89",
"output": "NO"
},
{
"input": "49 60\n0 1 2 5 7 8 9 10 11 12 13 14 15 16 17 18 20 21 23 25 26 27 28 29 30 31 32 33 34 36 38 39 40 41 42 43 44 46 47 48 49 50 51 52 53 54 55 58 59\n0 1 2 3 4 5 6 7 8 10 11 12 14 16 17 18 19 20 21 22 23 24 25 27 29 30 31 32 33 34 35 37 38 39 40 41 42 43 44 45 46 49 50 51 52 53 56 58 59",
"output": "NO"
},
{
"input": "49 97\n0 1 2 3 5 8 11 14 19 23 26 29 32 34 35 37 39 41 43 44 45 46 51 53 63 64 65 66 67 70 71 72 73 76 77 78 79 81 83 84 86 87 90 91 92 93 94 95 96\n0 3 4 5 6 7 8 9 10 11 12 13 16 18 21 24 29 33 36 39 42 44 45 47 49 51 53 54 55 56 61 63 73 74 75 76 77 80 81 82 83 86 87 88 89 91 93 94 96",
"output": "NO"
},
{
"input": "50 58\n0 1 2 3 5 6 7 8 10 11 12 13 14 15 16 17 18 19 21 22 23 24 25 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 50 54 55 56 57\n0 1 3 4 5 6 7 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 31 32 36 37 38 39 40 41 42 43 45 46 47 48 50 51 52 53 54 55 56 57",
"output": "NO"
},
{
"input": "5 10\n0 1 3 5 7\n0 1 2 4 7",
"output": "NO"
},
{
"input": "5 8\n0 2 4 6 7\n0 2 3 5 7",
"output": "NO"
}
] | 1,671,277,930 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 134 | 77 | 1,740,800 | n = input().split()
[n, l] = list(map(int, n))
l1 = input().split()
l1 = list(map(int, l1))
l2 = input().split()
l2 = list(map(int, l2))
def generator(l, f):
for i in range(len(l)-1):
l[i] = l[i+1]-l[i]
f -= l[i]
l[-1] = f
return l;
l1 = generator(l1, l)
l2 = generator(l2, l)
b = True
for i in range(n):
temp = l2[0]
for j in range(n-1):
l2[j] = l2[j+1]
l2[n-1] = temp
if l1 == l2 :
print("YES")
b = False
break
if b :
print("NO") | Title: Dasha and friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Running with barriers on the circle track is very popular in the country where Dasha lives, so no wonder that on her way to classes she saw the following situation:
The track is the circle with length *L*, in distinct points of which there are *n* barriers. Athlete always run the track in counterclockwise direction if you look on him from above. All barriers are located at integer distance from each other along the track.
Her friends the parrot Kefa and the leopard Sasha participated in competitions and each of them ran one lap. Each of the friends started from some integral point on the track. Both friends wrote the distance from their start along the track to each of the *n* barriers. Thus, each of them wrote *n* integers in the ascending order, each of them was between 0 and *L*<=-<=1, inclusively.
There are several tracks in the country, all of them have same length and same number of barriers, but the positions of the barriers can differ among different tracks. Now Dasha is interested if it is possible that Kefa and Sasha ran the same track or they participated on different tracks.
Write the program which will check that Kefa's and Sasha's tracks coincide (it means that one can be obtained from the other by changing the start position). Note that they always run the track in one direction — counterclockwise, if you look on a track from above.
Input Specification:
The first line contains two integers *n* and *L* (1<=≤<=*n*<=≤<=50, *n*<=≤<=*L*<=≤<=100) — the number of barriers on a track and its length.
The second line contains *n* distinct integers in the ascending order — the distance from Kefa's start to each barrier in the order of its appearance. All integers are in the range from 0 to *L*<=-<=1 inclusively.
The second line contains *n* distinct integers in the ascending order — the distance from Sasha's start to each barrier in the order of its overcoming. All integers are in the range from 0 to *L*<=-<=1 inclusively.
Output Specification:
Print "YES" (without quotes), if Kefa and Sasha ran the coinciding tracks (it means that the position of all barriers coincides, if they start running from the same points on the track). Otherwise print "NO" (without quotes).
Demo Input:
['3 8\n2 4 6\n1 5 7\n', '4 9\n2 3 5 8\n0 1 3 6\n', '2 4\n1 3\n1 2\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
The first test is analyzed in the statement. | ```python
n = input().split()
[n, l] = list(map(int, n))
l1 = input().split()
l1 = list(map(int, l1))
l2 = input().split()
l2 = list(map(int, l2))
def generator(l, f):
for i in range(len(l)-1):
l[i] = l[i+1]-l[i]
f -= l[i]
l[-1] = f
return l;
l1 = generator(l1, l)
l2 = generator(l2, l)
b = True
for i in range(n):
temp = l2[0]
for j in range(n-1):
l2[j] = l2[j+1]
l2[n-1] = temp
if l1 == l2 :
print("YES")
b = False
break
if b :
print("NO")
``` | 3 | |
514 | A | Chewbaсca and Number | PROGRAMMING | 1,200 | [
"greedy",
"implementation"
] | null | null | Luke Skywalker gave Chewbacca an integer number *x*. Chewbacca isn't good at numbers but he loves inverting digits in them. Inverting digit *t* means replacing it with digit 9<=-<=*t*.
Help Chewbacca to transform the initial number *x* to the minimum possible positive number by inverting some (possibly, zero) digits. The decimal representation of the final number shouldn't start with a zero. | The first line contains a single integer *x* (1<=≤<=*x*<=≤<=1018) — the number that Luke Skywalker gave to Chewbacca. | Print the minimum possible positive number that Chewbacca can obtain after inverting some digits. The number shouldn't contain leading zeroes. | [
"27\n",
"4545\n"
] | [
"22\n",
"4444\n"
] | none | 500 | [
{
"input": "27",
"output": "22"
},
{
"input": "4545",
"output": "4444"
},
{
"input": "1",
"output": "1"
},
{
"input": "9",
"output": "9"
},
{
"input": "8772",
"output": "1222"
},
{
"input": "81",
"output": "11"
},
{
"input": "71723447",
"output": "21223442"
},
{
"input": "91730629",
"output": "91230320"
},
{
"input": "420062703497",
"output": "420032203402"
},
{
"input": "332711047202",
"output": "332211042202"
},
{
"input": "3395184971407775",
"output": "3304114021402224"
},
{
"input": "8464062628894325",
"output": "1434032321104324"
},
{
"input": "164324828731963982",
"output": "134324121231033012"
},
{
"input": "384979173822804784",
"output": "314020123122104214"
},
{
"input": "41312150450968417",
"output": "41312140440031412"
},
{
"input": "2156",
"output": "2143"
},
{
"input": "1932",
"output": "1032"
},
{
"input": "5902",
"output": "4002"
},
{
"input": "5728",
"output": "4221"
},
{
"input": "8537",
"output": "1432"
},
{
"input": "55403857",
"output": "44403142"
},
{
"input": "270739",
"output": "220230"
},
{
"input": "28746918",
"output": "21243011"
},
{
"input": "10279211",
"output": "10220211"
},
{
"input": "40289679",
"output": "40210320"
},
{
"input": "545203238506",
"output": "444203231403"
},
{
"input": "461117063340",
"output": "431112033340"
},
{
"input": "658492686568",
"output": "341402313431"
},
{
"input": "857373361868",
"output": "142323331131"
},
{
"input": "429325660016",
"output": "420324330013"
},
{
"input": "9894448650287940",
"output": "9104441340212040"
},
{
"input": "6354510839296263",
"output": "3344410130203233"
},
{
"input": "6873575462224593",
"output": "3123424432224403"
},
{
"input": "4237951492601449",
"output": "4232041402301440"
},
{
"input": "2680352384836991",
"output": "2310342314133001"
},
{
"input": "606187734191890310",
"output": "303112234101100310"
},
{
"input": "351499943576823355",
"output": "341400043423123344"
},
{
"input": "180593481782177068",
"output": "110403411212122031"
},
{
"input": "999999999999999999",
"output": "900000000000000000"
},
{
"input": "1000000000000000000",
"output": "1000000000000000000"
},
{
"input": "9999",
"output": "9000"
},
{
"input": "99",
"output": "90"
},
{
"input": "9991",
"output": "9001"
}
] | 1,669,276,158 | 2,147,483,647 | Python 3 | OK | TESTS | 43 | 46 | 0 | n = input()
s = ""
for i in n:
if 9 - int(i) < int(i):
if 9 - int(i) == 0 and s == "":
s += i
else:
s += str(9 - int(i))
else:
s += i
print(s) | Title: Chewbaсca and Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luke Skywalker gave Chewbacca an integer number *x*. Chewbacca isn't good at numbers but he loves inverting digits in them. Inverting digit *t* means replacing it with digit 9<=-<=*t*.
Help Chewbacca to transform the initial number *x* to the minimum possible positive number by inverting some (possibly, zero) digits. The decimal representation of the final number shouldn't start with a zero.
Input Specification:
The first line contains a single integer *x* (1<=≤<=*x*<=≤<=1018) — the number that Luke Skywalker gave to Chewbacca.
Output Specification:
Print the minimum possible positive number that Chewbacca can obtain after inverting some digits. The number shouldn't contain leading zeroes.
Demo Input:
['27\n', '4545\n']
Demo Output:
['22\n', '4444\n']
Note:
none | ```python
n = input()
s = ""
for i in n:
if 9 - int(i) < int(i):
if 9 - int(i) == 0 and s == "":
s += i
else:
s += str(9 - int(i))
else:
s += i
print(s)
``` | 3 | |
9 | A | Die Roll | PROGRAMMING | 800 | [
"math",
"probabilities"
] | A. Die Roll | 1 | 64 | Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win. | The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls. | Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1». | [
"4 2\n"
] | [
"1/2\n"
] | Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | 0 | [
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
"output": "1/6"
},
{
"input": "2 1",
"output": "5/6"
},
{
"input": "2 2",
"output": "5/6"
},
{
"input": "2 3",
"output": "2/3"
},
{
"input": "2 4",
"output": "1/2"
},
{
"input": "2 5",
"output": "1/3"
},
{
"input": "2 6",
"output": "1/6"
},
{
"input": "3 1",
"output": "2/3"
},
{
"input": "3 2",
"output": "2/3"
},
{
"input": "3 3",
"output": "2/3"
},
{
"input": "3 4",
"output": "1/2"
},
{
"input": "3 5",
"output": "1/3"
},
{
"input": "3 6",
"output": "1/6"
},
{
"input": "4 1",
"output": "1/2"
},
{
"input": "4 3",
"output": "1/2"
},
{
"input": "4 4",
"output": "1/2"
},
{
"input": "4 5",
"output": "1/3"
},
{
"input": "4 6",
"output": "1/6"
},
{
"input": "5 1",
"output": "1/3"
},
{
"input": "5 2",
"output": "1/3"
},
{
"input": "5 3",
"output": "1/3"
},
{
"input": "5 4",
"output": "1/3"
},
{
"input": "5 5",
"output": "1/3"
},
{
"input": "5 6",
"output": "1/6"
},
{
"input": "6 1",
"output": "1/6"
},
{
"input": "6 2",
"output": "1/6"
},
{
"input": "6 3",
"output": "1/6"
},
{
"input": "6 4",
"output": "1/6"
},
{
"input": "6 5",
"output": "1/6"
},
{
"input": "6 6",
"output": "1/6"
}
] | 1,672,489,865 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 36 | 124 | 0 | s = ["1/1","5/6","2/3","1/2","1/3","1/6","0/1"]
t,w=map(int,input().split(" "))
ans = max(t , w)
print(s[ans-1]) | Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | ```python
s = ["1/1","5/6","2/3","1/2","1/3","1/6","0/1"]
t,w=map(int,input().split(" "))
ans = max(t , w)
print(s[ans-1])
``` | 3.938 |
12 | B | Correct Solution? | PROGRAMMING | 1,100 | [
"implementation",
"sortings"
] | B. Correct Solution? | 2 | 256 | One cold winter evening Alice and her older brother Bob was sitting at home near the fireplace and giving each other interesting problems to solve. When it was Alice's turn, she told the number *n* to Bob and said:
—Shuffle the digits in this number in order to obtain the smallest possible number without leading zeroes.
—No problem! — said Bob and immediately gave her an answer.
Alice said a random number, so she doesn't know whether Bob's answer is correct. Help her to find this out, because impatient brother is waiting for the verdict. | The first line contains one integer *n* (0<=≤<=*n*<=≤<=109) without leading zeroes. The second lines contains one integer *m* (0<=≤<=*m*<=≤<=109) — Bob's answer, possibly with leading zeroes. | Print OK if Bob's answer is correct and WRONG_ANSWER otherwise. | [
"3310\n1033\n",
"4\n5\n"
] | [
"OK\n",
"WRONG_ANSWER\n"
] | none | 0 | [
{
"input": "3310\n1033",
"output": "OK"
},
{
"input": "4\n5",
"output": "WRONG_ANSWER"
},
{
"input": "40\n04",
"output": "WRONG_ANSWER"
},
{
"input": "12\n12",
"output": "OK"
},
{
"input": "432\n234",
"output": "OK"
},
{
"input": "17109\n01179",
"output": "WRONG_ANSWER"
},
{
"input": "888\n888",
"output": "OK"
},
{
"input": "912\n9123",
"output": "WRONG_ANSWER"
},
{
"input": "0\n00",
"output": "WRONG_ANSWER"
},
{
"input": "11110\n1111",
"output": "WRONG_ANSWER"
},
{
"input": "7391\n1397",
"output": "WRONG_ANSWER"
},
{
"input": "201\n102",
"output": "OK"
},
{
"input": "111111111\n111111111",
"output": "OK"
},
{
"input": "32352320\n22203335",
"output": "WRONG_ANSWER"
},
{
"input": "1000000000\n1",
"output": "WRONG_ANSWER"
},
{
"input": "994321\n123499",
"output": "OK"
},
{
"input": "10101\n10101",
"output": "WRONG_ANSWER"
},
{
"input": "666\n0666",
"output": "WRONG_ANSWER"
},
{
"input": "2\n02",
"output": "WRONG_ANSWER"
},
{
"input": "313\n133",
"output": "OK"
},
{
"input": "987235645\n234556789",
"output": "OK"
},
{
"input": "90812\n010289",
"output": "WRONG_ANSWER"
},
{
"input": "123\n321",
"output": "WRONG_ANSWER"
},
{
"input": "707\n770",
"output": "WRONG_ANSWER"
},
{
"input": "77774444\n47474747",
"output": "WRONG_ANSWER"
},
{
"input": "1270\n1027",
"output": "OK"
},
{
"input": "320\n23",
"output": "WRONG_ANSWER"
},
{
"input": "123456789\n123456789",
"output": "OK"
},
{
"input": "918273645\n546372819",
"output": "WRONG_ANSWER"
},
{
"input": "300000003\n30000003",
"output": "WRONG_ANSWER"
},
{
"input": "0\n0",
"output": "OK"
},
{
"input": "0\n7",
"output": "WRONG_ANSWER"
}
] | 1,644,301,552 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 8 | 62 | 0 | n = int(input())
m = int(input())
o = str(n)
x = len(o)
list = []
for i in range(0, x):
list.append(o[i])
z = sorted(list)
if z[0] == '0':
z[0], z[1] = z[1], z[0]
strings = [str(integer) for integer in z]
a_string = "".join(strings)
an_integer = int(a_string)
if m == an_integer:
print("OK")
else:
print("WRONG_ANSWER")
| Title: Correct Solution?
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One cold winter evening Alice and her older brother Bob was sitting at home near the fireplace and giving each other interesting problems to solve. When it was Alice's turn, she told the number *n* to Bob and said:
—Shuffle the digits in this number in order to obtain the smallest possible number without leading zeroes.
—No problem! — said Bob and immediately gave her an answer.
Alice said a random number, so she doesn't know whether Bob's answer is correct. Help her to find this out, because impatient brother is waiting for the verdict.
Input Specification:
The first line contains one integer *n* (0<=≤<=*n*<=≤<=109) without leading zeroes. The second lines contains one integer *m* (0<=≤<=*m*<=≤<=109) — Bob's answer, possibly with leading zeroes.
Output Specification:
Print OK if Bob's answer is correct and WRONG_ANSWER otherwise.
Demo Input:
['3310\n1033\n', '4\n5\n']
Demo Output:
['OK\n', 'WRONG_ANSWER\n']
Note:
none | ```python
n = int(input())
m = int(input())
o = str(n)
x = len(o)
list = []
for i in range(0, x):
list.append(o[i])
z = sorted(list)
if z[0] == '0':
z[0], z[1] = z[1], z[0]
strings = [str(integer) for integer in z]
a_string = "".join(strings)
an_integer = int(a_string)
if m == an_integer:
print("OK")
else:
print("WRONG_ANSWER")
``` | -1 |
61 | A | Ultra-Fast Mathematician | PROGRAMMING | 800 | [
"implementation"
] | A. Ultra-Fast Mathematician | 2 | 256 | Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate. | There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. | Write one line — the corresponding answer. Do not omit the leading 0s. | [
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] | [
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] | none | 500 | [
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,694,899,082 | 2,147,483,647 | Python 3 | OK | TESTS | 102 | 46 | 0 | num1=(input())
num2=(input())
n1=list(num1)
n2=list(num2)
s=""
for i in range(len(n1)):
if n1[i]==n2[i]:
s=s+"0"
else:
s=s+"1"
print(s) | Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none | ```python
num1=(input())
num2=(input())
n1=list(num1)
n2=list(num2)
s=""
for i in range(len(n1)):
if n1[i]==n2[i]:
s=s+"0"
else:
s=s+"1"
print(s)
``` | 3.9885 |
37 | A | Towers | PROGRAMMING | 1,000 | [
"sortings"
] | A. Towers | 2 | 256 | Little Vasya has received a young builder’s kit. The kit consists of several wooden bars, the lengths of all of them are known. The bars can be put one on the top of the other if their lengths are the same.
Vasya wants to construct the minimal number of towers from the bars. Help Vasya to use the bars in the best way possible. | The first line contains an integer *N* (1<=≤<=*N*<=≤<=1000) — the number of bars at Vasya’s disposal. The second line contains *N* space-separated integers *l**i* — the lengths of the bars. All the lengths are natural numbers not exceeding 1000. | In one line output two numbers — the height of the largest tower and their total number. Remember that Vasya should use all the bars. | [
"3\n1 2 3\n",
"4\n6 5 6 7\n"
] | [
"1 3\n",
"2 3\n"
] | none | 500 | [
{
"input": "3\n1 2 3",
"output": "1 3"
},
{
"input": "4\n6 5 6 7",
"output": "2 3"
},
{
"input": "4\n3 2 1 1",
"output": "2 3"
},
{
"input": "4\n1 2 3 3",
"output": "2 3"
},
{
"input": "3\n20 22 36",
"output": "1 3"
},
{
"input": "25\n47 30 94 41 45 20 96 51 110 129 24 116 9 47 32 82 105 114 116 75 154 151 70 42 162",
"output": "2 23"
},
{
"input": "45\n802 664 442 318 318 827 417 878 711 291 231 414 807 553 657 392 279 202 386 606 465 655 658 112 887 15 25 502 95 44 679 775 942 609 209 871 31 234 4 231 150 110 22 823 193",
"output": "2 43"
},
{
"input": "63\n93 180 116 7 8 179 268 279 136 94 221 153 264 190 278 19 19 63 153 26 158 225 25 49 89 218 111 149 255 225 197 122 243 80 3 224 107 178 202 17 53 92 69 42 228 24 81 205 95 8 265 82 228 156 127 241 172 159 106 60 67 155 111",
"output": "2 57"
},
{
"input": "83\n246 535 994 33 390 927 321 97 223 922 812 705 79 80 977 457 476 636 511 137 6 360 815 319 717 674 368 551 714 628 278 713 761 553 184 414 623 753 428 214 581 115 439 61 677 216 772 592 187 603 658 310 439 559 870 376 109 321 189 337 277 26 70 734 796 907 979 693 570 227 345 650 737 633 701 914 134 403 972 940 371 6 642",
"output": "2 80"
},
{
"input": "105\n246 57 12 204 165 123 246 68 191 310 3 152 386 333 374 257 158 104 333 50 80 290 8 340 101 76 221 316 388 289 138 359 316 26 93 290 105 178 81 195 41 196 218 180 244 292 187 97 315 323 174 119 248 239 92 312 31 2 101 180 307 170 338 314 163 281 217 31 142 238 280 190 190 156 70 74 329 113 151 8 141 313 366 40 253 116 168 124 135 230 294 266 353 389 371 359 195 200 183 237 93 102 315 118 188",
"output": "2 92"
},
{
"input": "123\n112 277 170 247 252 115 157 293 256 143 196 90 12 164 164 42 8 223 167 109 175 232 239 111 148 51 9 254 93 32 268 162 231 91 47 162 161 191 195 145 247 292 129 199 230 94 144 217 18 205 176 20 143 198 121 243 211 262 230 277 195 255 108 290 220 275 158 2 286 200 60 267 278 207 123 150 123 116 131 13 12 226 33 244 30 275 263 45 158 192 254 149 242 176 62 224 221 288 250 160 155 225 132 143 276 293 218 145 197 175 33 129 79 206 210 192 222 262 190 52 274 243 233",
"output": "3 101"
},
{
"input": "5\n5 5 5 5 5",
"output": "5 1"
},
{
"input": "3\n1000 1000 1000",
"output": "3 1"
},
{
"input": "1\n1000",
"output": "1 1"
},
{
"input": "1\n1",
"output": "1 1"
},
{
"input": "5\n1 1000 1000 1000 1000",
"output": "4 2"
},
{
"input": "5\n1000 1000 1000 8 7",
"output": "3 3"
}
] | 1,697,179,735 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 61 | 154 | 0 |
from collections import Counter
n = int(input())
a = list(map(int,input().split()))
x = Counter(a)
print(max(x.values()),len(x))
| Title: Towers
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Vasya has received a young builder’s kit. The kit consists of several wooden bars, the lengths of all of them are known. The bars can be put one on the top of the other if their lengths are the same.
Vasya wants to construct the minimal number of towers from the bars. Help Vasya to use the bars in the best way possible.
Input Specification:
The first line contains an integer *N* (1<=≤<=*N*<=≤<=1000) — the number of bars at Vasya’s disposal. The second line contains *N* space-separated integers *l**i* — the lengths of the bars. All the lengths are natural numbers not exceeding 1000.
Output Specification:
In one line output two numbers — the height of the largest tower and their total number. Remember that Vasya should use all the bars.
Demo Input:
['3\n1 2 3\n', '4\n6 5 6 7\n']
Demo Output:
['1 3\n', '2 3\n']
Note:
none | ```python
from collections import Counter
n = int(input())
a = list(map(int,input().split()))
x = Counter(a)
print(max(x.values()),len(x))
``` | 3.9615 |
266 | A | Stones on the Table | PROGRAMMING | 800 | [
"implementation"
] | null | null | There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them. | The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stones on the table.
The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue. | Print a single integer — the answer to the problem. | [
"3\nRRG\n",
"5\nRRRRR\n",
"4\nBRBG\n"
] | [
"1\n",
"4\n",
"0\n"
] | none | 500 | [
{
"input": "3\nRRG",
"output": "1"
},
{
"input": "5\nRRRRR",
"output": "4"
},
{
"input": "4\nBRBG",
"output": "0"
},
{
"input": "1\nB",
"output": "0"
},
{
"input": "2\nBG",
"output": "0"
},
{
"input": "3\nBGB",
"output": "0"
},
{
"input": "4\nRBBR",
"output": "1"
},
{
"input": "5\nRGGBG",
"output": "1"
},
{
"input": "10\nGGBRBRGGRB",
"output": "2"
},
{
"input": "50\nGRBGGRBRGRBGGBBBBBGGGBBBBRBRGBRRBRGBBBRBBRRGBGGGRB",
"output": "18"
},
{
"input": "15\nBRRBRGGBBRRRRGR",
"output": "6"
},
{
"input": "20\nRRGBBRBRGRGBBGGRGRRR",
"output": "6"
},
{
"input": "25\nBBGBGRBGGBRRBGRRBGGBBRBRB",
"output": "6"
},
{
"input": "30\nGRGGGBGGRGBGGRGRBGBGBRRRRRRGRB",
"output": "9"
},
{
"input": "35\nGBBGBRGBBGGRBBGBRRGGRRRRRRRBRBBRRGB",
"output": "14"
},
{
"input": "40\nGBBRRGBGGGRGGGRRRRBRBGGBBGGGBGBBBBBRGGGG",
"output": "20"
},
{
"input": "45\nGGGBBRBBRRGRBBGGBGRBRGGBRBRGBRRGBGRRBGRGRBRRG",
"output": "11"
},
{
"input": "50\nRBGGBGGRBGRBBBGBBGRBBBGGGRBBBGBBBGRGGBGGBRBGBGRRGG",
"output": "17"
},
{
"input": "50\nGGGBBRGGGGGRRGGRBGGRGBBRBRRBGRGBBBGBRBGRGBBGRGGBRB",
"output": "16"
},
{
"input": "50\nGBGRGRRBRRRRRGGBBGBRRRBBBRBBBRRGRBBRGBRBGGRGRBBGGG",
"output": "19"
},
{
"input": "10\nGRRBRBRBGR",
"output": "1"
},
{
"input": "10\nBRBGBGRRBR",
"output": "1"
},
{
"input": "20\nGBGBGGRRRRGRBBGRGRGR",
"output": "5"
},
{
"input": "20\nRRGGRBBGBBRBGRRBRRBG",
"output": "6"
},
{
"input": "30\nBGBRGBBBGRGBBRGBGRBBBRGGRRGRRB",
"output": "8"
},
{
"input": "30\nBBBBGGBRBGBBGBGBGBGGGRGRRGGBBB",
"output": "11"
},
{
"input": "40\nGBRRGRBGBRRGBRGGGBRGBGBRGBBRRGRGGBBGBGBB",
"output": "9"
},
{
"input": "40\nBRGRGGRGGRBBRRRBRBBGGGRRGBGBBGRBBRGBRRGG",
"output": "13"
},
{
"input": "50\nRBGBGGRRGGRGGBGBGRRBGGBGBRRBBGBBGBBBGBBRBBRBRBRGRG",
"output": "13"
},
{
"input": "50\nRBRRGBGRRRBGRRBGRRGRBBRBBRRBRGGBRBRRBGGRBGGBRBRGRB",
"output": "12"
},
{
"input": "2\nBB",
"output": "1"
},
{
"input": "50\nRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "49"
},
{
"input": "50\nRRRRRRRRGRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR",
"output": "47"
},
{
"input": "50\nRRRRRRRRRRRRGGRRRRRRRRRBRRRRRRRRRRRRRRBBRRRRRRRRRR",
"output": "43"
}
] | 1,694,667,138 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 60 | 0 | n = int(input())
a = input()
c = 0
for i in range(1, len(a)):
if a[i-1] == a[i]:
del a[i]
c += 1
else:
pass
print(c) | Title: Stones on the Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stones on the table.
The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['3\nRRG\n', '5\nRRRRR\n', '4\nBRBG\n']
Demo Output:
['1\n', '4\n', '0\n']
Note:
none | ```python
n = int(input())
a = input()
c = 0
for i in range(1, len(a)):
if a[i-1] == a[i]:
del a[i]
c += 1
else:
pass
print(c)
``` | -1 | |
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,661,882,948 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 31 | 0 | n = input()
s1 = int(int(n[0])/int(n[4]))
s2 = int(int(n[2])/int(n[4]))
if int(n[0])%int(n[4])!=0 and int(n[0])>=int(n[4]):
s1+=1
if int(n[2])%int(n[4])!=0 and int(n[0])>=int(n[4]):
s2+=1
if (int(n[0])==1) or (int(n[2])==1):
s2-=1
s= s1+ s2
if s==0:
s=1
print(s)
| Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none | ```python
n = input()
s1 = int(int(n[0])/int(n[4]))
s2 = int(int(n[2])/int(n[4]))
if int(n[0])%int(n[4])!=0 and int(n[0])>=int(n[4]):
s1+=1
if int(n[2])%int(n[4])!=0 and int(n[0])>=int(n[4]):
s2+=1
if (int(n[0])==1) or (int(n[2])==1):
s2-=1
s= s1+ s2
if s==0:
s=1
print(s)
``` | 0 |
758 | A | Holiday Of Equality | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them. | The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen. | In the only line print the integer *S* — the minimum number of burles which are had to spend. | [
"5\n0 1 2 3 4\n",
"5\n1 1 0 1 1\n",
"3\n1 3 1\n",
"1\n12\n"
] | [
"10",
"1",
"4",
"0"
] | In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles. | 500 | [
{
"input": "5\n0 1 2 3 4",
"output": "10"
},
{
"input": "5\n1 1 0 1 1",
"output": "1"
},
{
"input": "3\n1 3 1",
"output": "4"
},
{
"input": "1\n12",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "14\n52518 718438 358883 462189 853171 592966 225788 46977 814826 295697 676256 561479 56545 764281",
"output": "5464380"
},
{
"input": "21\n842556 216391 427181 626688 775504 168309 851038 448402 880826 73697 593338 519033 135115 20128 424606 939484 846242 756907 377058 241543 29353",
"output": "9535765"
},
{
"input": "3\n1 3 2",
"output": "3"
},
{
"input": "3\n2 1 3",
"output": "3"
},
{
"input": "3\n2 3 1",
"output": "3"
},
{
"input": "3\n3 1 2",
"output": "3"
},
{
"input": "3\n3 2 1",
"output": "3"
},
{
"input": "1\n228503",
"output": "0"
},
{
"input": "2\n32576 550340",
"output": "517764"
},
{
"input": "3\n910648 542843 537125",
"output": "741328"
},
{
"input": "4\n751720 572344 569387 893618",
"output": "787403"
},
{
"input": "6\n433864 631347 597596 794426 713555 231193",
"output": "1364575"
},
{
"input": "9\n31078 645168 695751 126111 375934 150495 838412 434477 993107",
"output": "4647430"
},
{
"input": "30\n315421 772664 560686 654312 151528 356749 351486 707462 820089 226682 546700 136028 824236 842130 578079 337807 665903 764100 617900 822937 992759 591749 651310 742085 767695 695442 17967 515106 81059 186025",
"output": "13488674"
},
{
"input": "45\n908719 394261 815134 419990 926993 383792 772842 277695 527137 655356 684956 695716 273062 550324 106247 399133 442382 33076 462920 294674 846052 817752 421365 474141 290471 358990 109812 74492 543281 169434 919692 786809 24028 197184 310029 801476 699355 429672 51343 374128 776726 850380 293868 981569 550763",
"output": "21993384"
},
{
"input": "56\n100728 972537 13846 385421 756708 184642 259487 319707 376662 221694 675284 972837 499419 13846 38267 289898 901299 831197 954715 197515 514102 910423 127555 883934 362472 870788 538802 741008 973434 448124 391526 363321 947321 544618 68006 782313 955075 741981 815027 723297 585059 718114 700739 413489 454091 736144 308999 98065 3716 347323 9635 289003 986510 607065 60236 273351",
"output": "26984185"
},
{
"input": "70\n644488 5444 150441 714420 602059 335330 510670 196555 546346 740011 509449 850947 692874 524857 750434 952985 223744 374727 896124 753037 367352 679050 560202 172728 569291 778616 332116 286927 843598 372698 244906 498046 900681 709791 420904 724593 864493 813094 791377 39998 296710 625656 403891 579231 706693 984045 16901 574259 562265 761104 930361 256045 124461 538980 573508 372148 988722 108592 784354 55302 232524 277205 782251 299943 436488 743389 324618 742543 266915 99642",
"output": "32816391"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n1000000",
"output": "0"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "0"
},
{
"input": "100\n1000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "99000000"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1000000",
"output": "99000000"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "99000000"
},
{
"input": "3\n0 0 0",
"output": "0"
},
{
"input": "50\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "3\n5 0 0",
"output": "10"
},
{
"input": "5\n2 10 0 0 0",
"output": "38"
}
] | 1,662,067,142 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 41 | 77 | 0 | n = int(input())
wlfs = [int(i) for i in input().split()]
s = 0
wlfs.sort(reverse=True)
for i in wlfs:
s += wlfs[0] - i
print(s) | Title: Holiday Of Equality
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.
Input Specification:
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen.
Output Specification:
In the only line print the integer *S* — the minimum number of burles which are had to spend.
Demo Input:
['5\n0 1 2 3 4\n', '5\n1 1 0 1 1\n', '3\n1 3 1\n', '1\n12\n']
Demo Output:
['10', '1', '4', '0']
Note:
In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles. | ```python
n = int(input())
wlfs = [int(i) for i in input().split()]
s = 0
wlfs.sort(reverse=True)
for i in wlfs:
s += wlfs[0] - i
print(s)
``` | 3 | |
483 | A | Counterexample | PROGRAMMING | 1,100 | [
"brute force",
"implementation",
"math",
"number theory"
] | null | null | Your friend has recently learned about coprime numbers. A pair of numbers {*a*,<=*b*} is called coprime if the maximum number that divides both *a* and *b* is equal to one.
Your friend often comes up with different statements. He has recently supposed that if the pair (*a*,<=*b*) is coprime and the pair (*b*,<=*c*) is coprime, then the pair (*a*,<=*c*) is coprime.
You want to find a counterexample for your friend's statement. Therefore, your task is to find three distinct numbers (*a*,<=*b*,<=*c*), for which the statement is false, and the numbers meet the condition *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*.
More specifically, you need to find three numbers (*a*,<=*b*,<=*c*), such that *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*, pairs (*a*,<=*b*) and (*b*,<=*c*) are coprime, and pair (*a*,<=*c*) is not coprime. | The single line contains two positive space-separated integers *l*, *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018; *r*<=-<=*l*<=≤<=50). | Print three positive space-separated integers *a*, *b*, *c* — three distinct numbers (*a*,<=*b*,<=*c*) that form the counterexample. If there are several solutions, you are allowed to print any of them. The numbers must be printed in ascending order.
If the counterexample does not exist, print the single number -1. | [
"2 4\n",
"10 11\n",
"900000000000000009 900000000000000029\n"
] | [
"2 3 4\n",
"-1\n",
"900000000000000009 900000000000000010 900000000000000021\n"
] | In the first sample pair (2, 4) is not coprime and pairs (2, 3) and (3, 4) are.
In the second sample you cannot form a group of three distinct integers, so the answer is -1.
In the third sample it is easy to see that numbers 900000000000000009 and 900000000000000021 are divisible by three. | 500 | [
{
"input": "2 4",
"output": "2 3 4"
},
{
"input": "10 11",
"output": "-1"
},
{
"input": "900000000000000009 900000000000000029",
"output": "900000000000000009 900000000000000010 900000000000000021"
},
{
"input": "640097987171091791 640097987171091835",
"output": "640097987171091792 640097987171091793 640097987171091794"
},
{
"input": "19534350415104721 19534350415104725",
"output": "19534350415104722 19534350415104723 19534350415104724"
},
{
"input": "933700505788726243 933700505788726280",
"output": "933700505788726244 933700505788726245 933700505788726246"
},
{
"input": "1 3",
"output": "-1"
},
{
"input": "1 4",
"output": "2 3 4"
},
{
"input": "1 1",
"output": "-1"
},
{
"input": "266540997167959130 266540997167959164",
"output": "266540997167959130 266540997167959131 266540997167959132"
},
{
"input": "267367244641009850 267367244641009899",
"output": "267367244641009850 267367244641009851 267367244641009852"
},
{
"input": "268193483524125978 268193483524125993",
"output": "268193483524125978 268193483524125979 268193483524125980"
},
{
"input": "269019726702209402 269019726702209432",
"output": "269019726702209402 269019726702209403 269019726702209404"
},
{
"input": "269845965585325530 269845965585325576",
"output": "269845965585325530 269845965585325531 269845965585325532"
},
{
"input": "270672213058376250 270672213058376260",
"output": "270672213058376250 270672213058376251 270672213058376252"
},
{
"input": "271498451941492378 271498451941492378",
"output": "-1"
},
{
"input": "272324690824608506 272324690824608523",
"output": "272324690824608506 272324690824608507 272324690824608508"
},
{
"input": "273150934002691930 273150934002691962",
"output": "273150934002691930 273150934002691931 273150934002691932"
},
{
"input": "996517375802030516 996517375802030524",
"output": "996517375802030516 996517375802030517 996517375802030518"
},
{
"input": "997343614685146644 997343614685146694",
"output": "997343614685146644 997343614685146645 997343614685146646"
},
{
"input": "998169857863230068 998169857863230083",
"output": "998169857863230068 998169857863230069 998169857863230070"
},
{
"input": "998996101041313492 998996101041313522",
"output": "998996101041313492 998996101041313493 998996101041313494"
},
{
"input": "999822344219396916 999822344219396961",
"output": "999822344219396916 999822344219396917 999822344219396918"
},
{
"input": "648583102513043 648583102513053",
"output": "648583102513044 648583102513045 648583102513046"
},
{
"input": "266540997167959130 266540997167959131",
"output": "-1"
},
{
"input": "267367244641009850 267367244641009850",
"output": "-1"
},
{
"input": "268193483524125978 268193483524125979",
"output": "-1"
},
{
"input": "269019726702209402 269019726702209402",
"output": "-1"
},
{
"input": "269845965585325530 269845965585325530",
"output": "-1"
},
{
"input": "270672213058376250 270672213058376254",
"output": "270672213058376250 270672213058376251 270672213058376252"
},
{
"input": "271498451941492378 271498451941492379",
"output": "-1"
},
{
"input": "272324690824608506 272324690824608508",
"output": "272324690824608506 272324690824608507 272324690824608508"
},
{
"input": "273150934002691930 273150934002691931",
"output": "-1"
},
{
"input": "996517375802030516 996517375802030518",
"output": "996517375802030516 996517375802030517 996517375802030518"
},
{
"input": "997343614685146644 997343614685146644",
"output": "-1"
},
{
"input": "2147483647 2147483649",
"output": "-1"
},
{
"input": "3 5",
"output": "-1"
},
{
"input": "1 7",
"output": "2 3 4"
},
{
"input": "9 12",
"output": "9 11 12"
},
{
"input": "4 4",
"output": "-1"
},
{
"input": "11 13",
"output": "-1"
},
{
"input": "2 2",
"output": "-1"
}
] | 1,643,577,649 | 2,147,483,647 | PyPy 3 | OK | TESTS | 42 | 77 | 0 | def main():
l,r = map(int,input().split())
if r-l+1<3 :
print(-1)
exit()
if l%2==0 :
print(l,l+1,l+2)
exit()
if r-l+1>3 :
print(l+1,l+2,l+3)
exit()
print(-1)
main() | Title: Counterexample
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend has recently learned about coprime numbers. A pair of numbers {*a*,<=*b*} is called coprime if the maximum number that divides both *a* and *b* is equal to one.
Your friend often comes up with different statements. He has recently supposed that if the pair (*a*,<=*b*) is coprime and the pair (*b*,<=*c*) is coprime, then the pair (*a*,<=*c*) is coprime.
You want to find a counterexample for your friend's statement. Therefore, your task is to find three distinct numbers (*a*,<=*b*,<=*c*), for which the statement is false, and the numbers meet the condition *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*.
More specifically, you need to find three numbers (*a*,<=*b*,<=*c*), such that *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*, pairs (*a*,<=*b*) and (*b*,<=*c*) are coprime, and pair (*a*,<=*c*) is not coprime.
Input Specification:
The single line contains two positive space-separated integers *l*, *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018; *r*<=-<=*l*<=≤<=50).
Output Specification:
Print three positive space-separated integers *a*, *b*, *c* — three distinct numbers (*a*,<=*b*,<=*c*) that form the counterexample. If there are several solutions, you are allowed to print any of them. The numbers must be printed in ascending order.
If the counterexample does not exist, print the single number -1.
Demo Input:
['2 4\n', '10 11\n', '900000000000000009 900000000000000029\n']
Demo Output:
['2 3 4\n', '-1\n', '900000000000000009 900000000000000010 900000000000000021\n']
Note:
In the first sample pair (2, 4) is not coprime and pairs (2, 3) and (3, 4) are.
In the second sample you cannot form a group of three distinct integers, so the answer is -1.
In the third sample it is easy to see that numbers 900000000000000009 and 900000000000000021 are divisible by three. | ```python
def main():
l,r = map(int,input().split())
if r-l+1<3 :
print(-1)
exit()
if l%2==0 :
print(l,l+1,l+2)
exit()
if r-l+1>3 :
print(l+1,l+2,l+3)
exit()
print(-1)
main()
``` | 3 | |
984 | A | Game | PROGRAMMING | 800 | [
"sortings"
] | null | null | Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves. | The first line contains one integer $n$ ($1 \le n \le 1000$) — the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$). | Print one number that will be left on the board. | [
"3\n2 1 3\n",
"3\n2 2 2\n"
] | [
"2",
"2"
] | In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players. | 500 | [
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 2 2",
"output": "2"
},
{
"input": "9\n44 53 51 80 5 27 74 79 94",
"output": "53"
},
{
"input": "10\n38 82 23 37 96 4 81 60 67 86",
"output": "60"
},
{
"input": "10\n58 26 77 15 53 81 68 48 22 65",
"output": "53"
},
{
"input": "1\n124",
"output": "124"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n1 1 1000",
"output": "1"
},
{
"input": "2\n322 322",
"output": "322"
},
{
"input": "3\n9 92 12",
"output": "12"
},
{
"input": "3\n1 2 2",
"output": "2"
}
] | 1,601,717,451 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 2 | 202 | 2,150,400 | n = int(input())
arr = [int(i) for i in input().split()][:n]
arr = sorted(arr)
i=0
j = n-1
while(i<j):
arr.remove(arr[j])
j-=1
arr.remove(arr[i])
i+=1
print(arr[0]) | Title: Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players play a game.
Initially there are $n$ integers $a_1, a_2, \ldots, a_n$ written on the board. Each turn a player selects one number and erases it from the board. This continues until there is only one number left on the board, i. e. $n - 1$ turns are made. The first player makes the first move, then players alternate turns.
The first player wants to minimize the last number that would be left on the board, while the second player wants to maximize it.
You want to know what number will be left on the board after $n - 1$ turns if both players make optimal moves.
Input Specification:
The first line contains one integer $n$ ($1 \le n \le 1000$) — the number of numbers on the board.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^6$).
Output Specification:
Print one number that will be left on the board.
Demo Input:
['3\n2 1 3\n', '3\n2 2 2\n']
Demo Output:
['2', '2']
Note:
In the first sample, the first player erases $3$ and the second erases $1$. $2$ is left on the board.
In the second sample, $2$ is left on the board regardless of the actions of the players. | ```python
n = int(input())
arr = [int(i) for i in input().split()][:n]
arr = sorted(arr)
i=0
j = n-1
while(i<j):
arr.remove(arr[j])
j-=1
arr.remove(arr[i])
i+=1
print(arr[0])
``` | -1 | |
730 | J | Bottles | PROGRAMMING | 1,900 | [
"dp"
] | null | null | Nick has *n* bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda *a**i* and bottle volume *b**i* (*a**i*<=≤<=*b**i*).
Nick has decided to pour all remaining soda into minimal number of bottles, moreover he has to do it as soon as possible. Nick spends *x* seconds to pour *x* units of soda from one bottle to another.
Nick asks you to help him to determine *k* — the minimal number of bottles to store all remaining soda and *t* — the minimal time to pour soda into *k* bottles. A bottle can't store more soda than its volume. All remaining soda should be saved. | The first line contains positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of bottles.
The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is the amount of soda remaining in the *i*-th bottle.
The third line contains *n* positive integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=100), where *b**i* is the volume of the *i*-th bottle.
It is guaranteed that *a**i*<=≤<=*b**i* for any *i*. | The only line should contain two integers *k* and *t*, where *k* is the minimal number of bottles that can store all the soda and *t* is the minimal time to pour the soda into *k* bottles. | [
"4\n3 3 4 3\n4 7 6 5\n",
"2\n1 1\n100 100\n",
"5\n10 30 5 6 24\n10 41 7 8 24\n"
] | [
"2 6\n",
"1 1\n",
"3 11\n"
] | In the first example Nick can pour soda from the first bottle to the second bottle. It will take 3 seconds. After it the second bottle will contain 3 + 3 = 6 units of soda. Then he can pour soda from the fourth bottle to the second bottle and to the third bottle: one unit to the second and two units to the third. It will take 1 + 2 = 3 seconds. So, all the soda will be in two bottles and he will spend 3 + 3 = 6 seconds to do it. | 0 | [
{
"input": "4\n3 3 4 3\n4 7 6 5",
"output": "2 6"
},
{
"input": "2\n1 1\n100 100",
"output": "1 1"
},
{
"input": "5\n10 30 5 6 24\n10 41 7 8 24",
"output": "3 11"
},
{
"input": "1\n1\n100",
"output": "1 0"
},
{
"input": "1\n100\n100",
"output": "1 0"
},
{
"input": "1\n50\n100",
"output": "1 0"
},
{
"input": "10\n18 42 5 1 26 8 40 34 8 29\n18 71 21 67 38 13 99 37 47 76",
"output": "3 100"
},
{
"input": "20\n24 22 4 34 76 13 78 1 81 51 72 11 25 46 22 33 60 42 25 19\n40 81 10 34 84 16 90 38 99 81 100 19 79 65 26 80 62 47 76 47",
"output": "9 217"
},
{
"input": "30\n29 3 2 13 3 12 73 22 37 48 59 17 2 13 69 43 32 14 4 2 61 22 40 30 1 4 46 5 65 17\n55 3 3 92 25 27 97 40 55 74 91 31 7 33 72 62 61 40 16 2 70 61 67 72 8 5 48 9 75 84",
"output": "10 310"
},
{
"input": "40\n9 18 41 31 27 24 76 32 4 38 1 35 21 3 26 32 31 13 41 31 39 14 45 15 12 5 7 14 3 14 19 11 1 81 1 4 7 28 4 62\n70 21 95 63 66 30 100 42 4 80 83 39 34 6 27 55 72 38 43 48 81 53 54 30 63 23 9 59 3 83 83 95 1 81 30 40 35 58 8 66",
"output": "11 560"
},
{
"input": "50\n48 29 72 22 99 27 40 23 39 4 46 29 39 16 47 35 79 7 15 23 50 34 35 22 9 2 51 10 2 42 4 3 30 2 72 19 50 20 11 29 1 2 1 7 7 6 7 75 40 69\n81 36 76 26 100 41 99 39 52 73 83 51 54 86 73 49 79 27 83 90 100 40 49 81 22 54 85 21 26 79 36 96 73 10 98 31 65 39 89 39 1 32 5 20 71 39 87 80 60 86",
"output": "17 563"
},
{
"input": "60\n3 3 22 46 23 19 2 27 3 26 34 18 8 50 13 18 23 26 9 14 7 2 17 12 63 25 4 71 14 47 70 13 6 38 28 22 94 10 51 7 29 1 54 12 8 5 4 34 11 24 2 14 54 65 11 30 3 23 12 11\n4 54 69 97 45 53 2 41 4 74 78 66 85 59 19 38 82 28 11 41 15 43 41 43 77 77 50 75 46 66 97 93 50 44 69 22 94 23 61 27 44 1 56 25 31 63 8 37 23 57 6 17 54 68 14 40 43 31 31 60",
"output": "19 535"
},
{
"input": "70\n17 70 52 31 15 51 8 38 3 43 2 34 7 16 58 29 73 23 41 88 9 24 24 90 33 84 10 29 67 17 47 72 11 79 22 5 8 65 23 7 29 31 11 42 11 14 9 3 54 22 38 34 2 4 39 13 11 34 3 35 22 18 3 57 23 21 13 23 78 7\n18 72 58 55 87 56 9 39 60 79 74 82 9 39 66 32 89 25 46 95 26 31 28 94 36 96 19 37 77 61 50 82 22 81 37 9 11 96 33 12 90 74 11 42 88 86 24 3 85 31 82 81 3 7 69 47 27 51 49 98 33 40 5 94 83 35 21 24 89 49",
"output": "26 756"
},
{
"input": "80\n2 8 36 12 22 41 1 42 6 66 62 94 37 1 5 1 82 8 9 31 14 8 15 5 21 8 5 22 1 17 1 44 1 12 8 45 37 38 13 4 13 4 8 8 3 15 13 53 22 8 19 14 16 7 7 49 1 10 31 33 7 47 61 6 9 48 6 25 16 4 43 1 5 34 8 22 31 38 59 45\n33 90 47 22 28 67 4 44 13 76 65 94 40 8 12 21 88 15 74 37 37 22 19 53 91 26 88 99 1 61 3 75 2 14 8 96 41 76 13 96 41 44 66 48 40 17 41 60 48 9 62 46 56 46 31 63 6 84 68 43 7 88 62 36 52 92 23 27 46 87 52 9 50 44 33 30 33 63 79 72",
"output": "21 909"
},
{
"input": "90\n4 2 21 69 53 39 2 2 8 58 7 5 2 82 7 9 13 10 2 44 1 7 2 1 50 42 36 17 14 46 19 1 50 20 51 46 9 59 73 61 76 4 19 22 1 43 53 2 5 5 32 7 5 42 30 14 32 6 6 15 20 24 13 8 5 19 9 9 7 20 7 2 55 36 5 33 64 20 22 22 9 30 67 38 68 2 13 19 2 9\n48 4 39 85 69 70 11 42 65 77 61 6 60 84 67 15 99 12 2 84 51 17 10 3 50 45 57 53 20 52 64 72 74 44 80 83 70 61 82 81 88 17 22 53 1 44 66 21 10 84 39 11 5 77 93 74 90 17 83 85 70 36 28 87 6 48 22 23 100 22 97 64 96 89 52 49 95 93 34 37 18 69 69 43 83 70 14 54 2 30",
"output": "25 955"
},
{
"input": "10\n96 4 51 40 89 36 35 38 4 82\n99 8 56 42 94 46 35 43 4 84",
"output": "8 8"
},
{
"input": "20\n59 35 29 57 85 70 26 53 56 3 11 56 43 20 81 72 77 72 36 61\n67 53 80 69 100 71 30 63 60 3 20 56 75 23 97 80 81 85 49 80",
"output": "13 187"
},
{
"input": "30\n33 4 1 42 86 85 35 51 45 88 23 35 79 92 81 46 47 32 41 17 18 36 28 58 31 15 17 38 49 78\n36 4 1 49 86 86 43 51 64 93 24 42 82 98 92 47 56 41 41 25 20 53 32 61 53 26 20 38 49 98",
"output": "22 123"
},
{
"input": "40\n31 72 17 63 89 13 72 42 39 30 23 29 5 61 88 37 7 23 49 32 41 25 17 15 9 25 30 61 29 66 24 40 75 67 69 22 61 22 13 35\n32 73 20 68 98 13 74 79 41 33 27 85 5 68 95 44 9 24 95 36 45 26 20 31 10 53 37 72 51 84 24 59 80 75 74 22 72 27 13 39",
"output": "24 290"
},
{
"input": "50\n72 9 46 38 43 75 63 73 70 11 9 48 32 93 33 24 46 44 27 78 43 2 26 84 42 78 35 34 76 36 67 79 82 63 17 26 30 43 35 34 54 37 13 65 8 37 8 8 70 79\n96 19 54 54 44 75 66 80 71 12 9 54 38 95 39 25 48 52 39 86 44 2 27 99 54 99 35 44 80 36 86 93 98 73 27 30 39 43 80 34 61 38 13 69 9 37 8 9 75 97",
"output": "34 283"
},
{
"input": "60\n70 19 46 34 43 19 75 42 47 14 66 64 63 58 55 79 38 45 49 80 72 54 96 26 63 41 12 55 14 56 79 51 12 9 14 77 70 75 46 27 45 10 76 59 40 67 55 24 26 90 50 75 12 93 27 39 46 58 66 31\n73 23 48 49 53 23 76 62 65 14 67 89 66 71 59 90 40 47 68 82 81 61 96 48 99 53 13 60 21 63 83 75 15 12 16 80 74 87 66 31 45 12 76 61 45 88 55 32 28 90 50 75 12 94 29 51 57 85 84 38",
"output": "42 368"
},
{
"input": "70\n67 38 59 72 9 64 12 3 51 58 50 4 16 46 62 77 58 73 7 92 48 9 90 50 35 9 61 57 50 20 48 61 27 77 47 6 83 28 78 14 68 32 2 2 22 57 34 71 26 74 3 76 41 66 30 69 34 16 29 7 14 19 11 5 13 66 19 19 17 55\n69 41 84 91 10 77 12 7 70 74 55 7 30 63 66 79 89 88 10 93 89 15 91 81 41 26 65 67 55 37 73 94 34 94 47 6 90 31 100 25 69 33 2 3 43 97 37 95 35 85 3 78 50 86 30 73 34 21 32 13 21 32 11 5 13 80 23 20 17 58",
"output": "38 484"
},
{
"input": "80\n36 80 23 45 68 72 2 69 84 33 3 43 6 64 82 54 15 15 17 4 3 29 74 14 53 50 52 27 32 18 60 62 50 29 28 48 77 11 24 17 3 55 58 20 4 32 55 16 27 60 5 77 23 31 11 60 21 65 38 39 82 58 51 78 24 30 75 79 5 41 94 10 14 7 1 26 21 41 6 52\n37 93 24 46 99 74 2 93 86 33 3 44 6 71 88 65 15 19 24 4 3 40 82 14 62 81 56 30 33 30 62 62 70 29 31 53 78 13 27 31 3 65 61 20 5 41 58 25 27 61 6 87 26 31 13 62 25 71 44 45 82 75 62 95 24 44 82 94 6 50 94 10 15 15 1 29 35 60 8 68",
"output": "50 363"
},
{
"input": "10\n5 12 10 18 10 9 2 20 5 20\n70 91 36 94 46 15 10 73 55 43",
"output": "2 71"
},
{
"input": "20\n8 1 44 1 12 1 9 11 1 1 5 2 9 16 16 2 1 5 4 1\n88 2 80 33 55 3 74 61 17 11 11 16 42 81 88 14 4 81 60 10",
"output": "2 90"
},
{
"input": "30\n10 1 8 10 2 6 45 7 3 7 1 3 1 1 14 2 5 19 4 1 13 3 5 6 1 5 1 1 23 1\n98 4 43 41 56 58 85 51 47 55 20 85 93 12 49 15 95 72 20 4 68 24 16 97 21 52 18 69 89 15",
"output": "3 122"
},
{
"input": "40\n10 32 10 7 10 6 25 3 18 4 24 4 8 14 6 15 11 8 2 8 2 5 19 9 5 5 3 34 5 1 6 6 1 4 5 26 34 2 21 1\n35 66 54 11 58 68 75 12 69 94 80 33 23 48 45 66 94 53 25 53 83 30 64 49 69 84 73 85 26 41 10 65 23 56 58 93 58 7 100 7",
"output": "5 281"
},
{
"input": "50\n2 1 2 2 38 19 1 2 7 1 2 5 5 1 14 53 21 1 17 9 4 1 24 8 1 1 1 5 4 14 37 1 15 1 4 15 1 3 3 16 17 1 10 18 36 14 25 8 8 48\n45 24 8 12 83 37 6 20 88 9 10 11 28 9 60 98 76 20 84 95 15 45 74 48 37 2 46 34 99 57 94 70 31 22 11 88 58 25 20 73 64 64 81 80 59 64 92 31 43 89",
"output": "6 337"
},
{
"input": "60\n9 9 11 16 58 6 25 6 3 23 1 14 1 8 4 2 1 18 10 1 13 4 23 1 38 6 1 13 5 1 1 1 2 1 1 17 1 24 18 20 2 1 9 26 1 12 3 6 7 17 18 1 2 9 3 6 3 30 7 12\n47 82 78 52 99 51 90 23 58 49 2 98 100 60 25 60 6 69 79 6 91 47 69 18 99 46 30 51 11 3 42 17 33 61 14 81 16 76 72 94 13 5 51 88 26 43 80 31 26 70 93 76 18 67 25 86 60 81 40 38",
"output": "7 368"
},
{
"input": "70\n20 7 5 7 3 10 1 14 33 1 5 3 4 21 7 7 1 2 2 2 8 15 18 2 7 1 1 1 15 2 27 2 6 21 4 2 7 5 1 6 13 36 13 1 10 5 8 13 24 2 10 16 11 9 4 1 1 8 6 26 9 3 3 2 8 5 17 9 1 13\n85 36 76 36 65 24 37 56 78 42 33 13 29 93 31 38 1 59 71 31 28 55 70 14 33 9 1 5 41 22 86 41 92 89 88 10 39 54 6 32 58 82 49 22 62 44 29 19 54 12 59 54 51 80 66 16 22 74 8 68 35 34 24 8 22 14 55 76 32 75",
"output": "7 426"
},
{
"input": "80\n11 6 9 6 5 18 21 11 6 6 2 9 4 1 10 12 2 9 1 14 6 12 16 14 4 5 1 16 3 4 6 1 11 30 2 4 1 11 1 6 1 3 2 14 6 14 13 1 10 2 4 14 11 8 28 2 2 3 1 6 26 3 11 4 1 1 29 4 5 4 3 5 1 4 2 12 59 3 18 1\n94 43 36 86 12 75 50 80 55 14 5 97 17 25 28 86 51 56 17 88 48 40 31 39 51 58 4 75 70 30 11 8 61 88 10 25 35 46 31 51 20 79 22 54 19 67 31 89 42 70 30 37 35 78 95 31 31 51 31 50 54 90 63 27 6 2 92 80 48 9 27 33 61 63 30 38 95 46 86 45",
"output": "8 434"
},
{
"input": "90\n1 9 3 3 14 3 2 32 17 3 1 1 4 1 18 1 1 21 9 1 2 10 6 9 27 15 5 1 3 37 1 2 1 12 6 1 8 4 1 5 1 3 8 9 1 9 23 1 1 2 1 2 2 19 2 6 5 6 1 7 12 35 1 2 8 1 11 32 7 4 12 9 18 8 9 27 31 15 16 4 16 13 2 2 1 4 12 17 10 1\n8 52 13 56 42 40 8 98 64 47 84 11 12 1 97 8 8 66 35 4 6 62 22 38 68 57 50 28 28 88 7 57 9 81 14 37 71 57 33 24 2 21 54 58 58 27 79 3 55 13 2 95 17 97 61 22 28 85 78 72 68 80 12 41 98 18 35 70 40 22 98 85 51 70 79 100 68 29 73 45 89 64 53 6 16 29 73 53 24 69",
"output": "8 562"
},
{
"input": "32\n4 1 1 6 2 5 8 6 5 6 3 2 1 3 1 9 1 2 1 5 2 1 6 5 3 7 3 3 2 5 1 1\n8 1 3 6 4 7 9 8 6 8 10 2 5 3 2 10 1 10 9 5 4 1 8 7 8 7 4 10 4 6 9 2",
"output": "13 46"
},
{
"input": "38\n2 1 1 1 1 9 5 2 1 3 4 3 1 7 4 4 8 7 1 5 4 9 1 6 3 4 1 4 1 5 5 1 8 3 1 3 6 3\n2 1 6 2 9 10 6 2 1 5 4 6 1 7 4 6 10 8 8 6 4 10 1 6 4 4 6 4 4 8 5 2 10 7 3 5 6 3",
"output": "19 40"
},
{
"input": "35\n9 7 34 3 2 6 36 3 26 12 17 8 5 32 55 10 24 19 2 3 30 17 14 1 33 36 42 14 51 1 2 22 13 34 28\n9 9 55 17 16 12 37 14 27 58 51 16 10 37 69 15 43 26 14 60 86 34 54 1 37 50 58 18 92 66 7 24 25 92 30",
"output": "10 307"
},
{
"input": "35\n21 2 68 56 41 25 42 17 21 20 29 26 38 37 29 77 43 13 32 48 38 31 15 8 52 6 63 45 70 2 21 13 3 14 47\n46 83 100 87 59 95 47 33 56 60 38 76 63 75 60 92 65 43 56 94 70 80 46 40 64 6 83 50 75 19 52 66 13 88 62",
"output": "14 432"
},
{
"input": "69\n24 32 19 37 36 7 15 10 54 12 15 46 3 25 12 16 3 8 55 21 23 57 17 45 11 4 25 35 39 3 69 24 78 40 12 39 1 44 4 75 53 60 1 6 30 7 6 39 44 13 31 6 4 4 32 11 52 58 81 2 33 7 29 19 21 26 22 60 24\n57 56 50 64 40 58 31 20 81 14 43 64 48 38 56 71 58 26 98 92 52 88 71 93 11 20 79 39 56 7 92 54 88 58 19 85 12 71 4 87 78 90 29 18 89 13 86 71 100 24 65 95 46 8 91 35 62 66 96 36 80 24 81 58 53 86 89 67 73",
"output": "22 801"
},
{
"input": "63\n8 23 6 19 1 34 23 1 15 58 22 10 5 14 41 1 16 48 68 5 13 19 1 4 35 2 42 8 45 24 52 44 59 78 5 11 14 41 10 26 60 26 9 15 34 1 14 5 2 6 19 7 4 26 49 39 13 40 18 62 66 8 4\n17 25 39 45 2 44 40 1 82 68 80 27 7 58 90 20 100 80 79 21 53 62 2 11 51 98 78 55 48 37 89 74 83 91 64 30 20 50 24 74 81 94 33 64 56 28 57 9 27 50 81 34 18 33 53 61 39 89 44 77 86 40 89",
"output": "18 638"
},
{
"input": "73\n69 67 34 35 10 27 30 27 31 48 25 18 81 54 32 54 5 62 20 4 94 2 60 4 6 11 62 68 14 18 42 18 33 71 72 2 29 7 36 60 10 25 17 2 38 77 34 36 74 76 63 32 42 29 22 14 5 1 6 2 14 19 20 19 41 31 16 17 50 49 2 22 51\n73 70 58 54 10 71 59 35 91 61 52 65 90 70 37 80 12 94 78 34 97 4 62 95 10 11 93 100 14 38 56 42 96 96 84 71 69 43 50 79 11 83 95 76 39 79 61 42 89 90 71 62 43 38 39 21 5 40 27 13 21 73 30 46 47 34 23 22 57 59 6 25 72",
"output": "30 808"
},
{
"input": "90\n1 43 87 1 6 12 49 6 3 9 38 1 64 49 11 18 5 1 46 25 30 82 17 4 8 9 5 5 4 1 10 4 13 42 44 90 1 11 27 23 25 4 12 19 48 3 59 48 39 14 1 5 64 46 39 24 28 77 25 20 3 14 28 2 20 63 2 1 13 11 44 49 61 76 20 1 3 42 38 8 69 17 27 18 29 54 2 1 2 7\n8 96 91 1 11 20 83 34 41 88 54 4 65 82 48 60 62 18 76 74 75 89 87 8 11 32 67 7 5 1 92 88 57 92 76 95 35 58 68 23 30 25 12 31 85 5 89 84 71 23 1 5 76 56 57 57 76 94 33 34 66 20 54 5 22 69 2 19 28 62 74 88 91 86 30 6 3 48 80 10 84 20 44 37 81 100 12 3 6 8",
"output": "26 899"
},
{
"input": "85\n20 47 52 6 5 15 35 42 5 84 4 8 61 47 7 50 20 24 15 27 86 28 1 39 1 2 63 2 31 33 47 4 33 68 20 4 4 42 20 67 7 10 46 4 22 36 30 40 4 15 51 2 39 50 65 48 34 6 50 19 32 48 8 23 42 70 69 8 29 81 5 1 7 21 3 30 78 6 2 1 3 69 34 34 18\n74 64 89 61 5 17 75 43 13 87 30 51 93 54 7 76 44 44 98 77 86 97 1 41 1 3 69 3 80 87 67 6 90 100 31 5 7 46 99 67 9 44 56 7 39 39 55 80 80 33 77 9 89 79 86 53 49 49 72 87 43 84 24 23 43 94 74 17 54 96 28 64 14 42 91 60 87 69 20 1 30 95 44 50 20",
"output": "29 987"
},
{
"input": "81\n21 13 1 25 14 33 33 41 53 89 2 18 61 8 3 35 15 59 2 2 3 5 75 37 1 34 7 12 33 66 6 4 14 78 3 16 12 45 3 2 1 17 17 45 4 30 68 40 44 3 1 21 64 63 14 19 75 63 7 9 12 75 20 28 16 20 53 26 13 46 18 8 28 32 9 29 1 11 75 4 21\n45 90 21 31 36 68 71 47 59 89 61 32 98 67 7 53 90 86 6 28 4 83 93 62 8 56 18 35 33 92 36 37 23 98 44 21 23 79 10 4 2 18 48 87 29 86 79 74 45 3 6 23 79 71 17 39 88 73 50 15 13 92 33 47 83 48 73 33 15 63 43 14 90 72 9 95 1 22 83 20 29",
"output": "26 754"
},
{
"input": "2\n1 1\n1 1",
"output": "2 0"
},
{
"input": "1\n1\n1",
"output": "1 0"
},
{
"input": "1\n1\n2",
"output": "1 0"
},
{
"input": "2\n1 1\n1 100",
"output": "1 1"
},
{
"input": "2\n1 1\n100 1",
"output": "1 1"
},
{
"input": "86\n5 1 3 1 1 1 1 9 4 1 3 1 4 6 3 2 2 7 1 1 3 1 2 1 1 5 4 3 6 3 3 4 8 2 1 3 1 2 7 2 5 4 2 1 1 2 1 3 2 9 1 4 2 1 1 9 6 1 8 1 7 9 4 3 4 1 3 1 1 3 1 1 3 1 1 10 7 7 4 1 1 3 1 6 1 3\n10 2 5 7 1 4 7 9 4 7 3 1 5 6 3 8 4 10 5 1 9 3 4 2 1 5 7 4 7 7 7 5 9 5 3 3 6 4 7 2 9 7 3 4 2 3 1 5 6 9 10 4 8 10 10 9 7 8 10 1 7 10 10 7 8 5 8 2 1 4 1 2 3 8 1 10 9 7 4 2 1 3 4 9 2 3",
"output": "32 101"
},
{
"input": "90\n9 2 2 3 4 1 9 8 3 3 1 1 1 1 2 2 1 3 4 8 8 1 2 7 3 4 5 6 1 2 9 4 2 5 6 1 1 2 6 5 1 4 3 2 4 1 1 3 1 1 3 1 8 3 1 4 1 2 2 3 5 2 8 6 2 5 2 1 4 2 1 5 4 2 1 1 2 1 1 6 4 4 3 4 1 4 4 6 2 3\n10 6 2 3 10 1 10 10 6 4 1 3 6 1 2 5 3 7 7 9 9 2 3 8 3 4 9 7 8 4 10 7 8 10 9 5 1 4 6 5 1 9 10 4 6 4 1 3 3 1 6 1 9 4 1 6 4 5 5 10 7 9 9 10 4 5 2 1 4 2 1 7 6 5 3 9 2 5 1 8 6 4 6 10 1 7 5 9 6 4",
"output": "35 109"
},
{
"input": "33\n33 20 33 40 58 50 5 6 13 12 4 33 11 50 12 19 16 36 68 57 23 17 6 22 39 58 49 21 10 35 35 17 12\n62 22 53 44 66 60 97 7 33 18 10 59 33 77 55 63 91 86 87 86 27 62 65 53 46 69 64 63 10 53 52 23 24",
"output": "13 356"
},
{
"input": "83\n13 20 5 29 48 53 88 17 11 5 44 15 85 13 2 55 6 16 57 29 12 15 12 92 21 25 1 2 4 5 2 22 8 18 22 2 3 10 43 71 3 41 1 73 6 18 32 63 26 13 6 75 19 10 41 30 15 12 14 8 15 77 73 7 5 39 83 19 2 2 3 61 53 43 3 15 76 29 8 46 19 3 8\n54 34 15 58 50 67 100 43 30 15 46 26 94 75 2 58 85 38 68 98 83 51 82 100 61 27 5 5 41 89 17 34 10 48 48 4 15 13 71 75 4 44 2 82 18 82 59 96 26 13 66 95 81 33 85 45 16 92 41 37 85 78 83 17 7 72 83 38 69 24 18 76 71 66 3 66 78 31 73 72 43 89 49",
"output": "26 944"
},
{
"input": "70\n13 42 8 56 21 58 39 2 49 39 15 26 62 45 26 8 47 40 9 36 41 2 4 38 6 55 2 41 72 18 10 2 6 11 4 39 19 39 14 59 5 42 19 79 12 3 1 1 21 6 5 9 36 6 38 2 7 26 8 15 66 7 1 30 93 34 45 24 12 20\n26 56 25 60 26 79 99 7 68 92 99 32 81 48 39 97 49 95 18 82 59 4 99 41 10 63 43 54 76 97 73 7 17 43 4 84 35 86 20 63 8 59 87 80 34 3 8 13 49 55 14 11 68 8 41 33 14 39 43 31 89 13 7 88 93 51 84 73 26 30",
"output": "21 867"
},
{
"input": "77\n19 34 39 56 1 2 47 8 17 28 23 45 18 7 5 3 11 20 30 24 13 34 11 1 4 14 68 23 13 33 3 8 1 5 8 23 12 1 19 14 22 67 26 55 10 1 63 82 82 6 38 5 6 11 1 62 1 12 5 40 19 20 37 9 5 3 2 44 13 20 44 32 11 29 12 19 35\n28 41 43 68 1 36 57 13 84 89 26 92 47 19 7 94 79 75 74 42 32 44 46 23 96 46 82 86 91 33 25 11 12 68 22 31 89 14 81 32 50 94 27 66 50 39 98 90 91 11 69 6 45 19 15 74 22 31 7 92 23 98 88 32 8 4 2 51 79 69 70 43 16 60 29 20 98",
"output": "19 937"
},
{
"input": "77\n44 2 13 14 8 46 65 14 1 39 12 18 15 10 2 40 71 40 17 1 16 72 13 7 41 23 81 12 4 1 19 18 41 35 23 56 21 5 17 47 88 1 24 15 48 15 1 13 50 5 31 16 21 47 4 1 49 2 15 23 46 47 27 22 23 40 29 4 30 50 51 12 20 14 41 25 12\n57 16 72 59 28 80 74 19 4 60 52 52 97 20 5 69 84 66 63 38 50 79 24 84 58 92 99 36 38 97 66 79 41 48 26 95 28 38 28 72 95 71 30 15 63 17 7 69 90 29 89 40 21 83 73 24 51 14 15 74 100 88 74 27 46 61 38 4 32 52 52 51 47 51 81 75 19",
"output": "24 932"
}
] | 1,477,217,673 | 8,073 | Python 3 | WRONG_ANSWER | TESTS | 6 | 62 | 0 | n = int(input())
a, b = list(map(int, input().split())), list(map(int, input().split()))
have = sum(a)
k, c = 0, 0
for v in sorted(b)[::-1]:
c += v
k += 1
if c >= have:
break
t = sum(sorted(a)[:n-k])
print(k, t)
| Title: Bottles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nick has *n* bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda *a**i* and bottle volume *b**i* (*a**i*<=≤<=*b**i*).
Nick has decided to pour all remaining soda into minimal number of bottles, moreover he has to do it as soon as possible. Nick spends *x* seconds to pour *x* units of soda from one bottle to another.
Nick asks you to help him to determine *k* — the minimal number of bottles to store all remaining soda and *t* — the minimal time to pour soda into *k* bottles. A bottle can't store more soda than its volume. All remaining soda should be saved.
Input Specification:
The first line contains positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of bottles.
The second line contains *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is the amount of soda remaining in the *i*-th bottle.
The third line contains *n* positive integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=100), where *b**i* is the volume of the *i*-th bottle.
It is guaranteed that *a**i*<=≤<=*b**i* for any *i*.
Output Specification:
The only line should contain two integers *k* and *t*, where *k* is the minimal number of bottles that can store all the soda and *t* is the minimal time to pour the soda into *k* bottles.
Demo Input:
['4\n3 3 4 3\n4 7 6 5\n', '2\n1 1\n100 100\n', '5\n10 30 5 6 24\n10 41 7 8 24\n']
Demo Output:
['2 6\n', '1 1\n', '3 11\n']
Note:
In the first example Nick can pour soda from the first bottle to the second bottle. It will take 3 seconds. After it the second bottle will contain 3 + 3 = 6 units of soda. Then he can pour soda from the fourth bottle to the second bottle and to the third bottle: one unit to the second and two units to the third. It will take 1 + 2 = 3 seconds. So, all the soda will be in two bottles and he will spend 3 + 3 = 6 seconds to do it. | ```python
n = int(input())
a, b = list(map(int, input().split())), list(map(int, input().split()))
have = sum(a)
k, c = 0, 0
for v in sorted(b)[::-1]:
c += v
k += 1
if c >= have:
break
t = sum(sorted(a)[:n-k])
print(k, t)
``` | 0 | |
556 | A | Case of the Zeros and Ones | PROGRAMMING | 900 | [
"greedy"
] | null | null | Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result.
Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number. | First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has.
The second line contains the string of length *n* consisting only from zeros and ones. | Output the minimum length of the string that may remain after applying the described operations several times. | [
"4\n1100\n",
"5\n01010\n",
"8\n11101111\n"
] | [
"0\n",
"1\n",
"6\n"
] | In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 250 | [
{
"input": "4\n1100",
"output": "0"
},
{
"input": "5\n01010",
"output": "1"
},
{
"input": "8\n11101111",
"output": "6"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n00",
"output": "2"
},
{
"input": "2\n01",
"output": "0"
},
{
"input": "2\n10",
"output": "0"
},
{
"input": "2\n11",
"output": "2"
},
{
"input": "3\n001",
"output": "1"
},
{
"input": "6\n110110",
"output": "2"
},
{
"input": "7\n0000011",
"output": "3"
},
{
"input": "6\n110010",
"output": "0"
},
{
"input": "6\n110100",
"output": "0"
},
{
"input": "3\n100",
"output": "1"
},
{
"input": "6\n010111",
"output": "2"
},
{
"input": "8\n01011100",
"output": "0"
},
{
"input": "6\n001011",
"output": "0"
},
{
"input": "7\n1110000",
"output": "1"
},
{
"input": "9\n011111101",
"output": "5"
}
] | 1,679,192,580 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 78 | 409,600 | tamanho_string = int(input())
string = input()
conta_u = 0
conta_z = 0
for x in string:
if x == '1': conta_u += 1
if x == '0': conta_z += 1
if conta_u < conta_z:
tamanho_string -= conta_u * 2
else:
tamanho_string -= conta_z * 2
print(tamanho_string)
| Title: Case of the Zeros and Ones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result.
Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number.
Input Specification:
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has.
The second line contains the string of length *n* consisting only from zeros and ones.
Output Specification:
Output the minimum length of the string that may remain after applying the described operations several times.
Demo Input:
['4\n1100\n', '5\n01010\n', '8\n11101111\n']
Demo Output:
['0\n', '1\n', '6\n']
Note:
In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
tamanho_string = int(input())
string = input()
conta_u = 0
conta_z = 0
for x in string:
if x == '1': conta_u += 1
if x == '0': conta_z += 1
if conta_u < conta_z:
tamanho_string -= conta_u * 2
else:
tamanho_string -= conta_z * 2
print(tamanho_string)
``` | 3 | |
131 | A | cAPS lOCK | PROGRAMMING | 1,000 | [
"implementation",
"strings"
] | null | null | wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
- either it only contains uppercase letters; - or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged. | The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive. | Print the result of the given word's processing. | [
"cAPS\n",
"Lock\n"
] | [
"Caps",
"Lock\n"
] | none | 500 | [
{
"input": "cAPS",
"output": "Caps"
},
{
"input": "Lock",
"output": "Lock"
},
{
"input": "cAPSlOCK",
"output": "cAPSlOCK"
},
{
"input": "CAPs",
"output": "CAPs"
},
{
"input": "LoCK",
"output": "LoCK"
},
{
"input": "OOPS",
"output": "oops"
},
{
"input": "oops",
"output": "oops"
},
{
"input": "a",
"output": "A"
},
{
"input": "A",
"output": "a"
},
{
"input": "aA",
"output": "Aa"
},
{
"input": "Zz",
"output": "Zz"
},
{
"input": "Az",
"output": "Az"
},
{
"input": "zA",
"output": "Za"
},
{
"input": "AAA",
"output": "aaa"
},
{
"input": "AAa",
"output": "AAa"
},
{
"input": "AaR",
"output": "AaR"
},
{
"input": "Tdr",
"output": "Tdr"
},
{
"input": "aTF",
"output": "Atf"
},
{
"input": "fYd",
"output": "fYd"
},
{
"input": "dsA",
"output": "dsA"
},
{
"input": "fru",
"output": "fru"
},
{
"input": "hYBKF",
"output": "Hybkf"
},
{
"input": "XweAR",
"output": "XweAR"
},
{
"input": "mogqx",
"output": "mogqx"
},
{
"input": "eOhEi",
"output": "eOhEi"
},
{
"input": "nkdku",
"output": "nkdku"
},
{
"input": "zcnko",
"output": "zcnko"
},
{
"input": "lcccd",
"output": "lcccd"
},
{
"input": "vwmvg",
"output": "vwmvg"
},
{
"input": "lvchf",
"output": "lvchf"
},
{
"input": "IUNVZCCHEWENCHQQXQYPUJCRDZLUXCLJHXPHBXEUUGNXOOOPBMOBRIBHHMIRILYJGYYGFMTMFSVURGYHUWDRLQVIBRLPEVAMJQYO",
"output": "iunvzcchewenchqqxqypujcrdzluxcljhxphbxeuugnxooopbmobribhhmirilyjgyygfmtmfsvurgyhuwdrlqvibrlpevamjqyo"
},
{
"input": "OBHSZCAMDXEJWOZLKXQKIVXUUQJKJLMMFNBPXAEFXGVNSKQLJGXHUXHGCOTESIVKSFMVVXFVMTEKACRIWALAGGMCGFEXQKNYMRTG",
"output": "obhszcamdxejwozlkxqkivxuuqjkjlmmfnbpxaefxgvnskqljgxhuxhgcotesivksfmvvxfvmtekacriwalaggmcgfexqknymrtg"
},
{
"input": "IKJYZIKROIYUUCTHSVSKZTETNNOCMAUBLFJCEVANCADASMZRCNLBZPQRXESHEEMOMEPCHROSRTNBIDXYMEPJSIXSZQEBTEKKUHFS",
"output": "ikjyzikroiyuucthsvskztetnnocmaublfjcevancadasmzrcnlbzpqrxesheemomepchrosrtnbidxymepjsixszqebtekkuhfs"
},
{
"input": "cTKDZNWVYRTFPQLDAUUNSPKTDJTUPPFPRXRSINTVFVNNQNKXWUZUDHZBUSOKTABUEDQKUIVRTTVUREEOBJTSDKJKVEGFXVHXEYPE",
"output": "Ctkdznwvyrtfpqldauunspktdjtuppfprxrsintvfvnnqnkxwuzudhzbusoktabuedqkuivrttvureeobjtsdkjkvegfxvhxeype"
},
{
"input": "uCKJZRGZJCPPLEEYJTUNKOQSWGBMTBQEVPYFPIPEKRVYQNTDPANOIXKMPINNFUSZWCURGBDPYTEKBEKCPMVZPMWAOSHJYMGKOMBQ",
"output": "Uckjzrgzjcppleeyjtunkoqswgbmtbqevpyfpipekrvyqntdpanoixkmpinnfuszwcurgbdpytekbekcpmvzpmwaoshjymgkombq"
},
{
"input": "KETAXTSWAAOBKUOKUQREHIOMVMMRSAEWKGXZKRASwTVNSSFSNIWYNPSTMRADOADEEBURRHPOOBIEUIBGYDJCEKPNLEUCANZYJKMR",
"output": "KETAXTSWAAOBKUOKUQREHIOMVMMRSAEWKGXZKRASwTVNSSFSNIWYNPSTMRADOADEEBURRHPOOBIEUIBGYDJCEKPNLEUCANZYJKMR"
},
{
"input": "ZEKGDMWJPVUWFlNXRLUmWKLMMYSLRQQIBRWDPKWITUIMZYYKOEYGREKHHZRZZUFPVTNIHKGTCCTLOKSZITXXZDMPITHNZUIGDZLE",
"output": "ZEKGDMWJPVUWFlNXRLUmWKLMMYSLRQQIBRWDPKWITUIMZYYKOEYGREKHHZRZZUFPVTNIHKGTCCTLOKSZITXXZDMPITHNZUIGDZLE"
},
{
"input": "TcMbVPCFvnNkCEUUCIFLgBJeCOKuJhIGwXFrhAZjuAhBraMSchBfWwIuHAEbgJOFzGtxDLDXzDSaPCFujGGxgxdlHUIQYRrMFCgJ",
"output": "TcMbVPCFvnNkCEUUCIFLgBJeCOKuJhIGwXFrhAZjuAhBraMSchBfWwIuHAEbgJOFzGtxDLDXzDSaPCFujGGxgxdlHUIQYRrMFCgJ"
},
{
"input": "xFGqoLILNvxARKuIntPfeukFtMbvzDezKpPRAKkIoIvwqNXnehRVwkkXYvuRCeoieBaBfTjwsYhDeCLvBwktntyluoxCYVioXGdm",
"output": "xFGqoLILNvxARKuIntPfeukFtMbvzDezKpPRAKkIoIvwqNXnehRVwkkXYvuRCeoieBaBfTjwsYhDeCLvBwktntyluoxCYVioXGdm"
},
{
"input": "udvqolbxdwbkijwvhlyaelhynmnfgszbhgshlcwdkaibceqomzujndixuzivlsjyjqxzxodzbukxxhwwultvekdfntwpzlhhrIjm",
"output": "udvqolbxdwbkijwvhlyaelhynmnfgszbhgshlcwdkaibceqomzujndixuzivlsjyjqxzxodzbukxxhwwultvekdfntwpzlhhrIjm"
},
{
"input": "jgpwhetqqoncighgzbbaLwwwxkxivuwtokehrgprfgewzcwxkavwoflcgsgbhoeamzbefzoonwsyzisetoydrpufktzgbaycgaeg",
"output": "jgpwhetqqoncighgzbbaLwwwxkxivuwtokehrgprfgewzcwxkavwoflcgsgbhoeamzbefzoonwsyzisetoydrpufktzgbaycgaeg"
},
{
"input": "vyujsazdstbnkxeunedfbolicojzjpufgfemhtmdrswvmuhoivjvonacefqenbqudelmdegxqtbwezsbydmanzutvdgkgrjxzlnc",
"output": "vyujsazdstbnkxeunedfbolicojzjpufgfemhtmdrswvmuhoivjvonacefqenbqudelmdegxqtbwezsbydmanzutvdgkgrjxzlnc"
},
{
"input": "pivqnuqkaofcduvbttztjbuavrqwiqrwkfncmvatoxruelyoecnkpqraiahumiaiqeyjapbqyrsxcdgjbihivtqezvasfmzntdfv",
"output": "pivqnuqkaofcduvbttztjbuavrqwiqrwkfncmvatoxruelyoecnkpqraiahumiaiqeyjapbqyrsxcdgjbihivtqezvasfmzntdfv"
},
{
"input": "upvtbsxswbohxshdrbjxcungzquhuomgxwlryvshshsfvqbrxvcikbglietlpqkiwbhiqpmdwuqosbtdvyxekkaqiwrbsibpoect",
"output": "upvtbsxswbohxshdrbjxcungzquhuomgxwlryvshshsfvqbrxvcikbglietlpqkiwbhiqpmdwuqosbtdvyxekkaqiwrbsibpoect"
},
{
"input": "cTKDZNWVYRTFPQLDAUUNSPKTDJTUPPFPRXRSINTVFVNNQNKXWUZUDHZBUSOKTABUEDQKUIVRTTVUREEOBJTSDKJKVEGFXVHXEYPE",
"output": "Ctkdznwvyrtfpqldauunspktdjtuppfprxrsintvfvnnqnkxwuzudhzbusoktabuedqkuivrttvureeobjtsdkjkvegfxvhxeype"
},
{
"input": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu",
"output": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu"
},
{
"input": "aBACABa",
"output": "aBACABa"
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAaa",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAaa"
},
{
"input": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA"
},
{
"input": "dDDDDDDDDDDDDD",
"output": "Dddddddddddddd"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "z",
"output": "Z"
},
{
"input": "AZ",
"output": "az"
},
{
"input": "Z",
"output": "z"
},
{
"input": "aAAAA",
"output": "Aaaaa"
},
{
"input": "F",
"output": "f"
}
] | 1,698,749,446 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 31 | 0 |
s = str(input())
isAllUpperCase = 1
printed = 0
for i, item in enumerate(s):
if s.isupper():
print(s[0].lower() + s[1:].lower())
printed = 1
break
if i == 0 and printed == 0:
if item.islower():
for j in range(i+1,len(s)):
if s[j].islower():
isAllUpperCase = 0
break
if isAllUpperCase == 1:
print(s[0].upper()+s[1:].lower())
printed = 1
break
if s[0].isupper() and s[1:].islower() and printed == 0:
print(s)
break
| Title: cAPS lOCK
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
- either it only contains uppercase letters; - or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input Specification:
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output Specification:
Print the result of the given word's processing.
Demo Input:
['cAPS\n', 'Lock\n']
Demo Output:
['Caps', 'Lock\n']
Note:
none | ```python
s = str(input())
isAllUpperCase = 1
printed = 0
for i, item in enumerate(s):
if s.isupper():
print(s[0].lower() + s[1:].lower())
printed = 1
break
if i == 0 and printed == 0:
if item.islower():
for j in range(i+1,len(s)):
if s[j].islower():
isAllUpperCase = 0
break
if isAllUpperCase == 1:
print(s[0].upper()+s[1:].lower())
printed = 1
break
if s[0].isupper() and s[1:].islower() and printed == 0:
print(s)
break
``` | 0 | |
977 | A | Wrong Subtraction | PROGRAMMING | 800 | [
"implementation"
] | null | null | Little girl Tanya is learning how to decrease a number by one, but she does it wrong with a number consisting of two or more digits. Tanya subtracts one from a number by the following algorithm:
- if the last digit of the number is non-zero, she decreases the number by one; - if the last digit of the number is zero, she divides the number by 10 (i.e. removes the last digit).
You are given an integer number $n$. Tanya will subtract one from it $k$ times. Your task is to print the result after all $k$ subtractions.
It is guaranteed that the result will be positive integer number. | The first line of the input contains two integer numbers $n$ and $k$ ($2 \le n \le 10^9$, $1 \le k \le 50$) — the number from which Tanya will subtract and the number of subtractions correspondingly. | Print one integer number — the result of the decreasing $n$ by one $k$ times.
It is guaranteed that the result will be positive integer number. | [
"512 4\n",
"1000000000 9\n"
] | [
"50\n",
"1\n"
] | The first example corresponds to the following sequence: $512 \rightarrow 511 \rightarrow 510 \rightarrow 51 \rightarrow 50$. | 0 | [
{
"input": "512 4",
"output": "50"
},
{
"input": "1000000000 9",
"output": "1"
},
{
"input": "131203 11",
"output": "12"
},
{
"input": "999999999 50",
"output": "9999"
},
{
"input": "999999999 49",
"output": "99990"
},
{
"input": "131203 9",
"output": "130"
},
{
"input": "900000000 16",
"output": "1"
},
{
"input": "909090909 50",
"output": "3"
},
{
"input": "1001 2",
"output": "100"
},
{
"input": "5 2",
"output": "3"
},
{
"input": "2 1",
"output": "1"
}
] | 1,697,210,797 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 11 | 46 | 0 | a, b = input().split(" ")
n = int(a)
k = int(b)
while k > 0:
if n % 10 > 0:
n = n-1
k = k-1
else:
n = int(n/10)
k = k-1
print(n)
| Title: Wrong Subtraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little girl Tanya is learning how to decrease a number by one, but she does it wrong with a number consisting of two or more digits. Tanya subtracts one from a number by the following algorithm:
- if the last digit of the number is non-zero, she decreases the number by one; - if the last digit of the number is zero, she divides the number by 10 (i.e. removes the last digit).
You are given an integer number $n$. Tanya will subtract one from it $k$ times. Your task is to print the result after all $k$ subtractions.
It is guaranteed that the result will be positive integer number.
Input Specification:
The first line of the input contains two integer numbers $n$ and $k$ ($2 \le n \le 10^9$, $1 \le k \le 50$) — the number from which Tanya will subtract and the number of subtractions correspondingly.
Output Specification:
Print one integer number — the result of the decreasing $n$ by one $k$ times.
It is guaranteed that the result will be positive integer number.
Demo Input:
['512 4\n', '1000000000 9\n']
Demo Output:
['50\n', '1\n']
Note:
The first example corresponds to the following sequence: $512 \rightarrow 511 \rightarrow 510 \rightarrow 51 \rightarrow 50$. | ```python
a, b = input().split(" ")
n = int(a)
k = int(b)
while k > 0:
if n % 10 > 0:
n = n-1
k = k-1
else:
n = int(n/10)
k = k-1
print(n)
``` | 3 | |
25 | A | IQ test | PROGRAMMING | 1,300 | [
"brute force"
] | A. IQ test | 2 | 256 | Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness. | The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness. | Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order. | [
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] | [
"3\n",
"2\n"
] | none | 0 | [
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,675,179,699 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 62 | 0 | # -*- coding: utf-8 -*-
"""Untitled4.ipynb
Automatically generated by Colaboratory.
Original file is located at
https://colab.research.google.com/drive/1DjikK3UYlHIDWRNlyljbOv992ZPOXwfB
"""
n=int(input())
l=[int(i)%2 for i in input().split(' ')[:n]]
print(l)
if l[0]==l[-1]:
print(l.index(0)+1)
if l[0]==0:
print(l.index(1)+1)
else:
print(l.index(0))
else:
if l[0]==l[-2]:
print(len(l))
else:
print(1)
| Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none | ```python
# -*- coding: utf-8 -*-
"""Untitled4.ipynb
Automatically generated by Colaboratory.
Original file is located at
https://colab.research.google.com/drive/1DjikK3UYlHIDWRNlyljbOv992ZPOXwfB
"""
n=int(input())
l=[int(i)%2 for i in input().split(' ')[:n]]
print(l)
if l[0]==l[-1]:
print(l.index(0)+1)
if l[0]==0:
print(l.index(1)+1)
else:
print(l.index(0))
else:
if l[0]==l[-2]:
print(len(l))
else:
print(1)
``` | 0 |
862 | B | Mahmoud and Ehab and the bipartiteness | PROGRAMMING | 1,300 | [
"dfs and similar",
"graphs",
"trees"
] | null | null | Mahmoud and Ehab continue their adventures! As everybody in the evil land knows, Dr. Evil likes bipartite graphs, especially trees.
A tree is a connected acyclic graph. A bipartite graph is a graph, whose vertices can be partitioned into 2 sets in such a way, that for each edge (*u*,<=*v*) that belongs to the graph, *u* and *v* belong to different sets. You can find more formal definitions of a tree and a bipartite graph in the notes section below.
Dr. Evil gave Mahmoud and Ehab a tree consisting of *n* nodes and asked them to add edges to it in such a way, that the graph is still bipartite. Besides, after adding these edges the graph should be simple (doesn't contain loops or multiple edges). What is the maximum number of edges they can add?
A loop is an edge, which connects a node with itself. Graph doesn't contain multiple edges when for each pair of nodes there is no more than one edge between them. A cycle and a loop aren't the same . | The first line of input contains an integer *n* — the number of nodes in the tree (1<=≤<=*n*<=≤<=105).
The next *n*<=-<=1 lines contain integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*) — the description of the edges of the tree.
It's guaranteed that the given graph is a tree. | Output one integer — the maximum number of edges that Mahmoud and Ehab can add to the tree while fulfilling the conditions. | [
"3\n1 2\n1 3\n",
"5\n1 2\n2 3\n3 4\n4 5\n"
] | [
"0\n",
"2\n"
] | Tree definition: [https://en.wikipedia.org/wiki/Tree_(graph_theory)](https://en.wikipedia.org/wiki/Tree_(graph_theory))
Bipartite graph definition: [https://en.wikipedia.org/wiki/Bipartite_graph](https://en.wikipedia.org/wiki/Bipartite_graph)
In the first test case the only edge that can be added in such a way, that graph won't contain loops or multiple edges is (2, 3), but adding this edge will make the graph non-bipartite so the answer is 0.
In the second test case Mahmoud and Ehab can add edges (1, 4) and (2, 5). | 1,000 | [
{
"input": "3\n1 2\n1 3",
"output": "0"
},
{
"input": "5\n1 2\n2 3\n3 4\n4 5",
"output": "2"
},
{
"input": "10\n3 8\n6 2\n9 7\n10 1\n3 5\n1 3\n6 7\n5 4\n3 6",
"output": "16"
},
{
"input": "10\n7 6\n2 7\n4 1\n8 5\n9 4\n5 3\n8 7\n10 8\n10 4",
"output": "16"
},
{
"input": "10\n2 6\n3 7\n8 4\n4 10\n6 9\n9 7\n3 10\n1 2\n5 8",
"output": "16"
},
{
"input": "10\n6 9\n9 7\n9 4\n10 9\n9 1\n9 8\n9 2\n9 5\n3 9",
"output": "0"
},
{
"input": "2\n1 2",
"output": "0"
}
] | 1,506,779,693 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 62 | 0 | n = int(input())
s1 = {}
s2 = {}
pair = []
for i in range(n -1):
row = input().split(' ')
f = int(row[0]); t = int(row[1])
pair.append((f,t))
pair.sort()
for f,t in pair:
if f in s1:
s2[t] = True
elif f in s2:
s1[t] = True
elif t in s1:
s2[f] = True
elif t in s2:
s1[f] = True
else:
s1[f] = True; s2[t] = True
print(len(s1) * len(s2) - len(pair))
| Title: Mahmoud and Ehab and the bipartiteness
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mahmoud and Ehab continue their adventures! As everybody in the evil land knows, Dr. Evil likes bipartite graphs, especially trees.
A tree is a connected acyclic graph. A bipartite graph is a graph, whose vertices can be partitioned into 2 sets in such a way, that for each edge (*u*,<=*v*) that belongs to the graph, *u* and *v* belong to different sets. You can find more formal definitions of a tree and a bipartite graph in the notes section below.
Dr. Evil gave Mahmoud and Ehab a tree consisting of *n* nodes and asked them to add edges to it in such a way, that the graph is still bipartite. Besides, after adding these edges the graph should be simple (doesn't contain loops or multiple edges). What is the maximum number of edges they can add?
A loop is an edge, which connects a node with itself. Graph doesn't contain multiple edges when for each pair of nodes there is no more than one edge between them. A cycle and a loop aren't the same .
Input Specification:
The first line of input contains an integer *n* — the number of nodes in the tree (1<=≤<=*n*<=≤<=105).
The next *n*<=-<=1 lines contain integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*) — the description of the edges of the tree.
It's guaranteed that the given graph is a tree.
Output Specification:
Output one integer — the maximum number of edges that Mahmoud and Ehab can add to the tree while fulfilling the conditions.
Demo Input:
['3\n1 2\n1 3\n', '5\n1 2\n2 3\n3 4\n4 5\n']
Demo Output:
['0\n', '2\n']
Note:
Tree definition: [https://en.wikipedia.org/wiki/Tree_(graph_theory)](https://en.wikipedia.org/wiki/Tree_(graph_theory))
Bipartite graph definition: [https://en.wikipedia.org/wiki/Bipartite_graph](https://en.wikipedia.org/wiki/Bipartite_graph)
In the first test case the only edge that can be added in such a way, that graph won't contain loops or multiple edges is (2, 3), but adding this edge will make the graph non-bipartite so the answer is 0.
In the second test case Mahmoud and Ehab can add edges (1, 4) and (2, 5). | ```python
n = int(input())
s1 = {}
s2 = {}
pair = []
for i in range(n -1):
row = input().split(' ')
f = int(row[0]); t = int(row[1])
pair.append((f,t))
pair.sort()
for f,t in pair:
if f in s1:
s2[t] = True
elif f in s2:
s1[t] = True
elif t in s1:
s2[f] = True
elif t in s2:
s1[f] = True
else:
s1[f] = True; s2[t] = True
print(len(s1) * len(s2) - len(pair))
``` | 0 | |
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,634,788,161 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 62 | 4,198,400 | print("Domino piling")
input("Input Domino piling that is true:")
a = word()
print(a)
| Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
print("Domino piling")
input("Input Domino piling that is true:")
a = word()
print(a)
``` | -1 |
349 | A | Cinema Line | PROGRAMMING | 1,100 | [
"greedy",
"implementation"
] | null | null | The new "Die Hard" movie has just been released! There are *n* people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 ruble bill. A "Die Hard" ticket costs 25 rubles. Can the booking clerk sell a ticket to each person and give the change if he initially has no money and sells the tickets strictly in the order people follow in the line? | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of people in the line. The next line contains *n* integers, each of them equals 25, 50 or 100 — the values of the bills the people have. The numbers are given in the order from the beginning of the line (at the box office) to the end of the line. | Print "YES" (without the quotes) if the booking clerk can sell a ticket to each person and give the change. Otherwise print "NO". | [
"4\n25 25 50 50\n",
"2\n25 100\n",
"4\n50 50 25 25\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | none | 500 | [
{
"input": "4\n25 25 50 50",
"output": "YES"
},
{
"input": "2\n25 100",
"output": "NO"
},
{
"input": "4\n50 50 25 25",
"output": "NO"
},
{
"input": "3\n25 50 100",
"output": "NO"
},
{
"input": "10\n25 25 25 25 25 25 25 25 25 25",
"output": "YES"
},
{
"input": "10\n50 50 50 50 50 50 50 50 50 50",
"output": "NO"
},
{
"input": "10\n100 100 100 100 100 100 100 100 100 100",
"output": "NO"
},
{
"input": "7\n25 25 25 25 50 100 50",
"output": "YES"
},
{
"input": "2\n25 25",
"output": "YES"
},
{
"input": "3\n50 100 100",
"output": "NO"
},
{
"input": "3\n25 50 50",
"output": "NO"
},
{
"input": "2\n50 25",
"output": "NO"
},
{
"input": "14\n25 25 25 25 25 25 25 50 50 50 100 100 100 100",
"output": "NO"
},
{
"input": "5\n25 25 25 100 100",
"output": "NO"
},
{
"input": "7\n25 50 25 50 25 50 100",
"output": "NO"
},
{
"input": "5\n25 50 25 50 100",
"output": "NO"
},
{
"input": "5\n25 25 50 50 50",
"output": "NO"
},
{
"input": "5\n25 25 50 100 50",
"output": "NO"
},
{
"input": "5\n25 25 50 50 100",
"output": "NO"
},
{
"input": "1\n50",
"output": "NO"
}
] | 1,664,608,274 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 11 | 124 | 5,324,800 | input()
l=list(map(int, input().split(" ")))
bisa = True
uang =0
for e in l:
if e == 25:
uang += e
elif e == 50:
if uang < 25:
bisa=False
break
else:
uang -= 25
uang += 50
elif e == 100:
if uang < 75:
bisa=False
break
else:
uang += 100
uang -= 75
if bisa:
print ("YES")
else:
print ("NO") | Title: Cinema Line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The new "Die Hard" movie has just been released! There are *n* people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 ruble bill. A "Die Hard" ticket costs 25 rubles. Can the booking clerk sell a ticket to each person and give the change if he initially has no money and sells the tickets strictly in the order people follow in the line?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of people in the line. The next line contains *n* integers, each of them equals 25, 50 or 100 — the values of the bills the people have. The numbers are given in the order from the beginning of the line (at the box office) to the end of the line.
Output Specification:
Print "YES" (without the quotes) if the booking clerk can sell a ticket to each person and give the change. Otherwise print "NO".
Demo Input:
['4\n25 25 50 50\n', '2\n25 100\n', '4\n50 50 25 25\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none | ```python
input()
l=list(map(int, input().split(" ")))
bisa = True
uang =0
for e in l:
if e == 25:
uang += e
elif e == 50:
if uang < 25:
bisa=False
break
else:
uang -= 25
uang += 50
elif e == 100:
if uang < 75:
bisa=False
break
else:
uang += 100
uang -= 75
if bisa:
print ("YES")
else:
print ("NO")
``` | 0 | |
267 | A | Subtractions | PROGRAMMING | 900 | [
"math",
"number theory"
] | null | null | You've got two numbers. As long as they are both larger than zero, they go through the same operation: subtract the lesser number from the larger one. If they equal substract one number from the another. For example, one operation transforms pair (4,17) to pair (4,13), it transforms (5,5) to (0,5).
You've got some number of pairs (*a**i*,<=*b**i*). How many operations will be performed for each of them? | The first line contains the number of pairs *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000). Then follow *n* lines, each line contains a pair of positive integers *a**i*,<=*b**i* (1<=<=≤<=<=*a**i*,<=<=*b**i*<=<=≤<=<=109). | Print the sought number of operations for each pair on a single line. | [
"2\n4 17\n7 987654321\n"
] | [
"8\n141093479\n"
] | none | 500 | [
{
"input": "2\n4 17\n7 987654321",
"output": "8\n141093479"
},
{
"input": "10\n7 987654321\n7 987654321\n7 987654321\n7 987654321\n7 987654321\n7 987654321\n7 987654321\n7 987654321\n7 987654321\n7 987654321",
"output": "141093479\n141093479\n141093479\n141093479\n141093479\n141093479\n141093479\n141093479\n141093479\n141093479"
},
{
"input": "1\n536870912 32",
"output": "16777216"
},
{
"input": "20\n1000000000 999999999\n1000000000 999999999\n1000000000 999999999\n1000000000 999999999\n1000000000 999999999\n1000000000 999999999\n1000000000 999999999\n1000000000 999999999\n1000000000 999999999\n1000000000 999999999\n1000000000 999999999\n1000000000 999999999\n1000000000 999999999\n1000000000 999999999\n1000000000 999999999\n1000000000 999999999\n1000000000 999999999\n1000000000 999999999\n1000000000 999999999\n1000000000 999999999",
"output": "1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000\n1000000000"
},
{
"input": "3\n1000000000 1\n1000000000 1\n1 100000000",
"output": "1000000000\n1000000000\n100000000"
}
] | 1,596,252,727 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 124 | 6,963,200 | n = int(input())
i = 0
while i < n:
a = list(map(int, input().split()))
x = 0
while a[0] > 0 and a[1] > 0:
b = a[1] % a[0]
x += a[1]//a[0]
c = a[0]
a.clear()
a.append(c)
a.append(b)
a.sort()
print(x)
a = []
i += 1
| Title: Subtractions
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got two numbers. As long as they are both larger than zero, they go through the same operation: subtract the lesser number from the larger one. If they equal substract one number from the another. For example, one operation transforms pair (4,17) to pair (4,13), it transforms (5,5) to (0,5).
You've got some number of pairs (*a**i*,<=*b**i*). How many operations will be performed for each of them?
Input Specification:
The first line contains the number of pairs *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000). Then follow *n* lines, each line contains a pair of positive integers *a**i*,<=*b**i* (1<=<=≤<=<=*a**i*,<=<=*b**i*<=<=≤<=<=109).
Output Specification:
Print the sought number of operations for each pair on a single line.
Demo Input:
['2\n4 17\n7 987654321\n']
Demo Output:
['8\n141093479\n']
Note:
none | ```python
n = int(input())
i = 0
while i < n:
a = list(map(int, input().split()))
x = 0
while a[0] > 0 and a[1] > 0:
b = a[1] % a[0]
x += a[1]//a[0]
c = a[0]
a.clear()
a.append(c)
a.append(b)
a.sort()
print(x)
a = []
i += 1
``` | 3 | |
185 | A | Plant | PROGRAMMING | 1,300 | [
"math"
] | null | null | Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.
Help the dwarfs find out how many triangle plants that point "upwards" will be in *n* years. | The first line contains a single integer *n* (0<=≤<=*n*<=≤<=1018) — the number of full years when the plant grew.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier. | Print a single integer — the remainder of dividing the number of plants that will point "upwards" in *n* years by 1000000007 (109<=+<=7). | [
"1\n",
"2\n"
] | [
"3\n",
"10\n"
] | The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one. | 500 | [
{
"input": "1",
"output": "3"
},
{
"input": "2",
"output": "10"
},
{
"input": "385599124",
"output": "493875375"
},
{
"input": "989464295",
"output": "31966163"
},
{
"input": "376367012",
"output": "523204186"
},
{
"input": "529357306",
"output": "142578489"
},
{
"input": "782916801",
"output": "51174574"
},
{
"input": "74859961358140080",
"output": "478768275"
},
{
"input": "0",
"output": "1"
},
{
"input": "252509053898415171",
"output": "886314547"
},
{
"input": "760713016078377938",
"output": "79611270"
},
{
"input": "919845424847912644",
"output": "388845650"
},
{
"input": "585335721566249104",
"output": "301383716"
},
{
"input": "522842183413115087",
"output": "556012763"
},
{
"input": "148049062285906746",
"output": "913927498"
},
{
"input": "84324827171274022",
"output": "462535280"
},
{
"input": "354979172034763159",
"output": "239287993"
},
{
"input": "1312148742261680",
"output": "799725655"
},
{
"input": "269587448053313253",
"output": "536645997"
},
{
"input": "645762257531682045",
"output": "543988614"
},
{
"input": "615812227854199662",
"output": "357939938"
},
{
"input": "819875140559301751",
"output": "968653685"
},
{
"input": "349993003033420740",
"output": "709392758"
},
{
"input": "891351282398722856",
"output": "70758467"
},
{
"input": "563324730406715801",
"output": "353494903"
},
{
"input": "520974001002628386",
"output": "164118419"
},
{
"input": "666729339260489789",
"output": "784700006"
},
{
"input": "856674609788912527",
"output": "720540265"
},
{
"input": "791809296233191092",
"output": "369199735"
},
{
"input": "711066335916901717",
"output": "15590358"
},
{
"input": "931356501703211379",
"output": "239824013"
},
{
"input": "234122431978145893",
"output": "905163056"
},
{
"input": "1000000000000000000",
"output": "899770636"
},
{
"input": "3",
"output": "36"
},
{
"input": "4",
"output": "136"
},
{
"input": "5",
"output": "528"
},
{
"input": "6",
"output": "2080"
},
{
"input": "7",
"output": "8256"
},
{
"input": "8",
"output": "32896"
},
{
"input": "9",
"output": "131328"
},
{
"input": "10",
"output": "524800"
},
{
"input": "11",
"output": "2098176"
},
{
"input": "12",
"output": "8390656"
},
{
"input": "13",
"output": "33558528"
},
{
"input": "14",
"output": "134225920"
},
{
"input": "15",
"output": "536887296"
},
{
"input": "16",
"output": "147516402"
},
{
"input": "0",
"output": "1"
},
{
"input": "6265",
"output": "980996097"
}
] | 1,690,871,050 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 124 | 65,638,400 | '''
###### * User Profile : Keval_78
LinkedIn: https://www.linkedin.com/in/kevalpadsala78/
Github: https://github.com/Keval78
Leetcode: https://leetcode.com/Keval_78/
'''
import sys
# from os import path
# from math import log2, floor, ceil, sqrt, pow, gcd
# from random import random, randint, shuffle, choice
# from collections import Counter, defaultdict, deque
# from itertools import permutations, combinations
# from functools import reduce
# from heapq import heapify, heappop, heappush, heapreplace
# from bisect import bisect_left, bisect_right
def si(): return sys.stdin.readline().strip()
def ii(): return int(si())
def mi(ss=" "): return map(int, si().split(ss))
def msi(ss=" "): return map(str, si().split(ss))
def li(ss=" "): return list(mi(ss))
sys.setrecursionlimit(10 ** 5)
MOD, MOD2, INF = 10 ** 9 + 7, 998244353, float('inf')
MAX = 10**5+1
n = ii()
# 2^(n-1)*(1+2^n)-> Using Binary Exponentiation
ans = 1
while n > 0:
if n&1:
ans *= 2
ans %= MOD
ans *= ans
ans %= MOD
n >>= 1
ans = ((ans//2)*(1 + ans))%MOD
print(ans)
| Title: Plant
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.
Help the dwarfs find out how many triangle plants that point "upwards" will be in *n* years.
Input Specification:
The first line contains a single integer *n* (0<=≤<=*n*<=≤<=1018) — the number of full years when the plant grew.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Output Specification:
Print a single integer — the remainder of dividing the number of plants that will point "upwards" in *n* years by 1000000007 (109<=+<=7).
Demo Input:
['1\n', '2\n']
Demo Output:
['3\n', '10\n']
Note:
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one. | ```python
'''
###### * User Profile : Keval_78
LinkedIn: https://www.linkedin.com/in/kevalpadsala78/
Github: https://github.com/Keval78
Leetcode: https://leetcode.com/Keval_78/
'''
import sys
# from os import path
# from math import log2, floor, ceil, sqrt, pow, gcd
# from random import random, randint, shuffle, choice
# from collections import Counter, defaultdict, deque
# from itertools import permutations, combinations
# from functools import reduce
# from heapq import heapify, heappop, heappush, heapreplace
# from bisect import bisect_left, bisect_right
def si(): return sys.stdin.readline().strip()
def ii(): return int(si())
def mi(ss=" "): return map(int, si().split(ss))
def msi(ss=" "): return map(str, si().split(ss))
def li(ss=" "): return list(mi(ss))
sys.setrecursionlimit(10 ** 5)
MOD, MOD2, INF = 10 ** 9 + 7, 998244353, float('inf')
MAX = 10**5+1
n = ii()
# 2^(n-1)*(1+2^n)-> Using Binary Exponentiation
ans = 1
while n > 0:
if n&1:
ans *= 2
ans %= MOD
ans *= ans
ans %= MOD
n >>= 1
ans = ((ans//2)*(1 + ans))%MOD
print(ans)
``` | 0 | |
3 | A | Shortest path of the king | PROGRAMMING | 1,000 | [
"greedy",
"shortest paths"
] | A. Shortest path of the king | 1 | 64 | The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square *t*. As the king is not in habit of wasting his time, he wants to get from his current position *s* to square *t* in the least number of moves. Help him to do this.
In one move the king can get to the square that has a common side or a common vertex with the square the king is currently in (generally there are 8 different squares he can move to). | The first line contains the chessboard coordinates of square *s*, the second line — of square *t*.
Chessboard coordinates consist of two characters, the first one is a lowercase Latin letter (from a to h), the second one is a digit from 1 to 8. | In the first line print *n* — minimum number of the king's moves. Then in *n* lines print the moves themselves. Each move is described with one of the 8: L, R, U, D, LU, LD, RU or RD.
L, R, U, D stand respectively for moves left, right, up and down (according to the picture), and 2-letter combinations stand for diagonal moves. If the answer is not unique, print any of them. | [
"a8\nh1\n"
] | [
"7\nRD\nRD\nRD\nRD\nRD\nRD\nRD\n"
] | none | 0 | [
{
"input": "a8\nh1",
"output": "7\nRD\nRD\nRD\nRD\nRD\nRD\nRD"
},
{
"input": "b2\nb4",
"output": "2\nU\nU"
},
{
"input": "a5\na5",
"output": "0"
},
{
"input": "h1\nb2",
"output": "6\nLU\nL\nL\nL\nL\nL"
},
{
"input": "c5\nh2",
"output": "5\nRD\nRD\nRD\nR\nR"
},
{
"input": "e1\nf2",
"output": "1\nRU"
},
{
"input": "g4\nd2",
"output": "3\nLD\nLD\nL"
},
{
"input": "a8\nb2",
"output": "6\nRD\nD\nD\nD\nD\nD"
},
{
"input": "d4\nh2",
"output": "4\nRD\nRD\nR\nR"
},
{
"input": "c5\na2",
"output": "3\nLD\nLD\nD"
},
{
"input": "h5\nf8",
"output": "3\nLU\nLU\nU"
},
{
"input": "e6\nb6",
"output": "3\nL\nL\nL"
},
{
"input": "a6\ng4",
"output": "6\nRD\nRD\nR\nR\nR\nR"
},
{
"input": "f7\nc2",
"output": "5\nLD\nLD\nLD\nD\nD"
},
{
"input": "b7\nh8",
"output": "6\nRU\nR\nR\nR\nR\nR"
},
{
"input": "g7\nd6",
"output": "3\nLD\nL\nL"
},
{
"input": "c8\na3",
"output": "5\nLD\nLD\nD\nD\nD"
},
{
"input": "h8\nf1",
"output": "7\nLD\nLD\nD\nD\nD\nD\nD"
},
{
"input": "d1\nb7",
"output": "6\nLU\nLU\nU\nU\nU\nU"
},
{
"input": "a7\ne5",
"output": "4\nRD\nRD\nR\nR"
},
{
"input": "d6\nb1",
"output": "5\nLD\nLD\nD\nD\nD"
},
{
"input": "f5\ng5",
"output": "1\nR"
},
{
"input": "h4\nd1",
"output": "4\nLD\nLD\nLD\nL"
},
{
"input": "b3\na5",
"output": "2\nLU\nU"
},
{
"input": "d2\nf1",
"output": "2\nRD\nR"
},
{
"input": "f1\nc5",
"output": "4\nLU\nLU\nLU\nU"
},
{
"input": "a8\nh1",
"output": "7\nRD\nRD\nRD\nRD\nRD\nRD\nRD"
},
{
"input": "c7\ne5",
"output": "2\nRD\nRD"
},
{
"input": "e7\nb1",
"output": "6\nLD\nLD\nLD\nD\nD\nD"
},
{
"input": "g8\na8",
"output": "6\nL\nL\nL\nL\nL\nL"
},
{
"input": "g6\nf2",
"output": "4\nLD\nD\nD\nD"
},
{
"input": "g4\nc4",
"output": "4\nL\nL\nL\nL"
},
{
"input": "g2\na6",
"output": "6\nLU\nLU\nLU\nLU\nL\nL"
},
{
"input": "f8\nf8",
"output": "0"
},
{
"input": "f5\nd2",
"output": "3\nLD\nLD\nD"
}
] | 1,666,022,581 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 92 | 0 | #xet mA == mB: //dong
def check_in_row(colA,rowA,colB,rowB):
#number of king moves => n = abs(nA - nB)
numberOfKingMove = abs(colA - colB)
listMove = []
if colA > colB:
listMove = ['L'] * numberOfKingMove
else:
listMove = ['R'] * numberOfKingMove
return listMove
#xet nA == nB: // cot
def check_in_col(colA,rowA,colB,rowB):
##number of king moves => n = abs(mA - mB)
numberOfKingMove = abs(rowA - rowB)
listMove = []
if rowA > rowB:
listMove = ['U'] * numberOfKingMove
else:
listMove = ['D'] * numberOfKingMove
return listMove
#xet abs(nA - nB) == abs(mA- mB) // 2 duong cheo
def check_in_cross(colA,rowA,colB,rowB):
numberOfKingMove = abs(rowA - rowB)
listMove = []
if colA > colB:
if rowA > rowB:
listMove = ['LU'] * numberOfKingMove
else:
listMove = ['LD'] * numberOfKingMove
else:
if rowA > rowB:
listMove = ['RU'] * numberOfKingMove
else:
listMove = ['RD'] * numberOfKingMove
return listMove
#ko nam tren dong, cot, 2 duong cheo
def check_special_case(colA,rowA,colB,rowB):
leftUp = [-1, -1]
leftDown = [1, -1]
rightUp = [-1, 1]
rightDown = [1, 1]
temp = []
if rowA < rowB and colA > colB:
temp = leftDown
elif rowA < rowB and colA < colB:
temp = rightDown
elif rowA > rowB and colA > colB:
temp = leftUp
elif rowA > rowB and colA < colB:
temp = rightUp
tempRow = rowA
tempColumn = colA
while tempRow != rowB and tempColumn != colB:
tempRow += temp[0]
tempColumn += temp[1]
crossMoves = check_in_cross(colA, rowA, tempColumn, tempRow)
straintMoves = []
if tempRow == rowB:
straintMoves = check_in_row(tempColumn, tempRow, colB, rowB)
elif tempColumn == colB:
straintMoves = check_in_col(tempColumn, tempRow, colB, rowB)
return crossMoves + straintMoves
a = input()
b = input()
#tach ky tu dau tien => a => asci(a)%97 - 1 => lay dc gia tri cot
#tach ky tu thu 2 => 8 - 1 => lay dc gia tri dong
#co toa do diem bat dau A(mA,nA)
#co toa do diem ket thuc B(mB, nB)
colA = ord(a[0])%97
rowA = abs(int(a[1]) - 8)
colB = ord(b[0])%97
rowB = abs(int(b[1]) - 8)
result = check_special_case(colA,rowA,colB,rowB)
result_count = len(result)
print(result_count)
for move in result:
print(move) | Title: Shortest path of the king
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square *t*. As the king is not in habit of wasting his time, he wants to get from his current position *s* to square *t* in the least number of moves. Help him to do this.
In one move the king can get to the square that has a common side or a common vertex with the square the king is currently in (generally there are 8 different squares he can move to).
Input Specification:
The first line contains the chessboard coordinates of square *s*, the second line — of square *t*.
Chessboard coordinates consist of two characters, the first one is a lowercase Latin letter (from a to h), the second one is a digit from 1 to 8.
Output Specification:
In the first line print *n* — minimum number of the king's moves. Then in *n* lines print the moves themselves. Each move is described with one of the 8: L, R, U, D, LU, LD, RU or RD.
L, R, U, D stand respectively for moves left, right, up and down (according to the picture), and 2-letter combinations stand for diagonal moves. If the answer is not unique, print any of them.
Demo Input:
['a8\nh1\n']
Demo Output:
['7\nRD\nRD\nRD\nRD\nRD\nRD\nRD\n']
Note:
none | ```python
#xet mA == mB: //dong
def check_in_row(colA,rowA,colB,rowB):
#number of king moves => n = abs(nA - nB)
numberOfKingMove = abs(colA - colB)
listMove = []
if colA > colB:
listMove = ['L'] * numberOfKingMove
else:
listMove = ['R'] * numberOfKingMove
return listMove
#xet nA == nB: // cot
def check_in_col(colA,rowA,colB,rowB):
##number of king moves => n = abs(mA - mB)
numberOfKingMove = abs(rowA - rowB)
listMove = []
if rowA > rowB:
listMove = ['U'] * numberOfKingMove
else:
listMove = ['D'] * numberOfKingMove
return listMove
#xet abs(nA - nB) == abs(mA- mB) // 2 duong cheo
def check_in_cross(colA,rowA,colB,rowB):
numberOfKingMove = abs(rowA - rowB)
listMove = []
if colA > colB:
if rowA > rowB:
listMove = ['LU'] * numberOfKingMove
else:
listMove = ['LD'] * numberOfKingMove
else:
if rowA > rowB:
listMove = ['RU'] * numberOfKingMove
else:
listMove = ['RD'] * numberOfKingMove
return listMove
#ko nam tren dong, cot, 2 duong cheo
def check_special_case(colA,rowA,colB,rowB):
leftUp = [-1, -1]
leftDown = [1, -1]
rightUp = [-1, 1]
rightDown = [1, 1]
temp = []
if rowA < rowB and colA > colB:
temp = leftDown
elif rowA < rowB and colA < colB:
temp = rightDown
elif rowA > rowB and colA > colB:
temp = leftUp
elif rowA > rowB and colA < colB:
temp = rightUp
tempRow = rowA
tempColumn = colA
while tempRow != rowB and tempColumn != colB:
tempRow += temp[0]
tempColumn += temp[1]
crossMoves = check_in_cross(colA, rowA, tempColumn, tempRow)
straintMoves = []
if tempRow == rowB:
straintMoves = check_in_row(tempColumn, tempRow, colB, rowB)
elif tempColumn == colB:
straintMoves = check_in_col(tempColumn, tempRow, colB, rowB)
return crossMoves + straintMoves
a = input()
b = input()
#tach ky tu dau tien => a => asci(a)%97 - 1 => lay dc gia tri cot
#tach ky tu thu 2 => 8 - 1 => lay dc gia tri dong
#co toa do diem bat dau A(mA,nA)
#co toa do diem ket thuc B(mB, nB)
colA = ord(a[0])%97
rowA = abs(int(a[1]) - 8)
colB = ord(b[0])%97
rowB = abs(int(b[1]) - 8)
result = check_special_case(colA,rowA,colB,rowB)
result_count = len(result)
print(result_count)
for move in result:
print(move)
``` | 3.954 |
361 | B | Levko and Permutation | PROGRAMMING | 1,200 | [
"constructive algorithms",
"math",
"number theory"
] | null | null | Levko loves permutations very much. A permutation of length *n* is a sequence of distinct positive integers, each is at most *n*.
Let’s assume that value *gcd*(*a*,<=*b*) shows the greatest common divisor of numbers *a* and *b*. Levko assumes that element *p**i* of permutation *p*1,<=*p*2,<=... ,<=*p**n* is good if *gcd*(*i*,<=*p**i*)<=><=1. Levko considers a permutation beautiful, if it has exactly *k* good elements. Unfortunately, he doesn’t know any beautiful permutation. Your task is to help him to find at least one of them. | The single line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 0<=≤<=*k*<=≤<=*n*). | In a single line print either any beautiful permutation or -1, if such permutation doesn’t exist.
If there are multiple suitable permutations, you are allowed to print any of them. | [
"4 2\n",
"1 1\n"
] | [
"2 4 3 1",
"-1\n"
] | In the first sample elements 4 and 3 are good because *gcd*(2, 4) = 2 > 1 and *gcd*(3, 3) = 3 > 1. Elements 2 and 1 are not good because *gcd*(1, 2) = 1 and *gcd*(4, 1) = 1. As there are exactly 2 good elements, the permutation is beautiful.
The second sample has no beautiful permutations. | 1,000 | [
{
"input": "4 2",
"output": "2 1 3 4 "
},
{
"input": "1 1",
"output": "-1"
},
{
"input": "7 4",
"output": "3 1 2 4 5 6 7 "
},
{
"input": "10 9",
"output": "1 2 3 4 5 6 7 8 9 10 "
},
{
"input": "10000 5000",
"output": "5000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 15..."
},
{
"input": "7 0",
"output": "7 1 2 3 4 5 6 "
},
{
"input": "1 0",
"output": "1 "
},
{
"input": "7 7",
"output": "-1"
},
{
"input": "100000 47",
"output": "99953 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..."
},
{
"input": "100000 100000",
"output": "-1"
},
{
"input": "100000 43425",
"output": "56575 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..."
},
{
"input": "7 6",
"output": "1 2 3 4 5 6 7 "
},
{
"input": "100000 99999",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "47 46",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 "
},
{
"input": "5 0",
"output": "5 1 2 3 4 "
},
{
"input": "4 2",
"output": "2 1 3 4 "
},
{
"input": "1533 1052",
"output": "481 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154..."
},
{
"input": "81314 52747",
"output": "28567 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..."
},
{
"input": "17767 145",
"output": "17622 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..."
},
{
"input": "18168 7942",
"output": "10226 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..."
},
{
"input": "26593 15915",
"output": "10678 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..."
},
{
"input": "26593 8877",
"output": "17716 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..."
},
{
"input": "13852 12727",
"output": "1125 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 15..."
},
{
"input": "4 1",
"output": "3 1 2 4 "
},
{
"input": "8834 8834",
"output": "-1"
},
{
"input": "8485 8484",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "14564 14564",
"output": "-1"
},
{
"input": "8254 8253",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "81314 81312",
"output": "2 1 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "5795 5792",
"output": "3 1 2 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "6417 3",
"output": "6414 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 15..."
},
{
"input": "6896 0",
"output": "6896 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 15..."
},
{
"input": "6778 1",
"output": "6777 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 15..."
},
{
"input": "9448 1",
"output": "9447 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 15..."
},
{
"input": "5938 2",
"output": "5936 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 15..."
},
{
"input": "3072 0",
"output": "3072 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 15..."
},
{
"input": "8576 0",
"output": "8576 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 15..."
},
{
"input": "2 1",
"output": "1 2 "
},
{
"input": "4 4",
"output": "-1"
},
{
"input": "5 5",
"output": "-1"
},
{
"input": "2 2",
"output": "-1"
},
{
"input": "100000 1",
"output": "99999 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..."
},
{
"input": "100000 50000",
"output": "50000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..."
},
{
"input": "4 1",
"output": "3 1 2 4 "
},
{
"input": "100000 9999",
"output": "90001 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..."
},
{
"input": "100000 99000",
"output": "1000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 15..."
},
{
"input": "100000 12347",
"output": "87653 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 1..."
}
] | 1,613,627,593 | 2,147,483,647 | PyPy 3 | OK | TESTS | 47 | 170 | 7,987,200 | n,k=map(int,input().split())
if n==k:
print(-1)
else:
arr=[i for i in range(1,n+1)]
var=n-k
arr.remove(var)
arr.insert(0,var)
print(*arr)
| Title: Levko and Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Levko loves permutations very much. A permutation of length *n* is a sequence of distinct positive integers, each is at most *n*.
Let’s assume that value *gcd*(*a*,<=*b*) shows the greatest common divisor of numbers *a* and *b*. Levko assumes that element *p**i* of permutation *p*1,<=*p*2,<=... ,<=*p**n* is good if *gcd*(*i*,<=*p**i*)<=><=1. Levko considers a permutation beautiful, if it has exactly *k* good elements. Unfortunately, he doesn’t know any beautiful permutation. Your task is to help him to find at least one of them.
Input Specification:
The single line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 0<=≤<=*k*<=≤<=*n*).
Output Specification:
In a single line print either any beautiful permutation or -1, if such permutation doesn’t exist.
If there are multiple suitable permutations, you are allowed to print any of them.
Demo Input:
['4 2\n', '1 1\n']
Demo Output:
['2 4 3 1', '-1\n']
Note:
In the first sample elements 4 and 3 are good because *gcd*(2, 4) = 2 > 1 and *gcd*(3, 3) = 3 > 1. Elements 2 and 1 are not good because *gcd*(1, 2) = 1 and *gcd*(4, 1) = 1. As there are exactly 2 good elements, the permutation is beautiful.
The second sample has no beautiful permutations. | ```python
n,k=map(int,input().split())
if n==k:
print(-1)
else:
arr=[i for i in range(1,n+1)]
var=n-k
arr.remove(var)
arr.insert(0,var)
print(*arr)
``` | 3 | |
660 | A | Co-prime Array | PROGRAMMING | 1,200 | [
"greedy",
"implementation",
"math",
"number theory"
] | null | null | You are given an array of *n* elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers *a* and *b* are said to be co-prime if the only positive integer that divides both of them is 1. | The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the given array.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*. | Print integer *k* on the first line — the least number of elements needed to add to the array *a* to make it co-prime.
The second line should contain *n*<=+<=*k* integers *a**j* — the elements of the array *a* after adding *k* elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array *a* by adding *k* elements to it.
If there are multiple answers you can print any one of them. | [
"3\n2 7 28\n"
] | [
"1\n2 7 9 28\n"
] | none | 0 | [
{
"input": "3\n2 7 28",
"output": "1\n2 7 1 28"
},
{
"input": "1\n1",
"output": "0\n1"
},
{
"input": "1\n548",
"output": "0\n548"
},
{
"input": "1\n963837006",
"output": "0\n963837006"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "0\n1 1 1 1 1 1 1 1 1 1"
},
{
"input": "10\n26 723 970 13 422 968 875 329 234 983",
"output": "2\n26 723 970 13 422 1 968 875 1 329 234 983"
},
{
"input": "10\n319645572 758298525 812547177 459359946 355467212 304450522 807957797 916787906 239781206 242840396",
"output": "7\n319645572 1 758298525 1 812547177 1 459359946 1 355467212 1 304450522 807957797 916787906 1 239781206 1 242840396"
},
{
"input": "100\n1 1 1 1 2 1 1 1 1 1 2 2 1 1 2 1 2 1 1 1 2 1 1 2 1 2 1 1 2 2 2 1 1 2 1 1 1 2 2 2 1 1 1 2 1 2 2 1 2 1 1 2 2 1 2 1 2 1 2 2 1 1 1 2 1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 1 1 1 1 2 2 2 2 2 2 2 1 1 1 2 1 2 1",
"output": "19\n1 1 1 1 2 1 1 1 1 1 2 1 2 1 1 2 1 2 1 1 1 2 1 1 2 1 2 1 1 2 1 2 1 2 1 1 2 1 1 1 2 1 2 1 2 1 1 1 2 1 2 1 2 1 2 1 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 2 1 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 2 1 2 1"
},
{
"input": "100\n591 417 888 251 792 847 685 3 182 461 102 348 555 956 771 901 712 878 580 631 342 333 285 899 525 725 537 718 929 653 84 788 104 355 624 803 253 853 201 995 536 184 65 205 540 652 549 777 248 405 677 950 431 580 600 846 328 429 134 983 526 103 500 963 400 23 276 704 570 757 410 658 507 620 984 244 486 454 802 411 985 303 635 283 96 597 855 775 139 839 839 61 219 986 776 72 729 69 20 917",
"output": "38\n591 1 417 1 888 251 792 1 847 685 3 182 461 102 1 348 1 555 956 771 901 712 1 878 1 580 631 342 1 333 1 285 899 525 1 725 537 718 929 653 84 1 788 1 104 355 624 803 1 253 853 201 995 536 1 184 65 1 205 1 540 1 652 549 1 777 248 405 677 950 431 580 1 600 1 846 1 328 429 134 983 526 103 500 963 400 23 1 276 1 704 1 570 757 410 1 658 507 620 1 984 1 244 1 486 1 454 1 802 411 985 303 635 283 96 1 597 1 855 1 775 139 839 1 839 61 219 986 1 776 1 72 1 729 1 69 20 917"
},
{
"input": "5\n472882027 472882027 472882027 472882027 472882027",
"output": "4\n472882027 1 472882027 1 472882027 1 472882027 1 472882027"
},
{
"input": "2\n1000000000 1000000000",
"output": "1\n1000000000 1 1000000000"
},
{
"input": "2\n8 6",
"output": "1\n8 1 6"
},
{
"input": "3\n100000000 1000000000 1000000000",
"output": "2\n100000000 1 1000000000 1 1000000000"
},
{
"input": "5\n1 2 3 4 5",
"output": "0\n1 2 3 4 5"
},
{
"input": "20\n2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000 2 1000000000",
"output": "19\n2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000 1 2 1 1000000000"
},
{
"input": "2\n223092870 23",
"output": "1\n223092870 1 23"
},
{
"input": "2\n100000003 100000003",
"output": "1\n100000003 1 100000003"
},
{
"input": "2\n999999937 999999937",
"output": "1\n999999937 1 999999937"
},
{
"input": "4\n999 999999937 999999937 999",
"output": "1\n999 999999937 1 999999937 999"
},
{
"input": "2\n999999929 999999929",
"output": "1\n999999929 1 999999929"
},
{
"input": "2\n1049459 2098918",
"output": "1\n1049459 1 2098918"
},
{
"input": "2\n352229 704458",
"output": "1\n352229 1 704458"
},
{
"input": "2\n7293 4011",
"output": "1\n7293 1 4011"
},
{
"input": "2\n5565651 3999930",
"output": "1\n5565651 1 3999930"
},
{
"input": "2\n997 997",
"output": "1\n997 1 997"
},
{
"input": "3\n9994223 9994223 9994223",
"output": "2\n9994223 1 9994223 1 9994223"
},
{
"input": "2\n99999998 1000000000",
"output": "1\n99999998 1 1000000000"
},
{
"input": "3\n1000000000 1000000000 1000000000",
"output": "2\n1000000000 1 1000000000 1 1000000000"
},
{
"input": "2\n130471 130471",
"output": "1\n130471 1 130471"
},
{
"input": "3\n1000000000 2 2",
"output": "2\n1000000000 1 2 1 2"
},
{
"input": "2\n223092870 66526",
"output": "1\n223092870 1 66526"
},
{
"input": "14\n1000000000 1000000000 223092870 223092870 6 105 2 2 510510 510510 999999491 999999491 436077930 570018449",
"output": "10\n1000000000 1 1000000000 1 223092870 1 223092870 1 6 1 105 2 1 2 1 510510 1 510510 999999491 1 999999491 436077930 1 570018449"
},
{
"input": "2\n3996017 3996017",
"output": "1\n3996017 1 3996017"
},
{
"input": "2\n999983 999983",
"output": "1\n999983 1 999983"
},
{
"input": "2\n618575685 773990454",
"output": "1\n618575685 1 773990454"
},
{
"input": "3\n9699690 3 7",
"output": "1\n9699690 1 3 7"
},
{
"input": "2\n999999999 999999996",
"output": "1\n999999999 1 999999996"
},
{
"input": "2\n99999910 99999910",
"output": "1\n99999910 1 99999910"
},
{
"input": "12\n1000000000 1000000000 223092870 223092870 6 105 2 2 510510 510510 999999491 999999491",
"output": "9\n1000000000 1 1000000000 1 223092870 1 223092870 1 6 1 105 2 1 2 1 510510 1 510510 999999491 1 999999491"
},
{
"input": "3\n999999937 999999937 999999937",
"output": "2\n999999937 1 999999937 1 999999937"
},
{
"input": "2\n99839 99839",
"output": "1\n99839 1 99839"
},
{
"input": "3\n19999909 19999909 19999909",
"output": "2\n19999909 1 19999909 1 19999909"
},
{
"input": "4\n1 1000000000 1 1000000000",
"output": "0\n1 1000000000 1 1000000000"
},
{
"input": "2\n64006 64006",
"output": "1\n64006 1 64006"
},
{
"input": "2\n1956955 1956955",
"output": "1\n1956955 1 1956955"
},
{
"input": "3\n1 1000000000 1000000000",
"output": "1\n1 1000000000 1 1000000000"
},
{
"input": "2\n982451707 982451707",
"output": "1\n982451707 1 982451707"
},
{
"input": "2\n999999733 999999733",
"output": "1\n999999733 1 999999733"
},
{
"input": "3\n999999733 999999733 999999733",
"output": "2\n999999733 1 999999733 1 999999733"
},
{
"input": "2\n3257 3257",
"output": "1\n3257 1 3257"
},
{
"input": "2\n223092870 181598",
"output": "1\n223092870 1 181598"
},
{
"input": "3\n959919409 105935 105935",
"output": "2\n959919409 1 105935 1 105935"
},
{
"input": "2\n510510 510510",
"output": "1\n510510 1 510510"
},
{
"input": "3\n223092870 1000000000 1000000000",
"output": "2\n223092870 1 1000000000 1 1000000000"
},
{
"input": "14\n1000000000 2 1000000000 3 1000000000 6 1000000000 1000000000 15 1000000000 1000000000 1000000000 100000000 1000",
"output": "11\n1000000000 1 2 1 1000000000 3 1000000000 1 6 1 1000000000 1 1000000000 1 15 1 1000000000 1 1000000000 1 1000000000 1 100000000 1 1000"
},
{
"input": "7\n1 982451653 982451653 1 982451653 982451653 982451653",
"output": "3\n1 982451653 1 982451653 1 982451653 1 982451653 1 982451653"
},
{
"input": "2\n100000007 100000007",
"output": "1\n100000007 1 100000007"
},
{
"input": "3\n999999757 999999757 999999757",
"output": "2\n999999757 1 999999757 1 999999757"
},
{
"input": "3\n99999989 99999989 99999989",
"output": "2\n99999989 1 99999989 1 99999989"
},
{
"input": "5\n2 4 982451707 982451707 3",
"output": "2\n2 1 4 982451707 1 982451707 3"
},
{
"input": "2\n20000014 20000014",
"output": "1\n20000014 1 20000014"
},
{
"input": "2\n99999989 99999989",
"output": "1\n99999989 1 99999989"
},
{
"input": "2\n111546435 111546435",
"output": "1\n111546435 1 111546435"
},
{
"input": "2\n55288874 33538046",
"output": "1\n55288874 1 33538046"
},
{
"input": "5\n179424673 179424673 179424673 179424673 179424673",
"output": "4\n179424673 1 179424673 1 179424673 1 179424673 1 179424673"
},
{
"input": "2\n199999978 199999978",
"output": "1\n199999978 1 199999978"
},
{
"input": "2\n1000000000 2",
"output": "1\n1000000000 1 2"
},
{
"input": "3\n19999897 19999897 19999897",
"output": "2\n19999897 1 19999897 1 19999897"
},
{
"input": "2\n19999982 19999982",
"output": "1\n19999982 1 19999982"
},
{
"input": "2\n10000007 10000007",
"output": "1\n10000007 1 10000007"
},
{
"input": "3\n999999937 999999937 2",
"output": "1\n999999937 1 999999937 2"
},
{
"input": "5\n2017 2017 2017 2017 2017",
"output": "4\n2017 1 2017 1 2017 1 2017 1 2017"
},
{
"input": "2\n19999909 39999818",
"output": "1\n19999909 1 39999818"
},
{
"input": "2\n62615533 7919",
"output": "1\n62615533 1 7919"
},
{
"input": "5\n39989 39989 33 31 29",
"output": "1\n39989 1 39989 33 31 29"
},
{
"input": "2\n1000000000 100000",
"output": "1\n1000000000 1 100000"
},
{
"input": "2\n1938 10010",
"output": "1\n1938 1 10010"
},
{
"input": "2\n199999 199999",
"output": "1\n199999 1 199999"
},
{
"input": "2\n107273 107273",
"output": "1\n107273 1 107273"
},
{
"input": "3\n49999 49999 49999",
"output": "2\n49999 1 49999 1 49999"
},
{
"input": "2\n1999966 1999958",
"output": "1\n1999966 1 1999958"
},
{
"input": "2\n86020 300846",
"output": "1\n86020 1 300846"
},
{
"input": "2\n999999997 213",
"output": "1\n999999997 1 213"
},
{
"input": "2\n200000014 200000434",
"output": "1\n200000014 1 200000434"
}
] | 1,590,046,203 | 2,147,483,647 | PyPy 3 | OK | TESTS | 93 | 155 | 2,457,600 | from math import gcd
n=int(input())
k=0
l=list(map(int,input().split()))
p=[]
for i in range(n-1):
if gcd(l[i],l[i+1])>1:
p.append(l[i])
p.append(1)
k=k+1
else:
p.append(l[i])
p.append(l[-1])
print(k)
print(*p,sep=" ") | Title: Co-prime Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array of *n* elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers *a* and *b* are said to be co-prime if the only positive integer that divides both of them is 1.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the given array.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*.
Output Specification:
Print integer *k* on the first line — the least number of elements needed to add to the array *a* to make it co-prime.
The second line should contain *n*<=+<=*k* integers *a**j* — the elements of the array *a* after adding *k* elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array *a* by adding *k* elements to it.
If there are multiple answers you can print any one of them.
Demo Input:
['3\n2 7 28\n']
Demo Output:
['1\n2 7 9 28\n']
Note:
none | ```python
from math import gcd
n=int(input())
k=0
l=list(map(int,input().split()))
p=[]
for i in range(n-1):
if gcd(l[i],l[i+1])>1:
p.append(l[i])
p.append(1)
k=k+1
else:
p.append(l[i])
p.append(l[-1])
print(k)
print(*p,sep=" ")
``` | 3 | |
706 | B | Interesting drink | PROGRAMMING | 1,100 | [
"binary search",
"dp",
"implementation"
] | null | null | Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in *n* different shops in the city. It's known that the price of one bottle in the shop *i* is equal to *x**i* coins.
Vasiliy plans to buy his favorite drink for *q* consecutive days. He knows, that on the *i*-th day he will be able to spent *m**i* coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola". | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains *n* integers *x**i* (1<=≤<=*x**i*<=≤<=100<=000) — prices of the bottles of the drink in the *i*-th shop.
The third line contains a single integer *q* (1<=≤<=*q*<=≤<=100<=000) — the number of days Vasiliy plans to buy the drink.
Then follow *q* lines each containing one integer *m**i* (1<=≤<=*m**i*<=≤<=109) — the number of coins Vasiliy can spent on the *i*-th day. | Print *q* integers. The *i*-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the *i*-th day. | [
"5\n3 10 8 6 11\n4\n1\n10\n3\n11\n"
] | [
"0\n4\n1\n5\n"
] | On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop. | 1,000 | [
{
"input": "5\n3 10 8 6 11\n4\n1\n10\n3\n11",
"output": "0\n4\n1\n5"
},
{
"input": "5\n868 987 714 168 123\n10\n424\n192\n795\n873\n117\n914\n735\n158\n631\n471",
"output": "2\n2\n3\n4\n0\n4\n3\n1\n2\n2"
},
{
"input": "3\n435 482 309\n7\n245\n241\n909\n745\n980\n29\n521",
"output": "0\n0\n3\n3\n3\n0\n3"
},
{
"input": "1\n653\n9\n903\n980\n80\n770\n965\n874\n381\n657\n969",
"output": "1\n1\n0\n1\n1\n1\n0\n1\n1"
},
{
"input": "12\n35345 58181 32223 84621 35905 73863 99537 30666 67771 39229 36847 29038\n23\n55052824\n82504840\n35160556\n78141700\n73401989\n86305919\n39430705\n31939373\n23501765\n4406029\n61436920\n14295390\n34275309\n28028753\n85724689\n70158847\n2396455\n66994588\n84024224\n30175981\n65622319\n76517111\n76605341",
"output": "12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12"
},
{
"input": "4\n698 193 1584 76192\n4\n553640\n310518\n274280\n429192",
"output": "4\n4\n4\n4"
},
{
"input": "1\n1\n1\n1000000000",
"output": "1"
},
{
"input": "1\n1\n1\n100005",
"output": "1"
},
{
"input": "3\n1 2 3\n1\n100005",
"output": "3"
},
{
"input": "2\n1 2\n3\n100004\n100005\n100006",
"output": "2\n2\n2"
},
{
"input": "3\n1 1 1\n1\n1",
"output": "3"
},
{
"input": "1\n1\n1\n1",
"output": "1"
},
{
"input": "3\n1 2 3\n1\n100000",
"output": "3"
},
{
"input": "5\n1 1 1 1 10\n2\n1\n10",
"output": "4\n5"
},
{
"input": "5\n3 10 8 6 11\n1\n10000000",
"output": "5"
}
] | 1,699,410,028 | 2,147,483,647 | Python 3 | OK | TESTS | 103 | 811 | 13,619,200 | #罗景轩2300012610
import bisect
n = int(input())
l = sorted(list(map(int, input().split())))
q = int(input())
for _ in range(q):
print(bisect.bisect_right(l,int(input()))) | Title: Interesting drink
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in *n* different shops in the city. It's known that the price of one bottle in the shop *i* is equal to *x**i* coins.
Vasiliy plans to buy his favorite drink for *q* consecutive days. He knows, that on the *i*-th day he will be able to spent *m**i* coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains *n* integers *x**i* (1<=≤<=*x**i*<=≤<=100<=000) — prices of the bottles of the drink in the *i*-th shop.
The third line contains a single integer *q* (1<=≤<=*q*<=≤<=100<=000) — the number of days Vasiliy plans to buy the drink.
Then follow *q* lines each containing one integer *m**i* (1<=≤<=*m**i*<=≤<=109) — the number of coins Vasiliy can spent on the *i*-th day.
Output Specification:
Print *q* integers. The *i*-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the *i*-th day.
Demo Input:
['5\n3 10 8 6 11\n4\n1\n10\n3\n11\n']
Demo Output:
['0\n4\n1\n5\n']
Note:
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop. | ```python
#罗景轩2300012610
import bisect
n = int(input())
l = sorted(list(map(int, input().split())))
q = int(input())
for _ in range(q):
print(bisect.bisect_right(l,int(input())))
``` | 3 | |
492 | A | Vanya and Cubes | PROGRAMMING | 800 | [
"implementation"
] | null | null | Vanya got *n* cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1<=+<=2<==<=3 cubes, the third level must have 1<=+<=2<=+<=3<==<=6 cubes, and so on. Thus, the *i*-th level of the pyramid must have 1<=+<=2<=+<=...<=+<=(*i*<=-<=1)<=+<=*i* cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes. | The first line contains integer *n* (1<=≤<=*n*<=≤<=104) — the number of cubes given to Vanya. | Print the maximum possible height of the pyramid in the single line. | [
"1\n",
"25\n"
] | [
"1\n",
"4\n"
] | Illustration to the second sample: | 500 | [
{
"input": "1",
"output": "1"
},
{
"input": "25",
"output": "4"
},
{
"input": "2",
"output": "1"
},
{
"input": "4115",
"output": "28"
},
{
"input": "9894",
"output": "38"
},
{
"input": "7969",
"output": "35"
},
{
"input": "6560",
"output": "33"
},
{
"input": "4",
"output": "2"
},
{
"input": "3",
"output": "1"
},
{
"input": "5",
"output": "2"
},
{
"input": "19",
"output": "3"
},
{
"input": "20",
"output": "4"
},
{
"input": "9880",
"output": "38"
},
{
"input": "9879",
"output": "37"
},
{
"input": "7770",
"output": "35"
},
{
"input": "7769",
"output": "34"
},
{
"input": "2925",
"output": "25"
},
{
"input": "220",
"output": "10"
},
{
"input": "219",
"output": "9"
},
{
"input": "3046",
"output": "25"
},
{
"input": "7590",
"output": "34"
},
{
"input": "1014",
"output": "17"
},
{
"input": "7142",
"output": "34"
},
{
"input": "9999",
"output": "38"
},
{
"input": "10000",
"output": "38"
}
] | 1,697,990,918 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 8 | 46 | 0 | n = int(input())
s = 0
k = 0
while n>s:
k+= 1
s += k
n -= s
print(k) | Title: Vanya and Cubes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya got *n* cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1<=+<=2<==<=3 cubes, the third level must have 1<=+<=2<=+<=3<==<=6 cubes, and so on. Thus, the *i*-th level of the pyramid must have 1<=+<=2<=+<=...<=+<=(*i*<=-<=1)<=+<=*i* cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=104) — the number of cubes given to Vanya.
Output Specification:
Print the maximum possible height of the pyramid in the single line.
Demo Input:
['1\n', '25\n']
Demo Output:
['1\n', '4\n']
Note:
Illustration to the second sample: | ```python
n = int(input())
s = 0
k = 0
while n>s:
k+= 1
s += k
n -= s
print(k)
``` | 0 | |
755 | A | PolandBall and Hypothesis | PROGRAMMING | 800 | [
"brute force",
"graphs",
"math",
"number theory"
] | null | null | PolandBall is a young, clever Ball. He is interested in prime numbers. He has stated a following hypothesis: "There exists such a positive integer *n* that for each positive integer *m* number *n*·*m*<=+<=1 is a prime number".
Unfortunately, PolandBall is not experienced yet and doesn't know that his hypothesis is incorrect. Could you prove it wrong? Write a program that finds a counterexample for any *n*. | The only number in the input is *n* (1<=≤<=*n*<=≤<=1000) — number from the PolandBall's hypothesis. | Output such *m* that *n*·*m*<=+<=1 is not a prime number. Your answer will be considered correct if you output any suitable *m* such that 1<=≤<=*m*<=≤<=103. It is guaranteed the the answer exists. | [
"3\n",
"4\n"
] | [
"1",
"2"
] | A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.
For the first sample testcase, 3·1 + 1 = 4. We can output 1.
In the second sample testcase, 4·1 + 1 = 5. We cannot output 1 because 5 is prime. However, *m* = 2 is okay since 4·2 + 1 = 9, which is not a prime number. | 500 | [
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "2"
},
{
"input": "10",
"output": "2"
},
{
"input": "153",
"output": "1"
},
{
"input": "1000",
"output": "1"
},
{
"input": "1",
"output": "3"
},
{
"input": "2",
"output": "4"
},
{
"input": "5",
"output": "1"
},
{
"input": "6",
"output": "4"
},
{
"input": "7",
"output": "1"
},
{
"input": "8",
"output": "1"
},
{
"input": "9",
"output": "1"
},
{
"input": "11",
"output": "1"
},
{
"input": "998",
"output": "1"
},
{
"input": "996",
"output": "3"
},
{
"input": "36",
"output": "4"
},
{
"input": "210",
"output": "4"
},
{
"input": "270",
"output": "4"
},
{
"input": "306",
"output": "4"
},
{
"input": "330",
"output": "5"
},
{
"input": "336",
"output": "4"
},
{
"input": "600",
"output": "4"
},
{
"input": "726",
"output": "4"
},
{
"input": "988",
"output": "1"
},
{
"input": "12",
"output": "2"
},
{
"input": "987",
"output": "1"
},
{
"input": "13",
"output": "1"
},
{
"input": "986",
"output": "1"
},
{
"input": "14",
"output": "1"
},
{
"input": "985",
"output": "1"
},
{
"input": "15",
"output": "1"
},
{
"input": "984",
"output": "1"
},
{
"input": "16",
"output": "2"
},
{
"input": "983",
"output": "1"
},
{
"input": "17",
"output": "1"
},
{
"input": "982",
"output": "2"
},
{
"input": "18",
"output": "3"
},
{
"input": "981",
"output": "1"
},
{
"input": "19",
"output": "1"
},
{
"input": "980",
"output": "1"
},
{
"input": "20",
"output": "1"
},
{
"input": "979",
"output": "1"
},
{
"input": "21",
"output": "1"
},
{
"input": "978",
"output": "1"
},
{
"input": "22",
"output": "2"
},
{
"input": "977",
"output": "1"
},
{
"input": "23",
"output": "1"
},
{
"input": "976",
"output": "2"
},
{
"input": "24",
"output": "1"
},
{
"input": "975",
"output": "1"
},
{
"input": "25",
"output": "1"
},
{
"input": "2",
"output": "4"
},
{
"input": "6",
"output": "4"
},
{
"input": "996",
"output": "3"
},
{
"input": "8",
"output": "1"
},
{
"input": "998",
"output": "1"
},
{
"input": "330",
"output": "5"
},
{
"input": "18",
"output": "3"
},
{
"input": "12",
"output": "2"
},
{
"input": "24",
"output": "1"
},
{
"input": "999",
"output": "1"
},
{
"input": "726",
"output": "4"
},
{
"input": "876",
"output": "3"
}
] | 1,642,261,376 | 2,147,483,647 | Python 3 | OK | TESTS | 63 | 46 | 0 | n = int(input())
print(min(n+2, 999))
| Title: PolandBall and Hypothesis
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
PolandBall is a young, clever Ball. He is interested in prime numbers. He has stated a following hypothesis: "There exists such a positive integer *n* that for each positive integer *m* number *n*·*m*<=+<=1 is a prime number".
Unfortunately, PolandBall is not experienced yet and doesn't know that his hypothesis is incorrect. Could you prove it wrong? Write a program that finds a counterexample for any *n*.
Input Specification:
The only number in the input is *n* (1<=≤<=*n*<=≤<=1000) — number from the PolandBall's hypothesis.
Output Specification:
Output such *m* that *n*·*m*<=+<=1 is not a prime number. Your answer will be considered correct if you output any suitable *m* such that 1<=≤<=*m*<=≤<=103. It is guaranteed the the answer exists.
Demo Input:
['3\n', '4\n']
Demo Output:
['1', '2']
Note:
A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.
For the first sample testcase, 3·1 + 1 = 4. We can output 1.
In the second sample testcase, 4·1 + 1 = 5. We cannot output 1 because 5 is prime. However, *m* = 2 is okay since 4·2 + 1 = 9, which is not a prime number. | ```python
n = int(input())
print(min(n+2, 999))
``` | 3 | |
405 | A | Gravity Flip | PROGRAMMING | 900 | [
"greedy",
"implementation",
"sortings"
] | null | null | Little Chris is bored during his physics lessons (too easy), so he has built a toy box to keep himself occupied. The box is special, since it has the ability to change gravity.
There are *n* columns of toy cubes in the box arranged in a line. The *i*-th column contains *a**i* cubes. At first, the gravity in the box is pulling the cubes downwards. When Chris switches the gravity, it begins to pull all the cubes to the right side of the box. The figure shows the initial and final configurations of the cubes in the box: the cubes that have changed their position are highlighted with orange.
Given the initial configuration of the toy cubes in the box, find the amounts of cubes in each of the *n* columns after the gravity switch! | The first line of input contains an integer *n* (1<=≤<=*n*<=≤<=100), the number of the columns in the box. The next line contains *n* space-separated integer numbers. The *i*-th number *a**i* (1<=≤<=*a**i*<=≤<=100) denotes the number of cubes in the *i*-th column. | Output *n* integer numbers separated by spaces, where the *i*-th number is the amount of cubes in the *i*-th column after the gravity switch. | [
"4\n3 2 1 2\n",
"3\n2 3 8\n"
] | [
"1 2 2 3 \n",
"2 3 8 \n"
] | The first example case is shown on the figure. The top cube of the first column falls to the top of the last column; the top cube of the second column falls to the top of the third column; the middle cube of the first column falls to the top of the second column.
In the second example case the gravity switch does not change the heights of the columns. | 500 | [
{
"input": "4\n3 2 1 2",
"output": "1 2 2 3 "
},
{
"input": "3\n2 3 8",
"output": "2 3 8 "
},
{
"input": "5\n2 1 2 1 2",
"output": "1 1 2 2 2 "
},
{
"input": "1\n1",
"output": "1 "
},
{
"input": "2\n4 3",
"output": "3 4 "
},
{
"input": "6\n100 40 60 20 1 80",
"output": "1 20 40 60 80 100 "
},
{
"input": "10\n10 8 6 7 5 3 4 2 9 1",
"output": "1 2 3 4 5 6 7 8 9 10 "
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "1 2 3 4 5 6 7 8 9 10 "
},
{
"input": "100\n82 51 81 14 37 17 78 92 64 15 8 86 89 8 87 77 66 10 15 12 100 25 92 47 21 78 20 63 13 49 41 36 41 79 16 87 87 69 3 76 80 60 100 49 70 59 72 8 38 71 45 97 71 14 76 54 81 4 59 46 39 29 92 3 49 22 53 99 59 52 74 31 92 43 42 23 44 9 82 47 7 40 12 9 3 55 37 85 46 22 84 52 98 41 21 77 63 17 62 91",
"output": "3 3 3 4 7 8 8 8 9 9 10 12 12 13 14 14 15 15 16 17 17 20 21 21 22 22 23 25 29 31 36 37 37 38 39 40 41 41 41 42 43 44 45 46 46 47 47 49 49 49 51 52 52 53 54 55 59 59 59 60 62 63 63 64 66 69 70 71 71 72 74 76 76 77 77 78 78 79 80 81 81 82 82 84 85 86 87 87 87 89 91 92 92 92 92 97 98 99 100 100 "
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 "
},
{
"input": "10\n1 9 7 6 2 4 7 8 1 3",
"output": "1 1 2 3 4 6 7 7 8 9 "
},
{
"input": "20\n53 32 64 20 41 97 50 20 66 68 22 60 74 61 97 54 80 30 72 59",
"output": "20 20 22 30 32 41 50 53 54 59 60 61 64 66 68 72 74 80 97 97 "
},
{
"input": "30\n7 17 4 18 16 12 14 10 1 13 2 16 13 17 8 16 13 14 9 17 17 5 13 5 1 7 6 20 18 12",
"output": "1 1 2 4 5 5 6 7 7 8 9 10 12 12 13 13 13 13 14 14 16 16 16 17 17 17 17 18 18 20 "
},
{
"input": "40\n22 58 68 58 48 53 52 1 16 78 75 17 63 15 36 32 78 75 49 14 42 46 66 54 49 82 40 43 46 55 12 73 5 45 61 60 1 11 31 84",
"output": "1 1 5 11 12 14 15 16 17 22 31 32 36 40 42 43 45 46 46 48 49 49 52 53 54 55 58 58 60 61 63 66 68 73 75 75 78 78 82 84 "
},
{
"input": "70\n1 3 3 1 3 3 1 1 1 3 3 2 3 3 1 1 1 2 3 1 3 2 3 3 3 2 2 3 1 3 3 2 1 1 2 1 2 1 2 2 1 1 1 3 3 2 3 2 3 2 3 3 2 2 2 3 2 3 3 3 1 1 3 3 1 1 1 1 3 1",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 "
},
{
"input": "90\n17 75 51 30 100 5 50 95 51 73 66 5 7 76 43 49 23 55 3 24 95 79 10 11 44 93 17 99 53 66 82 66 63 76 19 4 51 71 75 43 27 5 24 19 48 7 91 15 55 21 7 6 27 10 2 91 64 58 18 21 16 71 90 88 21 20 6 6 95 85 11 7 40 65 52 49 92 98 46 88 17 48 85 96 77 46 100 34 67 52",
"output": "2 3 4 5 5 5 6 6 6 7 7 7 7 10 10 11 11 15 16 17 17 17 18 19 19 20 21 21 21 23 24 24 27 27 30 34 40 43 43 44 46 46 48 48 49 49 50 51 51 51 52 52 53 55 55 58 63 64 65 66 66 66 67 71 71 73 75 75 76 76 77 79 82 85 85 88 88 90 91 91 92 93 95 95 95 96 98 99 100 100 "
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 "
},
{
"input": "100\n1 1 1 1 2 1 1 1 1 1 2 2 1 1 2 1 2 1 1 1 2 1 1 2 1 2 1 1 2 2 2 1 1 2 1 1 1 2 2 2 1 1 1 2 1 2 2 1 2 1 1 2 2 1 2 1 2 1 2 2 1 1 1 2 1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 1 1 1 1 2 2 2 2 2 2 2 1 1 1 2 1 2 1",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "100\n2 1 1 1 3 2 3 3 2 3 3 1 3 3 1 3 3 1 1 1 2 3 1 2 3 1 2 3 3 1 3 1 1 2 3 2 3 3 2 3 3 1 2 2 1 2 3 2 3 2 2 1 1 3 1 3 2 1 3 1 3 1 3 1 1 3 3 3 2 3 2 2 2 2 1 3 3 3 1 2 1 2 3 2 1 3 1 3 2 1 3 1 2 1 2 3 1 3 2 3",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 "
},
{
"input": "100\n7 4 5 5 10 10 5 8 5 7 4 5 4 6 8 8 2 6 3 3 10 7 10 8 6 2 7 3 9 7 7 2 4 5 2 4 9 5 10 1 10 5 10 4 1 3 4 2 6 9 9 9 10 6 2 5 6 1 8 10 4 10 3 4 10 5 5 4 10 4 5 3 7 10 2 7 3 6 9 6 1 6 5 5 4 6 6 4 4 1 5 1 6 6 6 8 8 6 2 6",
"output": "1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 10 10 10 "
},
{
"input": "100\n12 10 5 11 13 12 14 13 7 15 15 12 13 19 12 18 14 10 10 3 1 10 16 11 19 8 10 15 5 10 12 16 11 13 11 15 14 12 16 8 11 8 15 2 18 2 14 13 15 20 8 8 4 12 14 7 10 3 9 1 7 19 6 7 2 14 8 20 7 17 18 20 3 18 18 9 6 10 4 1 4 19 9 13 3 3 12 11 11 20 8 2 13 6 7 12 1 4 17 3",
"output": "1 1 1 1 2 2 2 2 3 3 3 3 3 3 4 4 4 4 5 5 6 6 6 7 7 7 7 7 7 8 8 8 8 8 8 8 9 9 9 10 10 10 10 10 10 10 10 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12 12 13 13 13 13 13 13 13 14 14 14 14 14 14 15 15 15 15 15 15 16 16 16 17 17 18 18 18 18 18 19 19 19 19 20 20 20 20 "
},
{
"input": "100\n5 13 1 40 30 10 23 32 33 12 6 4 15 29 31 17 23 5 36 31 32 38 24 11 34 39 19 21 6 19 31 35 1 15 6 29 22 15 17 15 1 17 2 34 20 8 27 2 29 26 13 9 22 27 27 3 20 40 4 40 33 29 36 30 35 16 19 28 26 11 36 24 29 5 40 10 38 34 33 23 34 39 31 7 10 31 22 6 36 24 14 31 34 23 2 4 26 16 2 32",
"output": "1 1 1 2 2 2 2 3 4 4 4 5 5 5 6 6 6 6 7 8 9 10 10 10 11 11 12 13 13 14 15 15 15 15 16 16 17 17 17 19 19 19 20 20 21 22 22 22 23 23 23 23 24 24 24 26 26 26 27 27 27 28 29 29 29 29 29 30 30 31 31 31 31 31 31 32 32 32 33 33 33 34 34 34 34 34 35 35 36 36 36 36 38 38 39 39 40 40 40 40 "
},
{
"input": "100\n72 44 34 74 9 60 26 37 55 77 74 69 28 66 54 55 8 36 57 31 31 48 32 66 40 70 77 43 64 28 37 10 21 58 51 32 60 28 51 52 28 35 7 33 1 68 38 70 57 71 8 20 42 57 59 4 58 10 17 47 22 48 16 3 76 67 32 37 64 47 33 41 75 69 2 76 39 9 27 75 20 21 52 25 71 21 11 29 38 10 3 1 45 55 63 36 27 7 59 41",
"output": "1 1 2 3 3 4 7 7 8 8 9 9 10 10 10 11 16 17 20 20 21 21 21 22 25 26 27 27 28 28 28 28 29 31 31 32 32 32 33 33 34 35 36 36 37 37 37 38 38 39 40 41 41 42 43 44 45 47 47 48 48 51 51 52 52 54 55 55 55 57 57 57 58 58 59 59 60 60 63 64 64 66 66 67 68 69 69 70 70 71 71 72 74 74 75 75 76 76 77 77 "
},
{
"input": "100\n75 18 61 10 56 53 42 57 79 80 31 2 50 45 54 99 84 52 71 21 86 3 19 98 14 37 40 62 63 68 5 10 87 8 81 85 52 52 57 94 2 7 56 96 19 76 1 13 81 6 80 47 22 59 99 32 9 5 36 88 98 91 70 70 12 93 12 22 85 1 97 48 94 16 84 84 51 34 62 7 68 51 30 2 37 82 4 7 27 1 80 9 61 16 59 55 12 96 94 82",
"output": "1 1 1 2 2 2 3 4 5 5 6 7 7 7 8 9 9 10 10 12 12 12 13 14 16 16 18 19 19 21 22 22 27 30 31 32 34 36 37 37 40 42 45 47 48 50 51 51 52 52 52 53 54 55 56 56 57 57 59 59 61 61 62 62 63 68 68 70 70 71 75 76 79 80 80 80 81 81 82 82 84 84 84 85 85 86 87 88 91 93 94 94 94 96 96 97 98 98 99 99 "
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 "
},
{
"input": "100\n100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 "
},
{
"input": "100\n50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50",
"output": "50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 "
},
{
"input": "49\n1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97",
"output": "1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 "
},
{
"input": "30\n1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79 82 85 88",
"output": "1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79 82 85 88 "
},
{
"input": "100\n100 51 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 51 100 "
},
{
"input": "10\n100 90 80 70 60 50 40 30 20 10",
"output": "10 20 30 40 50 60 70 80 90 100 "
},
{
"input": "1\n10",
"output": "10 "
}
] | 1,694,515,069 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 46 | 0 | n = int(input())
b = input()
c = list(map(int, b.split()))
d = sorted(c)
for i in range(n-1):
print(d[i], end =' ')
print(d[n-1]) | Title: Gravity Flip
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Chris is bored during his physics lessons (too easy), so he has built a toy box to keep himself occupied. The box is special, since it has the ability to change gravity.
There are *n* columns of toy cubes in the box arranged in a line. The *i*-th column contains *a**i* cubes. At first, the gravity in the box is pulling the cubes downwards. When Chris switches the gravity, it begins to pull all the cubes to the right side of the box. The figure shows the initial and final configurations of the cubes in the box: the cubes that have changed their position are highlighted with orange.
Given the initial configuration of the toy cubes in the box, find the amounts of cubes in each of the *n* columns after the gravity switch!
Input Specification:
The first line of input contains an integer *n* (1<=≤<=*n*<=≤<=100), the number of the columns in the box. The next line contains *n* space-separated integer numbers. The *i*-th number *a**i* (1<=≤<=*a**i*<=≤<=100) denotes the number of cubes in the *i*-th column.
Output Specification:
Output *n* integer numbers separated by spaces, where the *i*-th number is the amount of cubes in the *i*-th column after the gravity switch.
Demo Input:
['4\n3 2 1 2\n', '3\n2 3 8\n']
Demo Output:
['1 2 2 3 \n', '2 3 8 \n']
Note:
The first example case is shown on the figure. The top cube of the first column falls to the top of the last column; the top cube of the second column falls to the top of the third column; the middle cube of the first column falls to the top of the second column.
In the second example case the gravity switch does not change the heights of the columns. | ```python
n = int(input())
b = input()
c = list(map(int, b.split()))
d = sorted(c)
for i in range(n-1):
print(d[i], end =' ')
print(d[n-1])
``` | 3 | |
578 | B | "Or" Game | PROGRAMMING | 1,700 | [
"brute force",
"greedy"
] | null | null | You are given *n* numbers *a*1,<=*a*2,<=...,<=*a**n*. You can perform at most *k* operations. For each operation you can multiply one of the numbers by *x*. We want to make as large as possible, where denotes the bitwise OR.
Find the maximum possible value of after performing at most *k* operations optimally. | The first line contains three integers *n*, *k* and *x* (1<=≤<=*n*<=≤<=200<=000, 1<=≤<=*k*<=≤<=10, 2<=≤<=*x*<=≤<=8).
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109). | Output the maximum value of a bitwise OR of sequence elements after performing operations. | [
"3 1 2\n1 1 1\n",
"4 2 3\n1 2 4 8\n"
] | [
"3\n",
"79\n"
] | For the first sample, any possible choice of doing one operation will result the same three numbers 1, 1, 2 so the result is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/1ee73b671ed4bc53f2f96ed1a85fd98388e1712b.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
For the second sample if we multiply 8 by 3 two times we'll get 72. In this case the numbers will become 1, 2, 4, 72 so the OR value will be 79 and is the largest possible result. | 500 | [
{
"input": "3 1 2\n1 1 1",
"output": "3"
},
{
"input": "4 2 3\n1 2 4 8",
"output": "79"
},
{
"input": "2 1 2\n12 9",
"output": "30"
},
{
"input": "2 1 2\n12 7",
"output": "31"
},
{
"input": "3 1 3\n3 2 0",
"output": "11"
},
{
"input": "5 10 8\n1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "1073741825000000000"
},
{
"input": "1 2 3\n612635770",
"output": "5513721930"
},
{
"input": "3 2 5\n0 2 3",
"output": "75"
},
{
"input": "2 1 8\n18 17",
"output": "154"
},
{
"input": "5 10 8\n0 0 0 0 0",
"output": "0"
},
{
"input": "1 1 2\n1",
"output": "2"
},
{
"input": "1 1 2\n0",
"output": "0"
},
{
"input": "3 2 6\n724148075 828984987 810015532",
"output": "29996605423"
},
{
"input": "3 1 2\n17 18 4",
"output": "54"
},
{
"input": "3 1 2\n4 17 18",
"output": "54"
},
{
"input": "2 2 2\n60 59",
"output": "252"
},
{
"input": "2 2 2\n9 10",
"output": "46"
},
{
"input": "3 1 2\n10 12 5",
"output": "31"
},
{
"input": "3 1 2\n20 17 8",
"output": "62"
},
{
"input": "3 1 2\n5 12 10",
"output": "31"
},
{
"input": "3 1 8\n10 17 18",
"output": "155"
},
{
"input": "3 1 2\n17 20 28",
"output": "62"
},
{
"input": "5 1 3\n1 5 13 8 16",
"output": "63"
}
] | 1,560,593,345 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 7 | 2,000 | 15,564,800 | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Fri Jun 14 21:25:50 2019
@author: pawangoyal
"""
input_1=input()
input_1=input_1.split()
n = int(input_1[0])
k = int(input_1[1])
x = int(input_1[2])
a=input()
a=a.split()
a=[int(x) for x in a]
maximum=0
for i in range(n):
b=a[:]
b[i]=b[i]*(x**k)
or_bit=0
for j in range(n):
or_bit = or_bit|b[j]
maximum=max(maximum,or_bit)
print(maximum)
| Title: "Or" Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given *n* numbers *a*1,<=*a*2,<=...,<=*a**n*. You can perform at most *k* operations. For each operation you can multiply one of the numbers by *x*. We want to make as large as possible, where denotes the bitwise OR.
Find the maximum possible value of after performing at most *k* operations optimally.
Input Specification:
The first line contains three integers *n*, *k* and *x* (1<=≤<=*n*<=≤<=200<=000, 1<=≤<=*k*<=≤<=10, 2<=≤<=*x*<=≤<=8).
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=109).
Output Specification:
Output the maximum value of a bitwise OR of sequence elements after performing operations.
Demo Input:
['3 1 2\n1 1 1\n', '4 2 3\n1 2 4 8\n']
Demo Output:
['3\n', '79\n']
Note:
For the first sample, any possible choice of doing one operation will result the same three numbers 1, 1, 2 so the result is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/1ee73b671ed4bc53f2f96ed1a85fd98388e1712b.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
For the second sample if we multiply 8 by 3 two times we'll get 72. In this case the numbers will become 1, 2, 4, 72 so the OR value will be 79 and is the largest possible result. | ```python
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Fri Jun 14 21:25:50 2019
@author: pawangoyal
"""
input_1=input()
input_1=input_1.split()
n = int(input_1[0])
k = int(input_1[1])
x = int(input_1[2])
a=input()
a=a.split()
a=[int(x) for x in a]
maximum=0
for i in range(n):
b=a[:]
b[i]=b[i]*(x**k)
or_bit=0
for j in range(n):
or_bit = or_bit|b[j]
maximum=max(maximum,or_bit)
print(maximum)
``` | 0 | |
978 | C | Letters | PROGRAMMING | 1,000 | [
"binary search",
"implementation",
"two pointers"
] | null | null | There are $n$ dormitories in Berland State University, they are numbered with integers from $1$ to $n$. Each dormitory consists of rooms, there are $a_i$ rooms in $i$-th dormitory. The rooms in $i$-th dormitory are numbered from $1$ to $a_i$.
A postman delivers letters. Sometimes there is no specific dormitory and room number in it on an envelope. Instead of it only a room number among all rooms of all $n$ dormitories is written on an envelope. In this case, assume that all the rooms are numbered from $1$ to $a_1 + a_2 + \dots + a_n$ and the rooms of the first dormitory go first, the rooms of the second dormitory go after them and so on.
For example, in case $n=2$, $a_1=3$ and $a_2=5$ an envelope can have any integer from $1$ to $8$ written on it. If the number $7$ is written on an envelope, it means that the letter should be delivered to the room number $4$ of the second dormitory.
For each of $m$ letters by the room number among all $n$ dormitories, determine the particular dormitory and the room number in a dormitory where this letter should be delivered. | The first line contains two integers $n$ and $m$ $(1 \le n, m \le 2 \cdot 10^{5})$ — the number of dormitories and the number of letters.
The second line contains a sequence $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^{10})$, where $a_i$ equals to the number of rooms in the $i$-th dormitory. The third line contains a sequence $b_1, b_2, \dots, b_m$ $(1 \le b_j \le a_1 + a_2 + \dots + a_n)$, where $b_j$ equals to the room number (among all rooms of all dormitories) for the $j$-th letter. All $b_j$ are given in increasing order. | Print $m$ lines. For each letter print two integers $f$ and $k$ — the dormitory number $f$ $(1 \le f \le n)$ and the room number $k$ in this dormitory $(1 \le k \le a_f)$ to deliver the letter. | [
"3 6\n10 15 12\n1 9 12 23 26 37\n",
"2 3\n5 10000000000\n5 6 9999999999\n"
] | [
"1 1\n1 9\n2 2\n2 13\n3 1\n3 12\n",
"1 5\n2 1\n2 9999999994\n"
] | In the first example letters should be delivered in the following order:
- the first letter in room $1$ of the first dormitory - the second letter in room $9$ of the first dormitory - the third letter in room $2$ of the second dormitory - the fourth letter in room $13$ of the second dormitory - the fifth letter in room $1$ of the third dormitory - the sixth letter in room $12$ of the third dormitory | 0 | [
{
"input": "3 6\n10 15 12\n1 9 12 23 26 37",
"output": "1 1\n1 9\n2 2\n2 13\n3 1\n3 12"
},
{
"input": "2 3\n5 10000000000\n5 6 9999999999",
"output": "1 5\n2 1\n2 9999999994"
},
{
"input": "1 1\n1\n1",
"output": "1 1"
},
{
"input": "5 15\n10 20 30 20 10\n1 6 10 11 15 30 31 54 60 61 76 80 81 84 90",
"output": "1 1\n1 6\n1 10\n2 1\n2 5\n2 20\n3 1\n3 24\n3 30\n4 1\n4 16\n4 20\n5 1\n5 4\n5 10"
},
{
"input": "1 10\n10\n1 2 3 4 5 6 7 8 9 10",
"output": "1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10"
},
{
"input": "5 8\n10 1 1 1 10\n9 10 11 12 13 14 15 23",
"output": "1 9\n1 10\n2 1\n3 1\n4 1\n5 1\n5 2\n5 10"
},
{
"input": "1 3\n10000\n1 4325 10000",
"output": "1 1\n1 4325\n1 10000"
},
{
"input": "4 18\n5 6 3 4\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18",
"output": "1 1\n1 2\n1 3\n1 4\n1 5\n2 1\n2 2\n2 3\n2 4\n2 5\n2 6\n3 1\n3 2\n3 3\n4 1\n4 2\n4 3\n4 4"
},
{
"input": "3 10\n1000000000 1000000000 1000000000\n543678543 567869543 1000000000 1000000001 1500000000 1999999999 2000000000 2000000001 2754432345 3000000000",
"output": "1 543678543\n1 567869543\n1 1000000000\n2 1\n2 500000000\n2 999999999\n2 1000000000\n3 1\n3 754432345\n3 1000000000"
}
] | 1,700,067,489 | 2,147,483,647 | Python 3 | OK | TESTS | 16 | 701 | 37,273,600 |
def solve(n,m,a,b):
cum =[]*len(a)
j=0
dor=0
room=0
for i in range(len(a)):
cum.append(a[i])
for i in range(1,len(a)):
cum[i]= cum[i]+cum[i-1]
for i in range(len(b)):
while b[i]>cum[j]:
j=j+1
dor = j+1
if j==0:
room=b[i]
else:
room = b[i] -cum[j-1]
print(dor,room)
n, m =input().split()
a =[int(_) for _ in input().split()]
b =[int(_) for _ in input().split()]
"""print(n,m)
print(a)
print(b)"""
solve(n,m,a,b)
| Title: Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are $n$ dormitories in Berland State University, they are numbered with integers from $1$ to $n$. Each dormitory consists of rooms, there are $a_i$ rooms in $i$-th dormitory. The rooms in $i$-th dormitory are numbered from $1$ to $a_i$.
A postman delivers letters. Sometimes there is no specific dormitory and room number in it on an envelope. Instead of it only a room number among all rooms of all $n$ dormitories is written on an envelope. In this case, assume that all the rooms are numbered from $1$ to $a_1 + a_2 + \dots + a_n$ and the rooms of the first dormitory go first, the rooms of the second dormitory go after them and so on.
For example, in case $n=2$, $a_1=3$ and $a_2=5$ an envelope can have any integer from $1$ to $8$ written on it. If the number $7$ is written on an envelope, it means that the letter should be delivered to the room number $4$ of the second dormitory.
For each of $m$ letters by the room number among all $n$ dormitories, determine the particular dormitory and the room number in a dormitory where this letter should be delivered.
Input Specification:
The first line contains two integers $n$ and $m$ $(1 \le n, m \le 2 \cdot 10^{5})$ — the number of dormitories and the number of letters.
The second line contains a sequence $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^{10})$, where $a_i$ equals to the number of rooms in the $i$-th dormitory. The third line contains a sequence $b_1, b_2, \dots, b_m$ $(1 \le b_j \le a_1 + a_2 + \dots + a_n)$, where $b_j$ equals to the room number (among all rooms of all dormitories) for the $j$-th letter. All $b_j$ are given in increasing order.
Output Specification:
Print $m$ lines. For each letter print two integers $f$ and $k$ — the dormitory number $f$ $(1 \le f \le n)$ and the room number $k$ in this dormitory $(1 \le k \le a_f)$ to deliver the letter.
Demo Input:
['3 6\n10 15 12\n1 9 12 23 26 37\n', '2 3\n5 10000000000\n5 6 9999999999\n']
Demo Output:
['1 1\n1 9\n2 2\n2 13\n3 1\n3 12\n', '1 5\n2 1\n2 9999999994\n']
Note:
In the first example letters should be delivered in the following order:
- the first letter in room $1$ of the first dormitory - the second letter in room $9$ of the first dormitory - the third letter in room $2$ of the second dormitory - the fourth letter in room $13$ of the second dormitory - the fifth letter in room $1$ of the third dormitory - the sixth letter in room $12$ of the third dormitory | ```python
def solve(n,m,a,b):
cum =[]*len(a)
j=0
dor=0
room=0
for i in range(len(a)):
cum.append(a[i])
for i in range(1,len(a)):
cum[i]= cum[i]+cum[i-1]
for i in range(len(b)):
while b[i]>cum[j]:
j=j+1
dor = j+1
if j==0:
room=b[i]
else:
room = b[i] -cum[j-1]
print(dor,room)
n, m =input().split()
a =[int(_) for _ in input().split()]
b =[int(_) for _ in input().split()]
"""print(n,m)
print(a)
print(b)"""
solve(n,m,a,b)
``` | 3 | |
598 | A | Tricky Sum | PROGRAMMING | 900 | [
"math"
] | null | null | In this problem you are to calculate the sum of all integers from 1 to *n*, but you should take all powers of two with minus in the sum.
For example, for *n*<==<=4 the sum is equal to <=-<=1<=-<=2<=+<=3<=-<=4<==<=<=-<=4, because 1, 2 and 4 are 20, 21 and 22 respectively.
Calculate the answer for *t* values of *n*. | The first line of the input contains a single integer *t* (1<=≤<=*t*<=≤<=100) — the number of values of *n* to be processed.
Each of next *t* lines contains a single integer *n* (1<=≤<=*n*<=≤<=109). | Print the requested sum for each of *t* integers *n* given in the input. | [
"2\n4\n1000000000\n"
] | [
"-4\n499999998352516354\n"
] | The answer for the first sample is explained in the statement. | 0 | [
{
"input": "2\n4\n1000000000",
"output": "-4\n499999998352516354"
},
{
"input": "10\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10",
"output": "-1\n-3\n0\n-4\n1\n7\n14\n6\n15\n25"
},
{
"input": "10\n10\n9\n47\n33\n99\n83\n62\n1\n100\n53",
"output": "25\n15\n1002\n435\n4696\n3232\n1827\n-1\n4796\n1305"
},
{
"input": "100\n901\n712\n3\n677\n652\n757\n963\n134\n205\n888\n847\n283\n591\n984\n1\n61\n540\n986\n950\n729\n104\n244\n500\n461\n251\n685\n631\n803\n526\n600\n1000\n899\n411\n219\n597\n342\n771\n348\n507\n775\n454\n102\n486\n333\n580\n431\n537\n355\n624\n23\n429\n276\n84\n704\n96\n536\n855\n653\n72\n718\n776\n658\n802\n777\n995\n285\n328\n405\n184\n555\n956\n410\n846\n853\n525\n983\n65\n549\n839\n929\n620\n725\n635\n303\n201\n878\n580\n139\n182\n69\n400\n788\n985\n792\n103\n248\n570\n839\n253\n417",
"output": "404305\n251782\n0\n227457\n210832\n284857\n462120\n8535\n20605\n392670\n357082\n39164\n172890\n482574\n-1\n1765\n144024\n484545\n449679\n264039\n5206\n29380\n124228\n105469\n31116\n232909\n197350\n320760\n136555\n178254\n498454\n402504\n83644\n23580\n176457\n57631\n295560\n59704\n127756\n298654\n102263\n4999\n117319\n54589\n166444\n92074\n142407\n62168\n192954\n214\n91213\n37204\n3316\n246114\n4402\n141870\n363894\n211485\n2374\n256075\n299430\n214765\n319957\n300207\n493464\n39733\n52934\n81193\n16510\n15..."
},
{
"input": "1\n16",
"output": "74"
},
{
"input": "60\n536870912\n536870911\n536870913\n1000000000\n999999999\n1\n2\n3\n4\n268435456\n268435455\n268435457\n536870912\n536870911\n536870913\n1000000000\n999999999\n1\n2\n3\n4\n268435456\n268435455\n268435457\n536870912\n536870911\n536870913\n1000000000\n999999999\n1\n2\n3\n4\n268435456\n268435455\n268435457\n536870912\n536870911\n536870913\n1000000000\n999999999\n1\n2\n3\n4\n268435456\n268435455\n268435457\n536870912\n536870911\n536870913\n1000000000\n999999999\n1\n2\n3\n4\n268435456\n268435455\n268435457",
"output": "144115186196807682\n144115186733678594\n144115186733678595\n499999998352516354\n499999997352516354\n-1\n-3\n0\n-4\n36028796079439874\n36028796347875330\n36028796347875331\n144115186196807682\n144115186733678594\n144115186733678595\n499999998352516354\n499999997352516354\n-1\n-3\n0\n-4\n36028796079439874\n36028796347875330\n36028796347875331\n144115186196807682\n144115186733678594\n144115186733678595\n499999998352516354\n499999997352516354\n-1\n-3\n0\n-4\n36028796079439874\n36028796347875330\n36028796347875..."
},
{
"input": "13\n1\n19\n31\n19\n19\n92\n74\n69\n32\n32\n91\n42\n73",
"output": "-1\n128\n434\n128\n128\n4024\n2521\n2161\n402\n402\n3932\n777\n2447"
},
{
"input": "1\n16383",
"output": "134176770"
},
{
"input": "16\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100",
"output": "5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908"
},
{
"input": "1\n414234",
"output": "85794061921"
},
{
"input": "1\n414232",
"output": "85793233454"
},
{
"input": "3\n414231\n414231\n414231",
"output": "85792819222\n85792819222\n85792819222"
},
{
"input": "1\n121",
"output": "7127"
}
] | 1,667,893,177 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 0 | 1,000 | 0 | import math
a=int(input())
while(int(a)>0):
s=int(input())
x=sum(range(0,s+1))
z=float(math.log(s))
za=float(math.log(2))
sa=float(z/za)
power=int(0)
for i in range (0,int(sa+1)):
power+=math.pow(2,i)
i+=1
power*=-2
print(int(x+power))
a-=1
| Title: Tricky Sum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem you are to calculate the sum of all integers from 1 to *n*, but you should take all powers of two with minus in the sum.
For example, for *n*<==<=4 the sum is equal to <=-<=1<=-<=2<=+<=3<=-<=4<==<=<=-<=4, because 1, 2 and 4 are 20, 21 and 22 respectively.
Calculate the answer for *t* values of *n*.
Input Specification:
The first line of the input contains a single integer *t* (1<=≤<=*t*<=≤<=100) — the number of values of *n* to be processed.
Each of next *t* lines contains a single integer *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
Print the requested sum for each of *t* integers *n* given in the input.
Demo Input:
['2\n4\n1000000000\n']
Demo Output:
['-4\n499999998352516354\n']
Note:
The answer for the first sample is explained in the statement. | ```python
import math
a=int(input())
while(int(a)>0):
s=int(input())
x=sum(range(0,s+1))
z=float(math.log(s))
za=float(math.log(2))
sa=float(z/za)
power=int(0)
for i in range (0,int(sa+1)):
power+=math.pow(2,i)
i+=1
power*=-2
print(int(x+power))
a-=1
``` | 0 | |
835 | A | Key races | PROGRAMMING | 800 | [
"math"
] | null | null | Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of *s* characters. The first participant types one character in *v*1 milliseconds and has ping *t*1 milliseconds. The second participant types one character in *v*2 milliseconds and has ping *t*2 milliseconds.
If connection ping (delay) is *t* milliseconds, the competition passes for a participant as follows:
1. Exactly after *t* milliseconds after the start of the competition the participant receives the text to be entered. 1. Right after that he starts to type it. 1. Exactly *t* milliseconds after he ends typing all the text, the site receives information about it.
The winner is the participant whose information on the success comes earlier. If the information comes from both participants at the same time, it is considered that there is a draw.
Given the length of the text and the information about participants, determine the result of the game. | The first line contains five integers *s*, *v*1, *v*2, *t*1, *t*2 (1<=≤<=*s*,<=*v*1,<=*v*2,<=*t*1,<=*t*2<=≤<=1000) — the number of characters in the text, the time of typing one character for the first participant, the time of typing one character for the the second participant, the ping of the first participant and the ping of the second participant. | If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship". | [
"5 1 2 1 2\n",
"3 3 1 1 1\n",
"4 5 3 1 5\n"
] | [
"First\n",
"Second\n",
"Friendship\n"
] | In the first example, information on the success of the first participant comes in 7 milliseconds, of the second participant — in 14 milliseconds. So, the first wins.
In the second example, information on the success of the first participant comes in 11 milliseconds, of the second participant — in 5 milliseconds. So, the second wins.
In the third example, information on the success of the first participant comes in 22 milliseconds, of the second participant — in 22 milliseconds. So, it is be a draw. | 500 | [
{
"input": "5 1 2 1 2",
"output": "First"
},
{
"input": "3 3 1 1 1",
"output": "Second"
},
{
"input": "4 5 3 1 5",
"output": "Friendship"
},
{
"input": "1000 1000 1000 1000 1000",
"output": "Friendship"
},
{
"input": "1 1 1 1 1",
"output": "Friendship"
},
{
"input": "8 8 1 1 1",
"output": "Second"
},
{
"input": "15 14 32 65 28",
"output": "First"
},
{
"input": "894 197 325 232 902",
"output": "First"
},
{
"input": "1 2 8 8 5",
"output": "Friendship"
},
{
"input": "37 261 207 1 1000",
"output": "Friendship"
},
{
"input": "29 344 406 900 1",
"output": "Friendship"
},
{
"input": "1 2 8 9 8",
"output": "First"
},
{
"input": "2 9 8 8 9",
"output": "Friendship"
},
{
"input": "213 480 811 134 745",
"output": "First"
},
{
"input": "2 313 856 964 421",
"output": "Friendship"
},
{
"input": "1 10 2 6 10",
"output": "Friendship"
},
{
"input": "2 7 6 2 3",
"output": "Friendship"
},
{
"input": "637 324 69 612 998",
"output": "Second"
},
{
"input": "13 849 819 723 918",
"output": "Friendship"
},
{
"input": "9 5 7 8 7",
"output": "First"
},
{
"input": "6 5 7 10 4",
"output": "Friendship"
},
{
"input": "61 464 623 89 548",
"output": "First"
},
{
"input": "641 31 29 161 802",
"output": "Friendship"
},
{
"input": "3 3 1 6 9",
"output": "Friendship"
},
{
"input": "2 3 9 8 2",
"output": "Friendship"
},
{
"input": "485 117 368 567 609",
"output": "First"
},
{
"input": "4 202 512 995 375",
"output": "Friendship"
},
{
"input": "424 41 41 909 909",
"output": "Friendship"
},
{
"input": "884 913 263 641 265",
"output": "Second"
},
{
"input": "12 462 8 311 327",
"output": "Second"
},
{
"input": "436 306 266 493 580",
"output": "Second"
},
{
"input": "69 1 2 1 2",
"output": "First"
}
] | 1,649,502,472 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 46 | 0 | s, v1, v2, t1, t2 = [int(i) for i in input().split()]
a = s*v1 + t1*2
b = s*v2 + t2*2
if a > b:
print("Second")
elif a < b:
print("First")
else:
print("Friendship") | Title: Key races
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two boys decided to compete in text typing on the site "Key races". During the competition, they have to type a text consisting of *s* characters. The first participant types one character in *v*1 milliseconds and has ping *t*1 milliseconds. The second participant types one character in *v*2 milliseconds and has ping *t*2 milliseconds.
If connection ping (delay) is *t* milliseconds, the competition passes for a participant as follows:
1. Exactly after *t* milliseconds after the start of the competition the participant receives the text to be entered. 1. Right after that he starts to type it. 1. Exactly *t* milliseconds after he ends typing all the text, the site receives information about it.
The winner is the participant whose information on the success comes earlier. If the information comes from both participants at the same time, it is considered that there is a draw.
Given the length of the text and the information about participants, determine the result of the game.
Input Specification:
The first line contains five integers *s*, *v*1, *v*2, *t*1, *t*2 (1<=≤<=*s*,<=*v*1,<=*v*2,<=*t*1,<=*t*2<=≤<=1000) — the number of characters in the text, the time of typing one character for the first participant, the time of typing one character for the the second participant, the ping of the first participant and the ping of the second participant.
Output Specification:
If the first participant wins, print "First". If the second participant wins, print "Second". In case of a draw print "Friendship".
Demo Input:
['5 1 2 1 2\n', '3 3 1 1 1\n', '4 5 3 1 5\n']
Demo Output:
['First\n', 'Second\n', 'Friendship\n']
Note:
In the first example, information on the success of the first participant comes in 7 milliseconds, of the second participant — in 14 milliseconds. So, the first wins.
In the second example, information on the success of the first participant comes in 11 milliseconds, of the second participant — in 5 milliseconds. So, the second wins.
In the third example, information on the success of the first participant comes in 22 milliseconds, of the second participant — in 22 milliseconds. So, it is be a draw. | ```python
s, v1, v2, t1, t2 = [int(i) for i in input().split()]
a = s*v1 + t1*2
b = s*v2 + t2*2
if a > b:
print("Second")
elif a < b:
print("First")
else:
print("Friendship")
``` | 3 | |
285 | B | Find Marble | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | Petya and Vasya are playing a game. Petya's got *n* non-transparent glasses, standing in a row. The glasses' positions are indexed with integers from 1 to *n* from left to right. Note that the positions are indexed but the glasses are not.
First Petya puts a marble under the glass in position *s*. Then he performs some (possibly zero) shuffling operations. One shuffling operation means moving the glass from the first position to position *p*1, the glass from the second position to position *p*2 and so on. That is, a glass goes from position *i* to position *p**i*. Consider all glasses are moving simultaneously during one shuffling operation. When the glasses are shuffled, the marble doesn't travel from one glass to another: it moves together with the glass it was initially been put in.
After all shuffling operations Petya shows Vasya that the ball has moved to position *t*. Vasya's task is to say what minimum number of shuffling operations Petya has performed or determine that Petya has made a mistake and the marble could not have got from position *s* to position *t*. | The first line contains three integers: *n*,<=*s*,<=*t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*s*,<=*t*<=≤<=*n*) — the number of glasses, the ball's initial and final position. The second line contains *n* space-separated integers: *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — the shuffling operation parameters. It is guaranteed that all *p**i*'s are distinct.
Note that *s* can equal *t*. | If the marble can move from position *s* to position *t*, then print on a single line a non-negative integer — the minimum number of shuffling operations, needed to get the marble to position *t*. If it is impossible, print number -1. | [
"4 2 1\n2 3 4 1\n",
"4 3 3\n4 1 3 2\n",
"4 3 4\n1 2 3 4\n",
"3 1 3\n2 1 3\n"
] | [
"3\n",
"0\n",
"-1\n",
"-1\n"
] | none | 1,000 | [
{
"input": "4 2 1\n2 3 4 1",
"output": "3"
},
{
"input": "4 3 3\n4 1 3 2",
"output": "0"
},
{
"input": "4 3 4\n1 2 3 4",
"output": "-1"
},
{
"input": "3 1 3\n2 1 3",
"output": "-1"
},
{
"input": "1 1 1\n1",
"output": "0"
},
{
"input": "10 6 7\n10 7 8 1 5 6 2 9 4 3",
"output": "-1"
},
{
"input": "10 3 6\n5 6 7 3 8 4 2 1 10 9",
"output": "3"
},
{
"input": "10 10 4\n4 2 6 9 5 3 8 1 10 7",
"output": "4"
},
{
"input": "100 90 57\n19 55 91 50 31 23 60 84 38 1 22 51 27 76 28 98 11 44 61 63 15 93 52 3 66 16 53 36 18 62 35 85 78 37 73 64 87 74 46 26 82 69 49 33 83 89 56 67 71 25 39 94 96 17 21 6 47 68 34 42 57 81 13 10 54 2 48 80 20 77 4 5 59 30 90 95 45 75 8 88 24 41 40 14 97 32 7 9 65 70 100 99 72 58 92 29 79 12 86 43",
"output": "-1"
},
{
"input": "100 11 20\n80 25 49 55 22 98 35 59 88 14 91 20 68 66 53 50 77 45 82 63 96 93 85 46 37 74 84 9 7 95 41 86 23 36 33 27 81 39 18 13 12 92 24 71 3 48 83 61 31 87 28 79 75 38 11 21 29 69 44 100 72 62 32 43 30 16 47 56 89 60 42 17 26 70 94 99 4 6 2 73 8 52 65 1 15 90 67 51 78 10 5 76 57 54 34 58 19 64 40 97",
"output": "26"
},
{
"input": "100 84 83\n30 67 53 89 94 54 92 17 26 57 15 5 74 85 10 61 18 70 91 75 14 11 93 41 25 78 88 81 20 51 35 4 62 1 97 39 68 52 47 77 64 3 2 72 60 80 8 83 65 98 21 22 45 7 58 31 43 38 90 99 49 87 55 36 29 6 37 23 66 76 59 79 40 86 63 44 82 32 48 16 50 100 28 96 46 12 27 13 24 9 19 84 73 69 71 42 56 33 34 95",
"output": "71"
},
{
"input": "100 6 93\n74 62 67 81 40 85 35 42 59 72 80 28 79 41 16 19 33 63 13 10 69 76 70 93 49 84 89 94 8 37 11 90 26 52 47 7 36 95 86 75 56 15 61 99 88 12 83 21 20 3 100 17 32 82 6 5 43 25 66 68 73 78 18 77 92 27 23 2 4 39 60 48 22 24 14 97 29 34 54 64 71 57 87 38 9 50 30 53 51 45 44 31 58 91 98 65 55 1 46 96",
"output": "-1"
},
{
"input": "100 27 56\n58 18 50 41 33 37 14 87 77 73 61 53 15 8 70 68 45 96 54 78 39 67 51 60 80 12 93 99 20 92 17 79 4 13 62 91 69 29 49 36 98 34 90 35 84 64 38 83 28 89 97 94 9 16 26 48 10 57 23 75 27 88 44 21 72 76 30 43 32 2 71 24 100 1 31 81 42 40 47 55 86 85 66 5 52 22 95 74 11 19 7 82 6 25 56 63 65 59 46 3",
"output": "20"
},
{
"input": "87 42 49\n45 55 24 44 56 72 74 23 4 7 37 67 22 6 58 76 40 36 3 20 26 87 64 75 49 70 62 42 31 1 80 33 25 59 78 27 32 2 41 61 66 28 19 85 15 69 52 77 50 14 16 34 18 43 73 83 11 39 29 9 35 13 81 54 79 21 60 46 71 57 12 17 5 47 38 30 10 84 53 63 68 8 51 65 48 86 82",
"output": "-1"
},
{
"input": "2 1 2\n1 2",
"output": "-1"
},
{
"input": "2 1 2\n2 1",
"output": "1"
},
{
"input": "2 2 2\n1 2",
"output": "0"
},
{
"input": "2 2 2\n2 1",
"output": "0"
},
{
"input": "2 1 1\n2 1",
"output": "0"
}
] | 1,562,849,698 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 342 | 7,372,800 | n, s, t=map(int,input().split())
l=list(map(int,input().split()))
x=s-1
count=0
done=0
while x!=(t-1):
if count!=n:
x=l[x]-1
count+=1
else:
done=1
break
if done==1:
print(-1)
else:
print(count) | Title: Find Marble
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya and Vasya are playing a game. Petya's got *n* non-transparent glasses, standing in a row. The glasses' positions are indexed with integers from 1 to *n* from left to right. Note that the positions are indexed but the glasses are not.
First Petya puts a marble under the glass in position *s*. Then he performs some (possibly zero) shuffling operations. One shuffling operation means moving the glass from the first position to position *p*1, the glass from the second position to position *p*2 and so on. That is, a glass goes from position *i* to position *p**i*. Consider all glasses are moving simultaneously during one shuffling operation. When the glasses are shuffled, the marble doesn't travel from one glass to another: it moves together with the glass it was initially been put in.
After all shuffling operations Petya shows Vasya that the ball has moved to position *t*. Vasya's task is to say what minimum number of shuffling operations Petya has performed or determine that Petya has made a mistake and the marble could not have got from position *s* to position *t*.
Input Specification:
The first line contains three integers: *n*,<=*s*,<=*t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*s*,<=*t*<=≤<=*n*) — the number of glasses, the ball's initial and final position. The second line contains *n* space-separated integers: *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — the shuffling operation parameters. It is guaranteed that all *p**i*'s are distinct.
Note that *s* can equal *t*.
Output Specification:
If the marble can move from position *s* to position *t*, then print on a single line a non-negative integer — the minimum number of shuffling operations, needed to get the marble to position *t*. If it is impossible, print number -1.
Demo Input:
['4 2 1\n2 3 4 1\n', '4 3 3\n4 1 3 2\n', '4 3 4\n1 2 3 4\n', '3 1 3\n2 1 3\n']
Demo Output:
['3\n', '0\n', '-1\n', '-1\n']
Note:
none | ```python
n, s, t=map(int,input().split())
l=list(map(int,input().split()))
x=s-1
count=0
done=0
while x!=(t-1):
if count!=n:
x=l[x]-1
count+=1
else:
done=1
break
if done==1:
print(-1)
else:
print(count)
``` | 3 | |
190 | B | Surrounded | PROGRAMMING | 1,800 | [
"geometry"
] | null | null | So, the Berland is at war with its eternal enemy Flatland again, and Vasya, an accountant, was assigned to fulfil his duty to the nation.
Right now the situation in Berland is dismal — their both cities are surrounded! The armies of flatlanders stand on the borders of circles, the circles' centers are in the surrounded cities. At any moment all points of the flatland ring can begin to move quickly in the direction of the city — that's the strategy the flatlanders usually follow when they besiege cities.
The berlanders are sure that they can repel the enemy's attack if they learn the exact time the attack starts. For that they need to construct a radar that would register any movement at the distance of at most *r* from it. Thus, we can install a radar at such point, that at least one point of the enemy ring will be in its detecting range (that is, at a distance of at most *r*). Then the radar can immediately inform about the enemy's attack.
Due to the newest technologies, we can place a radar at any point without any problems. But the problem is that the berlanders have the time to make only one radar. Besides, the larger the detection radius (*r*) is, the more the radar costs.
That's why Vasya's task (that is, your task) is to find the minimum possible detection radius for the radar. In other words, your task is to find the minimum radius *r* (*r*<=≥<=0) such, that a radar with radius *r* can be installed at some point and it can register the start of the movements of both flatland rings from that point.
In this problem you can consider the cities as material points, the attacking enemy rings - as circles with centers in the cities, the radar's detection range — as a disk (including the border) with the center at the point where the radar is placed. | The input files consist of two lines. Each line represents the city and the flatland ring that surrounds it as three space-separated integers *x**i*, *y**i*, *r**i* (|*x**i*|,<=|*y**i*|<=≤<=104; 1<=≤<=*r**i*<=≤<=104) — the city's coordinates and the distance from the city to the flatlanders, correspondingly.
It is guaranteed that the cities are located at different points. | Print a single real number — the minimum detection radius of the described radar. The answer is considered correct if the absolute or relative error does not exceed 10<=-<=6. | [
"0 0 1\n6 0 3\n",
"-10 10 3\n10 -10 3\n"
] | [
"1.000000000000000",
"11.142135623730951"
] | The figure below shows the answer to the first sample. In this sample the best decision is to put the radar at point with coordinates (2, 0).
The figure below shows the answer for the second sample. In this sample the best decision is to put the radar at point with coordinates (0, 0). | 1,000 | [
{
"input": "0 0 1\n6 0 3",
"output": "1.000000000000000"
},
{
"input": "-10 10 3\n10 -10 3",
"output": "11.142135623730951"
},
{
"input": "2 1 3\n8 9 5",
"output": "1.000000000000000"
},
{
"input": "0 0 1\n-10 -10 9",
"output": "2.071067811865475"
},
{
"input": "10000 -9268 1\n-9898 9000 10",
"output": "13500.519287710202000"
},
{
"input": "10000 10000 1\n-10000 -10000 1",
"output": "14141.135623730950000"
},
{
"input": "123 21 50\n10 100 1000",
"output": "406.061621719103360"
},
{
"input": "0 3278 2382\n2312 1 1111",
"output": "258.747677968983450"
},
{
"input": "3 4 5\n5 12 13",
"output": "0.000000000000000"
},
{
"input": "-2 7 5\n4 0 6",
"output": "0.000000000000000"
},
{
"input": "4 0 2\n6 -1 10",
"output": "2.881966011250105"
},
{
"input": "41 17 3\n71 -86 10",
"output": "47.140003728560643"
},
{
"input": "761 641 6\n506 -293 5",
"output": "478.592191632957450"
},
{
"input": "-5051 -7339 9\n-9030 755 8",
"output": "4501.080828635849700"
},
{
"input": "0 5 2\n8 -4 94",
"output": "39.979202710603850"
},
{
"input": "83 -64 85\n27 80 89",
"output": "0.000000000000000"
},
{
"input": "-655 -750 68\n905 -161 68",
"output": "765.744715125679250"
},
{
"input": "1055 -5271 60\n-2992 8832 38",
"output": "7287.089182936641900"
},
{
"input": "4 0 201\n-6 4 279",
"output": "33.614835192865499"
},
{
"input": "-34 -5 836\n52 -39 706",
"output": "18.761487913212431"
},
{
"input": "659 -674 277\n-345 -556 127",
"output": "303.455240352694320"
},
{
"input": "4763 2945 956\n3591 9812 180",
"output": "2915.147750239716500"
},
{
"input": "3 -7 5749\n1 -6 9750",
"output": "1999.381966011250100"
},
{
"input": "28 -63 2382\n43 -83 1364",
"output": "496.500000000000000"
},
{
"input": "315 -532 7813\n407 -157 2121",
"output": "2652.939776235497000"
},
{
"input": "-9577 9051 5276\n-4315 -1295 8453",
"output": "0.000000000000000"
},
{
"input": "-7 -10 1\n-4 3 1",
"output": "5.670832032063167"
},
{
"input": "-74 27 535\n18 84 1",
"output": "212.886692948961240"
},
{
"input": "-454 -721 72\n-33 279 911",
"output": "51.003686623418254"
},
{
"input": "-171 762 304\n-428 -85 523",
"output": "29.065814314662131"
},
{
"input": "192 -295 1386\n-54 -78 1",
"output": "528.483994683445640"
},
{
"input": "-5134 -9860 5513\n6291 -855 9034",
"output": "0.093506651303098"
},
{
"input": "6651 8200 610\n-9228 9387 10000",
"output": "2656.651995660197400"
},
{
"input": "6370 7728 933\n4595 3736 2748",
"output": "343.915768575204200"
},
{
"input": "-6 3 8\n7 2 1",
"output": "2.019202405202649"
},
{
"input": "0 -1 1\n1 -1 1",
"output": "0.000000000000000"
},
{
"input": "0 1 3\n1 -1 1",
"output": "0.000000000000000"
},
{
"input": "-2 0 1\n3 -2 1",
"output": "1.692582403567252"
},
{
"input": "-10000 42 10000\n10000 43 10000",
"output": "0.000012499999992"
},
{
"input": "103 104 5\n97 96 5",
"output": "0.000000000000000"
},
{
"input": "2587 4850 3327\n3278 -204 1774",
"output": "0.009605941526345"
},
{
"input": "826 4417 2901\n833 -2286 3802",
"output": "0.001827539409235"
},
{
"input": "1003 -5005 3399\n-6036 -1729 4365",
"output": "0.000032199896827"
}
] | 1,529,849,946 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 77 | 0 | x1, x2, r1 = [int(ch) for ch in input().split()]
y1, y2, r2 = [int(ch) for ch in input().split()]
printf((((x1-x2)**2 + (y1-y2)**2)**0.5 - r1 - r2) * 0.5) | Title: Surrounded
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
So, the Berland is at war with its eternal enemy Flatland again, and Vasya, an accountant, was assigned to fulfil his duty to the nation.
Right now the situation in Berland is dismal — their both cities are surrounded! The armies of flatlanders stand on the borders of circles, the circles' centers are in the surrounded cities. At any moment all points of the flatland ring can begin to move quickly in the direction of the city — that's the strategy the flatlanders usually follow when they besiege cities.
The berlanders are sure that they can repel the enemy's attack if they learn the exact time the attack starts. For that they need to construct a radar that would register any movement at the distance of at most *r* from it. Thus, we can install a radar at such point, that at least one point of the enemy ring will be in its detecting range (that is, at a distance of at most *r*). Then the radar can immediately inform about the enemy's attack.
Due to the newest technologies, we can place a radar at any point without any problems. But the problem is that the berlanders have the time to make only one radar. Besides, the larger the detection radius (*r*) is, the more the radar costs.
That's why Vasya's task (that is, your task) is to find the minimum possible detection radius for the radar. In other words, your task is to find the minimum radius *r* (*r*<=≥<=0) such, that a radar with radius *r* can be installed at some point and it can register the start of the movements of both flatland rings from that point.
In this problem you can consider the cities as material points, the attacking enemy rings - as circles with centers in the cities, the radar's detection range — as a disk (including the border) with the center at the point where the radar is placed.
Input Specification:
The input files consist of two lines. Each line represents the city and the flatland ring that surrounds it as three space-separated integers *x**i*, *y**i*, *r**i* (|*x**i*|,<=|*y**i*|<=≤<=104; 1<=≤<=*r**i*<=≤<=104) — the city's coordinates and the distance from the city to the flatlanders, correspondingly.
It is guaranteed that the cities are located at different points.
Output Specification:
Print a single real number — the minimum detection radius of the described radar. The answer is considered correct if the absolute or relative error does not exceed 10<=-<=6.
Demo Input:
['0 0 1\n6 0 3\n', '-10 10 3\n10 -10 3\n']
Demo Output:
['1.000000000000000', '11.142135623730951']
Note:
The figure below shows the answer to the first sample. In this sample the best decision is to put the radar at point with coordinates (2, 0).
The figure below shows the answer for the second sample. In this sample the best decision is to put the radar at point with coordinates (0, 0). | ```python
x1, x2, r1 = [int(ch) for ch in input().split()]
y1, y2, r2 = [int(ch) for ch in input().split()]
printf((((x1-x2)**2 + (y1-y2)**2)**0.5 - r1 - r2) * 0.5)
``` | -1 | |
523 | D | Statistics of Recompressing Videos | PROGRAMMING | 1,600 | [
"*special",
"data structures",
"implementation"
] | null | null | A social network for dogs called DH (DogHouse) has *k* special servers to recompress uploaded videos of cute cats. After each video is uploaded, it should be recompressed on one (any) of the servers, and only after that it can be saved in the social network.
We know that each server takes one second to recompress a one minute fragment. Thus, any server takes *m* seconds to recompress a *m* minute video.
We know the time when each of the *n* videos were uploaded to the network (in seconds starting from the moment all servers started working). All videos appear at different moments of time and they are recompressed in the order they appear. If some video appeared at time *s*, then its recompressing can start at that very moment, immediately. Some videos can await recompressing when all the servers are busy. In this case, as soon as a server is available, it immediately starts recompressing another video. The videos that await recompressing go in a queue. If by the moment the videos started being recompressed some servers are available, then any of them starts recompressing the video.
For each video find the moment it stops being recompressed. | The first line of the input contains integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=5·105) — the number of videos and servers, respectively.
Next *n* lines contain the descriptions of the videos as pairs of integers *s**i*,<=*m**i* (1<=≤<=*s**i*,<=*m**i*<=≤<=109), where *s**i* is the time in seconds when the *i*-th video appeared and *m**i* is its duration in minutes. It is guaranteed that all the *s**i*'s are distinct and the videos are given in the chronological order of upload, that is in the order of increasing *s**i*. | Print *n* numbers *e*1,<=*e*2,<=...,<=*e**n*, where *e**i* is the time in seconds after the servers start working, when the *i*-th video will be recompressed. | [
"3 2\n1 5\n2 5\n3 5\n",
"6 1\n1 1000000000\n2 1000000000\n3 1000000000\n4 1000000000\n5 1000000000\n6 3\n"
] | [
"6\n7\n11\n",
"1000000001\n2000000001\n3000000001\n4000000001\n5000000001\n5000000004\n"
] | none | 2,000 | [
{
"input": "3 2\n1 5\n2 5\n3 5",
"output": "6\n7\n11"
},
{
"input": "6 1\n1 1000000000\n2 1000000000\n3 1000000000\n4 1000000000\n5 1000000000\n6 3",
"output": "1000000001\n2000000001\n3000000001\n4000000001\n5000000001\n5000000004"
},
{
"input": "1 1\n1 1",
"output": "2"
},
{
"input": "1 1\n1000000000 10000",
"output": "1000010000"
},
{
"input": "10 6\n1 1\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1",
"output": "2\n3\n4\n5\n6\n7\n8\n9\n10\n11"
},
{
"input": "10 4\n1 1\n2 2\n3 1\n4 1\n5 1\n6 1\n7 1\n8 2\n9 1\n10 1",
"output": "2\n4\n4\n5\n6\n7\n8\n10\n10\n11"
},
{
"input": "10 2\n1 5650\n2 4753\n3 7632\n4 688\n5 8853\n6 284\n7 4659\n8 5650\n9 9768\n10 3905",
"output": "5651\n4755\n12387\n6339\n15192\n12671\n17330\n20842\n27098\n24747"
},
{
"input": "10 8\n1 5036\n7 9294\n8 6011\n10 8273\n11 9203\n12 7037\n14 383\n16 4568\n18 8136\n19 8288",
"output": "5037\n9301\n6019\n8283\n9214\n7049\n397\n4584\n8533\n12872"
},
{
"input": "10 2\n4 2\n7 2\n8 2\n9 1\n10 2\n12 2\n14 1\n15 2\n17 2\n19 1",
"output": "6\n9\n10\n10\n12\n14\n15\n17\n19\n20"
},
{
"input": "10 7\n195901104 7859\n265432683 5489\n290824505 5754\n346976046 4969\n406206484 8390\n522669517 6810\n800443397 4979\n839536223 1825\n918231479 8117\n941210310 1322",
"output": "195908963\n265438172\n290830259\n346981015\n406214874\n522676327\n800448376\n839538048\n918239596\n941211632"
},
{
"input": "10 4\n126995987 385321200\n195616854 752754110\n197489309 899442094\n285439286 247570387\n308620877 957032819\n428385669 227675453\n673115425 94614781\n766412355 105231165\n943783548 855684033\n994356572 699602107",
"output": "512317187\n948370964\n1096931403\n533009673\n1469350006\n760685126\n855299907\n960531072\n1804054997\n1693958679"
},
{
"input": "10 10\n999999991 1000000000\n999999992 1000000000\n999999993 1000000000\n999999994 1000000000\n999999995 1000000000\n999999996 1000000000\n999999997 1000000000\n999999998 1000000000\n999999999 1000000000\n1000000000 1000000000",
"output": "1999999991\n1999999992\n1999999993\n1999999994\n1999999995\n1999999996\n1999999997\n1999999998\n1999999999\n2000000000"
}
] | 1,426,357,753 | 12,552 | Python 3 | TIME_LIMIT_EXCEEDED | PRETESTS | 12 | 3,000 | 1,843,200 | __author__ = 'Gleb'
a=input().split()
n,k=int(a[0]),int(a[1])
serv=[]
for i in range(k):
serv.append(0)
for i in range(n):
a=input().split()
s,m=int(a[0]),int(a[1])
min=10000000000
for j in range(k):
if min>serv[j]:
min=serv[j]
mini=j
if min<s:
serv[mini]=s+m
else:
serv[mini]=serv[mini]+m
print(serv[mini]) | Title: Statistics of Recompressing Videos
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A social network for dogs called DH (DogHouse) has *k* special servers to recompress uploaded videos of cute cats. After each video is uploaded, it should be recompressed on one (any) of the servers, and only after that it can be saved in the social network.
We know that each server takes one second to recompress a one minute fragment. Thus, any server takes *m* seconds to recompress a *m* minute video.
We know the time when each of the *n* videos were uploaded to the network (in seconds starting from the moment all servers started working). All videos appear at different moments of time and they are recompressed in the order they appear. If some video appeared at time *s*, then its recompressing can start at that very moment, immediately. Some videos can await recompressing when all the servers are busy. In this case, as soon as a server is available, it immediately starts recompressing another video. The videos that await recompressing go in a queue. If by the moment the videos started being recompressed some servers are available, then any of them starts recompressing the video.
For each video find the moment it stops being recompressed.
Input Specification:
The first line of the input contains integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=5·105) — the number of videos and servers, respectively.
Next *n* lines contain the descriptions of the videos as pairs of integers *s**i*,<=*m**i* (1<=≤<=*s**i*,<=*m**i*<=≤<=109), where *s**i* is the time in seconds when the *i*-th video appeared and *m**i* is its duration in minutes. It is guaranteed that all the *s**i*'s are distinct and the videos are given in the chronological order of upload, that is in the order of increasing *s**i*.
Output Specification:
Print *n* numbers *e*1,<=*e*2,<=...,<=*e**n*, where *e**i* is the time in seconds after the servers start working, when the *i*-th video will be recompressed.
Demo Input:
['3 2\n1 5\n2 5\n3 5\n', '6 1\n1 1000000000\n2 1000000000\n3 1000000000\n4 1000000000\n5 1000000000\n6 3\n']
Demo Output:
['6\n7\n11\n', '1000000001\n2000000001\n3000000001\n4000000001\n5000000001\n5000000004\n']
Note:
none | ```python
__author__ = 'Gleb'
a=input().split()
n,k=int(a[0]),int(a[1])
serv=[]
for i in range(k):
serv.append(0)
for i in range(n):
a=input().split()
s,m=int(a[0]),int(a[1])
min=10000000000
for j in range(k):
if min>serv[j]:
min=serv[j]
mini=j
if min<s:
serv[mini]=s+m
else:
serv[mini]=serv[mini]+m
print(serv[mini])
``` | 0 | |
864 | A | Fair Game | PROGRAMMING | 1,000 | [
"implementation",
"sortings"
] | null | null | Petya and Vasya decided to play a game. They have *n* cards (*n* is an even number). A single integer is written on each card.
Before the game Petya will choose an integer and after that Vasya will choose another integer (different from the number that Petya chose). During the game each player takes all the cards with number he chose. For example, if Petya chose number 5 before the game he will take all cards on which 5 is written and if Vasya chose number 10 before the game he will take all cards on which 10 is written.
The game is considered fair if Petya and Vasya can take all *n* cards, and the number of cards each player gets is the same.
Determine whether Petya and Vasya can choose integer numbers before the game so that the game is fair. | The first line contains a single integer *n* (2<=≤<=*n*<=≤<=100) — number of cards. It is guaranteed that *n* is an even number.
The following *n* lines contain a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (one integer per line, 1<=≤<=*a**i*<=≤<=100) — numbers written on the *n* cards. | If it is impossible for Petya and Vasya to choose numbers in such a way that the game will be fair, print "NO" (without quotes) in the first line. In this case you should not print anything more.
In the other case print "YES" (without quotes) in the first line. In the second line print two distinct integers — number that Petya should choose and the number that Vasya should choose to make the game fair. If there are several solutions, print any of them. | [
"4\n11\n27\n27\n11\n",
"2\n6\n6\n",
"6\n10\n20\n30\n20\n10\n20\n",
"6\n1\n1\n2\n2\n3\n3\n"
] | [
"YES\n11 27\n",
"NO\n",
"NO\n",
"NO\n"
] | In the first example the game will be fair if, for example, Petya chooses number 11, and Vasya chooses number 27. Then the will take all cards — Petya will take cards 1 and 4, and Vasya will take cards 2 and 3. Thus, each of them will take exactly two cards.
In the second example fair game is impossible because the numbers written on the cards are equal, but the numbers that Petya and Vasya should choose should be distinct.
In the third example it is impossible to take all cards. Petya and Vasya can take at most five cards — for example, Petya can choose number 10 and Vasya can choose number 20. But for the game to be fair it is necessary to take 6 cards. | 500 | [
{
"input": "4\n11\n27\n27\n11",
"output": "YES\n11 27"
},
{
"input": "2\n6\n6",
"output": "NO"
},
{
"input": "6\n10\n20\n30\n20\n10\n20",
"output": "NO"
},
{
"input": "6\n1\n1\n2\n2\n3\n3",
"output": "NO"
},
{
"input": "2\n1\n100",
"output": "YES\n1 100"
},
{
"input": "2\n1\n1",
"output": "NO"
},
{
"input": "2\n100\n100",
"output": "NO"
},
{
"input": "14\n43\n43\n43\n43\n43\n43\n43\n43\n43\n43\n43\n43\n43\n43",
"output": "NO"
},
{
"input": "100\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32",
"output": "YES\n14 32"
},
{
"input": "2\n50\n100",
"output": "YES\n50 100"
},
{
"input": "2\n99\n100",
"output": "YES\n99 100"
},
{
"input": "4\n4\n4\n5\n5",
"output": "YES\n4 5"
},
{
"input": "10\n10\n10\n10\n10\n10\n23\n23\n23\n23\n23",
"output": "YES\n10 23"
},
{
"input": "20\n34\n34\n34\n34\n34\n34\n34\n34\n34\n34\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11",
"output": "YES\n11 34"
},
{
"input": "40\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30",
"output": "YES\n20 30"
},
{
"input": "58\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1",
"output": "YES\n1 100"
},
{
"input": "98\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99",
"output": "YES\n2 99"
},
{
"input": "100\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100",
"output": "YES\n1 100"
},
{
"input": "100\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2",
"output": "YES\n1 2"
},
{
"input": "100\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12",
"output": "YES\n12 49"
},
{
"input": "100\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94",
"output": "YES\n15 94"
},
{
"input": "100\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42",
"output": "YES\n33 42"
},
{
"input": "100\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35",
"output": "YES\n16 35"
},
{
"input": "100\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44",
"output": "YES\n33 44"
},
{
"input": "100\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98",
"output": "YES\n54 98"
},
{
"input": "100\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12",
"output": "YES\n12 81"
},
{
"input": "100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100",
"output": "NO"
},
{
"input": "100\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1",
"output": "NO"
},
{
"input": "40\n20\n20\n30\n30\n20\n20\n20\n30\n30\n20\n20\n30\n30\n30\n30\n20\n30\n30\n30\n30\n20\n20\n30\n30\n30\n20\n30\n20\n30\n20\n30\n20\n20\n20\n30\n20\n20\n20\n30\n30",
"output": "NO"
},
{
"input": "58\n100\n100\n100\n100\n100\n1\n1\n1\n1\n1\n1\n100\n100\n1\n100\n1\n100\n100\n1\n1\n100\n100\n1\n100\n1\n100\n100\n1\n1\n100\n1\n1\n1\n100\n1\n1\n1\n1\n100\n1\n100\n100\n100\n100\n100\n1\n1\n100\n100\n100\n100\n1\n100\n1\n1\n1\n1\n1",
"output": "NO"
},
{
"input": "98\n2\n99\n99\n99\n99\n2\n99\n99\n99\n2\n2\n99\n2\n2\n2\n2\n99\n99\n2\n99\n2\n2\n99\n99\n99\n99\n2\n2\n99\n2\n99\n99\n2\n2\n99\n2\n99\n2\n99\n2\n2\n2\n99\n2\n2\n2\n2\n99\n99\n99\n99\n2\n2\n2\n2\n2\n2\n2\n2\n99\n2\n99\n99\n2\n2\n99\n99\n99\n99\n99\n99\n99\n99\n2\n99\n2\n99\n2\n2\n2\n99\n99\n99\n99\n99\n99\n2\n99\n99\n2\n2\n2\n2\n2\n99\n99\n99\n2",
"output": "NO"
},
{
"input": "100\n100\n1\n100\n1\n1\n100\n1\n1\n1\n100\n100\n1\n100\n1\n100\n100\n1\n1\n1\n100\n1\n100\n1\n100\n100\n1\n100\n1\n100\n1\n1\n1\n1\n1\n100\n1\n100\n100\n100\n1\n100\n100\n1\n100\n1\n1\n100\n100\n100\n1\n100\n100\n1\n1\n100\n100\n1\n100\n1\n100\n1\n1\n100\n100\n100\n100\n100\n100\n1\n100\n100\n1\n100\n100\n1\n100\n1\n1\n1\n100\n100\n1\n100\n1\n100\n1\n1\n1\n1\n100\n1\n1\n100\n1\n100\n100\n1\n100\n1\n100",
"output": "NO"
},
{
"input": "100\n100\n100\n100\n1\n100\n1\n1\n1\n100\n1\n1\n1\n1\n100\n1\n100\n1\n100\n1\n100\n100\n100\n1\n100\n1\n1\n1\n100\n1\n1\n1\n1\n1\n100\n100\n1\n100\n1\n1\n100\n1\n1\n100\n1\n100\n100\n100\n1\n100\n100\n100\n1\n100\n1\n100\n100\n100\n1\n1\n100\n100\n100\n100\n1\n100\n36\n100\n1\n100\n1\n100\n100\n100\n1\n1\n1\n1\n1\n1\n1\n1\n1\n100\n1\n1\n100\n100\n100\n100\n100\n1\n100\n1\n100\n1\n1\n100\n100\n1\n100",
"output": "NO"
},
{
"input": "100\n2\n1\n1\n2\n2\n1\n1\n1\n1\n2\n1\n1\n1\n2\n2\n2\n1\n1\n1\n2\n1\n2\n2\n2\n2\n1\n1\n2\n1\n1\n2\n1\n27\n1\n1\n1\n2\n2\n2\n1\n2\n1\n2\n1\n1\n2\n2\n2\n2\n2\n2\n2\n2\n1\n2\n2\n2\n2\n1\n2\n1\n1\n1\n1\n1\n2\n1\n1\n1\n2\n2\n2\n2\n2\n2\n1\n1\n1\n1\n2\n2\n1\n2\n2\n1\n1\n1\n2\n1\n2\n2\n1\n1\n2\n1\n1\n1\n2\n2\n1",
"output": "NO"
},
{
"input": "100\n99\n99\n100\n99\n99\n100\n100\n100\n99\n100\n99\n99\n100\n99\n99\n99\n99\n99\n99\n100\n100\n100\n99\n100\n100\n99\n100\n99\n100\n100\n99\n100\n99\n99\n99\n100\n99\n10\n99\n100\n100\n100\n99\n100\n100\n100\n100\n100\n100\n100\n99\n100\n100\n100\n99\n99\n100\n99\n100\n99\n100\n100\n99\n99\n99\n99\n100\n99\n100\n100\n100\n100\n100\n100\n99\n99\n100\n100\n99\n99\n99\n99\n99\n99\n100\n99\n99\n100\n100\n99\n100\n99\n99\n100\n99\n99\n99\n99\n100\n100",
"output": "NO"
},
{
"input": "100\n29\n43\n43\n29\n43\n29\n29\n29\n43\n29\n29\n29\n29\n43\n29\n29\n29\n29\n43\n29\n29\n29\n43\n29\n29\n29\n43\n43\n43\n43\n43\n43\n29\n29\n43\n43\n43\n29\n43\n43\n43\n29\n29\n29\n43\n29\n29\n29\n43\n43\n43\n43\n29\n29\n29\n29\n43\n29\n43\n43\n29\n29\n43\n43\n29\n29\n95\n29\n29\n29\n43\n43\n29\n29\n29\n29\n29\n43\n43\n43\n43\n29\n29\n43\n43\n43\n43\n43\n43\n29\n43\n43\n43\n43\n43\n43\n29\n43\n29\n43",
"output": "NO"
},
{
"input": "100\n98\n98\n98\n88\n88\n88\n88\n98\n98\n88\n98\n88\n98\n88\n88\n88\n88\n88\n98\n98\n88\n98\n98\n98\n88\n88\n88\n98\n98\n88\n88\n88\n98\n88\n98\n88\n98\n88\n88\n98\n98\n98\n88\n88\n98\n98\n88\n88\n88\n88\n88\n98\n98\n98\n88\n98\n88\n88\n98\n98\n88\n98\n88\n88\n98\n88\n88\n98\n27\n88\n88\n88\n98\n98\n88\n88\n98\n98\n98\n98\n98\n88\n98\n88\n98\n98\n98\n98\n88\n88\n98\n88\n98\n88\n98\n98\n88\n98\n98\n88",
"output": "NO"
},
{
"input": "100\n50\n1\n1\n50\n50\n50\n50\n1\n50\n100\n50\n50\n50\n100\n1\n100\n1\n100\n50\n50\n50\n50\n50\n1\n50\n1\n100\n1\n1\n50\n100\n50\n50\n100\n50\n50\n100\n1\n50\n50\n100\n1\n1\n50\n1\n100\n50\n50\n100\n100\n1\n100\n1\n50\n100\n50\n50\n1\n1\n50\n100\n50\n100\n100\n100\n50\n50\n1\n1\n50\n100\n1\n50\n100\n100\n1\n50\n50\n50\n100\n50\n50\n100\n1\n50\n50\n50\n50\n1\n50\n50\n50\n50\n1\n50\n50\n100\n1\n50\n100",
"output": "NO"
},
{
"input": "100\n45\n45\n45\n45\n45\n45\n44\n44\n44\n43\n45\n44\n44\n45\n44\n44\n45\n44\n43\n44\n43\n43\n43\n45\n43\n45\n44\n45\n43\n44\n45\n45\n45\n45\n45\n45\n45\n45\n43\n45\n43\n43\n45\n44\n45\n45\n45\n44\n45\n45\n45\n45\n45\n45\n44\n43\n45\n45\n43\n44\n45\n45\n45\n45\n44\n45\n45\n45\n43\n43\n44\n44\n43\n45\n43\n45\n45\n45\n44\n44\n43\n43\n44\n44\n44\n43\n45\n43\n44\n43\n45\n43\n43\n45\n45\n44\n45\n43\n43\n45",
"output": "NO"
},
{
"input": "100\n12\n12\n97\n15\n97\n12\n15\n97\n12\n97\n12\n12\n97\n12\n15\n12\n12\n15\n12\n12\n97\n12\n12\n15\n15\n12\n97\n15\n12\n97\n15\n12\n12\n15\n15\n15\n97\n15\n97\n12\n12\n12\n12\n12\n97\n12\n97\n12\n15\n15\n12\n15\n12\n15\n12\n12\n12\n12\n12\n12\n12\n12\n97\n97\n12\n12\n97\n12\n97\n97\n15\n97\n12\n97\n97\n12\n12\n12\n97\n97\n15\n12\n12\n15\n12\n15\n97\n97\n12\n15\n12\n12\n97\n12\n15\n15\n15\n15\n12\n12",
"output": "NO"
},
{
"input": "12\n2\n3\n1\n3\n3\n1\n2\n1\n2\n1\n3\n2",
"output": "NO"
},
{
"input": "48\n99\n98\n100\n100\n99\n100\n99\n100\n100\n98\n99\n98\n98\n99\n98\n99\n98\n100\n100\n98\n100\n98\n99\n100\n98\n99\n98\n99\n99\n100\n98\n99\n99\n98\n100\n99\n98\n99\n98\n100\n100\n100\n99\n98\n99\n98\n100\n100",
"output": "NO"
},
{
"input": "4\n1\n3\n3\n3",
"output": "NO"
},
{
"input": "6\n1\n1\n1\n1\n2\n2",
"output": "NO"
},
{
"input": "4\n1\n1\n1\n2",
"output": "NO"
},
{
"input": "4\n1\n2\n2\n2",
"output": "NO"
},
{
"input": "4\n1\n2\n3\n4",
"output": "NO"
},
{
"input": "8\n1\n1\n2\n2\n3\n3\n4\n4",
"output": "NO"
},
{
"input": "4\n1\n3\n2\n4",
"output": "NO"
},
{
"input": "4\n10\n10\n10\n20",
"output": "NO"
},
{
"input": "4\n11\n12\n13\n13",
"output": "NO"
},
{
"input": "4\n1\n1\n1\n3",
"output": "NO"
},
{
"input": "6\n1\n1\n2\n2\n2\n2",
"output": "NO"
},
{
"input": "10\n1\n1\n2\n2\n2\n3\n3\n4\n4\n4",
"output": "NO"
}
] | 1,602,259,251 | 951 | PyPy 3 | OK | TESTS | 54 | 358 | 2,355,200 | import math as mt
import sys,string
input=sys.stdin.readline
L=lambda : list(map(int,input().split()))
Ls=lambda : list(input().split())
M=lambda : map(int,input().split())
I=lambda :int(input())
from collections import defaultdict
n=I()
d=defaultdict(int)
for i in range(n):
x=I()
d[x]+=1
if(len(d.keys())==2):
s=set([])
for i in d.keys():
s.add(d[i])
if(len(s)==1):
print("YES")
print(*list(d.keys()))
else:
print("NO")
else:
print("NO")
| Title: Fair Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya and Vasya decided to play a game. They have *n* cards (*n* is an even number). A single integer is written on each card.
Before the game Petya will choose an integer and after that Vasya will choose another integer (different from the number that Petya chose). During the game each player takes all the cards with number he chose. For example, if Petya chose number 5 before the game he will take all cards on which 5 is written and if Vasya chose number 10 before the game he will take all cards on which 10 is written.
The game is considered fair if Petya and Vasya can take all *n* cards, and the number of cards each player gets is the same.
Determine whether Petya and Vasya can choose integer numbers before the game so that the game is fair.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=100) — number of cards. It is guaranteed that *n* is an even number.
The following *n* lines contain a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (one integer per line, 1<=≤<=*a**i*<=≤<=100) — numbers written on the *n* cards.
Output Specification:
If it is impossible for Petya and Vasya to choose numbers in such a way that the game will be fair, print "NO" (without quotes) in the first line. In this case you should not print anything more.
In the other case print "YES" (without quotes) in the first line. In the second line print two distinct integers — number that Petya should choose and the number that Vasya should choose to make the game fair. If there are several solutions, print any of them.
Demo Input:
['4\n11\n27\n27\n11\n', '2\n6\n6\n', '6\n10\n20\n30\n20\n10\n20\n', '6\n1\n1\n2\n2\n3\n3\n']
Demo Output:
['YES\n11 27\n', 'NO\n', 'NO\n', 'NO\n']
Note:
In the first example the game will be fair if, for example, Petya chooses number 11, and Vasya chooses number 27. Then the will take all cards — Petya will take cards 1 and 4, and Vasya will take cards 2 and 3. Thus, each of them will take exactly two cards.
In the second example fair game is impossible because the numbers written on the cards are equal, but the numbers that Petya and Vasya should choose should be distinct.
In the third example it is impossible to take all cards. Petya and Vasya can take at most five cards — for example, Petya can choose number 10 and Vasya can choose number 20. But for the game to be fair it is necessary to take 6 cards. | ```python
import math as mt
import sys,string
input=sys.stdin.readline
L=lambda : list(map(int,input().split()))
Ls=lambda : list(input().split())
M=lambda : map(int,input().split())
I=lambda :int(input())
from collections import defaultdict
n=I()
d=defaultdict(int)
for i in range(n):
x=I()
d[x]+=1
if(len(d.keys())==2):
s=set([])
for i in d.keys():
s.add(d[i])
if(len(s)==1):
print("YES")
print(*list(d.keys()))
else:
print("NO")
else:
print("NO")
``` | 3 | |
29 | A | Spit Problem | PROGRAMMING | 1,000 | [
"brute force"
] | A. Spit Problem | 2 | 256 | In a Berland's zoo there is an enclosure with camels. It is known that camels like to spit. Bob watched these interesting animals for the whole day and registered in his notepad where each animal spitted. Now he wants to know if in the zoo there are two camels, which spitted at each other. Help him to solve this task.
The trajectory of a camel's spit is an arc, i.e. if the camel in position *x* spits *d* meters right, he can hit only the camel in position *x*<=+<=*d*, if such a camel exists. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the amount of camels in the zoo. Each of the following *n* lines contains two integers *x**i* and *d**i* (<=-<=104<=≤<=*x**i*<=≤<=104,<=1<=≤<=|*d**i*|<=≤<=2·104) — records in Bob's notepad. *x**i* is a position of the *i*-th camel, and *d**i* is a distance at which the *i*-th camel spitted. Positive values of *d**i* correspond to the spits right, negative values correspond to the spits left. No two camels may stand in the same position. | If there are two camels, which spitted at each other, output YES. Otherwise, output NO. | [
"2\n0 1\n1 -1\n",
"3\n0 1\n1 1\n2 -2\n",
"5\n2 -10\n3 10\n0 5\n5 -5\n10 1\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "2\n0 1\n1 -1",
"output": "YES"
},
{
"input": "3\n0 1\n1 1\n2 -2",
"output": "NO"
},
{
"input": "5\n2 -10\n3 10\n0 5\n5 -5\n10 1",
"output": "YES"
},
{
"input": "10\n-9897 -1144\n-4230 -6350\n2116 -3551\n-3635 4993\n3907 -9071\n-2362 4120\n-6542 984\n5807 3745\n7594 7675\n-5412 -6872",
"output": "NO"
},
{
"input": "11\n-1536 3809\n-2406 -8438\n-1866 395\n5636 -490\n-6867 -7030\n7525 3575\n-6796 2908\n3884 4629\n-2862 -6122\n-8984 6122\n7137 -326",
"output": "YES"
},
{
"input": "12\n-9765 1132\n-1382 -215\n-9405 7284\n-2040 3947\n-9360 3150\n6425 9386\n806 -2278\n-2121 -7284\n5663 -1608\n-8377 9297\n6245 708\n8470 6024",
"output": "YES"
},
{
"input": "15\n8122 -9991\n-4068 -3386\n8971 3731\n3458 5161\n-8700 7562\n2691 8735\n-1510 -3892\n5183 -3753\n-7018 6637\n-7454 3386\n-818 -6377\n6771 -8647\n-7357 -1246\n-6186 1922\n9889 -3627",
"output": "YES"
},
{
"input": "20\n-5264 6424\n-3664 -7459\n-2780 -9859\n-3317 6842\n5681 -8092\n1555 1904\n-6684 1414\n6593 -1253\n-5708 -1202\n335 1733\n-926 7579\n3459 -1904\n-4486 4006\n6201 3616\n2847 -5255\n8438 7057\n8171 6042\n-9102 3545\n7731 -233\n6264 6563",
"output": "YES"
},
{
"input": "30\n-398 -1774\n313 -6974\n2346 -4657\n8552 -9647\n-5265 1538\n8195 4864\n-5641 -5219\n-1394 8563\n-1190 1992\n-4669 -1156\n7574 256\n9206 -2414\n4140 -549\n-294 2169\n7029 -2871\n3808 -9799\n3141 5690\n4648 -2680\n-5990 9800\n-2299 1697\n6077 -7177\n-400 -9724\n-4644 -2392\n-2198 -9531\n-2105 9386\n-8165 -4201\n-1589 -7916\n2518 -7840\n4173 -6949\n-3368 -9943",
"output": "NO"
}
] | 1,673,686,543 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 92 | 0 | def __main__():
n = int(input())
a = []
b =[]
for i in range(n):
x, d = map(int, input().split())
a.append(x)
b.append(d)
for i in range(n):
for j in range(i+1,n):
if (a[i]+b[i]==a[j]) and (a[j]+b[j]==a[i]):
print('Yes')
return
print('No')
__main__() | Title: Spit Problem
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
In a Berland's zoo there is an enclosure with camels. It is known that camels like to spit. Bob watched these interesting animals for the whole day and registered in his notepad where each animal spitted. Now he wants to know if in the zoo there are two camels, which spitted at each other. Help him to solve this task.
The trajectory of a camel's spit is an arc, i.e. if the camel in position *x* spits *d* meters right, he can hit only the camel in position *x*<=+<=*d*, if such a camel exists.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the amount of camels in the zoo. Each of the following *n* lines contains two integers *x**i* and *d**i* (<=-<=104<=≤<=*x**i*<=≤<=104,<=1<=≤<=|*d**i*|<=≤<=2·104) — records in Bob's notepad. *x**i* is a position of the *i*-th camel, and *d**i* is a distance at which the *i*-th camel spitted. Positive values of *d**i* correspond to the spits right, negative values correspond to the spits left. No two camels may stand in the same position.
Output Specification:
If there are two camels, which spitted at each other, output YES. Otherwise, output NO.
Demo Input:
['2\n0 1\n1 -1\n', '3\n0 1\n1 1\n2 -2\n', '5\n2 -10\n3 10\n0 5\n5 -5\n10 1\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none | ```python
def __main__():
n = int(input())
a = []
b =[]
for i in range(n):
x, d = map(int, input().split())
a.append(x)
b.append(d)
for i in range(n):
for j in range(i+1,n):
if (a[i]+b[i]==a[j]) and (a[j]+b[j]==a[i]):
print('Yes')
return
print('No')
__main__()
``` | 3.977 |
377 | A | Maze | PROGRAMMING | 1,600 | [
"dfs and similar"
] | null | null | Pavel loves grid mazes. A grid maze is an *n*<=×<=*m* rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.
Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly *k* empty cells into walls so that all the remaining cells still formed a connected area. Help him. | The first line contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*<=≤<=500, 0<=≤<=*k*<=<<=*s*), where *n* and *m* are the maze's height and width, correspondingly, *k* is the number of walls Pavel wants to add and letter *s* represents the number of empty cells in the original maze.
Each of the next *n* lines contains *m* characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall. | Print *n* lines containing *m* characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").
It is guaranteed that a solution exists. If there are multiple solutions you can output any of them. | [
"3 4 2\n#..#\n..#.\n#...\n",
"5 4 5\n#...\n#.#.\n.#..\n...#\n.#.#\n"
] | [
"#.X#\nX.#.\n#...\n",
"#XXX\n#X#.\nX#..\n...#\n.#.#\n"
] | none | 500 | [
{
"input": "5 4 5\n#...\n#.#.\n.#..\n...#\n.#.#",
"output": "#XXX\n#X#.\nX#..\n...#\n.#.#"
},
{
"input": "3 3 2\n#.#\n...\n#.#",
"output": "#X#\nX..\n#.#"
},
{
"input": "7 7 18\n#.....#\n..#.#..\n.#...#.\n...#...\n.#...#.\n..#.#..\n#.....#",
"output": "#XXXXX#\nXX#X#X.\nX#XXX#.\nXXX#...\nX#...#.\nX.#.#..\n#.....#"
},
{
"input": "1 1 0\n.",
"output": "."
},
{
"input": "2 3 1\n..#\n#..",
"output": "X.#\n#.."
},
{
"input": "2 3 1\n#..\n..#",
"output": "#.X\n..#"
},
{
"input": "3 3 1\n...\n.#.\n..#",
"output": "...\n.#X\n..#"
},
{
"input": "3 3 1\n...\n.#.\n#..",
"output": "...\nX#.\n#.."
},
{
"input": "5 4 4\n#..#\n....\n.##.\n....\n#..#",
"output": "#XX#\nXX..\n.##.\n....\n#..#"
},
{
"input": "5 5 2\n.#..#\n..#.#\n#....\n##.#.\n###..",
"output": "X#..#\nX.#.#\n#....\n##.#.\n###.."
},
{
"input": "4 6 3\n#.....\n#.#.#.\n.#...#\n...#.#",
"output": "#.....\n#X#.#X\nX#...#\n...#.#"
},
{
"input": "7 5 4\n.....\n.#.#.\n#...#\n.#.#.\n.#...\n..#..\n....#",
"output": "X...X\nX#.#X\n#...#\n.#.#.\n.#...\n..#..\n....#"
},
{
"input": "16 14 19\n##############\n..############\n#.############\n#..###########\n....##########\n..############\n.#############\n.#.###########\n....##########\n###..#########\n##...#########\n###....#######\n###.##.......#\n###..###.#..#.\n###....#......\n#...#...##.###",
"output": "##############\nXX############\n#X############\n#XX###########\nXXXX##########\nXX############\nX#############\nX#.###########\nX...##########\n###..#########\n##...#########\n###....#######\n###.##.......#\n###..###.#..#.\n###...X#......\n#X..#XXX##.###"
},
{
"input": "10 17 32\n######.##########\n####.#.##########\n...#....#########\n.........########\n##.......########\n........#########\n#.....###########\n#################\n#################\n#################",
"output": "######X##########\n####X#X##########\nXXX#XXXX#########\nXXXXXXXXX########\n##XXX.XXX########\nXXXX...X#########\n#XX...###########\n#################\n#################\n#################"
},
{
"input": "16 10 38\n##########\n##########\n##########\n..########\n...#######\n...#######\n...#######\n....######\n.....####.\n......###.\n......##..\n.......#..\n.........#\n.........#\n.........#\n.........#",
"output": "##########\n##########\n##########\nXX########\nXXX#######\nXXX#######\nXXX#######\nXXXX######\nXXXXX####.\nXXXXX.###.\nXXXX..##..\nXXX....#..\nXXX......#\nXX.......#\nX........#\n.........#"
},
{
"input": "15 16 19\n########.....###\n########.....###\n############.###\n############.###\n############.###\n############.###\n############.###\n############.###\n############.###\n############.###\n.....#####.#..##\n................\n.#...........###\n###.########.###\n###.########.###",
"output": "########XXXXX###\n########XXXXX###\n############.###\n############.###\n############.###\n############.###\n############.###\n############.###\n############.###\n############.###\nXXXX.#####.#..##\nXXX.............\nX#...........###\n###.########.###\n###X########.###"
},
{
"input": "12 19 42\n.........##########\n...................\n.##.##############.\n..################.\n..#################\n..#################\n..#################\n..#################\n..#################\n..#################\n..##########.######\n.............######",
"output": "XXXXXXXXX##########\nXXXXXXXXXXXXXXXXXXX\nX##X##############X\nXX################X\nXX#################\nXX#################\nXX#################\nX.#################\nX.#################\n..#################\n..##########.######\n.............######"
},
{
"input": "3 5 1\n#...#\n..#..\n..#..",
"output": "#...#\n..#..\nX.#.."
},
{
"input": "4 5 10\n.....\n.....\n..#..\n..#..",
"output": "XXX..\nXXX..\nXX#..\nXX#.."
},
{
"input": "3 5 3\n.....\n..#..\n..#..",
"output": ".....\nX.#..\nXX#.."
},
{
"input": "3 5 1\n#....\n..#..\n..###",
"output": "#....\n..#.X\n..###"
},
{
"input": "4 5 1\n.....\n.##..\n..#..\n..###",
"output": ".....\n.##..\n..#.X\n..###"
},
{
"input": "3 5 2\n..#..\n..#..\n....#",
"output": "X.#..\nX.#..\n....#"
},
{
"input": "10 10 1\n##########\n##......##\n#..#..#..#\n#..####..#\n#######.##\n#######.##\n#..####..#\n#..#..#..#\n##......##\n##########",
"output": "##########\n##......##\n#..#..#..#\n#X.####..#\n#######.##\n#######.##\n#..####..#\n#..#..#..#\n##......##\n##########"
},
{
"input": "10 10 3\n..........\n.########.\n.########.\n.########.\n.########.\n.########.\n.#######..\n.#######..\n.####..###\n.......###",
"output": "..........\n.########.\n.########.\n.########.\n.########.\n.########.\n.#######X.\n.#######XX\n.####..###\n.......###"
},
{
"input": "5 7 10\n..#....\n..#.#..\n.##.#..\n..#.#..\n....#..",
"output": "XX#....\nXX#.#..\nX##.#..\nXX#.#..\nXXX.#.."
},
{
"input": "5 7 10\n..#....\n..#.##.\n.##.##.\n..#.#..\n....#..",
"output": "XX#....\nXX#.##.\nX##.##.\nXX#.#..\nXXX.#.."
},
{
"input": "10 10 1\n##########\n##..##..##\n#...##...#\n#.######.#\n#..####..#\n#..####..#\n#.######.#\n#........#\n##..##..##\n##########",
"output": "##########\n##.X##..##\n#...##...#\n#.######.#\n#..####..#\n#..####..#\n#.######.#\n#........#\n##..##..##\n##########"
},
{
"input": "4 5 1\n.....\n.###.\n..#..\n..#..",
"output": ".....\n.###.\n..#..\n.X#.."
},
{
"input": "2 5 2\n###..\n###..",
"output": "###X.\n###X."
},
{
"input": "2 5 3\n.....\n..#..",
"output": "X....\nXX#.."
},
{
"input": "12 12 3\n############\n#..........#\n#.########.#\n#.########.#\n#.########.#\n#.########.#\n#.########.#\n#.#######..#\n#.#######..#\n#.####..####\n#.......####\n############",
"output": "############\n#..........#\n#.########.#\n#.########.#\n#.########.#\n#.########.#\n#.########.#\n#.#######X.#\n#.#######XX#\n#.####..####\n#.......####\n############"
},
{
"input": "5 5 1\n.....\n.##..\n..###\n..###\n#####",
"output": ".....\n.##.X\n..###\n..###\n#####"
},
{
"input": "4 4 1\n....\n.#..\n..##\n..##",
"output": "....\n.#.X\n..##\n..##"
},
{
"input": "5 5 1\n....#\n.##..\n.##..\n...##\n...##",
"output": "....#\n.##..\n.##.X\n...##\n...##"
},
{
"input": "5 5 1\n.....\n.##..\n..###\n..###\n..###",
"output": ".....\n.##.X\n..###\n..###\n..###"
},
{
"input": "4 5 1\n#....\n#.#..\n..###\n..###",
"output": "#....\n#.#.X\n..###\n..###"
},
{
"input": "4 4 3\n....\n.#..\n..##\n..##",
"output": "...X\n.#XX\n..##\n..##"
},
{
"input": "4 7 6\n.......\n....#..\n.##.#..\n....#..",
"output": "X......\nX...#..\nX##.#..\nXXX.#.."
},
{
"input": "8 8 7\n........\n.##.....\n.#######\n..######\n..######\n..######\n..######\n..######",
"output": ".....XXX\n.##.XXXX\n.#######\n..######\n..######\n..######\n..######\n..######"
}
] | 1,636,621,629 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 8 | 514 | 118,681,600 |
def dfs(has,key,visited,out):
visited.add(key)
out.append(key)
for i in has[key]:
if i not in visited:
dfs(has, i, visited,out)
return out
grid = []
n,m,k = map(int,input().split())
for i in range(n):
s = str(input())
s = list(s)
grid.append(s)
'''
has = {}
for i in range(n):
for j in range(m):
if grid[i][j] == '.':
if i != 0:
if grid[i-1][j] == ".":
has[(i,j)].append((i-1,j))
else:
has[(i,j)] = []
has[(i,j)].append((i-1,j))
if i != n-1:
if grid[i+1][j] == ".":
if (i,j) in has:
has[(i,j)].append((i+1,j))
else:
has[(i,j)] = []
has[(i,j)].append((i+1,j))
if j != 0:
if grid[i][j-1] == ".":
if (i,j) in has:
has[(i,j)].append((i,j-1))
else:
has[(i,j)] = []
has[(i,j)].append((i,j-1))
if j != m-1:
if grid[i][j+1] == ".":
if (i,j) in has:
has[(i,j)].append((i,j+1))
else:
has[(i,j)] = []
has[(i,j)].append((i,j+1))
out = []
visited = set()
for key in has:
if key not in visited:
get = dfs(has,key,visited,out)
print(get)
'''
has = {}
for i in range(n):
count = 0
for j in range(m):
if grid[i][j] == '.':
if i != 0:
if grid[i-1][j] == ".":
count += 1
if i != n-1:
if grid[i+1][j] == ".":
if (i,j) in has:
has[(i,j)].append((i+1,j))
else:
has[(i,j)] = []
has[(i,j)].append((i+1,j))
count += 1
if j != 0:
if grid[i][j-1] == ".":
if (i,j) in has:
has[(i,j)].append((i,j-1))
else:
has[(i,j)] = []
has[(i,j)].append((i,j-1))
count += 1
if j != m-1:
if grid[i][j+1] == ".":
if (i,j) in has:
has[(i,j)].append((i,j+1))
else:
has[(i,j)] = []
has[(i,j)].append((i,j+1))
count += 1
if count == 0:
grid[i][j] = 'X'
k -= 1
if k:
for i in range(n):
for j in range(m):
if k and grid[i][j] == '.':
grid[i][j] = 'X'
k -= 1
for key in grid:
print("".join(key))
| Title: Maze
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pavel loves grid mazes. A grid maze is an *n*<=×<=*m* rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.
Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly *k* empty cells into walls so that all the remaining cells still formed a connected area. Help him.
Input Specification:
The first line contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*<=≤<=500, 0<=≤<=*k*<=<<=*s*), where *n* and *m* are the maze's height and width, correspondingly, *k* is the number of walls Pavel wants to add and letter *s* represents the number of empty cells in the original maze.
Each of the next *n* lines contains *m* characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.
Output Specification:
Print *n* lines containing *m* characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").
It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.
Demo Input:
['3 4 2\n#..#\n..#.\n#...\n', '5 4 5\n#...\n#.#.\n.#..\n...#\n.#.#\n']
Demo Output:
['#.X#\nX.#.\n#...\n', '#XXX\n#X#.\nX#..\n...#\n.#.#\n']
Note:
none | ```python
def dfs(has,key,visited,out):
visited.add(key)
out.append(key)
for i in has[key]:
if i not in visited:
dfs(has, i, visited,out)
return out
grid = []
n,m,k = map(int,input().split())
for i in range(n):
s = str(input())
s = list(s)
grid.append(s)
'''
has = {}
for i in range(n):
for j in range(m):
if grid[i][j] == '.':
if i != 0:
if grid[i-1][j] == ".":
has[(i,j)].append((i-1,j))
else:
has[(i,j)] = []
has[(i,j)].append((i-1,j))
if i != n-1:
if grid[i+1][j] == ".":
if (i,j) in has:
has[(i,j)].append((i+1,j))
else:
has[(i,j)] = []
has[(i,j)].append((i+1,j))
if j != 0:
if grid[i][j-1] == ".":
if (i,j) in has:
has[(i,j)].append((i,j-1))
else:
has[(i,j)] = []
has[(i,j)].append((i,j-1))
if j != m-1:
if grid[i][j+1] == ".":
if (i,j) in has:
has[(i,j)].append((i,j+1))
else:
has[(i,j)] = []
has[(i,j)].append((i,j+1))
out = []
visited = set()
for key in has:
if key not in visited:
get = dfs(has,key,visited,out)
print(get)
'''
has = {}
for i in range(n):
count = 0
for j in range(m):
if grid[i][j] == '.':
if i != 0:
if grid[i-1][j] == ".":
count += 1
if i != n-1:
if grid[i+1][j] == ".":
if (i,j) in has:
has[(i,j)].append((i+1,j))
else:
has[(i,j)] = []
has[(i,j)].append((i+1,j))
count += 1
if j != 0:
if grid[i][j-1] == ".":
if (i,j) in has:
has[(i,j)].append((i,j-1))
else:
has[(i,j)] = []
has[(i,j)].append((i,j-1))
count += 1
if j != m-1:
if grid[i][j+1] == ".":
if (i,j) in has:
has[(i,j)].append((i,j+1))
else:
has[(i,j)] = []
has[(i,j)].append((i,j+1))
count += 1
if count == 0:
grid[i][j] = 'X'
k -= 1
if k:
for i in range(n):
for j in range(m):
if k and grid[i][j] == '.':
grid[i][j] = 'X'
k -= 1
for key in grid:
print("".join(key))
``` | 0 | |
4 | A | Watermelon | PROGRAMMING | 800 | [
"brute force",
"math"
] | A. Watermelon | 1 | 64 | One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight. | The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys. | Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case. | [
"8\n"
] | [
"YES\n"
] | For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos). | 0 | [
{
"input": "8",
"output": "YES"
},
{
"input": "5",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "3",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "1",
"output": "NO"
},
{
"input": "7",
"output": "NO"
},
{
"input": "6",
"output": "YES"
},
{
"input": "10",
"output": "YES"
},
{
"input": "9",
"output": "NO"
},
{
"input": "53",
"output": "NO"
},
{
"input": "77",
"output": "NO"
},
{
"input": "32",
"output": "YES"
},
{
"input": "44",
"output": "YES"
},
{
"input": "98",
"output": "YES"
},
{
"input": "99",
"output": "NO"
},
{
"input": "90",
"output": "YES"
},
{
"input": "67",
"output": "NO"
},
{
"input": "100",
"output": "YES"
},
{
"input": "88",
"output": "YES"
}
] | 1,690,274,488 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 20 | 124 | 0 | x=int(input())
if x%2==0 and x>2:
print("Yes")
else:
print("No")
| Title: Watermelon
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
Input Specification:
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
Output Specification:
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
Demo Input:
['8\n']
Demo Output:
['YES\n']
Note:
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos). | ```python
x=int(input())
if x%2==0 and x>2:
print("Yes")
else:
print("No")
``` | 3.938 |
263 | A | Beautiful Matrix | PROGRAMMING | 800 | [
"implementation"
] | null | null | You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful. | The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one. | Print a single integer — the minimum number of moves needed to make the matrix beautiful. | [
"0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n",
"0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n"
] | [
"3\n",
"1\n"
] | none | 500 | [
{
"input": "0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1",
"output": "4"
},
{
"input": "0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0",
"output": "3"
}
] | 1,695,449,577 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 102,400 | c = [
[' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' ']
]
c[0][0], c[0][1], c[0][2], c[0][3], c[0][4] = input(" ").split()
c[1][0], c[1][1], c[1][2], c[1][3], c[1][4] = input(" ").split()
c[2][0], c[2][1], c[2][2], c[2][3], c[2][4] = input(" ").split()
c[3][0], c[3][1], c[3][2], c[3][3], c[3][4] = input(" ").split()
c[4][0], c[4][1], c[4][2], c[4][3], c[4][4] = input(" ").split()
m = 0
n = 0
a = 0
b = 0
number_step = 0
while a < 5:
while b < 5:
if int(c[a][b]) == 1:
m = a
n = b
break
else:
b += 1
a += 1
print(m)
print(n)
if m == 0 or m == 4:
number_step = number_step + 2
if m == 1 or m == 3:
number_step += 1
if n == 0 or n == 4:
number_step = number_step + 2
if n == 1 or n == 3:
number_step += 1
print(number_step) | Title: Beautiful Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
Input Specification:
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
Output Specification:
Print a single integer — the minimum number of moves needed to make the matrix beautiful.
Demo Input:
['0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n', '0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n']
Demo Output:
['3\n', '1\n']
Note:
none | ```python
c = [
[' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' '],
[' ', ' ', ' ', ' ', ' ']
]
c[0][0], c[0][1], c[0][2], c[0][3], c[0][4] = input(" ").split()
c[1][0], c[1][1], c[1][2], c[1][3], c[1][4] = input(" ").split()
c[2][0], c[2][1], c[2][2], c[2][3], c[2][4] = input(" ").split()
c[3][0], c[3][1], c[3][2], c[3][3], c[3][4] = input(" ").split()
c[4][0], c[4][1], c[4][2], c[4][3], c[4][4] = input(" ").split()
m = 0
n = 0
a = 0
b = 0
number_step = 0
while a < 5:
while b < 5:
if int(c[a][b]) == 1:
m = a
n = b
break
else:
b += 1
a += 1
print(m)
print(n)
if m == 0 or m == 4:
number_step = number_step + 2
if m == 1 or m == 3:
number_step += 1
if n == 0 or n == 4:
number_step = number_step + 2
if n == 1 or n == 3:
number_step += 1
print(number_step)
``` | 0 | |
611 | A | New Year and Days | PROGRAMMING | 900 | [
"implementation"
] | null | null | Today is Wednesday, the third day of the week. What's more interesting is that tomorrow is the last day of the year 2015.
Limak is a little polar bear. He enjoyed this year a lot. Now, he is so eager to the coming year 2016.
Limak wants to prove how responsible a bear he is. He is going to regularly save candies for the entire year 2016! He considers various saving plans. He can save one candy either on some fixed day of the week or on some fixed day of the month.
Limak chose one particular plan. He isn't sure how many candies he will save in the 2016 with his plan. Please, calculate it and tell him. | The only line of the input is in one of the following two formats:
- "*x* of week" where *x* (1<=≤<=*x*<=≤<=7) denotes the day of the week. The 1-st day is Monday and the 7-th one is Sunday. - "*x* of month" where *x* (1<=≤<=*x*<=≤<=31) denotes the day of the month. | Print one integer — the number of candies Limak will save in the year 2016. | [
"4 of week\n",
"30 of month\n"
] | [
"52\n",
"11\n"
] | Polar bears use the Gregorian calendar. It is the most common calendar and you likely use it too. You can read about it on Wikipedia if you want to – [https://en.wikipedia.org/wiki/Gregorian_calendar](https://en.wikipedia.org/wiki/Gregorian_calendar). The week starts with Monday.
In the first sample Limak wants to save one candy on each Thursday (the 4-th day of the week). There are 52 Thursdays in the 2016. Thus, he will save 52 candies in total.
In the second sample Limak wants to save one candy on the 30-th day of each month. There is the 30-th day in exactly 11 months in the 2016 — all months but February. It means that Limak will save 11 candies in total. | 500 | [
{
"input": "4 of week",
"output": "52"
},
{
"input": "30 of month",
"output": "11"
},
{
"input": "17 of month",
"output": "12"
},
{
"input": "31 of month",
"output": "7"
},
{
"input": "6 of week",
"output": "53"
},
{
"input": "1 of week",
"output": "52"
},
{
"input": "2 of week",
"output": "52"
},
{
"input": "3 of week",
"output": "52"
},
{
"input": "5 of week",
"output": "53"
},
{
"input": "7 of week",
"output": "52"
},
{
"input": "1 of month",
"output": "12"
},
{
"input": "2 of month",
"output": "12"
},
{
"input": "3 of month",
"output": "12"
},
{
"input": "4 of month",
"output": "12"
},
{
"input": "5 of month",
"output": "12"
},
{
"input": "6 of month",
"output": "12"
},
{
"input": "7 of month",
"output": "12"
},
{
"input": "8 of month",
"output": "12"
},
{
"input": "9 of month",
"output": "12"
},
{
"input": "10 of month",
"output": "12"
},
{
"input": "11 of month",
"output": "12"
},
{
"input": "12 of month",
"output": "12"
},
{
"input": "13 of month",
"output": "12"
},
{
"input": "14 of month",
"output": "12"
},
{
"input": "15 of month",
"output": "12"
},
{
"input": "16 of month",
"output": "12"
},
{
"input": "18 of month",
"output": "12"
},
{
"input": "19 of month",
"output": "12"
},
{
"input": "20 of month",
"output": "12"
},
{
"input": "21 of month",
"output": "12"
},
{
"input": "22 of month",
"output": "12"
},
{
"input": "23 of month",
"output": "12"
},
{
"input": "24 of month",
"output": "12"
},
{
"input": "25 of month",
"output": "12"
},
{
"input": "26 of month",
"output": "12"
},
{
"input": "27 of month",
"output": "12"
},
{
"input": "28 of month",
"output": "12"
},
{
"input": "29 of month",
"output": "12"
}
] | 1,451,488,325 | 424 | Python 3 | OK | TESTS | 38 | 62 | 0 | args = input().split()
num = int(args[0])
if args[2] == 'month':
if num <= 29:
print(12)
elif num == 30:
print(11)
else:
print(7)
else:
if (num == 5) or (num == 6):
print(53)
else:
print(52) | Title: New Year and Days
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today is Wednesday, the third day of the week. What's more interesting is that tomorrow is the last day of the year 2015.
Limak is a little polar bear. He enjoyed this year a lot. Now, he is so eager to the coming year 2016.
Limak wants to prove how responsible a bear he is. He is going to regularly save candies for the entire year 2016! He considers various saving plans. He can save one candy either on some fixed day of the week or on some fixed day of the month.
Limak chose one particular plan. He isn't sure how many candies he will save in the 2016 with his plan. Please, calculate it and tell him.
Input Specification:
The only line of the input is in one of the following two formats:
- "*x* of week" where *x* (1<=≤<=*x*<=≤<=7) denotes the day of the week. The 1-st day is Monday and the 7-th one is Sunday. - "*x* of month" where *x* (1<=≤<=*x*<=≤<=31) denotes the day of the month.
Output Specification:
Print one integer — the number of candies Limak will save in the year 2016.
Demo Input:
['4 of week\n', '30 of month\n']
Demo Output:
['52\n', '11\n']
Note:
Polar bears use the Gregorian calendar. It is the most common calendar and you likely use it too. You can read about it on Wikipedia if you want to – [https://en.wikipedia.org/wiki/Gregorian_calendar](https://en.wikipedia.org/wiki/Gregorian_calendar). The week starts with Monday.
In the first sample Limak wants to save one candy on each Thursday (the 4-th day of the week). There are 52 Thursdays in the 2016. Thus, he will save 52 candies in total.
In the second sample Limak wants to save one candy on the 30-th day of each month. There is the 30-th day in exactly 11 months in the 2016 — all months but February. It means that Limak will save 11 candies in total. | ```python
args = input().split()
num = int(args[0])
if args[2] == 'month':
if num <= 29:
print(12)
elif num == 30:
print(11)
else:
print(7)
else:
if (num == 5) or (num == 6):
print(53)
else:
print(52)
``` | 3 | |
994 | A | Fingerprints | PROGRAMMING | 800 | [
"implementation"
] | null | null | You are locked in a room with a door that has a keypad with 10 keys corresponding to digits from 0 to 9. To escape from the room, you need to enter a correct code. You also have a sequence of digits.
Some keys on the keypad have fingerprints. You believe the correct code is the longest not necessarily contiguous subsequence of the sequence you have that only contains digits with fingerprints on the corresponding keys. Find such code. | The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10$) representing the number of digits in the sequence you have and the number of keys on the keypad that have fingerprints.
The next line contains $n$ distinct space-separated integers $x_1, x_2, \ldots, x_n$ ($0 \le x_i \le 9$) representing the sequence.
The next line contains $m$ distinct space-separated integers $y_1, y_2, \ldots, y_m$ ($0 \le y_i \le 9$) — the keys with fingerprints. | In a single line print a space-separated sequence of integers representing the code. If the resulting sequence is empty, both printing nothing and printing a single line break is acceptable. | [
"7 3\n3 5 7 1 6 2 8\n1 2 7\n",
"4 4\n3 4 1 0\n0 1 7 9\n"
] | [
"7 1 2\n",
"1 0\n"
] | In the first example, the only digits with fingerprints are $1$, $2$ and $7$. All three of them appear in the sequence you know, $7$ first, then $1$ and then $2$. Therefore the output is 7 1 2. Note that the order is important, and shall be the same as the order in the original sequence.
In the second example digits $0$, $1$, $7$ and $9$ have fingerprints, however only $0$ and $1$ appear in the original sequence. $1$ appears earlier, so the output is 1 0. Again, the order is important. | 500 | [
{
"input": "7 3\n3 5 7 1 6 2 8\n1 2 7",
"output": "7 1 2"
},
{
"input": "4 4\n3 4 1 0\n0 1 7 9",
"output": "1 0"
},
{
"input": "9 4\n9 8 7 6 5 4 3 2 1\n2 4 6 8",
"output": "8 6 4 2"
},
{
"input": "10 5\n3 7 1 2 4 6 9 0 5 8\n4 3 0 7 9",
"output": "3 7 4 9 0"
},
{
"input": "10 10\n1 2 3 4 5 6 7 8 9 0\n4 5 6 7 1 2 3 0 9 8",
"output": "1 2 3 4 5 6 7 8 9 0"
},
{
"input": "1 1\n4\n4",
"output": "4"
},
{
"input": "3 7\n6 3 4\n4 9 0 1 7 8 6",
"output": "6 4"
},
{
"input": "10 1\n9 0 8 1 7 4 6 5 2 3\n0",
"output": "0"
},
{
"input": "5 10\n6 0 3 8 1\n3 1 0 5 4 7 2 8 9 6",
"output": "6 0 3 8 1"
},
{
"input": "8 2\n7 2 9 6 1 0 3 4\n6 3",
"output": "6 3"
},
{
"input": "5 4\n7 0 1 4 9\n0 9 5 3",
"output": "0 9"
},
{
"input": "10 1\n9 6 2 0 1 8 3 4 7 5\n6",
"output": "6"
},
{
"input": "10 2\n7 1 0 2 4 6 5 9 3 8\n3 2",
"output": "2 3"
},
{
"input": "5 9\n3 7 9 2 4\n3 8 4 5 9 6 1 0 2",
"output": "3 9 2 4"
},
{
"input": "10 6\n7 1 2 3 8 0 6 4 5 9\n1 5 8 2 3 6",
"output": "1 2 3 8 6 5"
},
{
"input": "8 2\n7 4 8 9 2 5 6 1\n6 4",
"output": "4 6"
},
{
"input": "10 2\n1 0 3 5 8 9 4 7 6 2\n0 3",
"output": "0 3"
},
{
"input": "7 6\n9 2 8 6 1 3 7\n4 2 0 3 1 8",
"output": "2 8 1 3"
},
{
"input": "1 6\n3\n6 8 2 4 5 3",
"output": "3"
},
{
"input": "1 8\n0\n9 2 4 8 1 5 0 7",
"output": "0"
},
{
"input": "6 9\n7 3 9 4 1 0\n9 1 5 8 0 6 2 7 4",
"output": "7 9 4 1 0"
},
{
"input": "10 2\n4 9 6 8 3 0 1 5 7 2\n0 1",
"output": "0 1"
},
{
"input": "10 5\n5 2 8 0 9 7 6 1 4 3\n9 6 4 1 2",
"output": "2 9 6 1 4"
},
{
"input": "6 3\n8 3 9 2 7 6\n5 4 3",
"output": "3"
},
{
"input": "4 10\n8 3 9 6\n4 9 6 2 7 0 8 1 3 5",
"output": "8 3 9 6"
},
{
"input": "1 2\n1\n1 0",
"output": "1"
},
{
"input": "3 6\n1 2 3\n4 5 6 1 2 3",
"output": "1 2 3"
},
{
"input": "1 2\n2\n1 2",
"output": "2"
},
{
"input": "1 10\n9\n0 1 2 3 4 5 6 7 8 9",
"output": "9"
}
] | 1,546,700,229 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 202 | 1,024,000 | import sys
input=sys.stdin.readline
n=int(input())
c=n//100
n%=100
c+=n//20
n%=20
c+=n//10
n%=10
c+=n//5
n%=5
c+=n
print(c)
| Title: Fingerprints
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are locked in a room with a door that has a keypad with 10 keys corresponding to digits from 0 to 9. To escape from the room, you need to enter a correct code. You also have a sequence of digits.
Some keys on the keypad have fingerprints. You believe the correct code is the longest not necessarily contiguous subsequence of the sequence you have that only contains digits with fingerprints on the corresponding keys. Find such code.
Input Specification:
The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10$) representing the number of digits in the sequence you have and the number of keys on the keypad that have fingerprints.
The next line contains $n$ distinct space-separated integers $x_1, x_2, \ldots, x_n$ ($0 \le x_i \le 9$) representing the sequence.
The next line contains $m$ distinct space-separated integers $y_1, y_2, \ldots, y_m$ ($0 \le y_i \le 9$) — the keys with fingerprints.
Output Specification:
In a single line print a space-separated sequence of integers representing the code. If the resulting sequence is empty, both printing nothing and printing a single line break is acceptable.
Demo Input:
['7 3\n3 5 7 1 6 2 8\n1 2 7\n', '4 4\n3 4 1 0\n0 1 7 9\n']
Demo Output:
['7 1 2\n', '1 0\n']
Note:
In the first example, the only digits with fingerprints are $1$, $2$ and $7$. All three of them appear in the sequence you know, $7$ first, then $1$ and then $2$. Therefore the output is 7 1 2. Note that the order is important, and shall be the same as the order in the original sequence.
In the second example digits $0$, $1$, $7$ and $9$ have fingerprints, however only $0$ and $1$ appear in the original sequence. $1$ appears earlier, so the output is 1 0. Again, the order is important. | ```python
import sys
input=sys.stdin.readline
n=int(input())
c=n//100
n%=100
c+=n//20
n%=20
c+=n//10
n%=10
c+=n//5
n%=5
c+=n
print(c)
``` | -1 | |
766 | B | Mahmoud and a Triangle | PROGRAMMING | 1,000 | [
"constructive algorithms",
"geometry",
"greedy",
"math",
"number theory",
"sortings"
] | null | null | Mahmoud has *n* line segments, the *i*-th of them has length *a**i*. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.
Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area. | The first line contains single integer *n* (3<=≤<=*n*<=≤<=105) — the number of line segments Mahmoud has.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the lengths of line segments Mahmoud has. | In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise. | [
"5\n1 5 3 2 4\n",
"3\n4 1 2\n"
] | [
"YES\n",
"NO\n"
] | For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle. | 1,000 | [
{
"input": "5\n1 5 3 2 4",
"output": "YES"
},
{
"input": "3\n4 1 2",
"output": "NO"
},
{
"input": "30\n197 75 517 39724 7906061 1153471 3 15166 168284 3019844 272293 316 16 24548 42 118 5792 5 9373 1866366 4886214 24 2206 712886 104005 1363 836 64273 440585 3576",
"output": "NO"
},
{
"input": "30\n229017064 335281886 247217656 670601882 743442492 615491486 544941439 911270108 474843964 803323771 177115397 62179276 390270885 754889875 881720571 902691435 154083299 328505383 761264351 182674686 94104683 357622370 573909964 320060691 33548810 247029007 812823597 946798893 813659359 710111761",
"output": "YES"
},
{
"input": "40\n740553458 532562042 138583675 75471987 487348843 476240280 972115023 103690894 546736371 915774563 35356828 819948191 138721993 24257926 761587264 767176616 608310208 78275645 386063134 227581756 672567198 177797611 87579917 941781518 274774331 843623616 981221615 630282032 118843963 749160513 354134861 132333165 405839062 522698334 29698277 541005920 856214146 167344951 398332403 68622974",
"output": "YES"
},
{
"input": "40\n155 1470176 7384 765965701 1075 4 561554 6227772 93 16304522 1744 662 3 292572860 19335 908613 42685804 347058 20 132560 3848974 69067081 58 2819 111752888 408 81925 30 11951 4564 251 26381275 473392832 50628 180819969 2378797 10076746 9 214492 31291",
"output": "NO"
},
{
"input": "3\n1 1000000000 1000000000",
"output": "YES"
},
{
"input": "4\n1 1000000000 1000000000 1000000000",
"output": "YES"
},
{
"input": "3\n1 1000000000 1",
"output": "NO"
},
{
"input": "5\n1 2 3 5 2",
"output": "YES"
},
{
"input": "41\n19 161 4090221 118757367 2 45361275 1562319 596751 140871 97 1844 310910829 10708344 6618115 698 1 87059 33 2527892 12703 73396090 17326460 3 368811 20550 813975131 10 53804 28034805 7847 2992 33254 1139 227930 965568 261 4846 503064297 192153458 57 431",
"output": "NO"
},
{
"input": "42\n4317083 530966905 202811311 104 389267 35 1203 18287479 125344279 21690 859122498 65 859122508 56790 1951 148683 457 1 22 2668100 8283 2 77467028 13405 11302280 47877251 328155592 35095 29589769 240574 4 10 1019123 6985189 629846 5118 169 1648973 91891 741 282 3159",
"output": "YES"
},
{
"input": "43\n729551585 11379 5931704 330557 1653 15529406 729551578 278663905 1 729551584 2683 40656510 29802 147 1400284 2 126260 865419 51 17 172223763 86 1 534861 450887671 32 234 25127103 9597697 48226 7034 389 204294 2265706 65783617 4343 3665990 626 78034 106440137 5 18421 1023",
"output": "YES"
},
{
"input": "44\n719528276 2 235 444692918 24781885 169857576 18164 47558 15316043 9465834 64879816 2234575 1631 853530 8 1001 621 719528259 84 6933 31 1 3615623 719528266 40097928 274835337 1381044 11225 2642 5850203 6 527506 18 104977753 76959 29393 49 4283 141 201482 380 1 124523 326015",
"output": "YES"
},
{
"input": "45\n28237 82 62327732 506757 691225170 5 970 4118 264024506 313192 367 14713577 73933 691225154 6660 599 691225145 3473403 51 427200630 1326718 2146678 100848386 1569 27 163176119 193562 10784 45687 819951 38520653 225 119620 1 3 691225169 691225164 17445 23807072 1 9093493 5620082 2542 139 14",
"output": "YES"
},
{
"input": "44\n165580141 21 34 55 1 89 144 17711 2 377 610 987 2584 13 5 4181 6765 10946 1597 8 28657 3 233 75025 121393 196418 317811 9227465 832040 1346269 2178309 3524578 5702887 1 14930352 102334155 24157817 39088169 63245986 701408733 267914296 433494437 514229 46368",
"output": "NO"
},
{
"input": "3\n1 1000000000 999999999",
"output": "NO"
},
{
"input": "5\n1 1 1 1 1",
"output": "YES"
},
{
"input": "10\n1 10 100 1000 10000 100000 1000000 10000000 100000000 1000000000",
"output": "NO"
},
{
"input": "5\n2 3 4 10 20",
"output": "YES"
},
{
"input": "6\n18 23 40 80 160 161",
"output": "YES"
},
{
"input": "4\n5 6 7 888",
"output": "YES"
},
{
"input": "9\n1 1 2 2 4 5 10 10 20",
"output": "YES"
},
{
"input": "7\n3 150 900 4 500 1500 5",
"output": "YES"
},
{
"input": "3\n2 2 3",
"output": "YES"
},
{
"input": "7\n1 2 100 200 250 1000000 2000000",
"output": "YES"
},
{
"input": "8\n2 3 5 5 5 6 6 13",
"output": "YES"
},
{
"input": "3\n2 3 4",
"output": "YES"
},
{
"input": "6\n1 1 1 4 5 100",
"output": "YES"
},
{
"input": "13\n1 2 3 5 8 13 22 34 55 89 144 233 377",
"output": "YES"
},
{
"input": "4\n2 3 4 8",
"output": "YES"
},
{
"input": "3\n5 6 7",
"output": "YES"
},
{
"input": "5\n1 4 5 6 1000000",
"output": "YES"
},
{
"input": "4\n5 6 7 20",
"output": "YES"
},
{
"input": "6\n1 1 1 1 1 65",
"output": "YES"
},
{
"input": "4\n3 4 5 100",
"output": "YES"
},
{
"input": "3\n2 4 5",
"output": "YES"
},
{
"input": "7\n1 1 1 1 1 10 1000",
"output": "YES"
},
{
"input": "4\n1 1 2 3",
"output": "NO"
},
{
"input": "11\n1 2 5 6 7 8 9 17 18 19 100",
"output": "YES"
},
{
"input": "4\n5 16 20 200",
"output": "YES"
},
{
"input": "5\n17 6 3 3 1",
"output": "YES"
},
{
"input": "3\n1 1 1",
"output": "YES"
},
{
"input": "6\n1 1 1 2 3 5",
"output": "YES"
},
{
"input": "4\n2 4 6 6",
"output": "YES"
},
{
"input": "9\n1 2 4 4 4 4 7 8 20",
"output": "YES"
},
{
"input": "9\n1 1 2 5 5 5 10 10 20",
"output": "YES"
},
{
"input": "7\n3 150 600 4 1700 6000 5",
"output": "YES"
},
{
"input": "5\n5761 20966 27841 28800 29399",
"output": "YES"
},
{
"input": "9\n1 2 3 6 7 10 11 12 24",
"output": "YES"
},
{
"input": "4\n1 2 1 1",
"output": "YES"
},
{
"input": "5\n1 1 2 3 4",
"output": "YES"
}
] | 1,642,693,349 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 8 | 78 | 7,884,800 | x=int(input())
y=list(map(int,input().split()))
y.sort()
for i in range(len(y)-2):
if i+y[-2]>y[-1]:
print('YES')
break
else:
print('NO') | Title: Mahmoud and a Triangle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mahmoud has *n* line segments, the *i*-th of them has length *a**i*. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.
Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area.
Input Specification:
The first line contains single integer *n* (3<=≤<=*n*<=≤<=105) — the number of line segments Mahmoud has.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the lengths of line segments Mahmoud has.
Output Specification:
In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise.
Demo Input:
['5\n1 5 3 2 4\n', '3\n4 1 2\n']
Demo Output:
['YES\n', 'NO\n']
Note:
For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle. | ```python
x=int(input())
y=list(map(int,input().split()))
y.sort()
for i in range(len(y)-2):
if i+y[-2]>y[-1]:
print('YES')
break
else:
print('NO')
``` | 0 | |
231 | A | Team | PROGRAMMING | 800 | [
"brute force",
"greedy"
] | null | null | One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. | The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. | Print a single integer — the number of problems the friends will implement on the contest. | [
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] | [
"2\n",
"1\n"
] | In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. | 500 | [
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
{
"input": "10\n0 1 0\n0 1 0\n1 1 0\n1 0 0\n0 0 1\n0 1 1\n1 1 1\n1 1 0\n0 0 0\n0 0 0",
"output": "4"
},
{
"input": "15\n0 1 0\n1 0 0\n1 1 0\n1 1 1\n0 1 0\n0 0 1\n1 0 1\n1 0 1\n1 0 1\n0 0 0\n1 1 1\n1 1 0\n0 1 1\n1 1 0\n1 1 1",
"output": "10"
},
{
"input": "50\n0 0 0\n0 1 1\n1 1 1\n0 1 0\n1 0 1\n1 1 1\n0 0 1\n1 0 0\n1 1 0\n1 0 1\n0 1 0\n0 0 1\n1 1 0\n0 1 0\n1 1 0\n0 0 0\n1 1 1\n1 0 1\n0 0 1\n1 1 0\n1 1 1\n0 1 1\n1 1 0\n0 0 0\n0 0 0\n1 1 1\n0 0 0\n1 1 1\n0 1 1\n0 0 1\n0 0 0\n0 0 0\n1 1 0\n1 1 0\n1 0 1\n1 0 0\n1 0 1\n1 0 1\n0 1 1\n1 1 0\n1 1 0\n0 1 0\n1 0 1\n0 0 0\n0 0 0\n0 0 0\n0 0 1\n1 1 1\n0 1 1\n1 0 1",
"output": "29"
},
{
"input": "1\n1 1 1",
"output": "1"
},
{
"input": "8\n0 0 0\n0 0 1\n0 0 0\n0 1 1\n1 0 0\n1 0 1\n1 1 0\n1 1 1",
"output": "4"
},
{
"input": "16\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1",
"output": "16"
}
] | 1,699,897,980 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 21 | 154 | 1,638,400 | def solve():
n = int(input())
ans = 0
for _ in range(n):
a, b, c = map(int, input().split())
if a+b+c >= 2:
ans += 1
print(ans)
solve() | Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer — the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. | ```python
def solve():
n = int(input())
ans = 0
for _ in range(n):
a, b, c = map(int, input().split())
if a+b+c >= 2:
ans += 1
print(ans)
solve()
``` | 3 | |
337 | A | Puzzles | PROGRAMMING | 900 | [
"greedy"
] | null | null | The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).
The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on.
Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*. | The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop. | Print a single integer — the least possible difference the teacher can obtain. | [
"4 6\n10 12 10 7 5 22\n"
] | [
"5\n"
] | Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5. | 500 | [
{
"input": "4 6\n10 12 10 7 5 22",
"output": "5"
},
{
"input": "2 2\n4 4",
"output": "0"
},
{
"input": "2 10\n4 5 6 7 8 9 10 11 12 12",
"output": "0"
},
{
"input": "4 5\n818 136 713 59 946",
"output": "759"
},
{
"input": "3 20\n446 852 783 313 549 965 40 88 86 617 479 118 768 34 47 826 366 957 463 903",
"output": "13"
},
{
"input": "2 25\n782 633 152 416 432 825 115 97 386 357 836 310 530 413 354 373 847 882 913 682 729 582 671 674 94",
"output": "3"
},
{
"input": "4 25\n226 790 628 528 114 64 239 279 619 39 894 763 763 847 525 93 882 697 999 643 650 244 159 884 190",
"output": "31"
},
{
"input": "2 50\n971 889 628 39 253 157 925 694 129 516 660 272 738 319 611 816 142 717 514 392 41 105 132 676 958 118 306 768 600 685 103 857 704 346 857 309 23 718 618 161 176 379 846 834 640 468 952 878 164 997",
"output": "0"
},
{
"input": "25 50\n582 146 750 905 313 509 402 21 488 512 32 898 282 64 579 869 37 996 377 929 975 697 666 837 311 205 116 992 533 298 648 268 54 479 792 595 152 69 267 417 184 433 894 603 988 712 24 414 301 176",
"output": "412"
},
{
"input": "49 50\n58 820 826 960 271 294 473 102 925 318 729 672 244 914 796 646 868 6 893 882 726 203 528 498 271 195 355 459 721 680 547 147 631 116 169 804 145 996 133 559 110 257 771 476 576 251 607 314 427 886",
"output": "938"
},
{
"input": "50 50\n374 573 323 744 190 806 485 247 628 336 491 606 702 321 991 678 337 579 86 240 993 208 668 686 855 205 363 177 719 249 896 919 782 434 59 647 787 996 286 216 636 212 546 903 958 559 544 126 608 993",
"output": "937"
},
{
"input": "6 50\n6 8 7 8 5 4 4 5 7 8 6 5 7 4 7 7 7 8 6 4 6 6 8 8 7 7 8 7 5 8 5 4 4 7 8 4 4 6 6 6 8 7 4 7 6 6 5 8 4 7",
"output": "0"
},
{
"input": "37 50\n14 5 11 17 8 20 19 16 20 11 17 20 16 9 14 14 13 18 11 20 8 8 8 5 19 17 6 18 10 20 9 7 12 6 14 17 4 4 10 13 7 4 11 6 20 19 12 12 15 19",
"output": "12"
},
{
"input": "40 50\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "0"
},
{
"input": "40 50\n17 20 43 26 41 37 14 8 30 35 30 24 43 8 42 9 41 50 41 35 27 32 35 43 28 36 31 16 5 7 23 16 14 29 8 39 12 16 36 18 49 39 33 37 38 6 6 27 23 17",
"output": "31"
},
{
"input": "2 2\n1000 4",
"output": "996"
},
{
"input": "2 3\n4 502 1000",
"output": "498"
},
{
"input": "3 3\n4 1000 4",
"output": "996"
}
] | 1,685,726,897 | 2,147,483,647 | Python 3 | OK | TESTS | 18 | 62 | 0 | a, b = input().split(), input().split()
uch = int(a[0])
paz = int(a[1])
b = [int(i) for i in b]
b.sort()
min = 1000000000000
for i in range(uch - 1, paz):
if b[i] - b[i - uch + 1] < min:
min = b[i] - b[i - uch + 1]
print(min) | Title: Puzzles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).
The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on.
Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.
Input Specification:
The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.
Output Specification:
Print a single integer — the least possible difference the teacher can obtain.
Demo Input:
['4 6\n10 12 10 7 5 22\n']
Demo Output:
['5\n']
Note:
Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5. | ```python
a, b = input().split(), input().split()
uch = int(a[0])
paz = int(a[1])
b = [int(i) for i in b]
b.sort()
min = 1000000000000
for i in range(uch - 1, paz):
if b[i] - b[i - uch + 1] < min:
min = b[i] - b[i - uch + 1]
print(min)
``` | 3 | |
177 | A1 | Good Matrix Elements | PROGRAMMING | 800 | [
"implementation"
] | null | null | The Smart Beaver from ABBYY got hooked on square matrices. Now he is busy studying an *n*<=×<=*n* size matrix, where *n* is odd. The Smart Beaver considers the following matrix elements good:
- Elements of the main diagonal. - Elements of the secondary diagonal. - Elements of the "middle" row — the row which has exactly rows above it and the same number of rows below it. - Elements of the "middle" column — the column that has exactly columns to the left of it and the same number of columns to the right of it.
Help the Smart Beaver count the sum of good elements of the given matrix. | The first line of input data contains a single odd integer *n*. Each of the next *n* lines contains *n* integers *a**ij* (0<=≤<=*a**ij*<=≤<=100) separated by single spaces — the elements of the given matrix.
The input limitations for getting 30 points are:
- 1<=≤<=*n*<=≤<=5
The input limitations for getting 100 points are:
- 1<=≤<=*n*<=≤<=101 | Print a single integer — the sum of good matrix elements. | [
"3\n1 2 3\n4 5 6\n7 8 9\n",
"5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n"
] | [
"45\n",
"17\n"
] | In the first sample all matrix elements will be good. Good elements in the second sample are shown on the figure. | 30 | [
{
"input": "3\n1 2 3\n4 5 6\n7 8 9",
"output": "45"
},
{
"input": "5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1",
"output": "17"
},
{
"input": "1\n3",
"output": "3"
},
{
"input": "5\n27 7 3 11 72\n19 49 68 19 59\n41 25 37 64 65\n8 39 96 62 90\n13 37 43 26 33",
"output": "756"
},
{
"input": "3\n19 7 16\n12 15 5\n15 15 5",
"output": "109"
},
{
"input": "3\n36 4 33\n11 46 32\n20 49 34",
"output": "265"
},
{
"input": "3\n79 91 74\n33 82 22\n18 28 54",
"output": "481"
},
{
"input": "5\n7 0 8 1 7\n5 1 1 0 4\n4 2 8 1 6\n1 2 3 2 7\n6 0 1 9 6",
"output": "65"
},
{
"input": "5\n27 20 28 11 17\n25 21 1 20 14\n14 22 28 1 6\n1 2 23 2 7\n6 0 1 29 6",
"output": "225"
},
{
"input": "5\n57 50 58 41 17\n25 21 1 50 44\n44 22 28 31 36\n31 32 23 32 37\n6 0 31 59 6",
"output": "495"
},
{
"input": "5\n57 80 28 41 47\n85 51 61 50 74\n44 82 28 31 36\n31 32 23 32 37\n66 60 31 59 6",
"output": "705"
},
{
"input": "5\n13 58 10 17 43\n61 73 100 0 9\n52 38 16 22 96\n11 4 14 67 62\n70 89 7 98 83",
"output": "708"
},
{
"input": "5\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "5\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "5\n0 1 0 1 0\n0 0 0 0 1\n0 0 5 0 0\n0 0 0 0 1\n0 0 0 0 0",
"output": "5"
},
{
"input": "5\n0 0 0 0 1\n0 1 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 1 0 0 0",
"output": "3"
},
{
"input": "3\n0 0 0\n0 0 0\n0 0 0",
"output": "0"
}
] | 1,617,457,288 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS1 | 3 | 92 | 0 | n= int(input())
a=[]
for i in range(n):
a.append([int(i) for i in input().split()])
s=0
for i in range(n):
s=s+a[i][i] + a[i][n-1-i]
k = int((n-1)/2) -1
for i in range(n):
s= s+a[k][i] + a[i][k]
s=s-3*a[k][k]
print(s)
| Title: Good Matrix Elements
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Smart Beaver from ABBYY got hooked on square matrices. Now he is busy studying an *n*<=×<=*n* size matrix, where *n* is odd. The Smart Beaver considers the following matrix elements good:
- Elements of the main diagonal. - Elements of the secondary diagonal. - Elements of the "middle" row — the row which has exactly rows above it and the same number of rows below it. - Elements of the "middle" column — the column that has exactly columns to the left of it and the same number of columns to the right of it.
Help the Smart Beaver count the sum of good elements of the given matrix.
Input Specification:
The first line of input data contains a single odd integer *n*. Each of the next *n* lines contains *n* integers *a**ij* (0<=≤<=*a**ij*<=≤<=100) separated by single spaces — the elements of the given matrix.
The input limitations for getting 30 points are:
- 1<=≤<=*n*<=≤<=5
The input limitations for getting 100 points are:
- 1<=≤<=*n*<=≤<=101
Output Specification:
Print a single integer — the sum of good matrix elements.
Demo Input:
['3\n1 2 3\n4 5 6\n7 8 9\n', '5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n']
Demo Output:
['45\n', '17\n']
Note:
In the first sample all matrix elements will be good. Good elements in the second sample are shown on the figure. | ```python
n= int(input())
a=[]
for i in range(n):
a.append([int(i) for i in input().split()])
s=0
for i in range(n):
s=s+a[i][i] + a[i][n-1-i]
k = int((n-1)/2) -1
for i in range(n):
s= s+a[k][i] + a[i][k]
s=s-3*a[k][k]
print(s)
``` | 0 | |
265 | B | Roadside Trees (Simplified Edition) | PROGRAMMING | 1,000 | [
"greedy",
"implementation"
] | null | null | Squirrel Liss loves nuts. There are *n* trees (numbered 1 to *n* from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree *i* is *h**i*. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
- Walk up or down one unit on a tree. - Eat a nut on the top of the current tree. - Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height *h* of the tree *i* (1<=≤<=*i*<=≤<=*n*<=-<=1), she jumps to height *h* of the tree *i*<=+<=1. This action can't be performed if *h*<=><=*h**i*<=+<=1.
Compute the minimal time (in seconds) required to eat all nuts. | The first line contains an integer *n* (1<=<=≤<=<=*n*<=≤<=105) — the number of trees.
Next *n* lines contains the height of trees: *i*-th line contains an integer *h**i* (1<=≤<=*h**i*<=≤<=104) — the height of the tree with the number *i*. | Print a single integer — the minimal time required to eat all nuts in seconds. | [
"2\n1\n2\n",
"5\n2\n1\n2\n1\n1\n"
] | [
"5\n",
"14\n"
] | none | 1,000 | [
{
"input": "2\n1\n2",
"output": "5"
},
{
"input": "5\n2\n1\n2\n1\n1",
"output": "14"
},
{
"input": "1\n1",
"output": "2"
}
] | 1,623,225,286 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 218 | 0 | temp = 0
c = 0
for i in range(int(input())):
n = int(input())
if i == 0:
c += n+1
temp = n
else:
if temp > n:
c += (temp-n)+1
c += n
else:
c += 1
c += (n-temp)+1
temp = n
print(c)
| Title: Roadside Trees (Simplified Edition)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Squirrel Liss loves nuts. There are *n* trees (numbered 1 to *n* from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree *i* is *h**i*. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
- Walk up or down one unit on a tree. - Eat a nut on the top of the current tree. - Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height *h* of the tree *i* (1<=≤<=*i*<=≤<=*n*<=-<=1), she jumps to height *h* of the tree *i*<=+<=1. This action can't be performed if *h*<=><=*h**i*<=+<=1.
Compute the minimal time (in seconds) required to eat all nuts.
Input Specification:
The first line contains an integer *n* (1<=<=≤<=<=*n*<=≤<=105) — the number of trees.
Next *n* lines contains the height of trees: *i*-th line contains an integer *h**i* (1<=≤<=*h**i*<=≤<=104) — the height of the tree with the number *i*.
Output Specification:
Print a single integer — the minimal time required to eat all nuts in seconds.
Demo Input:
['2\n1\n2\n', '5\n2\n1\n2\n1\n1\n']
Demo Output:
['5\n', '14\n']
Note:
none | ```python
temp = 0
c = 0
for i in range(int(input())):
n = int(input())
if i == 0:
c += n+1
temp = n
else:
if temp > n:
c += (temp-n)+1
c += n
else:
c += 1
c += (n-temp)+1
temp = n
print(c)
``` | 0 | |
214 | A | System of Equations | PROGRAMMING | 800 | [
"brute force"
] | null | null | Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
You should count, how many there are pairs of integers (*a*,<=*b*) (0<=≤<=*a*,<=*b*) which satisfy the system. | A single line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the parameters of the system. The numbers on the line are separated by a space. | On a single line print the answer to the problem. | [
"9 3\n",
"14 28\n",
"4 20\n"
] | [
"1\n",
"1\n",
"0\n"
] | In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair. | 500 | [
{
"input": "9 3",
"output": "1"
},
{
"input": "14 28",
"output": "1"
},
{
"input": "4 20",
"output": "0"
},
{
"input": "18 198",
"output": "1"
},
{
"input": "22 326",
"output": "1"
},
{
"input": "26 104",
"output": "1"
},
{
"input": "14 10",
"output": "0"
},
{
"input": "8 20",
"output": "0"
},
{
"input": "2 8",
"output": "0"
},
{
"input": "20 11",
"output": "0"
},
{
"input": "57 447",
"output": "1"
},
{
"input": "1 1",
"output": "2"
},
{
"input": "66 296",
"output": "1"
},
{
"input": "75 683",
"output": "1"
},
{
"input": "227 975",
"output": "1"
},
{
"input": "247 499",
"output": "1"
},
{
"input": "266 116",
"output": "1"
},
{
"input": "286 916",
"output": "1"
},
{
"input": "307 341",
"output": "1"
},
{
"input": "451 121",
"output": "1"
},
{
"input": "471 921",
"output": "1"
},
{
"input": "502 346",
"output": "1"
},
{
"input": "535 59",
"output": "1"
},
{
"input": "555 699",
"output": "1"
},
{
"input": "747 351",
"output": "1"
},
{
"input": "790 64",
"output": "1"
},
{
"input": "810 704",
"output": "1"
},
{
"input": "855 225",
"output": "1"
},
{
"input": "902 34",
"output": "1"
},
{
"input": "922 514",
"output": "1"
},
{
"input": "971 131",
"output": "1"
},
{
"input": "991 931",
"output": "1"
},
{
"input": "840 780",
"output": "0"
},
{
"input": "102 595",
"output": "0"
},
{
"input": "139 433",
"output": "0"
},
{
"input": "968 288",
"output": "0"
},
{
"input": "563 354",
"output": "0"
},
{
"input": "994 975",
"output": "0"
},
{
"input": "456 221",
"output": "0"
},
{
"input": "205 210",
"output": "0"
},
{
"input": "1 11",
"output": "0"
},
{
"input": "1000 1000",
"output": "0"
},
{
"input": "3 3",
"output": "0"
},
{
"input": "11 99",
"output": "0"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "11 1",
"output": "0"
},
{
"input": "6 6",
"output": "1"
},
{
"input": "100 452",
"output": "0"
},
{
"input": "420 380",
"output": "0"
},
{
"input": "31 31",
"output": "0"
},
{
"input": "2 6",
"output": "0"
},
{
"input": "112 134",
"output": "0"
},
{
"input": "13 13",
"output": "0"
},
{
"input": "1 571",
"output": "0"
}
] | 1,666,008,678 | 2,147,483,647 | Python 3 | OK | TESTS | 54 | 902 | 0 | n,m = map(int,input().split())
x = min(n,m)
li = []
for i in range(x+1):
x1= pow(i,2)
for j in range(x+1):
y1 = pow(j,2)
if(x1+j ==n and i+y1==m):
li.append((i,j))
print(len(li)) | Title: System of Equations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
You should count, how many there are pairs of integers (*a*,<=*b*) (0<=≤<=*a*,<=*b*) which satisfy the system.
Input Specification:
A single line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the parameters of the system. The numbers on the line are separated by a space.
Output Specification:
On a single line print the answer to the problem.
Demo Input:
['9 3\n', '14 28\n', '4 20\n']
Demo Output:
['1\n', '1\n', '0\n']
Note:
In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair. | ```python
n,m = map(int,input().split())
x = min(n,m)
li = []
for i in range(x+1):
x1= pow(i,2)
for j in range(x+1):
y1 = pow(j,2)
if(x1+j ==n and i+y1==m):
li.append((i,j))
print(len(li))
``` | 3 | |
845 | B | Luba And The Ticket | PROGRAMMING | 1,600 | [
"brute force",
"greedy",
"implementation"
] | null | null | Luba has a ticket consisting of 6 digits. In one move she can choose digit in any position and replace it with arbitrary digit. She wants to know the minimum number of digits she needs to replace in order to make the ticket lucky.
The ticket is considered lucky if the sum of first three digits equals to the sum of last three digits. | You are given a string consisting of 6 characters (all characters are digits from 0 to 9) — this string denotes Luba's ticket. The ticket can start with the digit 0. | Print one number — the minimum possible number of digits Luba needs to replace to make the ticket lucky. | [
"000000\n",
"123456\n",
"111000\n"
] | [
"0\n",
"2\n",
"1\n"
] | In the first example the ticket is already lucky, so the answer is 0.
In the second example Luba can replace 4 and 5 with zeroes, and the ticket will become lucky. It's easy to see that at least two replacements are required.
In the third example Luba can replace any zero with 3. It's easy to see that at least one replacement is required. | 0 | [
{
"input": "000000",
"output": "0"
},
{
"input": "123456",
"output": "2"
},
{
"input": "111000",
"output": "1"
},
{
"input": "120111",
"output": "0"
},
{
"input": "999999",
"output": "0"
},
{
"input": "199880",
"output": "1"
},
{
"input": "899889",
"output": "1"
},
{
"input": "899888",
"output": "1"
},
{
"input": "505777",
"output": "2"
},
{
"input": "999000",
"output": "3"
},
{
"input": "989010",
"output": "3"
},
{
"input": "651894",
"output": "1"
},
{
"input": "858022",
"output": "2"
},
{
"input": "103452",
"output": "1"
},
{
"input": "999801",
"output": "2"
},
{
"input": "999990",
"output": "1"
},
{
"input": "697742",
"output": "1"
},
{
"input": "242367",
"output": "2"
},
{
"input": "099999",
"output": "1"
},
{
"input": "198999",
"output": "1"
},
{
"input": "023680",
"output": "1"
},
{
"input": "999911",
"output": "2"
},
{
"input": "000990",
"output": "2"
},
{
"input": "117099",
"output": "1"
},
{
"input": "990999",
"output": "1"
},
{
"input": "000111",
"output": "1"
},
{
"input": "000444",
"output": "2"
},
{
"input": "202597",
"output": "2"
},
{
"input": "000333",
"output": "1"
},
{
"input": "030039",
"output": "1"
},
{
"input": "000009",
"output": "1"
},
{
"input": "006456",
"output": "1"
},
{
"input": "022995",
"output": "3"
},
{
"input": "999198",
"output": "1"
},
{
"input": "223456",
"output": "2"
},
{
"input": "333665",
"output": "2"
},
{
"input": "123986",
"output": "2"
},
{
"input": "599257",
"output": "1"
},
{
"input": "101488",
"output": "3"
},
{
"input": "111399",
"output": "2"
},
{
"input": "369009",
"output": "1"
},
{
"input": "024887",
"output": "2"
},
{
"input": "314347",
"output": "1"
},
{
"input": "145892",
"output": "1"
},
{
"input": "321933",
"output": "1"
},
{
"input": "100172",
"output": "1"
},
{
"input": "222455",
"output": "2"
},
{
"input": "317596",
"output": "1"
},
{
"input": "979245",
"output": "2"
},
{
"input": "000018",
"output": "1"
},
{
"input": "101389",
"output": "2"
},
{
"input": "123985",
"output": "2"
},
{
"input": "900000",
"output": "1"
},
{
"input": "132069",
"output": "1"
},
{
"input": "949256",
"output": "1"
},
{
"input": "123996",
"output": "2"
},
{
"input": "034988",
"output": "2"
},
{
"input": "320869",
"output": "2"
},
{
"input": "089753",
"output": "1"
},
{
"input": "335667",
"output": "2"
},
{
"input": "868580",
"output": "1"
},
{
"input": "958031",
"output": "2"
},
{
"input": "117999",
"output": "2"
},
{
"input": "000001",
"output": "1"
},
{
"input": "213986",
"output": "2"
},
{
"input": "123987",
"output": "3"
},
{
"input": "111993",
"output": "2"
},
{
"input": "642479",
"output": "1"
},
{
"input": "033788",
"output": "2"
},
{
"input": "766100",
"output": "2"
},
{
"input": "012561",
"output": "1"
},
{
"input": "111695",
"output": "2"
},
{
"input": "123689",
"output": "2"
},
{
"input": "944234",
"output": "1"
},
{
"input": "154999",
"output": "2"
},
{
"input": "333945",
"output": "1"
},
{
"input": "371130",
"output": "1"
},
{
"input": "977330",
"output": "2"
},
{
"input": "777544",
"output": "2"
},
{
"input": "111965",
"output": "2"
},
{
"input": "988430",
"output": "2"
},
{
"input": "123789",
"output": "3"
},
{
"input": "111956",
"output": "2"
},
{
"input": "444776",
"output": "2"
},
{
"input": "001019",
"output": "1"
},
{
"input": "011299",
"output": "2"
},
{
"input": "011389",
"output": "2"
},
{
"input": "999333",
"output": "2"
},
{
"input": "126999",
"output": "2"
},
{
"input": "744438",
"output": "0"
},
{
"input": "588121",
"output": "3"
},
{
"input": "698213",
"output": "2"
},
{
"input": "652858",
"output": "1"
},
{
"input": "989304",
"output": "3"
},
{
"input": "888213",
"output": "3"
},
{
"input": "969503",
"output": "2"
},
{
"input": "988034",
"output": "2"
},
{
"input": "889444",
"output": "2"
},
{
"input": "990900",
"output": "1"
},
{
"input": "301679",
"output": "2"
},
{
"input": "434946",
"output": "1"
},
{
"input": "191578",
"output": "2"
},
{
"input": "118000",
"output": "2"
},
{
"input": "636915",
"output": "0"
},
{
"input": "811010",
"output": "1"
},
{
"input": "822569",
"output": "1"
},
{
"input": "122669",
"output": "2"
},
{
"input": "010339",
"output": "2"
},
{
"input": "213698",
"output": "2"
},
{
"input": "895130",
"output": "2"
},
{
"input": "000900",
"output": "1"
},
{
"input": "191000",
"output": "2"
},
{
"input": "001000",
"output": "1"
},
{
"input": "080189",
"output": "2"
},
{
"input": "990000",
"output": "2"
},
{
"input": "201984",
"output": "2"
},
{
"input": "002667",
"output": "2"
},
{
"input": "877542",
"output": "2"
},
{
"input": "301697",
"output": "2"
},
{
"input": "211597",
"output": "2"
},
{
"input": "420337",
"output": "1"
},
{
"input": "024768",
"output": "2"
},
{
"input": "878033",
"output": "2"
},
{
"input": "788024",
"output": "2"
},
{
"input": "023869",
"output": "2"
},
{
"input": "466341",
"output": "1"
},
{
"input": "696327",
"output": "1"
},
{
"input": "779114",
"output": "2"
},
{
"input": "858643",
"output": "1"
},
{
"input": "011488",
"output": "3"
},
{
"input": "003669",
"output": "2"
},
{
"input": "202877",
"output": "3"
},
{
"input": "738000",
"output": "2"
},
{
"input": "567235",
"output": "2"
},
{
"input": "887321",
"output": "3"
},
{
"input": "401779",
"output": "2"
},
{
"input": "989473",
"output": "2"
},
{
"input": "004977",
"output": "3"
},
{
"input": "023778",
"output": "2"
},
{
"input": "809116",
"output": "1"
},
{
"input": "042762",
"output": "1"
},
{
"input": "777445",
"output": "2"
},
{
"input": "769302",
"output": "2"
},
{
"input": "023977",
"output": "2"
},
{
"input": "990131",
"output": "2"
}
] | 1,503,331,194 | 3,294 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | i = list(map(lambda c: int(c), input()))
print('input %s' % str(i))
d = sum([i[3]-i[0], i[4]-i[1], i[5]-i[2]])
print('difference: %i' % d)
if(d < 0):
print('reversing!')
i.reverse()
d *= -1
for z in range(3):
i[z] = 9 - i[z]
i = sorted(i)
i.reverse()
print('normalized: %s' % str(i))
ir = 0
for (ind, z) in enumerate(i):
print('reached: %d' % ir)
if ir >= d:
print(ind)
break
ir += z
| Title: Luba And The Ticket
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luba has a ticket consisting of 6 digits. In one move she can choose digit in any position and replace it with arbitrary digit. She wants to know the minimum number of digits she needs to replace in order to make the ticket lucky.
The ticket is considered lucky if the sum of first three digits equals to the sum of last three digits.
Input Specification:
You are given a string consisting of 6 characters (all characters are digits from 0 to 9) — this string denotes Luba's ticket. The ticket can start with the digit 0.
Output Specification:
Print one number — the minimum possible number of digits Luba needs to replace to make the ticket lucky.
Demo Input:
['000000\n', '123456\n', '111000\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the first example the ticket is already lucky, so the answer is 0.
In the second example Luba can replace 4 and 5 with zeroes, and the ticket will become lucky. It's easy to see that at least two replacements are required.
In the third example Luba can replace any zero with 3. It's easy to see that at least one replacement is required. | ```python
i = list(map(lambda c: int(c), input()))
print('input %s' % str(i))
d = sum([i[3]-i[0], i[4]-i[1], i[5]-i[2]])
print('difference: %i' % d)
if(d < 0):
print('reversing!')
i.reverse()
d *= -1
for z in range(3):
i[z] = 9 - i[z]
i = sorted(i)
i.reverse()
print('normalized: %s' % str(i))
ir = 0
for (ind, z) in enumerate(i):
print('reached: %d' % ir)
if ir >= d:
print(ind)
break
ir += z
``` | 0 | |
960 | B | Minimize the error | PROGRAMMING | 1,500 | [
"data structures",
"greedy",
"sortings"
] | null | null | You are given two arrays *A* and *B*, each of size *n*. The error, *E*, between these two arrays is defined . You have to perform exactly *k*1 operations on array *A* and exactly *k*2 operations on array *B*. In one operation, you have to choose one element of the array and increase or decrease it by 1.
Output the minimum possible value of error after *k*1 operations on array *A* and *k*2 operations on array *B* have been performed. | The first line contains three space-separated integers *n* (1<=≤<=*n*<=≤<=103), *k*1 and *k*2 (0<=≤<=*k*1<=+<=*k*2<=≤<=103, *k*1 and *k*2 are non-negative) — size of arrays and number of operations to perform on *A* and *B* respectively.
Second line contains *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=106<=≤<=*a**i*<=≤<=106) — array *A*.
Third line contains *n* space separated integers *b*1,<=*b*2,<=...,<=*b**n* (<=-<=106<=≤<=*b**i*<=≤<=106)— array *B*. | Output a single integer — the minimum possible value of after doing exactly *k*1 operations on array *A* and exactly *k*2 operations on array *B*. | [
"2 0 0\n1 2\n2 3\n",
"2 1 0\n1 2\n2 2\n",
"2 5 7\n3 4\n14 4\n"
] | [
"2",
"0",
"1"
] | In the first sample case, we cannot perform any operations on *A* or *B*. Therefore the minimum possible error *E* = (1 - 2)<sup class="upper-index">2</sup> + (2 - 3)<sup class="upper-index">2</sup> = 2.
In the second sample case, we are required to perform exactly one operation on *A*. In order to minimize error, we increment the first element of *A* by 1. Now, *A* = [2, 2]. The error is now *E* = (2 - 2)<sup class="upper-index">2</sup> + (2 - 2)<sup class="upper-index">2</sup> = 0. This is the minimum possible error obtainable.
In the third sample case, we can increase the first element of *A* to 8, using the all of the 5 moves available to us. Also, the first element of *B* can be reduced to 8 using the 6 of the 7 available moves. Now *A* = [8, 4] and *B* = [8, 4]. The error is now *E* = (8 - 8)<sup class="upper-index">2</sup> + (4 - 4)<sup class="upper-index">2</sup> = 0, but we are still left with 1 move for array *B*. Increasing the second element of *B* to 5 using the left move, we get *B* = [8, 5] and *E* = (8 - 8)<sup class="upper-index">2</sup> + (4 - 5)<sup class="upper-index">2</sup> = 1. | 1,000 | [
{
"input": "2 0 0\n1 2\n2 3",
"output": "2"
},
{
"input": "2 1 0\n1 2\n2 2",
"output": "0"
},
{
"input": "2 5 7\n3 4\n14 4",
"output": "1"
},
{
"input": "2 0 1\n1 2\n2 2",
"output": "0"
},
{
"input": "2 1 1\n0 0\n1 1",
"output": "0"
},
{
"input": "5 5 5\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "3 4 5\n1 2 3\n3 2 1",
"output": "1"
},
{
"input": "3 1000 0\n1 2 3\n-1000 -1000 -1000",
"output": "1341346"
},
{
"input": "10 300 517\n-6 -2 6 5 -3 8 9 -10 8 6\n5 -9 -2 6 1 4 6 -2 5 -3",
"output": "1"
},
{
"input": "10 819 133\n87 22 30 89 82 -97 -52 25 76 -22\n-20 95 21 25 2 -3 45 -7 -98 -56",
"output": "0"
},
{
"input": "10 10 580\n302 -553 -281 -299 -270 -890 -989 -749 -418 486\n735 330 6 725 -984 209 -855 -786 -502 967",
"output": "2983082"
},
{
"input": "10 403 187\n9691 -3200 3016 3540 -9475 8840 -4705 7940 6293 -2631\n-2288 9129 4067 696 -6754 9869 -5747 701 3344 -3426",
"output": "361744892"
},
{
"input": "10 561 439\n76639 67839 10670 -23 -18393 65114 46538 67596 86615 90480\n50690 620 -33631 -75857 75634 91321 -81662 -93668 -98557 -43621",
"output": "116776723778"
},
{
"input": "10 765 62\n-929885 -995154 254071 -370672 -435272 584846 -301610 -234118 -82557 743536\n-36327 439149 -977780 -821019 -585558 953598 -151943 140715 -311253 -383103",
"output": "6216649853365"
},
{
"input": "22 334 246\n-462653 -618002 4973 -348485 366658 192390 274752 200590 138367 779540 -661269 642587 113645 -110388 -604418 -491231 -933401 -219332 -603140 836439 167007 210226\n357638 -646669 -558432 -434313 -285851 -119233 323088 -512237 -729293 215256 39316 -984201 -209814 715016 -271932 796550 988227 -89577 67202 462973 -942079 -823339",
"output": "15389604923763"
},
{
"input": "1 1000 0\n1000000\n-1000000",
"output": "3996001000000"
},
{
"input": "1 1000 0\n1000000\n1000000",
"output": "0"
},
{
"input": "1 0 1000\n-1000000\n1000000",
"output": "3996001000000"
},
{
"input": "1 0 1000\n1000000\n1000000",
"output": "0"
},
{
"input": "1 5 7\n1\n2",
"output": "1"
},
{
"input": "2 1 3\n2 2\n2 2",
"output": "0"
},
{
"input": "3 3 0\n1 1 1\n1 1 1",
"output": "1"
},
{
"input": "1 0 6\n0\n0",
"output": "0"
},
{
"input": "2 1 1\n2 2\n0 0",
"output": "2"
},
{
"input": "3 1 4\n0 0 0\n1 5 6",
"output": "19"
},
{
"input": "1 1 2\n0\n8",
"output": "25"
},
{
"input": "2 2 0\n3 3\n3 3",
"output": "0"
}
] | 1,523,124,186 | 7,086 | Python 3 | WRONG_ANSWER | PRETESTS | 8 | 124 | 7,065,600 | n, k1, k2=map(int,input().split())
A=[int(x) for x in input().split()]
B=[int(x) for x in input().split()]
e=0
for i in range(k1+k2):
mx=max(A)
mx2=max(B)
a=A.index(mx)
sum1=abs(A[a]-B[a])
b=B.index(mx2)
sum2=abs(A[b]-B[b])
if sum2>=sum1:
if(B[b]>A[b]):
B[b]-=1
else:
A[b]-=1
else:
if A[a]>B[a]:
A[a]-=1
else:
B[a]-=1
#print('sum=',sum1,sum2)
for i in range(n):
e+=(A[i]-B[i])**2
print(e)
| Title: Minimize the error
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two arrays *A* and *B*, each of size *n*. The error, *E*, between these two arrays is defined . You have to perform exactly *k*1 operations on array *A* and exactly *k*2 operations on array *B*. In one operation, you have to choose one element of the array and increase or decrease it by 1.
Output the minimum possible value of error after *k*1 operations on array *A* and *k*2 operations on array *B* have been performed.
Input Specification:
The first line contains three space-separated integers *n* (1<=≤<=*n*<=≤<=103), *k*1 and *k*2 (0<=≤<=*k*1<=+<=*k*2<=≤<=103, *k*1 and *k*2 are non-negative) — size of arrays and number of operations to perform on *A* and *B* respectively.
Second line contains *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=106<=≤<=*a**i*<=≤<=106) — array *A*.
Third line contains *n* space separated integers *b*1,<=*b*2,<=...,<=*b**n* (<=-<=106<=≤<=*b**i*<=≤<=106)— array *B*.
Output Specification:
Output a single integer — the minimum possible value of after doing exactly *k*1 operations on array *A* and exactly *k*2 operations on array *B*.
Demo Input:
['2 0 0\n1 2\n2 3\n', '2 1 0\n1 2\n2 2\n', '2 5 7\n3 4\n14 4\n']
Demo Output:
['2', '0', '1']
Note:
In the first sample case, we cannot perform any operations on *A* or *B*. Therefore the minimum possible error *E* = (1 - 2)<sup class="upper-index">2</sup> + (2 - 3)<sup class="upper-index">2</sup> = 2.
In the second sample case, we are required to perform exactly one operation on *A*. In order to minimize error, we increment the first element of *A* by 1. Now, *A* = [2, 2]. The error is now *E* = (2 - 2)<sup class="upper-index">2</sup> + (2 - 2)<sup class="upper-index">2</sup> = 0. This is the minimum possible error obtainable.
In the third sample case, we can increase the first element of *A* to 8, using the all of the 5 moves available to us. Also, the first element of *B* can be reduced to 8 using the 6 of the 7 available moves. Now *A* = [8, 4] and *B* = [8, 4]. The error is now *E* = (8 - 8)<sup class="upper-index">2</sup> + (4 - 4)<sup class="upper-index">2</sup> = 0, but we are still left with 1 move for array *B*. Increasing the second element of *B* to 5 using the left move, we get *B* = [8, 5] and *E* = (8 - 8)<sup class="upper-index">2</sup> + (4 - 5)<sup class="upper-index">2</sup> = 1. | ```python
n, k1, k2=map(int,input().split())
A=[int(x) for x in input().split()]
B=[int(x) for x in input().split()]
e=0
for i in range(k1+k2):
mx=max(A)
mx2=max(B)
a=A.index(mx)
sum1=abs(A[a]-B[a])
b=B.index(mx2)
sum2=abs(A[b]-B[b])
if sum2>=sum1:
if(B[b]>A[b]):
B[b]-=1
else:
A[b]-=1
else:
if A[a]>B[a]:
A[a]-=1
else:
B[a]-=1
#print('sum=',sum1,sum2)
for i in range(n):
e+=(A[i]-B[i])**2
print(e)
``` | 0 | |
967 | A | Mind the Gap | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | These days Arkady works as an air traffic controller at a large airport. He controls a runway which is usually used for landings only. Thus, he has a schedule of planes that are landing in the nearest future, each landing lasts $1$ minute.
He was asked to insert one takeoff in the schedule. The takeoff takes $1$ minute itself, but for safety reasons there should be a time space between the takeoff and any landing of at least $s$ minutes from both sides.
Find the earliest time when Arkady can insert the takeoff. | The first line of input contains two integers $n$ and $s$ ($1 \le n \le 100$, $1 \le s \le 60$) — the number of landings on the schedule and the minimum allowed time (in minutes) between a landing and a takeoff.
Each of next $n$ lines contains two integers $h$ and $m$ ($0 \le h \le 23$, $0 \le m \le 59$) — the time, in hours and minutes, when a plane will land, starting from current moment (i. e. the current time is $0$ $0$). These times are given in increasing order. | Print two integers $h$ and $m$ — the hour and the minute from the current moment of the earliest time Arkady can insert the takeoff. | [
"6 60\n0 0\n1 20\n3 21\n5 0\n19 30\n23 40\n",
"16 50\n0 30\n1 20\n3 0\n4 30\n6 10\n7 50\n9 30\n11 10\n12 50\n14 30\n16 10\n17 50\n19 30\n21 10\n22 50\n23 59\n",
"3 17\n0 30\n1 0\n12 0\n"
] | [
"6 1\n",
"24 50\n",
"0 0\n"
] | In the first example note that there is not enough time between 1:20 and 3:21, because each landing and the takeoff take one minute.
In the second example there is no gaps in the schedule, so Arkady can only add takeoff after all landings. Note that it is possible that one should wait more than $24$ hours to insert the takeoff.
In the third example Arkady can insert the takeoff even between the first landing. | 500 | [
{
"input": "6 60\n0 0\n1 20\n3 21\n5 0\n19 30\n23 40",
"output": "6 1"
},
{
"input": "16 50\n0 30\n1 20\n3 0\n4 30\n6 10\n7 50\n9 30\n11 10\n12 50\n14 30\n16 10\n17 50\n19 30\n21 10\n22 50\n23 59",
"output": "24 50"
},
{
"input": "3 17\n0 30\n1 0\n12 0",
"output": "0 0"
},
{
"input": "24 60\n0 21\n2 21\n2 46\n3 17\n4 15\n5 43\n6 41\n7 50\n8 21\n9 8\n10 31\n10 45\n12 30\n14 8\n14 29\n14 32\n14 52\n15 16\n16 7\n16 52\n18 44\n20 25\n21 13\n22 7",
"output": "23 8"
},
{
"input": "20 60\n0 9\n0 19\n0 57\n2 42\n3 46\n3 47\n5 46\n8 1\n9 28\n9 41\n10 54\n12 52\n13 0\n14 49\n17 28\n17 39\n19 34\n20 52\n21 35\n23 22",
"output": "6 47"
},
{
"input": "57 20\n0 2\n0 31\n1 9\n1 42\n1 58\n2 4\n2 35\n2 49\n3 20\n3 46\n4 23\n4 52\n5 5\n5 39\n6 7\n6 48\n6 59\n7 8\n7 35\n8 10\n8 46\n8 53\n9 19\n9 33\n9 43\n10 18\n10 42\n11 0\n11 26\n12 3\n12 5\n12 30\n13 1\n13 38\n14 13\n14 54\n15 31\n16 5\n16 44\n17 18\n17 30\n17 58\n18 10\n18 34\n19 13\n19 49\n19 50\n19 59\n20 17\n20 23\n20 40\n21 18\n21 57\n22 31\n22 42\n22 56\n23 37",
"output": "23 58"
},
{
"input": "66 20\n0 16\n0 45\n0 58\n1 6\n1 19\n2 7\n2 9\n3 9\n3 25\n3 57\n4 38\n4 58\n5 21\n5 40\n6 16\n6 19\n6 58\n7 6\n7 26\n7 51\n8 13\n8 36\n8 55\n9 1\n9 15\n9 33\n10 12\n10 37\n11 15\n11 34\n12 8\n12 37\n12 55\n13 26\n14 0\n14 34\n14 36\n14 48\n15 23\n15 29\n15 43\n16 8\n16 41\n16 45\n17 5\n17 7\n17 15\n17 29\n17 46\n18 12\n18 19\n18 38\n18 57\n19 32\n19 58\n20 5\n20 40\n20 44\n20 50\n21 18\n21 49\n22 18\n22 47\n23 1\n23 38\n23 50",
"output": "1 40"
},
{
"input": "1 1\n0 0",
"output": "0 2"
},
{
"input": "10 1\n0 2\n0 4\n0 5\n0 8\n0 9\n0 11\n0 13\n0 16\n0 19\n0 21",
"output": "0 0"
},
{
"input": "10 1\n0 2\n0 5\n0 8\n0 11\n0 15\n0 17\n0 25\n0 28\n0 29\n0 32",
"output": "0 0"
},
{
"input": "15 20\n0 47\n2 24\n4 19\n4 34\n5 46\n8 15\n9 8\n10 28\n17 47\n17 52\n18 32\n19 50\n20 46\n20 50\n23 21",
"output": "0 0"
},
{
"input": "1 5\n1 0",
"output": "0 0"
},
{
"input": "24 60\n1 0\n2 0\n3 0\n4 0\n5 0\n6 0\n7 0\n8 0\n9 0\n10 0\n11 0\n12 0\n13 0\n14 0\n15 0\n16 0\n17 0\n18 0\n19 0\n20 0\n21 0\n22 0\n23 0\n23 59",
"output": "25 0"
},
{
"input": "1 30\n0 29",
"output": "1 0"
},
{
"input": "1 2\n3 0",
"output": "0 0"
},
{
"input": "16 60\n0 30\n1 20\n3 0\n4 30\n6 10\n7 50\n9 30\n11 10\n12 50\n14 30\n16 10\n17 50\n19 30\n21 10\n22 50\n23 59",
"output": "25 0"
},
{
"input": "1 5\n0 6",
"output": "0 0"
},
{
"input": "2 60\n0 59\n23 59",
"output": "2 0"
},
{
"input": "1 58\n0 1",
"output": "1 0"
},
{
"input": "25 60\n0 0\n1 0\n2 0\n3 0\n4 0\n5 0\n6 0\n7 0\n8 0\n9 0\n10 0\n11 0\n12 0\n13 0\n14 0\n15 0\n16 0\n17 0\n18 0\n19 0\n20 0\n21 0\n22 0\n23 0\n23 59",
"output": "25 0"
},
{
"input": "2 3\n0 3\n0 30",
"output": "0 7"
},
{
"input": "16 50\n0 30\n1 20\n3 0\n4 30\n6 10\n7 50\n9 30\n11 10\n12 50\n14 30\n16 10\n17 50\n19 30\n21 10\n22 50\n23 9",
"output": "24 0"
},
{
"input": "1 60\n2 0",
"output": "0 0"
},
{
"input": "2 60\n0 0\n5 0",
"output": "1 1"
},
{
"input": "1 30\n0 31",
"output": "0 0"
},
{
"input": "2 60\n0 59\n3 1",
"output": "2 0"
},
{
"input": "2 60\n0 59\n5 0",
"output": "2 0"
},
{
"input": "1 59\n0 0",
"output": "1 0"
},
{
"input": "3 25\n0 0\n1 0\n2 0",
"output": "0 26"
},
{
"input": "1 2\n2 3",
"output": "0 0"
}
] | 1,525,009,348 | 1,648 | Python 3 | OK | TESTS | 34 | 92 | 7,065,600 | rd = lambda: map(int, input().split())
n, s = rd()
a = [- s - 1]
for _ in range(n):
h, m = rd()
a.append(h * 60 + m)
x = 0
for i in range(1, len(a)):
if a[i - 1] + s + 1 <= x <= a[i] - s - 1:
break
x = a[i] + s + 1
print(x // 60, x % 60)
| Title: Mind the Gap
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
These days Arkady works as an air traffic controller at a large airport. He controls a runway which is usually used for landings only. Thus, he has a schedule of planes that are landing in the nearest future, each landing lasts $1$ minute.
He was asked to insert one takeoff in the schedule. The takeoff takes $1$ minute itself, but for safety reasons there should be a time space between the takeoff and any landing of at least $s$ minutes from both sides.
Find the earliest time when Arkady can insert the takeoff.
Input Specification:
The first line of input contains two integers $n$ and $s$ ($1 \le n \le 100$, $1 \le s \le 60$) — the number of landings on the schedule and the minimum allowed time (in minutes) between a landing and a takeoff.
Each of next $n$ lines contains two integers $h$ and $m$ ($0 \le h \le 23$, $0 \le m \le 59$) — the time, in hours and minutes, when a plane will land, starting from current moment (i. e. the current time is $0$ $0$). These times are given in increasing order.
Output Specification:
Print two integers $h$ and $m$ — the hour and the minute from the current moment of the earliest time Arkady can insert the takeoff.
Demo Input:
['6 60\n0 0\n1 20\n3 21\n5 0\n19 30\n23 40\n', '16 50\n0 30\n1 20\n3 0\n4 30\n6 10\n7 50\n9 30\n11 10\n12 50\n14 30\n16 10\n17 50\n19 30\n21 10\n22 50\n23 59\n', '3 17\n0 30\n1 0\n12 0\n']
Demo Output:
['6 1\n', '24 50\n', '0 0\n']
Note:
In the first example note that there is not enough time between 1:20 and 3:21, because each landing and the takeoff take one minute.
In the second example there is no gaps in the schedule, so Arkady can only add takeoff after all landings. Note that it is possible that one should wait more than $24$ hours to insert the takeoff.
In the third example Arkady can insert the takeoff even between the first landing. | ```python
rd = lambda: map(int, input().split())
n, s = rd()
a = [- s - 1]
for _ in range(n):
h, m = rd()
a.append(h * 60 + m)
x = 0
for i in range(1, len(a)):
if a[i - 1] + s + 1 <= x <= a[i] - s - 1:
break
x = a[i] + s + 1
print(x // 60, x % 60)
``` | 3 | |
25 | D | Roads not only in Berland | PROGRAMMING | 1,900 | [
"dsu",
"graphs",
"trees"
] | D. Roads not only in Berland | 2 | 256 | Berland Government decided to improve relations with neighboring countries. First of all, it was decided to build new roads so that from each city of Berland and neighboring countries it became possible to reach all the others. There are *n* cities in Berland and neighboring countries in total and exactly *n*<=-<=1 two-way roads. Because of the recent financial crisis, the Berland Government is strongly pressed for money, so to build a new road it has to close some of the existing ones. Every day it is possible to close one existing road and immediately build a new one. Your task is to determine how many days would be needed to rebuild roads so that from each city it became possible to reach all the others, and to draw a plan of closure of old roads and building of new ones. | The first line contains integer *n* (2<=≤<=*n*<=≤<=1000) — amount of cities in Berland and neighboring countries. Next *n*<=-<=1 lines contain the description of roads. Each road is described by two space-separated integers *a**i*, *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*) — pair of cities, which the road connects. It can't be more than one road between a pair of cities. No road connects the city with itself. | Output the answer, number *t* — what is the least amount of days needed to rebuild roads so that from each city it became possible to reach all the others. Then output *t* lines — the plan of closure of old roads and building of new ones. Each line should describe one day in the format i j u v — it means that road between cities i and j became closed and a new road between cities u and v is built. Cities are numbered from 1. If the answer is not unique, output any. | [
"2\n1 2\n",
"7\n1 2\n2 3\n3 1\n4 5\n5 6\n6 7\n"
] | [
"0\n",
"1\n3 1 3 7\n"
] | none | 0 | [
{
"input": "2\n1 2",
"output": "0"
},
{
"input": "7\n1 2\n2 3\n3 1\n4 5\n5 6\n6 7",
"output": "1\n3 1 3 7"
},
{
"input": "3\n3 2\n1 2",
"output": "0"
},
{
"input": "3\n3 1\n3 2",
"output": "0"
},
{
"input": "4\n1 4\n3 1\n3 4",
"output": "1\n3 4 2 4"
},
{
"input": "5\n4 1\n4 3\n5 3\n2 4",
"output": "0"
},
{
"input": "6\n5 2\n5 3\n1 4\n3 1\n5 6",
"output": "0"
},
{
"input": "10\n5 9\n8 5\n7 6\n7 9\n3 9\n2 1\n7 2\n3 6\n7 1",
"output": "2\n3 6 1 4\n7 1 4 10"
},
{
"input": "21\n7 15\n13 1\n14 3\n4 10\n2 3\n16 18\n17 20\n16 20\n8 4\n3 12\n2 17\n13 11\n16 1\n13 2\n13 5\n8 9\n6 14\n3 17\n16 9\n13 8",
"output": "3\n13 2 9 15\n3 17 15 19\n13 8 19 21"
},
{
"input": "39\n6 13\n15 39\n10 35\n31 28\n4 21\n12 39\n3 7\n3 13\n6 1\n5 14\n36 28\n12 15\n18 38\n30 29\n19 34\n36 16\n20 22\n8 13\n38 32\n26 39\n21 37\n1 7\n15 27\n12 26\n8 3\n6 14\n29 2\n25 23\n32 21\n5 16\n32 25\n6 8\n13 10\n23 30\n34 37\n29 33\n28 14\n36 5",
"output": "7\n12 15 9 11\n1 7 11 17\n12 26 17 22\n8 3 22 24\n6 8 24 27\n28 14 27 33\n36 5 33 35"
},
{
"input": "60\n17 34\n46 22\n50 44\n46 33\n41 21\n31 33\n48 6\n38 19\n35 60\n2 24\n49 29\n7 53\n34 1\n19 55\n32 1\n31 42\n27 28\n4 53\n6 50\n21 34\n1 10\n12 36\n54 8\n16 13\n51 43\n45 51\n54 20\n13 53\n34 33\n49 33\n51 11\n59 34\n15 5\n59 28\n30 39\n13 30\n58 4\n34 14\n3 9\n19 34\n4 18\n26 48\n56 54\n3 43\n57 25\n3 41\n35 3\n48 44\n19 13\n54 1\n23 31\n59 47\n5 1\n46 40\n6 26\n20 25\n37 5\n17 24\n20 52",
"output": "2\n48 44 36 44\n6 26 44 52"
}
] | 1,631,679,513 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 154 | 7,475,200 | import bisect
import heapq
import math
import collections
import sys
import copy
from functools import reduce
import decimal
from io import BytesIO, IOBase
import os
import itertools
import functools
from types import GeneratorType
sys.setrecursionlimit(10 ** 9)
decimal.getcontext().rounding = decimal.ROUND_HALF_UP
graphDict = collections.defaultdict
queue = collections.deque
################## pypy deep recursion handling ##############
# Author = @pajenegod
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
to = f(*args, **kwargs)
if stack:
return to
else:
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
return to
to = stack[-1].send(to)
return wrappedfunc
################## Graphs ###################
class Graphs:
def __init__(self):
self.graph = graphDict(set)
def add_edge(self, u, v):
self.graph[u].add(v)
self.graph[v].add(u)
def dfs_utility(self, nodes, visited_nodes):
visited_nodes.add(nodes)
for neighbour in self.graph[nodes]:
if neighbour not in visited_nodes:
self.dfs_utility(neighbour, visited_nodes)
def dfs(self, node):
Visited = set()
self.dfs_utility(i, Visited)
def bfs(self, node, f_node):
count = float("inf")
visited = set()
level = 0
if node not in visited:
queue.append([node, level])
visited.add(node)
flag = 0
while queue:
parent = queue.popleft()
if parent[0] == f_node:
flag = 1
count = min(count, parent[1])
level = parent[1] + 1
for item in self.graph[parent[0]]:
if item not in visited:
queue.append([item, level])
visited.add(item)
return count if flag else -1
################### Tree Implementaion ##############
class Tree:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
def inorder(node, lis):
if node:
inorder(node.left, lis)
lis.append(node.data)
inorder(node.right, lis)
return lis
def leaf_node_sum(root):
if root is None:
return 0
if root.left is None and root.right is None:
return root.data
return leaf_node_sum(root.left) + leaf_node_sum(root.right)
def hight(root):
if root is None:
return -1
if root.left is None and root.right is None:
return 0
return max(hight(root.left), hight(root.right)) + 1
################## Union Find #######################
class DisjointSets:
def __init__(self, size: int) -> None:
self.parents = [-1 for _ in range(size)]
self.sizes = [1 for _ in range(size)]
self._count = size
# finds the "representative" node in a's component
@bootstrap
def find(self, x: int) -> int:
if self.parents[x] == -1:
yield x
self.parents[x] = yield self.find(self.parents[x])
yield self.parents[x]
# returns whether the merge changed connectivity
def union(self, x: int, y: int) -> bool:
x_root = self.find(x)
y_root = self.find(y)
if x_root == y_root:
return False
self._count -= 1
if self.sizes[x_root] < self.sizes[y_root]:
x_root, y_root = y_root, x_root
self.parents[y_root] = x_root
self.sizes[x_root] += self.sizes[y_root]
return True
# returns whether two nodes are in the same connected component
def connected(self, x: int, y: int) -> bool:
return self.find(x) == self.find(y)
def count(self):
return self._count
def add_roads(self):
return set(self.parents)
#################################################
def rounding(n):
return int(decimal.Decimal(f'{n}').to_integral_value())
def factors(n):
return set(reduce(list.__add__,
([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0), []))
def p_sum(array):
return list(itertools.accumulate(array))
def diophantine(a: int, b: int, c: int):
d, x, y = extended_gcd(a, b)
r = c // d
return r * x, r * y
@bootstrap
def extended_gcd(a: int, b: int):
if b == 0:
d, x, y = a, 1, 0
else:
(d, p, q) = yield extended_gcd(b, a % b)
x = q
y = p - q * (a // b)
yield d, x, y
################################ <fast I/O> ###########################################
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self, **kwargs):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
#############################################<I/O Region >##############################################
def inp():
return sys.stdin.readline().strip()
def map_inp(v_type):
return map(v_type, inp().split())
def list_inp(v_type):
return list(map_inp(v_type))
######################################## Solution ####################################
n = int(inp())
roads = DisjointSets(n)
delete = []
for i in range(n - 1):
a, b = map_inp(int)
if roads.connected(a-1, b-1):
delete.append([a, b])
continue
roads.union(a-1, b-1)
print(len(delete))
new = list(roads.add_roads())
for i in range(len(delete)):
print(delete[i][0], delete[i][1], new[i % len(new)]+1, new[(i + 1) % len(new)]+1)
| Title: Roads not only in Berland
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Berland Government decided to improve relations with neighboring countries. First of all, it was decided to build new roads so that from each city of Berland and neighboring countries it became possible to reach all the others. There are *n* cities in Berland and neighboring countries in total and exactly *n*<=-<=1 two-way roads. Because of the recent financial crisis, the Berland Government is strongly pressed for money, so to build a new road it has to close some of the existing ones. Every day it is possible to close one existing road and immediately build a new one. Your task is to determine how many days would be needed to rebuild roads so that from each city it became possible to reach all the others, and to draw a plan of closure of old roads and building of new ones.
Input Specification:
The first line contains integer *n* (2<=≤<=*n*<=≤<=1000) — amount of cities in Berland and neighboring countries. Next *n*<=-<=1 lines contain the description of roads. Each road is described by two space-separated integers *a**i*, *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*) — pair of cities, which the road connects. It can't be more than one road between a pair of cities. No road connects the city with itself.
Output Specification:
Output the answer, number *t* — what is the least amount of days needed to rebuild roads so that from each city it became possible to reach all the others. Then output *t* lines — the plan of closure of old roads and building of new ones. Each line should describe one day in the format i j u v — it means that road between cities i and j became closed and a new road between cities u and v is built. Cities are numbered from 1. If the answer is not unique, output any.
Demo Input:
['2\n1 2\n', '7\n1 2\n2 3\n3 1\n4 5\n5 6\n6 7\n']
Demo Output:
['0\n', '1\n3 1 3 7\n']
Note:
none | ```python
import bisect
import heapq
import math
import collections
import sys
import copy
from functools import reduce
import decimal
from io import BytesIO, IOBase
import os
import itertools
import functools
from types import GeneratorType
sys.setrecursionlimit(10 ** 9)
decimal.getcontext().rounding = decimal.ROUND_HALF_UP
graphDict = collections.defaultdict
queue = collections.deque
################## pypy deep recursion handling ##############
# Author = @pajenegod
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
to = f(*args, **kwargs)
if stack:
return to
else:
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
return to
to = stack[-1].send(to)
return wrappedfunc
################## Graphs ###################
class Graphs:
def __init__(self):
self.graph = graphDict(set)
def add_edge(self, u, v):
self.graph[u].add(v)
self.graph[v].add(u)
def dfs_utility(self, nodes, visited_nodes):
visited_nodes.add(nodes)
for neighbour in self.graph[nodes]:
if neighbour not in visited_nodes:
self.dfs_utility(neighbour, visited_nodes)
def dfs(self, node):
Visited = set()
self.dfs_utility(i, Visited)
def bfs(self, node, f_node):
count = float("inf")
visited = set()
level = 0
if node not in visited:
queue.append([node, level])
visited.add(node)
flag = 0
while queue:
parent = queue.popleft()
if parent[0] == f_node:
flag = 1
count = min(count, parent[1])
level = parent[1] + 1
for item in self.graph[parent[0]]:
if item not in visited:
queue.append([item, level])
visited.add(item)
return count if flag else -1
################### Tree Implementaion ##############
class Tree:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
def inorder(node, lis):
if node:
inorder(node.left, lis)
lis.append(node.data)
inorder(node.right, lis)
return lis
def leaf_node_sum(root):
if root is None:
return 0
if root.left is None and root.right is None:
return root.data
return leaf_node_sum(root.left) + leaf_node_sum(root.right)
def hight(root):
if root is None:
return -1
if root.left is None and root.right is None:
return 0
return max(hight(root.left), hight(root.right)) + 1
################## Union Find #######################
class DisjointSets:
def __init__(self, size: int) -> None:
self.parents = [-1 for _ in range(size)]
self.sizes = [1 for _ in range(size)]
self._count = size
# finds the "representative" node in a's component
@bootstrap
def find(self, x: int) -> int:
if self.parents[x] == -1:
yield x
self.parents[x] = yield self.find(self.parents[x])
yield self.parents[x]
# returns whether the merge changed connectivity
def union(self, x: int, y: int) -> bool:
x_root = self.find(x)
y_root = self.find(y)
if x_root == y_root:
return False
self._count -= 1
if self.sizes[x_root] < self.sizes[y_root]:
x_root, y_root = y_root, x_root
self.parents[y_root] = x_root
self.sizes[x_root] += self.sizes[y_root]
return True
# returns whether two nodes are in the same connected component
def connected(self, x: int, y: int) -> bool:
return self.find(x) == self.find(y)
def count(self):
return self._count
def add_roads(self):
return set(self.parents)
#################################################
def rounding(n):
return int(decimal.Decimal(f'{n}').to_integral_value())
def factors(n):
return set(reduce(list.__add__,
([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0), []))
def p_sum(array):
return list(itertools.accumulate(array))
def diophantine(a: int, b: int, c: int):
d, x, y = extended_gcd(a, b)
r = c // d
return r * x, r * y
@bootstrap
def extended_gcd(a: int, b: int):
if b == 0:
d, x, y = a, 1, 0
else:
(d, p, q) = yield extended_gcd(b, a % b)
x = q
y = p - q * (a // b)
yield d, x, y
################################ <fast I/O> ###########################################
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self, **kwargs):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
#############################################<I/O Region >##############################################
def inp():
return sys.stdin.readline().strip()
def map_inp(v_type):
return map(v_type, inp().split())
def list_inp(v_type):
return list(map_inp(v_type))
######################################## Solution ####################################
n = int(inp())
roads = DisjointSets(n)
delete = []
for i in range(n - 1):
a, b = map_inp(int)
if roads.connected(a-1, b-1):
delete.append([a, b])
continue
roads.union(a-1, b-1)
print(len(delete))
new = list(roads.add_roads())
for i in range(len(delete)):
print(delete[i][0], delete[i][1], new[i % len(new)]+1, new[(i + 1) % len(new)]+1)
``` | 0 |
962 | A | Equator | PROGRAMMING | 1,300 | [
"implementation"
] | null | null | Polycarp has created his own training plan to prepare for the programming contests. He will train for $n$ days, all days are numbered from $1$ to $n$, beginning from the first.
On the $i$-th day Polycarp will necessarily solve $a_i$ problems. One evening Polycarp plans to celebrate the equator. He will celebrate it on the first evening of such a day that from the beginning of the training and to this day inclusive he will solve half or more of all the problems.
Determine the index of day when Polycarp will celebrate the equator. | The first line contains a single integer $n$ ($1 \le n \le 200\,000$) — the number of days to prepare for the programming contests.
The second line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10\,000$), where $a_i$ equals to the number of problems, which Polycarp will solve on the $i$-th day. | Print the index of the day when Polycarp will celebrate the equator. | [
"4\n1 3 2 1\n",
"6\n2 2 2 2 2 2\n"
] | [
"2\n",
"3\n"
] | In the first example Polycarp will celebrate the equator on the evening of the second day, because up to this day (inclusive) he will solve $4$ out of $7$ scheduled problems on four days of the training.
In the second example Polycarp will celebrate the equator on the evening of the third day, because up to this day (inclusive) he will solve $6$ out of $12$ scheduled problems on six days of the training. | 0 | [
{
"input": "4\n1 3 2 1",
"output": "2"
},
{
"input": "6\n2 2 2 2 2 2",
"output": "3"
},
{
"input": "1\n10000",
"output": "1"
},
{
"input": "3\n2 1 1",
"output": "1"
},
{
"input": "2\n1 3",
"output": "2"
},
{
"input": "4\n2 1 1 3",
"output": "3"
},
{
"input": "3\n1 1 3",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "2"
},
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n2 1 2",
"output": "2"
},
{
"input": "5\n1 2 4 3 5",
"output": "4"
},
{
"input": "5\n2 2 2 4 3",
"output": "4"
},
{
"input": "4\n1 2 3 1",
"output": "3"
},
{
"input": "6\n7 3 10 7 3 11",
"output": "4"
},
{
"input": "2\n3 4",
"output": "2"
},
{
"input": "5\n1 1 1 1 1",
"output": "3"
},
{
"input": "4\n1 3 2 3",
"output": "3"
},
{
"input": "2\n2 3",
"output": "2"
},
{
"input": "3\n32 10 23",
"output": "2"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "4"
},
{
"input": "3\n1 2 4",
"output": "3"
},
{
"input": "6\n3 3 3 2 4 4",
"output": "4"
},
{
"input": "9\n1 1 1 1 1 1 1 1 1",
"output": "5"
},
{
"input": "5\n1 3 3 1 1",
"output": "3"
},
{
"input": "4\n1 1 1 2",
"output": "3"
},
{
"input": "4\n1 2 1 3",
"output": "3"
},
{
"input": "3\n2 2 1",
"output": "2"
},
{
"input": "4\n2 3 3 3",
"output": "3"
},
{
"input": "4\n3 2 3 3",
"output": "3"
},
{
"input": "4\n2 1 1 1",
"output": "2"
},
{
"input": "3\n2 1 4",
"output": "3"
},
{
"input": "2\n6 7",
"output": "2"
},
{
"input": "4\n3 3 4 3",
"output": "3"
},
{
"input": "4\n1 1 2 5",
"output": "4"
},
{
"input": "4\n1 8 7 3",
"output": "3"
},
{
"input": "6\n2 2 2 2 2 3",
"output": "4"
},
{
"input": "3\n2 2 5",
"output": "3"
},
{
"input": "4\n1 1 2 1",
"output": "3"
},
{
"input": "5\n1 1 2 2 3",
"output": "4"
},
{
"input": "5\n9 5 3 4 8",
"output": "3"
},
{
"input": "3\n3 3 1",
"output": "2"
},
{
"input": "4\n1 2 2 2",
"output": "3"
},
{
"input": "3\n1 3 5",
"output": "3"
},
{
"input": "4\n1 1 3 6",
"output": "4"
},
{
"input": "6\n1 2 1 1 1 1",
"output": "3"
},
{
"input": "3\n3 1 3",
"output": "2"
},
{
"input": "5\n3 4 5 1 2",
"output": "3"
},
{
"input": "11\n1 1 1 1 1 1 1 1 1 1 1",
"output": "6"
},
{
"input": "5\n3 1 2 5 2",
"output": "4"
},
{
"input": "4\n1 1 1 4",
"output": "4"
},
{
"input": "4\n2 6 1 10",
"output": "4"
},
{
"input": "4\n2 2 3 2",
"output": "3"
},
{
"input": "4\n4 2 2 1",
"output": "2"
},
{
"input": "6\n1 1 1 1 1 4",
"output": "5"
},
{
"input": "3\n3 2 2",
"output": "2"
},
{
"input": "6\n1 3 5 1 7 4",
"output": "5"
},
{
"input": "5\n1 2 4 8 16",
"output": "5"
},
{
"input": "5\n1 2 4 4 4",
"output": "4"
},
{
"input": "6\n4 2 1 2 3 1",
"output": "3"
},
{
"input": "4\n3 2 1 5",
"output": "3"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "3\n2 4 7",
"output": "3"
},
{
"input": "5\n1 1 1 1 3",
"output": "4"
},
{
"input": "3\n3 1 5",
"output": "3"
},
{
"input": "4\n1 2 3 7",
"output": "4"
},
{
"input": "3\n1 4 6",
"output": "3"
},
{
"input": "4\n2 1 2 2",
"output": "3"
},
{
"input": "2\n4 5",
"output": "2"
},
{
"input": "5\n1 2 1 2 1",
"output": "3"
},
{
"input": "3\n2 3 6",
"output": "3"
},
{
"input": "6\n1 1 4 1 1 5",
"output": "4"
},
{
"input": "5\n2 2 2 2 1",
"output": "3"
},
{
"input": "2\n5 6",
"output": "2"
},
{
"input": "4\n2 2 1 4",
"output": "3"
},
{
"input": "5\n2 2 3 4 4",
"output": "4"
},
{
"input": "4\n3 1 1 2",
"output": "2"
},
{
"input": "5\n3 4 1 4 5",
"output": "4"
},
{
"input": "4\n1 3 1 6",
"output": "4"
},
{
"input": "5\n1 1 1 2 2",
"output": "4"
},
{
"input": "4\n1 4 2 4",
"output": "3"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 8",
"output": "9"
},
{
"input": "4\n1 4 5 1",
"output": "3"
},
{
"input": "5\n1 1 1 1 5",
"output": "5"
},
{
"input": "4\n1 3 4 1",
"output": "3"
},
{
"input": "4\n2 2 2 3",
"output": "3"
},
{
"input": "4\n2 3 2 4",
"output": "3"
},
{
"input": "5\n2 2 1 2 2",
"output": "3"
},
{
"input": "3\n4 3 2",
"output": "2"
},
{
"input": "3\n6 5 2",
"output": "2"
},
{
"input": "69\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "35"
},
{
"input": "6\n1 1 1 1 1 2",
"output": "4"
},
{
"input": "5\n1 2 5 4 5",
"output": "4"
},
{
"input": "2\n9 10",
"output": "2"
},
{
"input": "3\n1 1 5",
"output": "3"
},
{
"input": "4\n3 4 3 5",
"output": "3"
},
{
"input": "4\n1 4 3 3",
"output": "3"
},
{
"input": "4\n7 1 3 4",
"output": "2"
},
{
"input": "3\n100 100 1",
"output": "2"
},
{
"input": "4\n5 2 2 2",
"output": "2"
}
] | 1,523,536,241 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | n = int(input())
a = list(map(int, input().split()))
h = sum(a)//2
for i in range(n);
if sum(a[:i+1])>=h:
print(i+1)
break | Title: Equator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has created his own training plan to prepare for the programming contests. He will train for $n$ days, all days are numbered from $1$ to $n$, beginning from the first.
On the $i$-th day Polycarp will necessarily solve $a_i$ problems. One evening Polycarp plans to celebrate the equator. He will celebrate it on the first evening of such a day that from the beginning of the training and to this day inclusive he will solve half or more of all the problems.
Determine the index of day when Polycarp will celebrate the equator.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 200\,000$) — the number of days to prepare for the programming contests.
The second line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10\,000$), where $a_i$ equals to the number of problems, which Polycarp will solve on the $i$-th day.
Output Specification:
Print the index of the day when Polycarp will celebrate the equator.
Demo Input:
['4\n1 3 2 1\n', '6\n2 2 2 2 2 2\n']
Demo Output:
['2\n', '3\n']
Note:
In the first example Polycarp will celebrate the equator on the evening of the second day, because up to this day (inclusive) he will solve $4$ out of $7$ scheduled problems on four days of the training.
In the second example Polycarp will celebrate the equator on the evening of the third day, because up to this day (inclusive) he will solve $6$ out of $12$ scheduled problems on six days of the training. | ```python
n = int(input())
a = list(map(int, input().split()))
h = sum(a)//2
for i in range(n);
if sum(a[:i+1])>=h:
print(i+1)
break
``` | -1 | |
556 | A | Case of the Zeros and Ones | PROGRAMMING | 900 | [
"greedy"
] | null | null | Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result.
Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number. | First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has.
The second line contains the string of length *n* consisting only from zeros and ones. | Output the minimum length of the string that may remain after applying the described operations several times. | [
"4\n1100\n",
"5\n01010\n",
"8\n11101111\n"
] | [
"0\n",
"1\n",
"6\n"
] | In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 250 | [
{
"input": "4\n1100",
"output": "0"
},
{
"input": "5\n01010",
"output": "1"
},
{
"input": "8\n11101111",
"output": "6"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n00",
"output": "2"
},
{
"input": "2\n01",
"output": "0"
},
{
"input": "2\n10",
"output": "0"
},
{
"input": "2\n11",
"output": "2"
},
{
"input": "3\n001",
"output": "1"
},
{
"input": "6\n110110",
"output": "2"
},
{
"input": "7\n0000011",
"output": "3"
},
{
"input": "6\n110010",
"output": "0"
},
{
"input": "6\n110100",
"output": "0"
},
{
"input": "3\n100",
"output": "1"
},
{
"input": "6\n010111",
"output": "2"
},
{
"input": "8\n01011100",
"output": "0"
},
{
"input": "6\n001011",
"output": "0"
},
{
"input": "7\n1110000",
"output": "1"
},
{
"input": "9\n011111101",
"output": "5"
}
] | 1,693,651,934 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 11 | 1,000 | 14,643,200 | j=int(input())
m=list(input())
i=1
while i<len(m):
if m[i-1]!=m[i]:
del m[i-1:i+1]
i-=2
i+=1
print(len(m))
| Title: Case of the Zeros and Ones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result.
Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number.
Input Specification:
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has.
The second line contains the string of length *n* consisting only from zeros and ones.
Output Specification:
Output the minimum length of the string that may remain after applying the described operations several times.
Demo Input:
['4\n1100\n', '5\n01010\n', '8\n11101111\n']
Demo Output:
['0\n', '1\n', '6\n']
Note:
In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
j=int(input())
m=list(input())
i=1
while i<len(m):
if m[i-1]!=m[i]:
del m[i-1:i+1]
i-=2
i+=1
print(len(m))
``` | 0 | |
318 | A | Even Odds | PROGRAMMING | 900 | [
"math"
] | null | null | Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first *n*. He writes down the following sequence of numbers: firstly all odd integers from 1 to *n* (in ascending order), then all even integers from 1 to *n* (also in ascending order). Help our hero to find out which number will stand at the position number *k*. | The only line of input contains integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier. | Print the number that will stand at the position number *k* after Volodya's manipulations. | [
"10 3\n",
"7 7\n"
] | [
"5",
"6"
] | In the first sample Volodya's sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5. | 500 | [
{
"input": "10 3",
"output": "5"
},
{
"input": "7 7",
"output": "6"
},
{
"input": "7 1",
"output": "1"
},
{
"input": "7 2",
"output": "3"
},
{
"input": "8 5",
"output": "2"
},
{
"input": "8 3",
"output": "5"
},
{
"input": "8 4",
"output": "7"
},
{
"input": "1000000000000 500000000001",
"output": "2"
},
{
"input": "999999999997 499999999999",
"output": "999999999997"
},
{
"input": "999999999999 999999999999",
"output": "999999999998"
},
{
"input": "1000000000000 1",
"output": "1"
},
{
"input": "999999999999 1",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1000000000000 1000000000000",
"output": "1000000000000"
},
{
"input": "1000000000000 500000000000",
"output": "999999999999"
},
{
"input": "1000000000000 499999999999",
"output": "999999999997"
},
{
"input": "999999999997 499999999998",
"output": "999999999995"
},
{
"input": "619234238 556154835",
"output": "493075432"
},
{
"input": "38151981 36650624",
"output": "35149266"
},
{
"input": "680402465 442571217",
"output": "204739968"
},
{
"input": "109135284 9408714",
"output": "18817427"
},
{
"input": "603701841 56038951",
"output": "112077901"
},
{
"input": "356764822 321510177",
"output": "286255532"
},
{
"input": "284911189 142190783",
"output": "284381565"
},
{
"input": "91028405 61435545",
"output": "31842684"
}
] | 1,693,015,235 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 62 | 0 | import math
def my_solution(n, k):
mid = math.ceil(n/2)
if k <= mid:
x = k + (k-1)
elif k > mid:
x = (k - (mid+1)+1) * 2
print(x)
n, k = map(int, input().split())
my_solution(n, k) | Title: Even Odds
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first *n*. He writes down the following sequence of numbers: firstly all odd integers from 1 to *n* (in ascending order), then all even integers from 1 to *n* (also in ascending order). Help our hero to find out which number will stand at the position number *k*.
Input Specification:
The only line of input contains integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print the number that will stand at the position number *k* after Volodya's manipulations.
Demo Input:
['10 3\n', '7 7\n']
Demo Output:
['5', '6']
Note:
In the first sample Volodya's sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5. | ```python
import math
def my_solution(n, k):
mid = math.ceil(n/2)
if k <= mid:
x = k + (k-1)
elif k > mid:
x = (k - (mid+1)+1) * 2
print(x)
n, k = map(int, input().split())
my_solution(n, k)
``` | 3 | |
571 | A | Lengthening Sticks | PROGRAMMING | 2,100 | [
"combinatorics",
"implementation",
"math"
] | null | null | You are given three sticks with positive integer lengths of *a*,<=*b*, and *c* centimeters. You can increase length of some of them by some positive integer number of centimeters (different sticks can be increased by a different length), but in total by at most *l* centimeters. In particular, it is allowed not to increase the length of any stick.
Determine the number of ways to increase the lengths of some sticks so that you can form from them a non-degenerate (that is, having a positive area) triangle. Two ways are considered different, if the length of some stick is increased by different number of centimeters in them. | The single line contains 4 integers *a*,<=*b*,<=*c*,<=*l* (1<=≤<=*a*,<=*b*,<=*c*<=≤<=3·105, 0<=≤<=*l*<=≤<=3·105). | Print a single integer — the number of ways to increase the sizes of the sticks by the total of at most *l* centimeters, so that you can make a non-degenerate triangle from it. | [
"1 1 1 2\n",
"1 2 3 1\n",
"10 2 1 7\n"
] | [
"4\n",
"2\n",
"0\n"
] | In the first sample test you can either not increase any stick or increase any two sticks by 1 centimeter.
In the second sample test you can increase either the first or the second stick by one centimeter. Note that the triangle made from the initial sticks is degenerate and thus, doesn't meet the conditions. | 750 | [
{
"input": "1 1 1 2",
"output": "4"
},
{
"input": "1 2 3 1",
"output": "2"
},
{
"input": "10 2 1 7",
"output": "0"
},
{
"input": "1 2 1 5",
"output": "20"
},
{
"input": "10 15 17 10",
"output": "281"
},
{
"input": "5 5 5 10000",
"output": "41841675001"
},
{
"input": "5 7 30 100",
"output": "71696"
},
{
"input": "5 5 5 300000",
"output": "1125157500250001"
},
{
"input": "4 2 5 28",
"output": "1893"
},
{
"input": "2 7 8 4",
"output": "25"
},
{
"input": "85 50 17 89",
"output": "68620"
},
{
"input": "17 28 19 5558",
"output": "7396315389"
},
{
"input": "5276 8562 1074 8453",
"output": "49093268246"
},
{
"input": "9133 7818 3682 82004",
"output": "38306048676255"
},
{
"input": "81780 54799 231699 808",
"output": "0"
},
{
"input": "53553 262850 271957 182759",
"output": "834977070873802"
},
{
"input": "300000 300000 300000 300000",
"output": "4500090000549998"
},
{
"input": "1 1 300000 300000",
"output": "599999"
},
{
"input": "300000 300000 1 300000",
"output": "2250045000350001"
},
{
"input": "300000 300000 1 24234",
"output": "1186319275394"
},
{
"input": "1 1 1 300000",
"output": "1125022500250001"
},
{
"input": "3 5 7 300000",
"output": "1125157499050009"
},
{
"input": "63 5 52 78",
"output": "46502"
},
{
"input": "2 42 49 93",
"output": "72542"
},
{
"input": "61 100 3 8502",
"output": "27050809786"
},
{
"input": "30 918 702 591",
"output": "14315560"
},
{
"input": "98406 37723 3 257918",
"output": "1154347569149860"
},
{
"input": "552 250082 77579 278985",
"output": "596240712378446"
},
{
"input": "183808 8 8 294771",
"output": "622921327009564"
},
{
"input": "2958 4133 233463 259655",
"output": "65797591388150"
},
{
"input": "300000 200000 100000 1",
"output": "2"
},
{
"input": "300000 200000 100000 0",
"output": "0"
},
{
"input": "100000 300000 100000 100000",
"output": "0"
},
{
"input": "100000 300000 100000 100001",
"output": "100002"
},
{
"input": "100000 300000 100000 100002",
"output": "200005"
},
{
"input": "100000 300000 100000 100003",
"output": "400012"
},
{
"input": "100000 300000 100000 100010",
"output": "3000195"
},
{
"input": "100000 300000 100000 100100",
"output": "255131325"
},
{
"input": "100000 300000 199999 0",
"output": "0"
},
{
"input": "100000 300000 200001 0",
"output": "1"
},
{
"input": "1 1 1 300000",
"output": "1125022500250001"
},
{
"input": "3 1 29 1",
"output": "0"
},
{
"input": "300000 300000 300000 300000",
"output": "4500090000549998"
}
] | 1,487,937,677 | 2,147,483,647 | Python 3 | OK | TESTS | 43 | 701 | 4,608,000 | #in the name of god
#Mr_Rubick
a,b,c,l=map(int, input().split())
cnt=(l+3)*(l+2)*(l+1)//3
for i in (a,b,c):
s=2*i-a-b-c
for x in range(max(0,-s),l+1):
m = min(s+x,l-x)
cnt-=(m+1)*(m+2)
print(cnt//2)
| Title: Lengthening Sticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given three sticks with positive integer lengths of *a*,<=*b*, and *c* centimeters. You can increase length of some of them by some positive integer number of centimeters (different sticks can be increased by a different length), but in total by at most *l* centimeters. In particular, it is allowed not to increase the length of any stick.
Determine the number of ways to increase the lengths of some sticks so that you can form from them a non-degenerate (that is, having a positive area) triangle. Two ways are considered different, if the length of some stick is increased by different number of centimeters in them.
Input Specification:
The single line contains 4 integers *a*,<=*b*,<=*c*,<=*l* (1<=≤<=*a*,<=*b*,<=*c*<=≤<=3·105, 0<=≤<=*l*<=≤<=3·105).
Output Specification:
Print a single integer — the number of ways to increase the sizes of the sticks by the total of at most *l* centimeters, so that you can make a non-degenerate triangle from it.
Demo Input:
['1 1 1 2\n', '1 2 3 1\n', '10 2 1 7\n']
Demo Output:
['4\n', '2\n', '0\n']
Note:
In the first sample test you can either not increase any stick or increase any two sticks by 1 centimeter.
In the second sample test you can increase either the first or the second stick by one centimeter. Note that the triangle made from the initial sticks is degenerate and thus, doesn't meet the conditions. | ```python
#in the name of god
#Mr_Rubick
a,b,c,l=map(int, input().split())
cnt=(l+3)*(l+2)*(l+1)//3
for i in (a,b,c):
s=2*i-a-b-c
for x in range(max(0,-s),l+1):
m = min(s+x,l-x)
cnt-=(m+1)*(m+2)
print(cnt//2)
``` | 3 | |
69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,513,343,544 | 2,147,483,647 | Python 3 | OK | TESTS | 81 | 124 | 5,529,600 | n = int(input())
a = []
for i in range(n):
a.append([int(j) for j in input().split()])
s = 0
k = 0
for i in range(len(a)):
s = s + a[i][k]
if i == len(a):
k = k + 1
if s == 0:
print("YES")
else:
print("NO") | Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
n = int(input())
a = []
for i in range(n):
a.append([int(j) for j in input().split()])
s = 0
k = 0
for i in range(len(a)):
s = s + a[i][k]
if i == len(a):
k = k + 1
if s == 0:
print("YES")
else:
print("NO")
``` | 3.9587 |
816 | B | Karen and Coffee | PROGRAMMING | 1,400 | [
"binary search",
"data structures",
"implementation"
] | null | null | To stay woke and attentive during classes, Karen needs some coffee!
Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally acclaimed "The Art of the Covfefe".
She knows *n* coffee recipes. The *i*-th recipe suggests that coffee should be brewed between *l**i* and *r**i* degrees, inclusive, to achieve the optimal taste.
Karen thinks that a temperature is admissible if at least *k* recipes recommend it.
Karen has a rather fickle mind, and so she asks *q* questions. In each question, given that she only wants to prepare coffee with a temperature between *a* and *b*, inclusive, can you tell her how many admissible integer temperatures fall within the range? | The first line of input contains three integers, *n*, *k* (1<=≤<=*k*<=≤<=*n*<=≤<=200000), and *q* (1<=≤<=*q*<=≤<=200000), the number of recipes, the minimum number of recipes a certain temperature must be recommended by to be admissible, and the number of questions Karen has, respectively.
The next *n* lines describe the recipes. Specifically, the *i*-th line among these contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=200000), describing that the *i*-th recipe suggests that the coffee be brewed between *l**i* and *r**i* degrees, inclusive.
The next *q* lines describe the questions. Each of these lines contains *a* and *b*, (1<=≤<=*a*<=≤<=*b*<=≤<=200000), describing that she wants to know the number of admissible integer temperatures between *a* and *b* degrees, inclusive. | For each question, output a single integer on a line by itself, the number of admissible integer temperatures between *a* and *b* degrees, inclusive. | [
"3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100\n",
"2 1 1\n1 1\n200000 200000\n90 100\n"
] | [
"3\n3\n0\n4\n",
"0\n"
] | In the first test case, Karen knows 3 recipes.
1. The first one recommends brewing the coffee between 91 and 94 degrees, inclusive. 1. The second one recommends brewing the coffee between 92 and 97 degrees, inclusive. 1. The third one recommends brewing the coffee between 97 and 99 degrees, inclusive.
A temperature is admissible if at least 2 recipes recommend it.
She asks 4 questions.
In her first question, she wants to know the number of admissible integer temperatures between 92 and 94 degrees, inclusive. There are 3: 92, 93 and 94 degrees are all admissible.
In her second question, she wants to know the number of admissible integer temperatures between 93 and 97 degrees, inclusive. There are 3: 93, 94 and 97 degrees are all admissible.
In her third question, she wants to know the number of admissible integer temperatures between 95 and 96 degrees, inclusive. There are none.
In her final question, she wants to know the number of admissible integer temperatures between 90 and 100 degrees, inclusive. There are 4: 92, 93, 94 and 97 degrees are all admissible.
In the second test case, Karen knows 2 recipes.
1. The first one, "wikiHow to make Cold Brew Coffee", recommends brewing the coffee at exactly 1 degree. 1. The second one, "What good is coffee that isn't brewed at at least 36.3306 times the temperature of the surface of the sun?", recommends brewing the coffee at exactly 200000 degrees.
A temperature is admissible if at least 1 recipe recommends it.
In her first and only question, she wants to know the number of admissible integer temperatures that are actually reasonable. There are none. | 1,000 | [
{
"input": "3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100",
"output": "3\n3\n0\n4"
},
{
"input": "2 1 1\n1 1\n200000 200000\n90 100",
"output": "0"
},
{
"input": "1 1 1\n1 1\n1 1",
"output": "1"
},
{
"input": "1 1 1\n200000 200000\n200000 200000",
"output": "1"
}
] | 1,633,967,515 | 2,147,483,647 | PyPy 3 | OK | TESTS | 45 | 920 | 31,641,600 | from sys import stdin
input = stdin.readline
def answer():
prefix = []
count = 0
for i in range(size + 1):
if(s[i] >= k):count += 1
prefix.append(count)
for i in range(q):
l , r = map(int,input().split())
print(prefix[r] - prefix[l - 1])
for T in range(1):
n , k , q = map(int,input().split())
size = 2 * (10**5) + 1
s = [0]*(size + 1)
for i in range(n):
l , r = map(int,input().split())
s[l] += 1
s[r + 1] -= 1
for i in range(1 , size + 1):
s[i] += s[i - 1]
answer()
| Title: Karen and Coffee
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
To stay woke and attentive during classes, Karen needs some coffee!
Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally acclaimed "The Art of the Covfefe".
She knows *n* coffee recipes. The *i*-th recipe suggests that coffee should be brewed between *l**i* and *r**i* degrees, inclusive, to achieve the optimal taste.
Karen thinks that a temperature is admissible if at least *k* recipes recommend it.
Karen has a rather fickle mind, and so she asks *q* questions. In each question, given that she only wants to prepare coffee with a temperature between *a* and *b*, inclusive, can you tell her how many admissible integer temperatures fall within the range?
Input Specification:
The first line of input contains three integers, *n*, *k* (1<=≤<=*k*<=≤<=*n*<=≤<=200000), and *q* (1<=≤<=*q*<=≤<=200000), the number of recipes, the minimum number of recipes a certain temperature must be recommended by to be admissible, and the number of questions Karen has, respectively.
The next *n* lines describe the recipes. Specifically, the *i*-th line among these contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=200000), describing that the *i*-th recipe suggests that the coffee be brewed between *l**i* and *r**i* degrees, inclusive.
The next *q* lines describe the questions. Each of these lines contains *a* and *b*, (1<=≤<=*a*<=≤<=*b*<=≤<=200000), describing that she wants to know the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
Output Specification:
For each question, output a single integer on a line by itself, the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
Demo Input:
['3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100\n', '2 1 1\n1 1\n200000 200000\n90 100\n']
Demo Output:
['3\n3\n0\n4\n', '0\n']
Note:
In the first test case, Karen knows 3 recipes.
1. The first one recommends brewing the coffee between 91 and 94 degrees, inclusive. 1. The second one recommends brewing the coffee between 92 and 97 degrees, inclusive. 1. The third one recommends brewing the coffee between 97 and 99 degrees, inclusive.
A temperature is admissible if at least 2 recipes recommend it.
She asks 4 questions.
In her first question, she wants to know the number of admissible integer temperatures between 92 and 94 degrees, inclusive. There are 3: 92, 93 and 94 degrees are all admissible.
In her second question, she wants to know the number of admissible integer temperatures between 93 and 97 degrees, inclusive. There are 3: 93, 94 and 97 degrees are all admissible.
In her third question, she wants to know the number of admissible integer temperatures between 95 and 96 degrees, inclusive. There are none.
In her final question, she wants to know the number of admissible integer temperatures between 90 and 100 degrees, inclusive. There are 4: 92, 93, 94 and 97 degrees are all admissible.
In the second test case, Karen knows 2 recipes.
1. The first one, "wikiHow to make Cold Brew Coffee", recommends brewing the coffee at exactly 1 degree. 1. The second one, "What good is coffee that isn't brewed at at least 36.3306 times the temperature of the surface of the sun?", recommends brewing the coffee at exactly 200000 degrees.
A temperature is admissible if at least 1 recipe recommends it.
In her first and only question, she wants to know the number of admissible integer temperatures that are actually reasonable. There are none. | ```python
from sys import stdin
input = stdin.readline
def answer():
prefix = []
count = 0
for i in range(size + 1):
if(s[i] >= k):count += 1
prefix.append(count)
for i in range(q):
l , r = map(int,input().split())
print(prefix[r] - prefix[l - 1])
for T in range(1):
n , k , q = map(int,input().split())
size = 2 * (10**5) + 1
s = [0]*(size + 1)
for i in range(n):
l , r = map(int,input().split())
s[l] += 1
s[r + 1] -= 1
for i in range(1 , size + 1):
s[i] += s[i - 1]
answer()
``` | 3 | |
322 | B | Ciel and Flowers | PROGRAMMING | 1,600 | [
"combinatorics",
"math"
] | null | null | Fox Ciel has some flowers: *r* red flowers, *g* green flowers and *b* blue flowers. She wants to use these flowers to make several bouquets. There are 4 types of bouquets:
- To make a "red bouquet", it needs 3 red flowers. - To make a "green bouquet", it needs 3 green flowers. - To make a "blue bouquet", it needs 3 blue flowers. - To make a "mixing bouquet", it needs 1 red, 1 green and 1 blue flower.
Help Fox Ciel to find the maximal number of bouquets she can make. | The first line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=109) — the number of red, green and blue flowers. | Print the maximal number of bouquets Fox Ciel can make. | [
"3 6 9\n",
"4 4 4\n",
"0 0 0\n"
] | [
"6\n",
"4\n",
"0\n"
] | In test case 1, we can make 1 red bouquet, 2 green bouquets and 3 blue bouquets.
In test case 2, we can make 1 red, 1 green, 1 blue and 1 mixing bouquet. | 1,000 | [
{
"input": "3 6 9",
"output": "6"
},
{
"input": "4 4 4",
"output": "4"
},
{
"input": "0 0 0",
"output": "0"
},
{
"input": "0 3 6",
"output": "3"
},
{
"input": "7 8 9",
"output": "7"
},
{
"input": "8 8 9",
"output": "8"
},
{
"input": "15 3 999",
"output": "339"
},
{
"input": "32 62 92",
"output": "62"
},
{
"input": "123456789 123456789 123456789",
"output": "123456789"
},
{
"input": "3 5 5",
"output": "4"
},
{
"input": "666806767 385540591 357848286",
"output": "470065214"
},
{
"input": "80010646 727118126 817880463",
"output": "541669744"
},
{
"input": "829651016 732259171 572879931",
"output": "711596705"
},
{
"input": "242854896 442432924 180395753",
"output": "288561190"
},
{
"input": "139978911 5123031 935395222",
"output": "360165721"
},
{
"input": "553182792 10264076 395427398",
"output": "319624755"
},
{
"input": "597790453 720437830 855459575",
"output": "724562619"
},
{
"input": "494914467 356982656 757942689",
"output": "536613270"
},
{
"input": "908118348 67156409 217974865",
"output": "397749873"
},
{
"input": "952726009 629846517 972974334",
"output": "851848953"
},
{
"input": "775140200 616574841 630329230",
"output": "674014756"
},
{
"input": "524780569 326748594 90361407",
"output": "313963523"
},
{
"input": "937984449 184405994 992844522",
"output": "705078321"
},
{
"input": "835108464 525983528 452876698",
"output": "604656229"
},
{
"input": "879716125 531124573 207876166",
"output": "539572288"
},
{
"input": "292920005 241298326 667908343",
"output": "400708891"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1000000000"
},
{
"input": "1000000000 999999999 999999998",
"output": "999999998"
},
{
"input": "999999998 999999998 999999999",
"output": "999999998"
},
{
"input": "0 1 1",
"output": "0"
},
{
"input": "0 1000000000 0",
"output": "333333333"
},
{
"input": "0 1 0",
"output": "0"
},
{
"input": "1 0 0",
"output": "0"
},
{
"input": "0 2 2",
"output": "0"
},
{
"input": "3 3 5",
"output": "3"
},
{
"input": "2 2 0",
"output": "0"
},
{
"input": "0 5 5",
"output": "2"
},
{
"input": "2 0 11",
"output": "3"
},
{
"input": "9 9 7",
"output": "8"
},
{
"input": "65 30 74",
"output": "56"
}
] | 1,613,734,939 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 124 | 307,200 | r,g,b = map(int,input().split())
a,b,c = sorted([r,g,b])
x=y=z=0
x = a + (b-a)//3 + (c-a)//3
if a>0:
y = a-1 + (b-a+1)//3 + (c-a+1)//3
if a>1:
z = a-2 + (b-a+2)//3 + (c-a+2)//3
print(max(x,y,z)) | Title: Ciel and Flowers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel has some flowers: *r* red flowers, *g* green flowers and *b* blue flowers. She wants to use these flowers to make several bouquets. There are 4 types of bouquets:
- To make a "red bouquet", it needs 3 red flowers. - To make a "green bouquet", it needs 3 green flowers. - To make a "blue bouquet", it needs 3 blue flowers. - To make a "mixing bouquet", it needs 1 red, 1 green and 1 blue flower.
Help Fox Ciel to find the maximal number of bouquets she can make.
Input Specification:
The first line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=109) — the number of red, green and blue flowers.
Output Specification:
Print the maximal number of bouquets Fox Ciel can make.
Demo Input:
['3 6 9\n', '4 4 4\n', '0 0 0\n']
Demo Output:
['6\n', '4\n', '0\n']
Note:
In test case 1, we can make 1 red bouquet, 2 green bouquets and 3 blue bouquets.
In test case 2, we can make 1 red, 1 green, 1 blue and 1 mixing bouquet. | ```python
r,g,b = map(int,input().split())
a,b,c = sorted([r,g,b])
x=y=z=0
x = a + (b-a)//3 + (c-a)//3
if a>0:
y = a-1 + (b-a+1)//3 + (c-a+1)//3
if a>1:
z = a-2 + (b-a+2)//3 + (c-a+2)//3
print(max(x,y,z))
``` | 3 | |
71 | A | Way Too Long Words | PROGRAMMING | 800 | [
"strings"
] | A. Way Too Long Words | 1 | 256 | Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. | Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. | [
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] | [
"word\nl10n\ni18n\np43s\n"
] | none | 500 | [
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,674,140,895 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 31 | 0 | n=int(input())
k=[]
for _ in range(n):
a=input()
if len(a)>10:
a=a[0]+str(len(a)-2)+a[-1]
k.append(a)
else:
k.append(a)
for i in k:
print(i)
| Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none | ```python
n=int(input())
k=[]
for _ in range(n):
a=input()
if len(a)>10:
a=a[0]+str(len(a)-2)+a[-1]
k.append(a)
else:
k.append(a)
for i in k:
print(i)
``` | 3.9845 |
466 | A | Cheap Travel | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | Ann has recently started commuting by subway. We know that a one ride subway ticket costs *a* rubles. Besides, Ann found out that she can buy a special ticket for *m* rides (she can buy it several times). It costs *b* rubles. Ann did the math; she will need to use subway *n* times. Help Ann, tell her what is the minimum sum of money she will have to spend to make *n* rides? | The single line contains four space-separated integers *n*, *m*, *a*, *b* (1<=≤<=*n*,<=*m*,<=*a*,<=*b*<=≤<=1000) — the number of rides Ann has planned, the number of rides covered by the *m* ride ticket, the price of a one ride ticket and the price of an *m* ride ticket. | Print a single integer — the minimum sum in rubles that Ann will need to spend. | [
"6 2 1 2\n",
"5 2 2 3\n"
] | [
"6\n",
"8\n"
] | In the first sample one of the optimal solutions is: each time buy a one ride ticket. There are other optimal solutions. For example, buy three *m* ride tickets. | 500 | [
{
"input": "6 2 1 2",
"output": "6"
},
{
"input": "5 2 2 3",
"output": "8"
},
{
"input": "10 3 5 1",
"output": "4"
},
{
"input": "1000 1 1000 1000",
"output": "1000000"
},
{
"input": "1000 3 1000 1000",
"output": "334000"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "10 2 1 1",
"output": "5"
},
{
"input": "1 1000 1 2",
"output": "1"
},
{
"input": "1 1000 3 2",
"output": "2"
},
{
"input": "10 3 1 2",
"output": "7"
},
{
"input": "995 1 2 1",
"output": "995"
},
{
"input": "556 2 16 15",
"output": "4170"
},
{
"input": "477 2 16 14",
"output": "3346"
},
{
"input": "101 110 1 100",
"output": "100"
},
{
"input": "9 3 3 10",
"output": "27"
},
{
"input": "100 8 10 1",
"output": "13"
},
{
"input": "6 4 1 3",
"output": "5"
},
{
"input": "8 5 2 8",
"output": "14"
},
{
"input": "1000 2 1 1000",
"output": "1000"
}
] | 1,692,254,042 | 2,147,483,647 | Python 3 | OK | TESTS | 19 | 46 | 0 | n , a , b , c = map(int , input().split())
if (c / a) < b:
if b * (n % a) < c:
print(c * (n // a) + b * (n % a))
else:
print(c * (n // a + 1))
else:
print(n * b) | Title: Cheap Travel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ann has recently started commuting by subway. We know that a one ride subway ticket costs *a* rubles. Besides, Ann found out that she can buy a special ticket for *m* rides (she can buy it several times). It costs *b* rubles. Ann did the math; she will need to use subway *n* times. Help Ann, tell her what is the minimum sum of money she will have to spend to make *n* rides?
Input Specification:
The single line contains four space-separated integers *n*, *m*, *a*, *b* (1<=≤<=*n*,<=*m*,<=*a*,<=*b*<=≤<=1000) — the number of rides Ann has planned, the number of rides covered by the *m* ride ticket, the price of a one ride ticket and the price of an *m* ride ticket.
Output Specification:
Print a single integer — the minimum sum in rubles that Ann will need to spend.
Demo Input:
['6 2 1 2\n', '5 2 2 3\n']
Demo Output:
['6\n', '8\n']
Note:
In the first sample one of the optimal solutions is: each time buy a one ride ticket. There are other optimal solutions. For example, buy three *m* ride tickets. | ```python
n , a , b , c = map(int , input().split())
if (c / a) < b:
if b * (n % a) < c:
print(c * (n // a) + b * (n % a))
else:
print(c * (n // a + 1))
else:
print(n * b)
``` | 3 | |
79 | A | Bus Game | PROGRAMMING | 1,200 | [
"greedy"
] | A. Bus Game | 2 | 256 | After Fox Ciel won an onsite round of a programming contest, she took a bus to return to her castle. The fee of the bus was 220 yen. She met Rabbit Hanako in the bus. They decided to play the following game because they got bored in the bus.
- Initially, there is a pile that contains *x* 100-yen coins and *y* 10-yen coins. - They take turns alternatively. Ciel takes the first turn. - In each turn, they must take exactly 220 yen from the pile. In Ciel's turn, if there are multiple ways to take 220 yen, she will choose the way that contains the maximal number of 100-yen coins. In Hanako's turn, if there are multiple ways to take 220 yen, she will choose the way that contains the maximal number of 10-yen coins. - If Ciel or Hanako can't take exactly 220 yen from the pile, she loses.
Determine the winner of the game. | The first line contains two integers *x* (0<=≤<=*x*<=≤<=106) and *y* (0<=≤<=*y*<=≤<=106), separated by a single space. | If Ciel wins, print "Ciel". Otherwise, print "Hanako". | [
"2 2\n",
"3 22\n"
] | [
"Ciel\n",
"Hanako\n"
] | In the first turn (Ciel's turn), she will choose 2 100-yen coins and 2 10-yen coins. In the second turn (Hanako's turn), she will choose 1 100-yen coin and 12 10-yen coins. In the third turn (Ciel's turn), she can't pay exactly 220 yen, so Ciel will lose. | 500 | [
{
"input": "2 2",
"output": "Ciel"
},
{
"input": "3 22",
"output": "Hanako"
},
{
"input": "0 22",
"output": "Ciel"
},
{
"input": "1000 1000",
"output": "Ciel"
},
{
"input": "0 0",
"output": "Hanako"
},
{
"input": "0 21",
"output": "Hanako"
},
{
"input": "1 11",
"output": "Hanako"
},
{
"input": "1 12",
"output": "Ciel"
},
{
"input": "2 1",
"output": "Hanako"
},
{
"input": "2 23",
"output": "Ciel"
},
{
"input": "2 24",
"output": "Hanako"
},
{
"input": "3 1",
"output": "Hanako"
},
{
"input": "3 2",
"output": "Ciel"
},
{
"input": "3 13",
"output": "Ciel"
},
{
"input": "3 14",
"output": "Hanako"
},
{
"input": "4 1",
"output": "Hanako"
},
{
"input": "4 2",
"output": "Ciel"
},
{
"input": "4 25",
"output": "Hanako"
},
{
"input": "4 26",
"output": "Ciel"
},
{
"input": "5 1",
"output": "Hanako"
},
{
"input": "5 2",
"output": "Ciel"
},
{
"input": "5 15",
"output": "Hanako"
},
{
"input": "5 16",
"output": "Ciel"
},
{
"input": "5 23",
"output": "Ciel"
},
{
"input": "5 24",
"output": "Hanako"
},
{
"input": "6 1",
"output": "Hanako"
},
{
"input": "6 2",
"output": "Ciel"
},
{
"input": "6 13",
"output": "Ciel"
},
{
"input": "6 14",
"output": "Hanako"
},
{
"input": "6 23",
"output": "Ciel"
},
{
"input": "6 24",
"output": "Hanako"
},
{
"input": "7 1",
"output": "Hanako"
},
{
"input": "7 2",
"output": "Ciel"
},
{
"input": "7 13",
"output": "Ciel"
},
{
"input": "7 14",
"output": "Hanako"
},
{
"input": "7 25",
"output": "Hanako"
},
{
"input": "7 26",
"output": "Ciel"
},
{
"input": "8 1",
"output": "Hanako"
},
{
"input": "8 2",
"output": "Ciel"
},
{
"input": "8 15",
"output": "Hanako"
},
{
"input": "8 16",
"output": "Ciel"
},
{
"input": "8 25",
"output": "Hanako"
},
{
"input": "8 26",
"output": "Ciel"
},
{
"input": "9 1",
"output": "Hanako"
},
{
"input": "9 2",
"output": "Ciel"
},
{
"input": "9 15",
"output": "Hanako"
},
{
"input": "9 16",
"output": "Ciel"
},
{
"input": "9 23",
"output": "Ciel"
},
{
"input": "9 24",
"output": "Hanako"
},
{
"input": "10 12",
"output": "Ciel"
},
{
"input": "10 13",
"output": "Ciel"
},
{
"input": "10 22",
"output": "Ciel"
},
{
"input": "10 23",
"output": "Ciel"
},
{
"input": "11 12",
"output": "Ciel"
},
{
"input": "11 13",
"output": "Ciel"
},
{
"input": "11 24",
"output": "Hanako"
},
{
"input": "11 25",
"output": "Hanako"
},
{
"input": "12 14",
"output": "Hanako"
},
{
"input": "12 15",
"output": "Hanako"
},
{
"input": "12 24",
"output": "Hanako"
},
{
"input": "12 25",
"output": "Hanako"
},
{
"input": "0 1000000",
"output": "Hanako"
},
{
"input": "1000000 0",
"output": "Hanako"
},
{
"input": "1000000 1000000",
"output": "Ciel"
},
{
"input": "178087 42116",
"output": "Ciel"
},
{
"input": "378897 104123",
"output": "Ciel"
},
{
"input": "61207 166129",
"output": "Hanako"
},
{
"input": "743519 228136",
"output": "Ciel"
},
{
"input": "425829 771644",
"output": "Ciel"
},
{
"input": "626640 833651",
"output": "Ciel"
},
{
"input": "308950 895657",
"output": "Hanako"
},
{
"input": "991262 957664",
"output": "Ciel"
},
{
"input": "192071 19670",
"output": "Hanako"
},
{
"input": "874382 81677",
"output": "Hanako"
},
{
"input": "202081 745873",
"output": "Hanako"
},
{
"input": "233663 723781",
"output": "Hanako"
},
{
"input": "783744 701689",
"output": "Hanako"
},
{
"input": "333825 679597",
"output": "Hanako"
},
{
"input": "365407 657504",
"output": "Hanako"
},
{
"input": "915488 635412",
"output": "Hanako"
},
{
"input": "947070 613320",
"output": "Hanako"
},
{
"input": "497151 591228",
"output": "Hanako"
},
{
"input": "528732 87635",
"output": "Ciel"
},
{
"input": "78813 65543",
"output": "Hanako"
},
{
"input": "6 4",
"output": "Hanako"
},
{
"input": "3 5",
"output": "Ciel"
},
{
"input": "4 24",
"output": "Hanako"
},
{
"input": "2 14",
"output": "Ciel"
},
{
"input": "2 26",
"output": "Hanako"
},
{
"input": "5 26",
"output": "Ciel"
},
{
"input": "5 5",
"output": "Hanako"
},
{
"input": "359 479",
"output": "Hanako"
},
{
"input": "1 20",
"output": "Ciel"
},
{
"input": "11 0",
"output": "Hanako"
},
{
"input": "3 0",
"output": "Hanako"
},
{
"input": "0 30",
"output": "Ciel"
},
{
"input": "0 35",
"output": "Ciel"
},
{
"input": "100 99",
"output": "Ciel"
},
{
"input": "3 4",
"output": "Ciel"
}
] | 1,519,805,433 | 2,147,483,647 | Python 3 | OK | TESTS | 99 | 186 | 5,734,400 | x, y = map(int, input().split())
c = min(x // 2, y // 24)
x -= 2 * c
y -= 24 * c
c = min(x // 3, y // 14)
x -= 3 * c
y -= 14 * c
c = min(x // 4, y // 4)
x -= 4 * c
y -= 4 * c
c = min(x, y // 34)
x -= 1 * c
y -= 34 * c
c = y // 44
y -= 44 * c
if y >= 2 and 10 * x + y >= 22:
print('Ciel')
else:
print('Hanako') | Title: Bus Game
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
After Fox Ciel won an onsite round of a programming contest, she took a bus to return to her castle. The fee of the bus was 220 yen. She met Rabbit Hanako in the bus. They decided to play the following game because they got bored in the bus.
- Initially, there is a pile that contains *x* 100-yen coins and *y* 10-yen coins. - They take turns alternatively. Ciel takes the first turn. - In each turn, they must take exactly 220 yen from the pile. In Ciel's turn, if there are multiple ways to take 220 yen, she will choose the way that contains the maximal number of 100-yen coins. In Hanako's turn, if there are multiple ways to take 220 yen, she will choose the way that contains the maximal number of 10-yen coins. - If Ciel or Hanako can't take exactly 220 yen from the pile, she loses.
Determine the winner of the game.
Input Specification:
The first line contains two integers *x* (0<=≤<=*x*<=≤<=106) and *y* (0<=≤<=*y*<=≤<=106), separated by a single space.
Output Specification:
If Ciel wins, print "Ciel". Otherwise, print "Hanako".
Demo Input:
['2 2\n', '3 22\n']
Demo Output:
['Ciel\n', 'Hanako\n']
Note:
In the first turn (Ciel's turn), she will choose 2 100-yen coins and 2 10-yen coins. In the second turn (Hanako's turn), she will choose 1 100-yen coin and 12 10-yen coins. In the third turn (Ciel's turn), she can't pay exactly 220 yen, so Ciel will lose. | ```python
x, y = map(int, input().split())
c = min(x // 2, y // 24)
x -= 2 * c
y -= 24 * c
c = min(x // 3, y // 14)
x -= 3 * c
y -= 14 * c
c = min(x // 4, y // 4)
x -= 4 * c
y -= 4 * c
c = min(x, y // 34)
x -= 1 * c
y -= 34 * c
c = y // 44
y -= 44 * c
if y >= 2 and 10 * x + y >= 22:
print('Ciel')
else:
print('Hanako')
``` | 3.942819 |
735 | C | Tennis Championship | PROGRAMMING | 1,600 | [
"combinatorics",
"constructive algorithms",
"greedy",
"math"
] | null | null | Famous Brazil city Rio de Janeiro holds a tennis tournament and Ostap Bender doesn't want to miss this event. There will be *n* players participating, and the tournament will follow knockout rules from the very first game. That means, that if someone loses a game he leaves the tournament immediately.
Organizers are still arranging tournament grid (i.e. the order games will happen and who is going to play with whom) but they have already fixed one rule: two players can play against each other only if the number of games one of them has already played differs by no more than one from the number of games the other one has already played. Of course, both players had to win all their games in order to continue participating in the tournament.
Tournament hasn't started yet so the audience is a bit bored. Ostap decided to find out what is the maximum number of games the winner of the tournament can take part in (assuming the rule above is used). However, it is unlikely he can deal with this problem without your help. | The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=1018) — the number of players to participate in the tournament. | Print the maximum number of games in which the winner of the tournament can take part. | [
"2\n",
"3\n",
"4\n",
"10\n"
] | [
"1\n",
"2\n",
"2\n",
"4\n"
] | In all samples we consider that player number 1 is the winner.
In the first sample, there would be only one game so the answer is 1.
In the second sample, player 1 can consequently beat players 2 and 3.
In the third sample, player 1 can't play with each other player as after he plays with players 2 and 3 he can't play against player 4, as he has 0 games played, while player 1 already played 2. Thus, the answer is 2 and to achieve we make pairs (1, 2) and (3, 4) and then clash the winners. | 1,750 | [
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "2"
},
{
"input": "10",
"output": "4"
},
{
"input": "1000",
"output": "14"
},
{
"input": "2500",
"output": "15"
},
{
"input": "690000",
"output": "27"
},
{
"input": "3000000000",
"output": "45"
},
{
"input": "123456789123456789",
"output": "81"
},
{
"input": "5",
"output": "3"
},
{
"input": "143",
"output": "9"
},
{
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"output": "10"
},
{
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"output": "10"
},
{
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"output": "10"
},
{
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"output": "11"
},
{
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"output": "11"
},
{
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"output": "84"
},
{
"input": "679891637638612258",
"output": "85"
},
{
"input": "679891637638612259",
"output": "85"
},
{
"input": "1000000000000000000",
"output": "85"
},
{
"input": "10235439547",
"output": "47"
},
{
"input": "1240723548",
"output": "43"
},
{
"input": "92353046212453",
"output": "66"
},
{
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"output": "68"
},
{
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"output": "62"
},
{
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"output": "62"
},
{
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"output": "63"
},
{
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"output": "56"
},
{
"input": "793056859214355",
"output": "70"
},
{
"input": "982045466234565",
"output": "71"
},
{
"input": "126743950353465",
"output": "67"
},
{
"input": "12405430465",
"output": "47"
},
{
"input": "10238439257768",
"output": "61"
},
{
"input": "1728493055346",
"output": "58"
},
{
"input": "927553829046",
"output": "56"
},
{
"input": "62735129403",
"output": "51"
},
{
"input": "71624823950223",
"output": "65"
},
{
"input": "8902353464851212",
"output": "75"
},
{
"input": "61824012598535",
"output": "65"
},
{
"input": "1294902504603347",
"output": "71"
},
{
"input": "6",
"output": "3"
},
{
"input": "7",
"output": "3"
},
{
"input": "8",
"output": "4"
},
{
"input": "9",
"output": "4"
},
{
"input": "11",
"output": "4"
},
{
"input": "12",
"output": "4"
},
{
"input": "13",
"output": "5"
},
{
"input": "14",
"output": "5"
},
{
"input": "15",
"output": "5"
},
{
"input": "16",
"output": "5"
},
{
"input": "17",
"output": "5"
},
{
"input": "18",
"output": "5"
},
{
"input": "19",
"output": "5"
},
{
"input": "20",
"output": "5"
},
{
"input": "21",
"output": "6"
},
{
"input": "22",
"output": "6"
},
{
"input": "23",
"output": "6"
},
{
"input": "355687428096000",
"output": "69"
},
{
"input": "576460752303423488",
"output": "84"
},
{
"input": "32212254719",
"output": "49"
},
{
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"output": "63"
},
{
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"output": "56"
},
{
"input": "262406072477",
"output": "54"
}
] | 1,537,506,832 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 4 | 139 | 0 | from math import *
n=int(input())
ans=ceil(log(n)/log(2))
print(ans)
| Title: Tennis Championship
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Famous Brazil city Rio de Janeiro holds a tennis tournament and Ostap Bender doesn't want to miss this event. There will be *n* players participating, and the tournament will follow knockout rules from the very first game. That means, that if someone loses a game he leaves the tournament immediately.
Organizers are still arranging tournament grid (i.e. the order games will happen and who is going to play with whom) but they have already fixed one rule: two players can play against each other only if the number of games one of them has already played differs by no more than one from the number of games the other one has already played. Of course, both players had to win all their games in order to continue participating in the tournament.
Tournament hasn't started yet so the audience is a bit bored. Ostap decided to find out what is the maximum number of games the winner of the tournament can take part in (assuming the rule above is used). However, it is unlikely he can deal with this problem without your help.
Input Specification:
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=1018) — the number of players to participate in the tournament.
Output Specification:
Print the maximum number of games in which the winner of the tournament can take part.
Demo Input:
['2\n', '3\n', '4\n', '10\n']
Demo Output:
['1\n', '2\n', '2\n', '4\n']
Note:
In all samples we consider that player number 1 is the winner.
In the first sample, there would be only one game so the answer is 1.
In the second sample, player 1 can consequently beat players 2 and 3.
In the third sample, player 1 can't play with each other player as after he plays with players 2 and 3 he can't play against player 4, as he has 0 games played, while player 1 already played 2. Thus, the answer is 2 and to achieve we make pairs (1, 2) and (3, 4) and then clash the winners. | ```python
from math import *
n=int(input())
ans=ceil(log(n)/log(2))
print(ans)
``` | 0 | |
405 | A | Gravity Flip | PROGRAMMING | 900 | [
"greedy",
"implementation",
"sortings"
] | null | null | Little Chris is bored during his physics lessons (too easy), so he has built a toy box to keep himself occupied. The box is special, since it has the ability to change gravity.
There are *n* columns of toy cubes in the box arranged in a line. The *i*-th column contains *a**i* cubes. At first, the gravity in the box is pulling the cubes downwards. When Chris switches the gravity, it begins to pull all the cubes to the right side of the box. The figure shows the initial and final configurations of the cubes in the box: the cubes that have changed their position are highlighted with orange.
Given the initial configuration of the toy cubes in the box, find the amounts of cubes in each of the *n* columns after the gravity switch! | The first line of input contains an integer *n* (1<=≤<=*n*<=≤<=100), the number of the columns in the box. The next line contains *n* space-separated integer numbers. The *i*-th number *a**i* (1<=≤<=*a**i*<=≤<=100) denotes the number of cubes in the *i*-th column. | Output *n* integer numbers separated by spaces, where the *i*-th number is the amount of cubes in the *i*-th column after the gravity switch. | [
"4\n3 2 1 2\n",
"3\n2 3 8\n"
] | [
"1 2 2 3 \n",
"2 3 8 \n"
] | The first example case is shown on the figure. The top cube of the first column falls to the top of the last column; the top cube of the second column falls to the top of the third column; the middle cube of the first column falls to the top of the second column.
In the second example case the gravity switch does not change the heights of the columns. | 500 | [
{
"input": "4\n3 2 1 2",
"output": "1 2 2 3 "
},
{
"input": "3\n2 3 8",
"output": "2 3 8 "
},
{
"input": "5\n2 1 2 1 2",
"output": "1 1 2 2 2 "
},
{
"input": "1\n1",
"output": "1 "
},
{
"input": "2\n4 3",
"output": "3 4 "
},
{
"input": "6\n100 40 60 20 1 80",
"output": "1 20 40 60 80 100 "
},
{
"input": "10\n10 8 6 7 5 3 4 2 9 1",
"output": "1 2 3 4 5 6 7 8 9 10 "
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "1 2 3 4 5 6 7 8 9 10 "
},
{
"input": "100\n82 51 81 14 37 17 78 92 64 15 8 86 89 8 87 77 66 10 15 12 100 25 92 47 21 78 20 63 13 49 41 36 41 79 16 87 87 69 3 76 80 60 100 49 70 59 72 8 38 71 45 97 71 14 76 54 81 4 59 46 39 29 92 3 49 22 53 99 59 52 74 31 92 43 42 23 44 9 82 47 7 40 12 9 3 55 37 85 46 22 84 52 98 41 21 77 63 17 62 91",
"output": "3 3 3 4 7 8 8 8 9 9 10 12 12 13 14 14 15 15 16 17 17 20 21 21 22 22 23 25 29 31 36 37 37 38 39 40 41 41 41 42 43 44 45 46 46 47 47 49 49 49 51 52 52 53 54 55 59 59 59 60 62 63 63 64 66 69 70 71 71 72 74 76 76 77 77 78 78 79 80 81 81 82 82 84 85 86 87 87 87 89 91 92 92 92 92 97 98 99 100 100 "
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 "
},
{
"input": "10\n1 9 7 6 2 4 7 8 1 3",
"output": "1 1 2 3 4 6 7 7 8 9 "
},
{
"input": "20\n53 32 64 20 41 97 50 20 66 68 22 60 74 61 97 54 80 30 72 59",
"output": "20 20 22 30 32 41 50 53 54 59 60 61 64 66 68 72 74 80 97 97 "
},
{
"input": "30\n7 17 4 18 16 12 14 10 1 13 2 16 13 17 8 16 13 14 9 17 17 5 13 5 1 7 6 20 18 12",
"output": "1 1 2 4 5 5 6 7 7 8 9 10 12 12 13 13 13 13 14 14 16 16 16 17 17 17 17 18 18 20 "
},
{
"input": "40\n22 58 68 58 48 53 52 1 16 78 75 17 63 15 36 32 78 75 49 14 42 46 66 54 49 82 40 43 46 55 12 73 5 45 61 60 1 11 31 84",
"output": "1 1 5 11 12 14 15 16 17 22 31 32 36 40 42 43 45 46 46 48 49 49 52 53 54 55 58 58 60 61 63 66 68 73 75 75 78 78 82 84 "
},
{
"input": "70\n1 3 3 1 3 3 1 1 1 3 3 2 3 3 1 1 1 2 3 1 3 2 3 3 3 2 2 3 1 3 3 2 1 1 2 1 2 1 2 2 1 1 1 3 3 2 3 2 3 2 3 3 2 2 2 3 2 3 3 3 1 1 3 3 1 1 1 1 3 1",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 "
},
{
"input": "90\n17 75 51 30 100 5 50 95 51 73 66 5 7 76 43 49 23 55 3 24 95 79 10 11 44 93 17 99 53 66 82 66 63 76 19 4 51 71 75 43 27 5 24 19 48 7 91 15 55 21 7 6 27 10 2 91 64 58 18 21 16 71 90 88 21 20 6 6 95 85 11 7 40 65 52 49 92 98 46 88 17 48 85 96 77 46 100 34 67 52",
"output": "2 3 4 5 5 5 6 6 6 7 7 7 7 10 10 11 11 15 16 17 17 17 18 19 19 20 21 21 21 23 24 24 27 27 30 34 40 43 43 44 46 46 48 48 49 49 50 51 51 51 52 52 53 55 55 58 63 64 65 66 66 66 67 71 71 73 75 75 76 76 77 79 82 85 85 88 88 90 91 91 92 93 95 95 95 96 98 99 100 100 "
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 "
},
{
"input": "100\n1 1 1 1 2 1 1 1 1 1 2 2 1 1 2 1 2 1 1 1 2 1 1 2 1 2 1 1 2 2 2 1 1 2 1 1 1 2 2 2 1 1 1 2 1 2 2 1 2 1 1 2 2 1 2 1 2 1 2 2 1 1 1 2 1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 1 1 1 1 2 2 2 2 2 2 2 1 1 1 2 1 2 1",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "100\n2 1 1 1 3 2 3 3 2 3 3 1 3 3 1 3 3 1 1 1 2 3 1 2 3 1 2 3 3 1 3 1 1 2 3 2 3 3 2 3 3 1 2 2 1 2 3 2 3 2 2 1 1 3 1 3 2 1 3 1 3 1 3 1 1 3 3 3 2 3 2 2 2 2 1 3 3 3 1 2 1 2 3 2 1 3 1 3 2 1 3 1 2 1 2 3 1 3 2 3",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 "
},
{
"input": "100\n7 4 5 5 10 10 5 8 5 7 4 5 4 6 8 8 2 6 3 3 10 7 10 8 6 2 7 3 9 7 7 2 4 5 2 4 9 5 10 1 10 5 10 4 1 3 4 2 6 9 9 9 10 6 2 5 6 1 8 10 4 10 3 4 10 5 5 4 10 4 5 3 7 10 2 7 3 6 9 6 1 6 5 5 4 6 6 4 4 1 5 1 6 6 6 8 8 6 2 6",
"output": "1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 10 10 10 "
},
{
"input": "100\n12 10 5 11 13 12 14 13 7 15 15 12 13 19 12 18 14 10 10 3 1 10 16 11 19 8 10 15 5 10 12 16 11 13 11 15 14 12 16 8 11 8 15 2 18 2 14 13 15 20 8 8 4 12 14 7 10 3 9 1 7 19 6 7 2 14 8 20 7 17 18 20 3 18 18 9 6 10 4 1 4 19 9 13 3 3 12 11 11 20 8 2 13 6 7 12 1 4 17 3",
"output": "1 1 1 1 2 2 2 2 3 3 3 3 3 3 4 4 4 4 5 5 6 6 6 7 7 7 7 7 7 8 8 8 8 8 8 8 9 9 9 10 10 10 10 10 10 10 10 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12 12 13 13 13 13 13 13 13 14 14 14 14 14 14 15 15 15 15 15 15 16 16 16 17 17 18 18 18 18 18 19 19 19 19 20 20 20 20 "
},
{
"input": "100\n5 13 1 40 30 10 23 32 33 12 6 4 15 29 31 17 23 5 36 31 32 38 24 11 34 39 19 21 6 19 31 35 1 15 6 29 22 15 17 15 1 17 2 34 20 8 27 2 29 26 13 9 22 27 27 3 20 40 4 40 33 29 36 30 35 16 19 28 26 11 36 24 29 5 40 10 38 34 33 23 34 39 31 7 10 31 22 6 36 24 14 31 34 23 2 4 26 16 2 32",
"output": "1 1 1 2 2 2 2 3 4 4 4 5 5 5 6 6 6 6 7 8 9 10 10 10 11 11 12 13 13 14 15 15 15 15 16 16 17 17 17 19 19 19 20 20 21 22 22 22 23 23 23 23 24 24 24 26 26 26 27 27 27 28 29 29 29 29 29 30 30 31 31 31 31 31 31 32 32 32 33 33 33 34 34 34 34 34 35 35 36 36 36 36 38 38 39 39 40 40 40 40 "
},
{
"input": "100\n72 44 34 74 9 60 26 37 55 77 74 69 28 66 54 55 8 36 57 31 31 48 32 66 40 70 77 43 64 28 37 10 21 58 51 32 60 28 51 52 28 35 7 33 1 68 38 70 57 71 8 20 42 57 59 4 58 10 17 47 22 48 16 3 76 67 32 37 64 47 33 41 75 69 2 76 39 9 27 75 20 21 52 25 71 21 11 29 38 10 3 1 45 55 63 36 27 7 59 41",
"output": "1 1 2 3 3 4 7 7 8 8 9 9 10 10 10 11 16 17 20 20 21 21 21 22 25 26 27 27 28 28 28 28 29 31 31 32 32 32 33 33 34 35 36 36 37 37 37 38 38 39 40 41 41 42 43 44 45 47 47 48 48 51 51 52 52 54 55 55 55 57 57 57 58 58 59 59 60 60 63 64 64 66 66 67 68 69 69 70 70 71 71 72 74 74 75 75 76 76 77 77 "
},
{
"input": "100\n75 18 61 10 56 53 42 57 79 80 31 2 50 45 54 99 84 52 71 21 86 3 19 98 14 37 40 62 63 68 5 10 87 8 81 85 52 52 57 94 2 7 56 96 19 76 1 13 81 6 80 47 22 59 99 32 9 5 36 88 98 91 70 70 12 93 12 22 85 1 97 48 94 16 84 84 51 34 62 7 68 51 30 2 37 82 4 7 27 1 80 9 61 16 59 55 12 96 94 82",
"output": "1 1 1 2 2 2 3 4 5 5 6 7 7 7 8 9 9 10 10 12 12 12 13 14 16 16 18 19 19 21 22 22 27 30 31 32 34 36 37 37 40 42 45 47 48 50 51 51 52 52 52 53 54 55 56 56 57 57 59 59 61 61 62 62 63 68 68 70 70 71 75 76 79 80 80 80 81 81 82 82 84 84 84 85 85 86 87 88 91 93 94 94 94 96 96 97 98 98 99 99 "
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 "
},
{
"input": "100\n100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 "
},
{
"input": "100\n50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50",
"output": "50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 "
},
{
"input": "49\n1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97",
"output": "1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 "
},
{
"input": "30\n1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79 82 85 88",
"output": "1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79 82 85 88 "
},
{
"input": "100\n100 51 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 51 100 "
},
{
"input": "10\n100 90 80 70 60 50 40 30 20 10",
"output": "10 20 30 40 50 60 70 80 90 100 "
},
{
"input": "1\n10",
"output": "10 "
}
] | 1,694,673,149 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 46 | 0 | a=int(input())
b=list(map(int,input().split()))
c=""
for i in range(0,len(b)):
a=max(b)
b.remove(a)
c=str(a)+" "+c
print(c) | Title: Gravity Flip
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Chris is bored during his physics lessons (too easy), so he has built a toy box to keep himself occupied. The box is special, since it has the ability to change gravity.
There are *n* columns of toy cubes in the box arranged in a line. The *i*-th column contains *a**i* cubes. At first, the gravity in the box is pulling the cubes downwards. When Chris switches the gravity, it begins to pull all the cubes to the right side of the box. The figure shows the initial and final configurations of the cubes in the box: the cubes that have changed their position are highlighted with orange.
Given the initial configuration of the toy cubes in the box, find the amounts of cubes in each of the *n* columns after the gravity switch!
Input Specification:
The first line of input contains an integer *n* (1<=≤<=*n*<=≤<=100), the number of the columns in the box. The next line contains *n* space-separated integer numbers. The *i*-th number *a**i* (1<=≤<=*a**i*<=≤<=100) denotes the number of cubes in the *i*-th column.
Output Specification:
Output *n* integer numbers separated by spaces, where the *i*-th number is the amount of cubes in the *i*-th column after the gravity switch.
Demo Input:
['4\n3 2 1 2\n', '3\n2 3 8\n']
Demo Output:
['1 2 2 3 \n', '2 3 8 \n']
Note:
The first example case is shown on the figure. The top cube of the first column falls to the top of the last column; the top cube of the second column falls to the top of the third column; the middle cube of the first column falls to the top of the second column.
In the second example case the gravity switch does not change the heights of the columns. | ```python
a=int(input())
b=list(map(int,input().split()))
c=""
for i in range(0,len(b)):
a=max(b)
b.remove(a)
c=str(a)+" "+c
print(c)
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | The last stage of Football World Cup is played using the play-off system.
There are *n* teams left in this stage, they are enumerated from 1 to *n*. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids *a* and *b* can meet. | The only line contains three integers *n*, *a* and *b* (2<=≤<=*n*<=≤<=256, 1<=≤<=*a*,<=*b*<=≤<=*n*) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that *n* is such that in each round an even number of team advance, and that *a* and *b* are not equal. | In the only line print "Final!" (without quotes), if teams *a* and *b* can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams *a* and *b* can meet. The round are enumerated from 1. | [
"4 1 2\n",
"8 2 6\n",
"8 7 5\n"
] | [
"1\n",
"Final!\n",
"2\n"
] | In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round. | 0 | [
{
"input": "4 1 2",
"output": "1"
},
{
"input": "8 2 6",
"output": "Final!"
},
{
"input": "8 7 5",
"output": "2"
},
{
"input": "128 30 98",
"output": "Final!"
},
{
"input": "256 128 256",
"output": "Final!"
},
{
"input": "256 2 127",
"output": "7"
},
{
"input": "2 1 2",
"output": "Final!"
},
{
"input": "2 2 1",
"output": "Final!"
},
{
"input": "4 1 3",
"output": "Final!"
},
{
"input": "4 1 4",
"output": "Final!"
},
{
"input": "4 2 1",
"output": "1"
},
{
"input": "4 2 3",
"output": "Final!"
},
{
"input": "4 2 4",
"output": "Final!"
},
{
"input": "4 3 1",
"output": "Final!"
},
{
"input": "4 3 2",
"output": "Final!"
},
{
"input": "4 3 4",
"output": "1"
},
{
"input": "4 4 1",
"output": "Final!"
},
{
"input": "4 4 2",
"output": "Final!"
},
{
"input": "4 4 3",
"output": "1"
},
{
"input": "8 8 7",
"output": "1"
},
{
"input": "8 8 5",
"output": "2"
},
{
"input": "8 8 1",
"output": "Final!"
},
{
"input": "16 4 3",
"output": "1"
},
{
"input": "16 2 4",
"output": "2"
},
{
"input": "16 14 11",
"output": "3"
},
{
"input": "16 3 11",
"output": "Final!"
},
{
"input": "32 10 9",
"output": "1"
},
{
"input": "32 25 28",
"output": "2"
},
{
"input": "32 22 18",
"output": "3"
},
{
"input": "32 17 25",
"output": "4"
},
{
"input": "32 18 3",
"output": "Final!"
},
{
"input": "64 40 39",
"output": "1"
},
{
"input": "64 60 58",
"output": "2"
},
{
"input": "64 34 37",
"output": "3"
},
{
"input": "64 26 24",
"output": "4"
},
{
"input": "64 50 43",
"output": "5"
},
{
"input": "64 17 42",
"output": "Final!"
},
{
"input": "128 116 115",
"output": "1"
},
{
"input": "128 35 33",
"output": "2"
},
{
"input": "128 61 59",
"output": "3"
},
{
"input": "128 116 123",
"output": "4"
},
{
"input": "128 17 15",
"output": "5"
},
{
"input": "128 124 77",
"output": "6"
},
{
"input": "128 4 80",
"output": "Final!"
},
{
"input": "256 224 223",
"output": "1"
},
{
"input": "256 24 22",
"output": "2"
},
{
"input": "256 199 196",
"output": "3"
},
{
"input": "256 148 159",
"output": "4"
},
{
"input": "256 178 166",
"output": "5"
},
{
"input": "256 75 97",
"output": "6"
},
{
"input": "256 185 200",
"output": "7"
},
{
"input": "256 3 238",
"output": "Final!"
},
{
"input": "256 128 129",
"output": "Final!"
},
{
"input": "256 255 129",
"output": "7"
},
{
"input": "256 255 128",
"output": "Final!"
},
{
"input": "256 129 256",
"output": "7"
},
{
"input": "128 98 69",
"output": "6"
},
{
"input": "128 47 83",
"output": "Final!"
},
{
"input": "16 2 3",
"output": "2"
},
{
"input": "64 32 30",
"output": "2"
},
{
"input": "8 4 5",
"output": "Final!"
},
{
"input": "8 7 8",
"output": "1"
},
{
"input": "8 2 3",
"output": "2"
},
{
"input": "8 2 5",
"output": "Final!"
}
] | 1,565,283,130 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 109 | 0 | from math import log
n,a,b=(int(i) for i in input().split())
z=int(log(abs(b-a+1), 2))
print("Final!" if 2**z==n else z)
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The last stage of Football World Cup is played using the play-off system.
There are *n* teams left in this stage, they are enumerated from 1 to *n*. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids *a* and *b* can meet.
Input Specification:
The only line contains three integers *n*, *a* and *b* (2<=≤<=*n*<=≤<=256, 1<=≤<=*a*,<=*b*<=≤<=*n*) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that *n* is such that in each round an even number of team advance, and that *a* and *b* are not equal.
Output Specification:
In the only line print "Final!" (without quotes), if teams *a* and *b* can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams *a* and *b* can meet. The round are enumerated from 1.
Demo Input:
['4 1 2\n', '8 2 6\n', '8 7 5\n']
Demo Output:
['1\n', 'Final!\n', '2\n']
Note:
In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round. | ```python
from math import log
n,a,b=(int(i) for i in input().split())
z=int(log(abs(b-a+1), 2))
print("Final!" if 2**z==n else z)
``` | 0 | |
177 | B1 | Rectangular Game | PROGRAMMING | 1,000 | [
"number theory"
] | null | null | The Smart Beaver from ABBYY decided to have a day off. But doing nothing the whole day turned out to be too boring, and he decided to play a game with pebbles. Initially, the Beaver has *n* pebbles. He arranges them in *a* equal rows, each row has *b* pebbles (*a*<=><=1). Note that the Beaver must use all the pebbles he has, i. e. *n*<==<=*a*·*b*.
Once the Smart Beaver has arranged the pebbles, he takes back any of the resulting rows (that is, *b* pebbles) and discards all other pebbles. Then he arranges all his pebbles again (possibly choosing other values of *a* and *b*) and takes back one row, and so on. The game continues until at some point the Beaver ends up with exactly one pebble.
The game process can be represented as a finite sequence of integers *c*1,<=...,<=*c**k*, where:
- *c*1<==<=*n* - *c**i*<=+<=1 is the number of pebbles that the Beaver ends up with after the *i*-th move, that is, the number of pebbles in a row after some arrangement of *c**i* pebbles (1<=≤<=*i*<=<<=*k*). Note that *c**i*<=><=*c**i*<=+<=1. - *c**k*<==<=1
The result of the game is the sum of numbers *c**i*. You are given *n*. Find the maximum possible result of the game. | The single line of the input contains a single integer *n* — the initial number of pebbles the Smart Beaver has.
The input limitations for getting 30 points are:
- 2<=≤<=*n*<=≤<=50
The input limitations for getting 100 points are:
- 2<=≤<=*n*<=≤<=109 | Print a single number — the maximum possible result of the game. | [
"10\n",
"8\n"
] | [
"16\n",
"15\n"
] | Consider the first example (*c*<sub class="lower-index">1</sub> = 10). The possible options for the game development are:
- Arrange the pebbles in 10 rows, one pebble per row. Then *c*<sub class="lower-index">2</sub> = 1, and the game ends after the first move with the result of 11. - Arrange the pebbles in 5 rows, two pebbles per row. Then *c*<sub class="lower-index">2</sub> = 2, and the game continues. During the second move we have two pebbles which can be arranged in a unique way (remember that you are not allowed to put all the pebbles in the same row!) — 2 rows, one pebble per row. *c*<sub class="lower-index">3</sub> = 1, and the game ends with the result of 13. - Finally, arrange the pebbles in two rows, five pebbles per row. The same logic leads us to *c*<sub class="lower-index">2</sub> = 5, *c*<sub class="lower-index">3</sub> = 1, and the game ends with the result of 16 — the maximum possible result. | 30 | [
{
"input": "10",
"output": "16"
},
{
"input": "8",
"output": "15"
},
{
"input": "4",
"output": "7"
},
{
"input": "36",
"output": "67"
},
{
"input": "32",
"output": "63"
},
{
"input": "46",
"output": "70"
},
{
"input": "6",
"output": "10"
},
{
"input": "13",
"output": "14"
},
{
"input": "9",
"output": "13"
},
{
"input": "18",
"output": "31"
},
{
"input": "20",
"output": "36"
},
{
"input": "37",
"output": "38"
},
{
"input": "42",
"output": "71"
},
{
"input": "50",
"output": "81"
},
{
"input": "48",
"output": "94"
},
{
"input": "33",
"output": "45"
},
{
"input": "49",
"output": "57"
},
{
"input": "34",
"output": "52"
},
{
"input": "35",
"output": "43"
},
{
"input": "37",
"output": "38"
},
{
"input": "39",
"output": "53"
},
{
"input": "42",
"output": "71"
},
{
"input": "14",
"output": "22"
},
{
"input": "2",
"output": "3"
},
{
"input": "3",
"output": "4"
},
{
"input": "5",
"output": "6"
}
] | 1,630,517,364 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS1 | 0 | 62 | 6,963,200 | t=int(input())
max_result=t
def smallestDivisor(n):
# if divisible by 2
if (n % 2 == 0):
return 2;
# iterate from 3 to sqrt(n)
i = 3;
while(i * i <= n):
if (n % i == 0):
return i;
i += 2;
return n;
while t!=1:
s=smallestDivisor(t)
t=t/s
max_result+=t
print(max_result) | Title: Rectangular Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Smart Beaver from ABBYY decided to have a day off. But doing nothing the whole day turned out to be too boring, and he decided to play a game with pebbles. Initially, the Beaver has *n* pebbles. He arranges them in *a* equal rows, each row has *b* pebbles (*a*<=><=1). Note that the Beaver must use all the pebbles he has, i. e. *n*<==<=*a*·*b*.
Once the Smart Beaver has arranged the pebbles, he takes back any of the resulting rows (that is, *b* pebbles) and discards all other pebbles. Then he arranges all his pebbles again (possibly choosing other values of *a* and *b*) and takes back one row, and so on. The game continues until at some point the Beaver ends up with exactly one pebble.
The game process can be represented as a finite sequence of integers *c*1,<=...,<=*c**k*, where:
- *c*1<==<=*n* - *c**i*<=+<=1 is the number of pebbles that the Beaver ends up with after the *i*-th move, that is, the number of pebbles in a row after some arrangement of *c**i* pebbles (1<=≤<=*i*<=<<=*k*). Note that *c**i*<=><=*c**i*<=+<=1. - *c**k*<==<=1
The result of the game is the sum of numbers *c**i*. You are given *n*. Find the maximum possible result of the game.
Input Specification:
The single line of the input contains a single integer *n* — the initial number of pebbles the Smart Beaver has.
The input limitations for getting 30 points are:
- 2<=≤<=*n*<=≤<=50
The input limitations for getting 100 points are:
- 2<=≤<=*n*<=≤<=109
Output Specification:
Print a single number — the maximum possible result of the game.
Demo Input:
['10\n', '8\n']
Demo Output:
['16\n', '15\n']
Note:
Consider the first example (*c*<sub class="lower-index">1</sub> = 10). The possible options for the game development are:
- Arrange the pebbles in 10 rows, one pebble per row. Then *c*<sub class="lower-index">2</sub> = 1, and the game ends after the first move with the result of 11. - Arrange the pebbles in 5 rows, two pebbles per row. Then *c*<sub class="lower-index">2</sub> = 2, and the game continues. During the second move we have two pebbles which can be arranged in a unique way (remember that you are not allowed to put all the pebbles in the same row!) — 2 rows, one pebble per row. *c*<sub class="lower-index">3</sub> = 1, and the game ends with the result of 13. - Finally, arrange the pebbles in two rows, five pebbles per row. The same logic leads us to *c*<sub class="lower-index">2</sub> = 5, *c*<sub class="lower-index">3</sub> = 1, and the game ends with the result of 16 — the maximum possible result. | ```python
t=int(input())
max_result=t
def smallestDivisor(n):
# if divisible by 2
if (n % 2 == 0):
return 2;
# iterate from 3 to sqrt(n)
i = 3;
while(i * i <= n):
if (n % i == 0):
return i;
i += 2;
return n;
while t!=1:
s=smallestDivisor(t)
t=t/s
max_result+=t
print(max_result)
``` | 0 | |
777 | A | Shell Game | PROGRAMMING | 1,000 | [
"constructive algorithms",
"implementation",
"math"
] | null | null | Bomboslav likes to look out of the window in his room and watch lads outside playing famous shell game. The game is played by two persons: operator and player. Operator takes three similar opaque shells and places a ball beneath one of them. Then he shuffles the shells by swapping some pairs and the player has to guess the current position of the ball.
Bomboslav noticed that guys are not very inventive, so the operator always swaps the left shell with the middle one during odd moves (first, third, fifth, etc.) and always swaps the middle shell with the right one during even moves (second, fourth, etc.).
Let's number shells from 0 to 2 from left to right. Thus the left shell is assigned number 0, the middle shell is 1 and the right shell is 2. Bomboslav has missed the moment when the ball was placed beneath the shell, but he knows that exactly *n* movements were made by the operator and the ball was under shell *x* at the end. Now he wonders, what was the initial position of the ball? | The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=2·109) — the number of movements made by the operator.
The second line contains a single integer *x* (0<=≤<=*x*<=≤<=2) — the index of the shell where the ball was found after *n* movements. | Print one integer from 0 to 2 — the index of the shell where the ball was initially placed. | [
"4\n2\n",
"1\n1\n"
] | [
"1\n",
"0\n"
] | In the first sample, the ball was initially placed beneath the middle shell and the operator completed four movements.
1. During the first move operator swapped the left shell and the middle shell. The ball is now under the left shell. 1. During the second move operator swapped the middle shell and the right one. The ball is still under the left shell. 1. During the third move operator swapped the left shell and the middle shell again. The ball is again in the middle. 1. Finally, the operators swapped the middle shell and the right shell. The ball is now beneath the right shell. | 500 | [
{
"input": "4\n2",
"output": "1"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "2\n2",
"output": "0"
},
{
"input": "3\n1",
"output": "1"
},
{
"input": "3\n2",
"output": "0"
},
{
"input": "3\n0",
"output": "2"
},
{
"input": "2000000000\n0",
"output": "1"
},
{
"input": "2\n0",
"output": "1"
},
{
"input": "2\n1",
"output": "2"
},
{
"input": "4\n0",
"output": "2"
},
{
"input": "4\n1",
"output": "0"
},
{
"input": "5\n0",
"output": "0"
},
{
"input": "5\n1",
"output": "2"
},
{
"input": "5\n2",
"output": "1"
},
{
"input": "6\n0",
"output": "0"
},
{
"input": "6\n1",
"output": "1"
},
{
"input": "6\n2",
"output": "2"
},
{
"input": "7\n0",
"output": "1"
},
{
"input": "7\n1",
"output": "0"
},
{
"input": "7\n2",
"output": "2"
},
{
"input": "100000\n0",
"output": "2"
},
{
"input": "100000\n1",
"output": "0"
},
{
"input": "100000\n2",
"output": "1"
},
{
"input": "99999\n1",
"output": "1"
},
{
"input": "99998\n1",
"output": "2"
},
{
"input": "99997\n1",
"output": "0"
},
{
"input": "99996\n1",
"output": "1"
},
{
"input": "99995\n1",
"output": "2"
},
{
"input": "1999999995\n0",
"output": "2"
},
{
"input": "1999999995\n1",
"output": "1"
},
{
"input": "1999999995\n2",
"output": "0"
},
{
"input": "1999999996\n0",
"output": "2"
},
{
"input": "1999999996\n1",
"output": "0"
},
{
"input": "1999999996\n2",
"output": "1"
},
{
"input": "1999999997\n0",
"output": "0"
},
{
"input": "1999999997\n1",
"output": "2"
},
{
"input": "1999999997\n2",
"output": "1"
},
{
"input": "1999999998\n0",
"output": "0"
},
{
"input": "1999999998\n1",
"output": "1"
},
{
"input": "1999999998\n2",
"output": "2"
},
{
"input": "1999999999\n0",
"output": "1"
},
{
"input": "1999999999\n1",
"output": "0"
},
{
"input": "1999999999\n2",
"output": "2"
},
{
"input": "2000000000\n1",
"output": "2"
},
{
"input": "2000000000\n2",
"output": "0"
},
{
"input": "1234567890\n0",
"output": "0"
},
{
"input": "1234567890\n1",
"output": "1"
},
{
"input": "1234567890\n2",
"output": "2"
},
{
"input": "123456789\n0",
"output": "2"
},
{
"input": "123456789\n1",
"output": "1"
},
{
"input": "123456789\n2",
"output": "0"
},
{
"input": "123456790\n0",
"output": "2"
},
{
"input": "12\n2",
"output": "2"
},
{
"input": "32\n1",
"output": "2"
},
{
"input": "20\n2",
"output": "0"
},
{
"input": "10\n1",
"output": "0"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "76994383\n1",
"output": "0"
},
{
"input": "25\n2",
"output": "2"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "12\n0",
"output": "0"
},
{
"input": "150\n2",
"output": "2"
},
{
"input": "15\n0",
"output": "2"
},
{
"input": "21\n2",
"output": "0"
},
{
"input": "18\n2",
"output": "2"
},
{
"input": "8\n2",
"output": "0"
},
{
"input": "10\n0",
"output": "2"
},
{
"input": "16\n0",
"output": "2"
}
] | 1,627,132,919 | 2,147,483,647 | PyPy 3 | OK | TESTS | 68 | 109 | 20,172,800 | import math
def main_function():
n = int(input())
initial_position = int(input())
rem = n % 6
last_move = n % 2
if last_move == 0:
first_move = False
else:
first_move = True
f = [0, 0, 0]
f[initial_position] = 1
for i in range(rem):
if first_move:
f[0], f[1] = f[1], f[0]
else:
f[1], f[2] = f[2], f[1]
first_move = not first_move
for i in range(len(f)):
if f[i] == 1:
print(i)
main_function() | Title: Shell Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bomboslav likes to look out of the window in his room and watch lads outside playing famous shell game. The game is played by two persons: operator and player. Operator takes three similar opaque shells and places a ball beneath one of them. Then he shuffles the shells by swapping some pairs and the player has to guess the current position of the ball.
Bomboslav noticed that guys are not very inventive, so the operator always swaps the left shell with the middle one during odd moves (first, third, fifth, etc.) and always swaps the middle shell with the right one during even moves (second, fourth, etc.).
Let's number shells from 0 to 2 from left to right. Thus the left shell is assigned number 0, the middle shell is 1 and the right shell is 2. Bomboslav has missed the moment when the ball was placed beneath the shell, but he knows that exactly *n* movements were made by the operator and the ball was under shell *x* at the end. Now he wonders, what was the initial position of the ball?
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=2·109) — the number of movements made by the operator.
The second line contains a single integer *x* (0<=≤<=*x*<=≤<=2) — the index of the shell where the ball was found after *n* movements.
Output Specification:
Print one integer from 0 to 2 — the index of the shell where the ball was initially placed.
Demo Input:
['4\n2\n', '1\n1\n']
Demo Output:
['1\n', '0\n']
Note:
In the first sample, the ball was initially placed beneath the middle shell and the operator completed four movements.
1. During the first move operator swapped the left shell and the middle shell. The ball is now under the left shell. 1. During the second move operator swapped the middle shell and the right one. The ball is still under the left shell. 1. During the third move operator swapped the left shell and the middle shell again. The ball is again in the middle. 1. Finally, the operators swapped the middle shell and the right shell. The ball is now beneath the right shell. | ```python
import math
def main_function():
n = int(input())
initial_position = int(input())
rem = n % 6
last_move = n % 2
if last_move == 0:
first_move = False
else:
first_move = True
f = [0, 0, 0]
f[initial_position] = 1
for i in range(rem):
if first_move:
f[0], f[1] = f[1], f[0]
else:
f[1], f[2] = f[2], f[1]
first_move = not first_move
for i in range(len(f)):
if f[i] == 1:
print(i)
main_function()
``` | 3 | |
629 | A | Far Relative’s Birthday Cake | PROGRAMMING | 800 | [
"brute force",
"combinatorics",
"constructive algorithms",
"implementation"
] | null | null | Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird!
The cake is a *n*<=×<=*n* square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be?
Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column. | In the first line of the input, you are given a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the side of the cake.
Then follow *n* lines, each containing *n* characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'. | Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column. | [
"3\n.CC\nC..\nC.C\n",
"4\nCC..\nC..C\n.CC.\n.CC.\n"
] | [
"4\n",
"9\n"
] | If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are:
1. (1, 2) and (1, 3) 1. (3, 1) and (3, 3) 1. (2, 1) and (3, 1) 1. (1, 3) and (3, 3) | 500 | [
{
"input": "3\n.CC\nC..\nC.C",
"output": "4"
},
{
"input": "4\nCC..\nC..C\n.CC.\n.CC.",
"output": "9"
},
{
"input": "5\n.CCCC\nCCCCC\n.CCC.\nCC...\n.CC.C",
"output": "46"
},
{
"input": "7\n.CC..CC\nCC.C..C\nC.C..C.\nC...C.C\nCCC.CCC\n.CC...C\n.C.CCC.",
"output": "84"
},
{
"input": "8\n..C....C\nC.CCC.CC\n.C..C.CC\nCC......\nC..C..CC\nC.C...C.\nC.C..C..\nC...C.C.",
"output": "80"
},
{
"input": "9\n.C...CCCC\nC.CCCC...\n....C..CC\n.CC.CCC..\n.C.C..CC.\nC...C.CCC\nCCC.C...C\nCCCC....C\n..C..C..C",
"output": "144"
},
{
"input": "10\n..C..C.C..\n..CC..C.CC\n.C.C...C.C\n..C.CC..CC\n....C..C.C\n...C..C..C\nCC.CC....C\n..CCCC.C.C\n..CC.CCC..\nCCCC..C.CC",
"output": "190"
},
{
"input": "11\nC.CC...C.CC\nCC.C....C.C\n.....C..CCC\n....C.CC.CC\nC..C..CC...\nC...C...C..\nCC..CCC.C.C\n..C.CC.C..C\nC...C.C..CC\n.C.C..CC..C\n.C.C.CC.C..",
"output": "228"
},
{
"input": "21\n...CCC.....CC..C..C.C\n..CCC...CC...CC.CCC.C\n....C.C.C..CCC..C.C.C\n....CCC..C..C.CC.CCC.\n...CCC.C..C.C.....CCC\n.CCC.....CCC..C...C.C\nCCCC.C...CCC.C...C.CC\nC..C...C.CCC..CC..C..\nC...CC..C.C.CC..C.CC.\nCC..CCCCCCCCC..C....C\n.C..CCCC.CCCC.CCC...C\nCCC...CCC...CCC.C..C.\n.CCCCCCCC.CCCC.CC.C..\n.C.C..C....C.CCCCCC.C\n...C...C.CCC.C.CC..C.\nCCC...CC..CC...C..C.C\n.CCCCC...C.C..C.CC.C.\n..CCC.C.C..CCC.CCC...\n..C..C.C.C.....CC.C..\n.CC.C...C.CCC.C....CC\n...C..CCCC.CCC....C..",
"output": "2103"
},
{
"input": "20\nC.C.CCC.C....C.CCCCC\nC.CC.C..CCC....CCCC.\n.CCC.CC...CC.CCCCCC.\n.C...CCCC..C....CCC.\n.C..CCCCCCC.C.C.....\nC....C.C..CCC.C..CCC\n...C.C.CC..CC..CC...\nC...CC.C.CCCCC....CC\n.CC.C.CCC....C.CCC.C\nCC...CC...CC..CC...C\nC.C..CC.C.CCCC.C.CC.\n..CCCCC.C.CCC..CCCC.\n....C..C..C.CC...C.C\nC..CCC..CC..C.CC..CC\n...CC......C.C..C.C.\nCC.CCCCC.CC.CC...C.C\n.C.CC..CC..CCC.C.CCC\nC..C.CC....C....C...\n..CCC..CCC...CC..C.C\n.C.CCC.CCCCCCCCC..CC",
"output": "2071"
},
{
"input": "17\nCCC..C.C....C.C.C\n.C.CC.CC...CC..C.\n.CCCC.CC.C..CCC.C\n...CCC.CC.CCC.C.C\nCCCCCCCC..C.CC.CC\n...C..C....C.CC.C\nCC....CCC...C.CC.\n.CC.C.CC..C......\n.CCCCC.C.CC.CCCCC\n..CCCC...C..CC..C\nC.CC.C.CC..C.C.C.\nC..C..C..CCC.C...\n.C..CCCC..C......\n.CC.C...C..CC.CC.\nC..C....CC...CC..\nC.CC.CC..C.C..C..\nCCCC...C.C..CCCC.",
"output": "1160"
},
{
"input": "15\nCCCC.C..CCC....\nCCCCCC.CC.....C\n...C.CC.C.C.CC.\nCCCCCCC..C..C..\nC..CCC..C.CCCC.\n.CC..C.C.C.CC.C\n.C.C..C..C.C..C\n...C...C..CCCC.\n.....C.C..CC...\nCC.C.C..CC.C..C\n..CCCCC..CCC...\nCC.CC.C..CC.CCC\n..CCC...CC.C..C\nCC..C.C..CCC..C\n.C.C....CCC...C",
"output": "789"
},
{
"input": "1\n.",
"output": "0"
},
{
"input": "3\n.CC\nC..\nC.C",
"output": "4"
},
{
"input": "13\nC.C...C.C.C..\nCC.CCCC.CC..C\n.C.CCCCC.CC..\nCCCC..C...C..\n...CC.C.C...C\n.CC.CCC...CC.\nCC.CCCCCC....\n.C...C..CC..C\nCCCC.CC...C..\n.C.CCC..C.CC.\n..C...CC..C.C\n..C.CCC..CC.C\n.C...CCC.CC.C",
"output": "529"
},
{
"input": "16\n.C.C.C.C.C...C.C\n..C..C.CCCCCC...\n..C.C.C.C..C..C.\n.CC....C.CCC..C.\n.C.CCC..C....CCC\nCC..C.CC..C.C.CC\n...C..C..CC..CC.\n.CCC..C.CC.C.C..\n.CC.C..........C\nC...C....CC..C..\nC.CCC.C..C..C...\n.CCCCCCCCCCCC..C\n..C.C.CC.CC.CCC.\nCC..C.C....C..CC\nC.CCC..C..C.C.CC\n.C.CCC.CC..CCC.C",
"output": "874"
},
{
"input": "2\nCC\nCC",
"output": "4"
},
{
"input": "3\nC..\nC..\nC..",
"output": "3"
}
] | 1,628,698,651 | 2,147,483,647 | Python 3 | OK | TESTS | 48 | 77 | 6,963,200 | def fact(n):
res = 1
for i in range(2,n+1):
res *= i
return res
def comb(n,r):
if n < r:
return 0
return fact(n)/(fact(n-r)*fact(r))
n = int(input())
cop = n
cake = []
for i in range(n):
cake.append(input())
rows = []
cols = []
for i in range(n):
rows.append(cake[i].count('C'))
c = 0
while cop>0:
col = 0
for i in range(n):
if cake[i][c] == 'C':
col += 1
c += 1
cop -= 1
cols.append(col)
result = 0
for i in range(n):
result += comb(rows[i],2)
result += comb(cols[i],2)
print(int(result)) | Title: Far Relative’s Birthday Cake
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird!
The cake is a *n*<=×<=*n* square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be?
Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column.
Input Specification:
In the first line of the input, you are given a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the side of the cake.
Then follow *n* lines, each containing *n* characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'.
Output Specification:
Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column.
Demo Input:
['3\n.CC\nC..\nC.C\n', '4\nCC..\nC..C\n.CC.\n.CC.\n']
Demo Output:
['4\n', '9\n']
Note:
If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are:
1. (1, 2) and (1, 3) 1. (3, 1) and (3, 3) 1. (2, 1) and (3, 1) 1. (1, 3) and (3, 3) | ```python
def fact(n):
res = 1
for i in range(2,n+1):
res *= i
return res
def comb(n,r):
if n < r:
return 0
return fact(n)/(fact(n-r)*fact(r))
n = int(input())
cop = n
cake = []
for i in range(n):
cake.append(input())
rows = []
cols = []
for i in range(n):
rows.append(cake[i].count('C'))
c = 0
while cop>0:
col = 0
for i in range(n):
if cake[i][c] == 'C':
col += 1
c += 1
cop -= 1
cols.append(col)
result = 0
for i in range(n):
result += comb(rows[i],2)
result += comb(cols[i],2)
print(int(result))
``` | 3 | |
858 | D | Polycarp's phone book | PROGRAMMING | 1,600 | [
"data structures",
"implementation",
"sortings"
] | null | null | There are *n* phone numbers in Polycarp's contacts on his phone. Each number is a 9-digit integer, starting with a digit different from 0. All the numbers are distinct.
There is the latest version of Berdroid OS installed on Polycarp's phone. If some number is entered, is shows up all the numbers in the contacts for which there is a substring equal to the entered sequence of digits. For example, is there are three phone numbers in Polycarp's contacts: 123456789, 100000000 and 100123456, then:
- if he enters 00 two numbers will show up: 100000000 and 100123456, - if he enters 123 two numbers will show up 123456789 and 100123456, - if he enters 01 there will be only one number 100123456.
For each of the phone numbers in Polycarp's contacts, find the minimum in length sequence of digits such that if Polycarp enters this sequence, Berdroid shows this only phone number. | The first line contains single integer *n* (1<=≤<=*n*<=≤<=70000) — the total number of phone contacts in Polycarp's contacts.
The phone numbers follow, one in each line. Each number is a positive 9-digit integer starting with a digit from 1 to 9. All the numbers are distinct. | Print exactly *n* lines: the *i*-th of them should contain the shortest non-empty sequence of digits, such that if Polycarp enters it, the Berdroid OS shows up only the *i*-th number from the contacts. If there are several such sequences, print any of them. | [
"3\n123456789\n100000000\n100123456\n",
"4\n123456789\n193456789\n134567819\n934567891\n"
] | [
"9\n000\n01\n",
"2\n193\n81\n91\n"
] | none | 2,000 | [
{
"input": "3\n123456789\n100000000\n100123456",
"output": "9\n000\n01"
},
{
"input": "4\n123456789\n193456789\n134567819\n934567891",
"output": "2\n193\n81\n91"
},
{
"input": "1\n167038488",
"output": "4"
},
{
"input": "5\n115830748\n403459907\n556271610\n430358099\n413961410",
"output": "15\n40\n2\n35\n14"
},
{
"input": "5\n139127034\n452751056\n193432721\n894001929\n426470953",
"output": "39\n05\n32\n8\n53"
},
{
"input": "5\n343216531\n914073407\n420246472\n855857272\n801664978",
"output": "32\n07\n46\n27\n78"
},
{
"input": "5\n567323818\n353474649\n468171032\n989223926\n685081078",
"output": "67\n35\n03\n26\n78"
},
{
"input": "5\n774610315\n325796798\n989859836\n707706423\n310546337",
"output": "61\n32\n89\n23\n37"
},
{
"input": "10\n181033039\n210698534\n971006898\n391227170\n323096464\n560766866\n377374442\n654389922\n544146403\n779261493",
"output": "18\n53\n97\n27\n09\n07\n42\n99\n41\n93"
},
{
"input": "10\n197120216\n680990683\n319631438\n442393410\n888300189\n170777450\n164487872\n487350759\n651751346\n652859411",
"output": "12\n09\n43\n42\n01\n74\n72\n73\n46\n11"
},
{
"input": "10\n302417715\n621211824\n474451896\n961495400\n633841010\n839982537\n797812119\n510708100\n770758643\n228046084",
"output": "15\n62\n47\n14\n33\n98\n97\n100\n43\n60"
},
{
"input": "10\n506504092\n561611075\n265260859\n557114891\n838578824\n985006846\n456984731\n856424964\n658005674\n666280709",
"output": "40\n75\n60\n55\n78\n68\n31\n96\n74\n66"
},
{
"input": "10\n510613599\n931933224\n693094490\n508960931\n313762868\n396027639\n164098962\n749880019\n709024305\n498545812",
"output": "61\n22\n44\n08\n68\n39\n64\n88\n43\n12"
},
{
"input": "3\n638631659\n929648227\n848163730",
"output": "5\n2\n0"
},
{
"input": "4\n898855826\n343430636\n210120107\n467957087",
"output": "98\n3\n1\n57"
},
{
"input": "5\n202080398\n357502772\n269676952\n711559315\n111366203",
"output": "8\n57\n26\n93\n62"
},
{
"input": "9\n111111111\n111111110\n111111100\n111111000\n111110000\n111100000\n111000000\n110000000\n100000000",
"output": "111111111\n111111110\n111111100\n111111000\n111110000\n111100000\n111000000\n110000000\n00000000"
}
] | 1,508,006,176 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 2 | 4,000 | 3,379,200 | n=int(input())
m=[]
for i in range(n):
m.append(input())
p=[]
for y in m:
d=y
for i in range(len(y)+1):
for j in range(i+1,len(y)+1):
c=0
for x in m:
if y[i:j] in x:
c+=1
if (c==1) and (len(y[i:j])<len(d)):
d=y[i:j]
p.append(d)
[print(i) for i in p] | Title: Polycarp's phone book
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* phone numbers in Polycarp's contacts on his phone. Each number is a 9-digit integer, starting with a digit different from 0. All the numbers are distinct.
There is the latest version of Berdroid OS installed on Polycarp's phone. If some number is entered, is shows up all the numbers in the contacts for which there is a substring equal to the entered sequence of digits. For example, is there are three phone numbers in Polycarp's contacts: 123456789, 100000000 and 100123456, then:
- if he enters 00 two numbers will show up: 100000000 and 100123456, - if he enters 123 two numbers will show up 123456789 and 100123456, - if he enters 01 there will be only one number 100123456.
For each of the phone numbers in Polycarp's contacts, find the minimum in length sequence of digits such that if Polycarp enters this sequence, Berdroid shows this only phone number.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=70000) — the total number of phone contacts in Polycarp's contacts.
The phone numbers follow, one in each line. Each number is a positive 9-digit integer starting with a digit from 1 to 9. All the numbers are distinct.
Output Specification:
Print exactly *n* lines: the *i*-th of them should contain the shortest non-empty sequence of digits, such that if Polycarp enters it, the Berdroid OS shows up only the *i*-th number from the contacts. If there are several such sequences, print any of them.
Demo Input:
['3\n123456789\n100000000\n100123456\n', '4\n123456789\n193456789\n134567819\n934567891\n']
Demo Output:
['9\n000\n01\n', '2\n193\n81\n91\n']
Note:
none | ```python
n=int(input())
m=[]
for i in range(n):
m.append(input())
p=[]
for y in m:
d=y
for i in range(len(y)+1):
for j in range(i+1,len(y)+1):
c=0
for x in m:
if y[i:j] in x:
c+=1
if (c==1) and (len(y[i:j])<len(d)):
d=y[i:j]
p.append(d)
[print(i) for i in p]
``` | 0 | |
599 | A | Patrick and Shopping | PROGRAMMING | 800 | [
"implementation"
] | null | null | Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled. | The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops. | Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house. | [
"10 20 30\n",
"1 1 5\n"
] | [
"60\n",
"4\n"
] | The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house. | 500 | [
{
"input": "10 20 30",
"output": "60"
},
{
"input": "1 1 5",
"output": "4"
},
{
"input": "100 33 34",
"output": "134"
},
{
"input": "777 777 777",
"output": "2331"
},
{
"input": "2 2 8",
"output": "8"
},
{
"input": "12 34 56",
"output": "92"
},
{
"input": "789 101112 131415",
"output": "203802"
},
{
"input": "27485716 99999999 35182",
"output": "55041796"
},
{
"input": "1 293548 5",
"output": "12"
},
{
"input": "12059 259855 5874875",
"output": "543828"
},
{
"input": "46981 105809 585858",
"output": "305580"
},
{
"input": "9889 1221 2442",
"output": "7326"
},
{
"input": "100500 200600 300700",
"output": "601800"
},
{
"input": "318476 318476 318476",
"output": "955428"
},
{
"input": "23985 3353 75633",
"output": "54676"
},
{
"input": "120 1298 2222",
"output": "2836"
},
{
"input": "98437 23487 666672",
"output": "243848"
},
{
"input": "100000000 100000000 100000000",
"output": "300000000"
},
{
"input": "2 5 2",
"output": "8"
},
{
"input": "1 1000 1",
"output": "4"
},
{
"input": "1 100000000 1",
"output": "4"
}
] | 1,599,405,238 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 7 | 109 | 307,200 | #A. Patrick and Shopping
Data = [int (x) for x in input().split()]
d1 = Data[0]
d2 = Data[1]
d3 = Data[2]
#Route 1: H->S1->S2
#From S1 to S2
if d3 <= d1 + d2:
ToS2 = d3
else:
ToS2 = d1 + d2
#From S2 to home
if d1 + d3 <= d2:
ToHome = d1 + d3
else:
R1ToHome = d2
R1TotalD = d1 + ToS2 + R1ToHome
#Route 2: H->S2->S1
#From S2 to S1
if d3 <= d2 + d1:
ToS1 = d3
else:
ToS1 = d2 + d1
#From S2 to home
if d1 <= d2 + d3:
R2ToHome = d1
else:
R2ToHome = d2 + d3
R2TotalD = d2 + ToS1 + R2ToHome
if R1TotalD <= R2TotalD:
print (R1TotalD)
else:
print (R2TotalD)
| Title: Patrick and Shopping
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.
Input Specification:
The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.
Output Specification:
Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.
Demo Input:
['10 20 30\n', '1 1 5\n']
Demo Output:
['60\n', '4\n']
Note:
The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house. | ```python
#A. Patrick and Shopping
Data = [int (x) for x in input().split()]
d1 = Data[0]
d2 = Data[1]
d3 = Data[2]
#Route 1: H->S1->S2
#From S1 to S2
if d3 <= d1 + d2:
ToS2 = d3
else:
ToS2 = d1 + d2
#From S2 to home
if d1 + d3 <= d2:
ToHome = d1 + d3
else:
R1ToHome = d2
R1TotalD = d1 + ToS2 + R1ToHome
#Route 2: H->S2->S1
#From S2 to S1
if d3 <= d2 + d1:
ToS1 = d3
else:
ToS1 = d2 + d1
#From S2 to home
if d1 <= d2 + d3:
R2ToHome = d1
else:
R2ToHome = d2 + d3
R2TotalD = d2 + ToS1 + R2ToHome
if R1TotalD <= R2TotalD:
print (R1TotalD)
else:
print (R2TotalD)
``` | -1 | |
386 | A | Second-Price Auction | PROGRAMMING | 800 | [
"implementation"
] | null | null | In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different. | The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder. | The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based. | [
"2\n5 7\n",
"3\n10 2 8\n",
"6\n3 8 2 9 4 14\n"
] | [
"2 5\n",
"1 8\n",
"6 9\n"
] | none | 500 | [
{
"input": "2\n5 7",
"output": "2 5"
},
{
"input": "3\n10 2 8",
"output": "1 8"
},
{
"input": "6\n3 8 2 9 4 14",
"output": "6 9"
},
{
"input": "4\n4707 7586 4221 5842",
"output": "2 5842"
},
{
"input": "5\n3304 4227 4869 6937 6002",
"output": "4 6002"
},
{
"input": "6\n5083 3289 7708 5362 9031 7458",
"output": "5 7708"
},
{
"input": "7\n9038 6222 3392 1706 3778 1807 2657",
"output": "1 6222"
},
{
"input": "8\n7062 2194 4481 3864 7470 1814 8091 733",
"output": "7 7470"
},
{
"input": "9\n2678 5659 9199 2628 7906 7496 4524 2663 3408",
"output": "3 7906"
},
{
"input": "2\n3458 1504",
"output": "1 1504"
},
{
"input": "50\n9237 3904 407 9052 6657 9229 9752 3888 7732 2512 4614 1055 2355 7108 6506 6849 2529 8862 159 8630 7906 7941 960 8470 333 8659 54 9475 3163 5625 6393 6814 2656 3388 169 7918 4881 8468 9983 6281 6340 280 5108 2996 101 7617 3313 8172 326 1991",
"output": "39 9752"
},
{
"input": "100\n2515 3324 7975 6171 4240 1217 4829 5203 8603 6900 3031 4699 4732 6070 4221 3228 6497 7359 9130 4346 4619 1109 3945 5442 3271 16 9711 2045 6410 2301 3406 8125 6003 1892 1260 9661 3940 6692 4708 7027 4930 6925 1979 5361 4263 3144 867 8639 6230 5562 9714 3676 4231 3347 4733 4920 4881 3431 1059 7313 8912 3038 9308 72 9583 7009 3034 7425 2398 6141 3245 2495 2933 6710 8289 9806 1226 8393 7349 6462 1618 9613 3546 6012 2964 9995 1578 210 2123 4874 1252 8625 348 8020 803 7244 9080 5088 706 2602",
"output": "86 9806"
},
{
"input": "2\n2 1",
"output": "1 1"
},
{
"input": "2\n1 2",
"output": "2 1"
},
{
"input": "3\n10 20 30",
"output": "3 20"
},
{
"input": "3\n10 30 20",
"output": "2 20"
},
{
"input": "3\n20 10 30",
"output": "3 20"
},
{
"input": "3\n20 30 10",
"output": "2 20"
},
{
"input": "3\n30 10 20",
"output": "1 20"
},
{
"input": "3\n30 20 10",
"output": "1 20"
},
{
"input": "2\n1 10000",
"output": "2 1"
},
{
"input": "2\n10000 999",
"output": "1 999"
},
{
"input": "3\n3 4 1",
"output": "2 3"
},
{
"input": "6\n1 2 24 6 5 7",
"output": "3 7"
}
] | 1,627,572,826 | 2,147,483,647 | Python 3 | OK | TESTS | 42 | 77 | 6,963,200 | # Second-Price Auction
n = int(input())
p = list(map(int,input().split()))
y = max(p)
x = p.index(y)+1
p.remove(y)
print(x,max(p))
| Title: Second-Price Auction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different.
Input Specification:
The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder.
Output Specification:
The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based.
Demo Input:
['2\n5 7\n', '3\n10 2 8\n', '6\n3 8 2 9 4 14\n']
Demo Output:
['2 5\n', '1 8\n', '6 9\n']
Note:
none | ```python
# Second-Price Auction
n = int(input())
p = list(map(int,input().split()))
y = max(p)
x = p.index(y)+1
p.remove(y)
print(x,max(p))
``` | 3 | |
463 | B | Caisa and Pylons | PROGRAMMING | 1,100 | [
"brute force",
"implementation",
"math"
] | null | null | Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (*n*<=+<=1) pylons numbered from 0 to *n* in this game. The pylon with number 0 has zero height, the pylon with number *i* (*i*<=><=0) has height *h**i*. The goal of the game is to reach *n*-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as *k*) to the next one (its number will be *k*<=+<=1). When the player have made such a move, its energy increases by *h**k*<=-<=*h**k*<=+<=1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game? | The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *h*1, *h*2,<=..., *h**n* (1<=<=≤<=<=*h**i*<=<=≤<=<=105) representing the heights of the pylons. | Print a single number representing the minimum number of dollars paid by Caisa. | [
"5\n3 4 3 2 4\n",
"3\n4 4 4\n"
] | [
"4\n",
"4\n"
] | In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon. | 1,000 | [
{
"input": "5\n3 4 3 2 4",
"output": "4"
},
{
"input": "3\n4 4 4",
"output": "4"
},
{
"input": "99\n1401 2019 1748 3785 3236 3177 3443 3772 2138 1049 353 908 310 2388 1322 88 2160 2783 435 2248 1471 706 2468 2319 3156 3506 2794 1999 1983 2519 2597 3735 537 344 3519 3772 3872 2961 3895 2010 10 247 3269 671 2986 942 758 1146 77 1545 3745 1547 2250 2565 217 1406 2070 3010 3404 404 1528 2352 138 2065 3047 3656 2188 2919 2616 2083 1280 2977 2681 548 4000 1667 1489 1109 3164 1565 2653 3260 3463 903 1824 3679 2308 245 2689 2063 648 568 766 785 2984 3812 440 1172 2730",
"output": "4000"
},
{
"input": "68\n477 1931 3738 3921 2306 1823 3328 2057 661 3993 2967 3520 171 1739 1525 1817 209 3475 1902 2666 518 3283 3412 3040 3383 2331 1147 1460 1452 1800 1327 2280 82 1416 2200 2388 3238 1879 796 250 1872 114 121 2042 1853 1645 211 2061 1472 2464 726 1989 1746 489 1380 1128 2819 2527 2939 622 678 265 2902 1111 2032 1453 3850 1621",
"output": "3993"
},
{
"input": "30\n30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "30"
},
{
"input": "3\n3 2 1",
"output": "3"
},
{
"input": "1\n69",
"output": "69"
}
] | 1,628,683,296 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 140 | 14,438,400 | n = int(input())
h = list(map(int, input().split()))
h.insert(0, 0)
e = 0
c = 0
for i in range(n):
t = h[i] - h[i + 1]
if t + e < 0:
c += abs(t + e)
e = 0
else:
e += t
print(c)
| Title: Caisa and Pylons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (*n*<=+<=1) pylons numbered from 0 to *n* in this game. The pylon with number 0 has zero height, the pylon with number *i* (*i*<=><=0) has height *h**i*. The goal of the game is to reach *n*-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as *k*) to the next one (its number will be *k*<=+<=1). When the player have made such a move, its energy increases by *h**k*<=-<=*h**k*<=+<=1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *h*1, *h*2,<=..., *h**n* (1<=<=≤<=<=*h**i*<=<=≤<=<=105) representing the heights of the pylons.
Output Specification:
Print a single number representing the minimum number of dollars paid by Caisa.
Demo Input:
['5\n3 4 3 2 4\n', '3\n4 4 4\n']
Demo Output:
['4\n', '4\n']
Note:
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon. | ```python
n = int(input())
h = list(map(int, input().split()))
h.insert(0, 0)
e = 0
c = 0
for i in range(n):
t = h[i] - h[i + 1]
if t + e < 0:
c += abs(t + e)
e = 0
else:
e += t
print(c)
``` | 3 | |
34 | A | Reconnaissance 2 | PROGRAMMING | 800 | [
"implementation"
] | A. Reconnaissance 2 | 2 | 256 | *n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit. | The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction. | Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle. | [
"5\n10 12 13 15 10\n",
"4\n10 20 30 40\n"
] | [
"5 1\n",
"1 2\n"
] | none | 500 | [
{
"input": "5\n10 12 13 15 10",
"output": "5 1"
},
{
"input": "4\n10 20 30 40",
"output": "1 2"
},
{
"input": "6\n744 359 230 586 944 442",
"output": "2 3"
},
{
"input": "5\n826 747 849 687 437",
"output": "1 2"
},
{
"input": "5\n999 999 993 969 999",
"output": "1 2"
},
{
"input": "5\n4 24 6 1 15",
"output": "3 4"
},
{
"input": "2\n511 32",
"output": "1 2"
},
{
"input": "3\n907 452 355",
"output": "2 3"
},
{
"input": "4\n303 872 764 401",
"output": "4 1"
},
{
"input": "10\n684 698 429 694 956 812 594 170 937 764",
"output": "1 2"
},
{
"input": "20\n646 840 437 946 640 564 936 917 487 752 844 734 468 969 674 646 728 642 514 695",
"output": "7 8"
},
{
"input": "30\n996 999 998 984 989 1000 996 993 1000 983 992 999 999 1000 979 992 987 1000 996 1000 1000 989 981 996 995 999 999 989 999 1000",
"output": "12 13"
},
{
"input": "50\n93 27 28 4 5 78 59 24 19 134 31 128 118 36 90 32 32 1 44 32 33 13 31 10 12 25 38 50 25 12 4 22 28 53 48 83 4 25 57 31 71 24 8 7 28 86 23 80 101 58",
"output": "16 17"
},
{
"input": "88\n1000 1000 1000 1000 1000 998 998 1000 1000 1000 1000 999 999 1000 1000 1000 999 1000 997 999 997 1000 999 998 1000 999 1000 1000 1000 999 1000 999 999 1000 1000 999 1000 999 1000 1000 998 1000 1000 1000 998 998 1000 1000 999 1000 1000 1000 1000 1000 1000 1000 998 1000 1000 1000 999 1000 1000 999 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 998 1000 1000 1000 998 1000 1000 998 1000 999 1000 1000 1000 1000",
"output": "1 2"
},
{
"input": "99\n4 4 21 6 5 3 13 2 6 1 3 4 1 3 1 9 11 1 6 17 4 5 20 4 1 9 5 11 3 4 14 1 3 3 1 4 3 5 27 1 1 2 10 7 11 4 19 7 11 6 11 13 3 1 10 7 2 1 16 1 9 4 29 13 2 12 14 2 21 1 9 8 26 12 12 5 2 14 7 8 8 8 9 4 12 2 6 6 7 16 8 14 2 10 20 15 3 7 4",
"output": "1 2"
},
{
"input": "100\n713 572 318 890 577 657 646 146 373 783 392 229 455 871 20 593 573 336 26 381 280 916 907 732 820 713 111 840 570 446 184 711 481 399 788 647 492 15 40 530 549 506 719 782 126 20 778 996 712 761 9 74 812 418 488 175 103 585 900 3 604 521 109 513 145 708 990 361 682 827 791 22 596 780 596 385 450 643 158 496 876 975 319 783 654 895 891 361 397 81 682 899 347 623 809 557 435 279 513 438",
"output": "86 87"
},
{
"input": "100\n31 75 86 68 111 27 22 22 26 30 54 163 107 75 160 122 14 23 17 26 27 20 43 58 59 71 21 148 9 32 43 91 133 286 132 70 90 156 84 14 77 93 23 18 13 72 18 131 33 28 72 175 30 86 249 20 14 208 28 57 63 199 6 10 24 30 62 267 43 479 60 28 138 1 45 3 19 47 7 166 116 117 50 140 28 14 95 85 93 43 61 15 2 70 10 51 7 95 9 25",
"output": "7 8"
},
{
"input": "100\n896 898 967 979 973 709 961 968 806 967 896 967 826 975 936 903 986 856 851 931 852 971 786 837 949 978 686 936 952 909 965 749 908 916 943 973 983 975 939 886 964 928 960 976 907 788 994 773 949 871 947 980 945 985 726 981 887 943 907 990 931 874 840 867 948 951 961 904 888 901 976 967 994 921 828 970 972 722 755 970 860 855 914 869 714 899 969 978 898 862 642 939 904 936 819 934 884 983 955 964",
"output": "1 2"
},
{
"input": "100\n994 927 872 970 815 986 952 996 965 1000 877 986 978 999 950 990 936 997 993 960 921 860 895 869 943 998 983 968 973 953 999 990 995 871 853 979 973 963 953 938 997 989 993 964 960 973 946 975 1000 962 920 746 989 957 904 965 920 979 966 961 1000 993 975 952 846 971 991 979 985 969 984 973 956 1000 952 778 983 974 956 927 995 997 980 997 1000 970 960 970 988 983 947 904 935 972 1000 863 992 996 932 967",
"output": "81 82"
},
{
"input": "100\n48 108 63 21 27 8 49 21 75 8 24 42 149 18 8 28 21 18 25 35 59 70 59 33 40 1 67 34 120 82 4 115 72 87 3 15 15 63 37 12 40 27 83 14 38 20 14 58 93 10 31 3 39 6 197 77 54 16 31 146 9 49 14 8 77 82 5 11 80 116 8 61 50 24 7 103 29 11 3 3 1 12 46 24 21 131 39 29 36 2 107 40 16 99 31 41 29 48 17 17",
"output": "36 37"
}
] | 1,615,355,884 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 186 | 0 | n = int(input())
arr = list(map(int,input().split()))
min1=arr[n-1]-arr[0]
c=0
d=n-1
for i in range(len(arr)-1):
p = abs(arr[i+1]-arr[i])
if p<min1:
c = i+1
d = i
min1 = p
print(d+1,c+1)
| Title: Reconnaissance 2
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
*n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
Input Specification:
The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction.
Output Specification:
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
Demo Input:
['5\n10 12 13 15 10\n', '4\n10 20 30 40\n']
Demo Output:
['5 1\n', '1 2\n']
Note:
none | ```python
n = int(input())
arr = list(map(int,input().split()))
min1=arr[n-1]-arr[0]
c=0
d=n-1
for i in range(len(arr)-1):
p = abs(arr[i+1]-arr[i])
if p<min1:
c = i+1
d = i
min1 = p
print(d+1,c+1)
``` | 0 |
672 | A | Summer Camp | PROGRAMMING | 800 | [
"implementation"
] | null | null | Every year, hundreds of people come to summer camps, they learn new algorithms and solve hard problems.
This is your first year at summer camp, and you are asked to solve the following problem. All integers starting with 1 are written in one line. The prefix of these line is "123456789101112131415...". Your task is to print the *n*-th digit of this string (digits are numbered starting with 1. | The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the position of the digit you need to print. | Print the *n*-th digit of the line. | [
"3\n",
"11\n"
] | [
"3\n",
"0\n"
] | In the first sample the digit at position 3 is '3', as both integers 1 and 2 consist on one digit.
In the second sample, the digit at position 11 is '0', it belongs to the integer 10. | 500 | [
{
"input": "3",
"output": "3"
},
{
"input": "11",
"output": "0"
},
{
"input": "12",
"output": "1"
},
{
"input": "13",
"output": "1"
},
{
"input": "29",
"output": "9"
},
{
"input": "30",
"output": "2"
},
{
"input": "1000",
"output": "3"
},
{
"input": "999",
"output": "9"
},
{
"input": "100",
"output": "5"
},
{
"input": "123",
"output": "6"
},
{
"input": "8",
"output": "8"
},
{
"input": "157",
"output": "3"
},
{
"input": "289",
"output": "1"
},
{
"input": "179",
"output": "4"
},
{
"input": "942",
"output": "0"
},
{
"input": "879",
"output": "9"
},
{
"input": "394",
"output": "1"
},
{
"input": "423",
"output": "7"
},
{
"input": "952",
"output": "3"
},
{
"input": "121",
"output": "5"
},
{
"input": "613",
"output": "2"
},
{
"input": "945",
"output": "1"
},
{
"input": "270",
"output": "6"
},
{
"input": "781",
"output": "2"
},
{
"input": "453",
"output": "7"
},
{
"input": "171",
"output": "0"
},
{
"input": "643",
"output": "2"
},
{
"input": "570",
"output": "6"
},
{
"input": "750",
"output": "6"
},
{
"input": "500",
"output": "0"
},
{
"input": "2",
"output": "2"
},
{
"input": "1",
"output": "1"
},
{
"input": "108",
"output": "5"
},
{
"input": "500",
"output": "0"
},
{
"input": "189",
"output": "9"
},
{
"input": "491",
"output": "0"
},
{
"input": "191",
"output": "0"
}
] | 1,598,471,096 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 108 | 307,200 | n = int(input())
arr = list(map(str, range(n + 5)))
st = ""
for i in arr:
st += i
print(st[n]) | Title: Summer Camp
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Every year, hundreds of people come to summer camps, they learn new algorithms and solve hard problems.
This is your first year at summer camp, and you are asked to solve the following problem. All integers starting with 1 are written in one line. The prefix of these line is "123456789101112131415...". Your task is to print the *n*-th digit of this string (digits are numbered starting with 1.
Input Specification:
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the position of the digit you need to print.
Output Specification:
Print the *n*-th digit of the line.
Demo Input:
['3\n', '11\n']
Demo Output:
['3\n', '0\n']
Note:
In the first sample the digit at position 3 is '3', as both integers 1 and 2 consist on one digit.
In the second sample, the digit at position 11 is '0', it belongs to the integer 10. | ```python
n = int(input())
arr = list(map(str, range(n + 5)))
st = ""
for i in arr:
st += i
print(st[n])
``` | 3 | |
875 | A | Classroom Watch | PROGRAMMING | 1,200 | [
"brute force",
"math"
] | null | null | Eighth-grader Vova is on duty today in the class. After classes, he went into the office to wash the board, and found on it the number *n*. He asked what is this number and the teacher of mathematics Inna Petrovna answered Vova that *n* is the answer to the arithmetic task for first-graders. In the textbook, a certain positive integer *x* was given. The task was to add *x* to the sum of the digits of the number *x* written in decimal numeral system.
Since the number *n* on the board was small, Vova quickly guessed which *x* could be in the textbook. Now he wants to get a program which will search for arbitrary values of the number *n* for all suitable values of *x* or determine that such *x* does not exist. Write such a program for Vova. | The first line contains integer *n* (1<=≤<=*n*<=≤<=109). | In the first line print one integer *k* — number of different values of *x* satisfying the condition.
In next *k* lines print these values in ascending order. | [
"21\n",
"20\n"
] | [
"1\n15\n",
"0\n"
] | In the first test case *x* = 15 there is only one variant: 15 + 1 + 5 = 21.
In the second test case there are no such *x*. | 500 | [
{
"input": "21",
"output": "1\n15"
},
{
"input": "20",
"output": "0"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "1\n1"
},
{
"input": "3",
"output": "0"
},
{
"input": "100000001",
"output": "2\n99999937\n100000000"
},
{
"input": "1000000000",
"output": "1\n999999932"
},
{
"input": "999999979",
"output": "2\n999999899\n999999908"
},
{
"input": "9",
"output": "0"
},
{
"input": "10",
"output": "1\n5"
},
{
"input": "11",
"output": "1\n10"
},
{
"input": "39",
"output": "1\n33"
},
{
"input": "66",
"output": "1\n60"
},
{
"input": "75",
"output": "0"
},
{
"input": "100",
"output": "1\n86"
},
{
"input": "101",
"output": "2\n91\n100"
},
{
"input": "2014",
"output": "2\n1988\n2006"
},
{
"input": "999999994",
"output": "0"
}
] | 1,644,172,891 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 1,000 | 0 | number_x = int(input())
def sum_digit(number):
total = int(number)
for i in range(len(number)):
total += int(number[i])
return total
lista = []
for i in range(number_x):
if sum_digit(str(i)) == number_x:
lista.append(i)
print(len(lista))
for i in lista:
print(i) | Title: Classroom Watch
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Eighth-grader Vova is on duty today in the class. After classes, he went into the office to wash the board, and found on it the number *n*. He asked what is this number and the teacher of mathematics Inna Petrovna answered Vova that *n* is the answer to the arithmetic task for first-graders. In the textbook, a certain positive integer *x* was given. The task was to add *x* to the sum of the digits of the number *x* written in decimal numeral system.
Since the number *n* on the board was small, Vova quickly guessed which *x* could be in the textbook. Now he wants to get a program which will search for arbitrary values of the number *n* for all suitable values of *x* or determine that such *x* does not exist. Write such a program for Vova.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
In the first line print one integer *k* — number of different values of *x* satisfying the condition.
In next *k* lines print these values in ascending order.
Demo Input:
['21\n', '20\n']
Demo Output:
['1\n15\n', '0\n']
Note:
In the first test case *x* = 15 there is only one variant: 15 + 1 + 5 = 21.
In the second test case there are no such *x*. | ```python
number_x = int(input())
def sum_digit(number):
total = int(number)
for i in range(len(number)):
total += int(number[i])
return total
lista = []
for i in range(number_x):
if sum_digit(str(i)) == number_x:
lista.append(i)
print(len(lista))
for i in lista:
print(i)
``` | 0 | |
600 | D | Area of Two Circles' Intersection | PROGRAMMING | 2,000 | [
"geometry"
] | null | null | You are given two circles. Find the area of their intersection. | The first line contains three integers *x*1,<=*y*1,<=*r*1 (<=-<=109<=≤<=*x*1,<=*y*1<=≤<=109,<=1<=≤<=*r*1<=≤<=109) — the position of the center and the radius of the first circle.
The second line contains three integers *x*2,<=*y*2,<=*r*2 (<=-<=109<=≤<=*x*2,<=*y*2<=≤<=109,<=1<=≤<=*r*2<=≤<=109) — the position of the center and the radius of the second circle. | Print the area of the intersection of the circles. The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6. | [
"0 0 4\n6 0 4\n",
"0 0 5\n11 0 5\n"
] | [
"7.25298806364175601379\n",
"0.00000000000000000000\n"
] | none | 0 | [
{
"input": "0 0 4\n6 0 4",
"output": "7.25298806364175601379"
},
{
"input": "0 0 5\n11 0 5",
"output": "0.00000000000000000000"
},
{
"input": "0 0 10\n9 0 1",
"output": "3.14159265358979311600"
},
{
"input": "0 0 2\n2 2 2",
"output": "2.28318530717958647659"
},
{
"input": "0 0 10\n5 0 5",
"output": "78.53981633974482789995"
},
{
"input": "-9 8 7\n-9 8 5",
"output": "78.53981633974482789995"
},
{
"input": "-60 -85 95\n-69 -94 95",
"output": "25936.37843115316246844770"
},
{
"input": "159 111 998\n161 121 1023",
"output": "3129038.84934604830277748988"
},
{
"input": "6008 8591 6693\n5310 8351 7192",
"output": "138921450.46886559338599909097"
},
{
"input": "-13563 -6901 22958\n-19316 -16534 18514",
"output": "868466038.83295116270892322063"
},
{
"input": "-875463 79216 524620\n-891344 76571 536598",
"output": "862534134678.47474157810211181641"
},
{
"input": "-8907963 -8149654 8808560\n-8893489 -8125053 8830600",
"output": "243706233220003.66226196289062500000"
},
{
"input": "-56452806 56199829 45467742\n-56397667 56292048 45489064",
"output": "6487743741270471.46582031250000000000"
},
{
"input": "-11786939 388749051 844435993\n-11696460 388789113 844535886",
"output": "2240182216213578196.25000000000000000000"
},
{
"input": "-944341103 -3062765 891990581\n-943884414 -3338765 891882754",
"output": "2498325849744150942.00000000000000000000"
},
{
"input": "808468733 166975547 650132512\n807140196 169714842 655993403",
"output": "1327864139649690571.00000000000000000000"
},
{
"input": "-16 -107 146\n75 25 19",
"output": "75.73941676175987183783"
},
{
"input": "468534418 -876402362 779510\n392125478 -856995174 1",
"output": "0.00000000000000000000"
},
{
"input": "368831644 125127030 959524552\n690900461 -368007601 1000000000",
"output": "1877639096067727828.75000000000000000000"
},
{
"input": "638572730 86093565 553198855\n-151099010 -5582761 1000000000",
"output": "648156847022339121.87500000000000000000"
},
{
"input": "567845488 379750385 112902105\n567845488 379750385 112902105",
"output": "40045521256826535.57031250000000000000"
},
{
"input": "817163584 -145230792 164258581\n826720200 -149804696 98",
"output": "30171.85584507637308604444"
},
{
"input": "-812130546 -209199732 799576707\n-728169661 -278950375 4385",
"output": "60407250.40157159973750822246"
},
{
"input": "-36140638 -933845433 250828868\n90789911 -245130908 328547",
"output": "0.00000000000000000000"
},
{
"input": "34537868 -531411810 591044372\n34536968 -531411968 58",
"output": "10568.31768667606404221715"
},
{
"input": "-410889750 -716765873 303980004\n-410889749 -716765874 7",
"output": "153.93804002589986268390"
},
{
"input": "-304 -310 476\n120 -294 1",
"output": "3.14159265358979311600"
},
{
"input": "-999999999 0 1000000000\n999999999 0 1000000000",
"output": "119256.95877838134765625000"
},
{
"input": "-1000000000 0 1000000000\n999999999 0 1000000000",
"output": "42163.70213317871093750000"
},
{
"input": "-99999999 0 100000000\n99999999 0 100000000",
"output": "37712.36160683631896972656"
},
{
"input": "-999999999 0 1000000000\n999999999 1 1000000000",
"output": "119256.95874786376953125000"
},
{
"input": "-1000000000 0 999999999\n999999997 0 999999999",
"output": "42163.70211410522460937500"
},
{
"input": "0 1000000000 1\n0 0 1000000000",
"output": "1.57079632649338855020"
},
{
"input": "10000000 0 10000001\n-10000000 0 10000000",
"output": "4216.37028734199702739716"
},
{
"input": "1000000000 0 1000000000\n-999999999 1 1000000000",
"output": "42163.70212173461914062500"
},
{
"input": "44721 999999999 400000000\n0 0 600000000",
"output": "0.00188343226909637451"
},
{
"input": "-1000000000 1 1000000000\n999999998 0 1000000000",
"output": "119256.95874786376953125000"
},
{
"input": "0 0 500000000\n431276 999999907 500000000",
"output": "0.33492207527160644531"
},
{
"input": "1000000000 0 1000000000\n-999999998 -87334 1000000000",
"output": "1199.53919601440429687500"
},
{
"input": "0 0 10\n0 0 25",
"output": "314.15926535897931159980"
},
{
"input": "0 0 1000000000\n707106781 707106781 1",
"output": "2.09224628662147114737"
},
{
"input": "100 10 10\n100 20 10",
"output": "122.83696986087568455565"
},
{
"input": "1000000000 0 1000000000\n-999999998 -88334 1000000000",
"output": "461.20431423187255859375"
},
{
"input": "0 0 999999999\n1000000000 0 2",
"output": "2.45673939563023624650"
},
{
"input": "-99999999 0 100000000\n99999999 1 100000000",
"output": "37712.36153602600097656250"
},
{
"input": "1000000000 0 1000000000\n-999999999 60333 1000000000",
"output": "1138.08371162414550781250"
},
{
"input": "1000000000 0 1000000000\n-999999999 58333 1000000000",
"output": "2432.73669052124023437500"
},
{
"input": "1000000000 0 1000000000\n-999999998 -85334 1000000000",
"output": "3207.25725555419921875000"
},
{
"input": "0 0 1000000000\n999999999 1 2",
"output": "10.10963121370591567653"
},
{
"input": "0 0 1000000000\n999999998 0 3",
"output": "25.17685179846658691770"
},
{
"input": "141 9999 5000\n0 0 5000",
"output": "0.04272695172407026121"
},
{
"input": "-1000000000 0 1000000000\n999999998 0 1000000000",
"output": "119256.95877838134765625000"
},
{
"input": "0 0 10\n1 0 10",
"output": "294.16760182010623145277"
},
{
"input": "0 0 1000000000\n707106782 707106781 2",
"output": "4.52465731000908907454"
}
] | 1,672,085,277 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 5 | 77 | 2,867,200 | import math
x1,y1,r1 = list(map(int,input().split()))
x2,y2,r2 = list(map(int,input().split()))
d = math.sqrt( (x1 - x2)**2 + (y1 - y2) ** 2)
error = 1e-13
if ((r1 + r2) - d < error):
print(0)
else:
angle1 = math.acos( (d**2 + r1 **2 - r2**2) / (2 * d * r1 * 1.0))
angle2 = math.acos( (d**2 + r2 **2 - r1**2) / (2 * d * r2 * 1.0))
# print(angle1, angle2)
area = angle1 * r1 * r1 + angle2 * r2 * r2
area -= 0.5 * r1 * r1 * math.sin(2.0 * angle1)
area -= 0.5 * r2 * r2 * math.sin(2.0 * angle2)
print(area) | Title: Area of Two Circles' Intersection
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two circles. Find the area of their intersection.
Input Specification:
The first line contains three integers *x*1,<=*y*1,<=*r*1 (<=-<=109<=≤<=*x*1,<=*y*1<=≤<=109,<=1<=≤<=*r*1<=≤<=109) — the position of the center and the radius of the first circle.
The second line contains three integers *x*2,<=*y*2,<=*r*2 (<=-<=109<=≤<=*x*2,<=*y*2<=≤<=109,<=1<=≤<=*r*2<=≤<=109) — the position of the center and the radius of the second circle.
Output Specification:
Print the area of the intersection of the circles. The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6.
Demo Input:
['0 0 4\n6 0 4\n', '0 0 5\n11 0 5\n']
Demo Output:
['7.25298806364175601379\n', '0.00000000000000000000\n']
Note:
none | ```python
import math
x1,y1,r1 = list(map(int,input().split()))
x2,y2,r2 = list(map(int,input().split()))
d = math.sqrt( (x1 - x2)**2 + (y1 - y2) ** 2)
error = 1e-13
if ((r1 + r2) - d < error):
print(0)
else:
angle1 = math.acos( (d**2 + r1 **2 - r2**2) / (2 * d * r1 * 1.0))
angle2 = math.acos( (d**2 + r2 **2 - r1**2) / (2 * d * r2 * 1.0))
# print(angle1, angle2)
area = angle1 * r1 * r1 + angle2 * r2 * r2
area -= 0.5 * r1 * r1 * math.sin(2.0 * angle1)
area -= 0.5 * r2 * r2 * math.sin(2.0 * angle2)
print(area)
``` | -1 | |
920 | E | Connected Components? | PROGRAMMING | 2,100 | [
"data structures",
"dfs and similar",
"dsu",
"graphs"
] | null | null | You are given an undirected graph consisting of *n* vertices and edges. Instead of giving you the edges that exist in the graph, we give you *m* unordered pairs (*x*,<=*y*) such that there is no edge between *x* and *y*, and if some pair of vertices is not listed in the input, then there is an edge between these vertices.
You have to find the number of connected components in the graph and the size of each component. A connected component is a set of vertices *X* such that for every two vertices from this set there exists at least one path in the graph connecting these vertices, but adding any other vertex to *X* violates this rule. | The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=200000, ).
Then *m* lines follow, each containing a pair of integers *x* and *y* (1<=≤<=*x*,<=*y*<=≤<=*n*, *x*<=≠<=*y*) denoting that there is no edge between *x* and *y*. Each pair is listed at most once; (*x*,<=*y*) and (*y*,<=*x*) are considered the same (so they are never listed in the same test). If some pair of vertices is not listed in the input, then there exists an edge between those vertices. | Firstly print *k* — the number of connected components in this graph.
Then print *k* integers — the sizes of components. You should output these integers in non-descending order. | [
"5 5\n1 2\n3 4\n3 2\n4 2\n2 5\n"
] | [
"2\n1 4 "
] | none | 0 | [
{
"input": "5 5\n1 2\n3 4\n3 2\n4 2\n2 5",
"output": "2\n1 4 "
},
{
"input": "8 15\n2 1\n4 5\n2 4\n3 4\n2 5\n3 5\n2 6\n3 6\n5 6\n4 6\n2 7\n3 8\n2 8\n3 7\n6 7",
"output": "1\n8 "
},
{
"input": "12 58\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 10\n1 11\n1 12\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n2 11\n2 12\n3 4\n3 5\n3 6\n3 7\n3 8\n3 9\n3 10\n3 11\n3 12\n4 5\n4 6\n4 8\n4 11\n4 12\n5 6\n5 7\n5 8\n5 9\n5 10\n5 11\n6 7\n6 8\n6 9\n6 10\n6 11\n6 12\n7 8\n7 9\n7 10\n7 11\n7 12\n8 9\n8 10\n8 11\n9 10\n9 11\n9 12\n10 12",
"output": "4\n1 1 1 9 "
},
{
"input": "5 7\n1 2\n2 3\n3 4\n1 5\n2 5\n3 5\n4 5",
"output": "2\n1 4 "
},
{
"input": "6 10\n1 2\n1 3\n1 4\n1 6\n2 3\n2 4\n2 5\n3 5\n3 6\n4 6",
"output": "1\n6 "
},
{
"input": "8 23\n1 2\n1 4\n1 6\n1 8\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n3 4\n3 5\n3 6\n3 7\n3 8\n4 5\n4 6\n4 7\n5 6\n5 7\n5 8\n6 8\n7 8",
"output": "3\n1 2 5 "
},
{
"input": "4 3\n2 1\n3 1\n4 2",
"output": "1\n4 "
},
{
"input": "6 9\n1 2\n1 4\n1 5\n2 3\n2 5\n2 6\n3 5\n4 6\n5 6",
"output": "1\n6 "
},
{
"input": "2 0",
"output": "1\n2 "
},
{
"input": "8 18\n1 4\n1 6\n1 7\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n3 4\n3 8\n4 7\n5 6\n5 7\n5 8\n6 7\n6 8\n7 8",
"output": "1\n8 "
},
{
"input": "4 3\n1 2\n3 1\n4 3",
"output": "1\n4 "
},
{
"input": "8 23\n2 7\n7 5\n8 6\n8 2\n6 3\n3 5\n8 1\n8 4\n8 3\n3 4\n1 2\n2 6\n5 2\n6 4\n7 6\n6 5\n7 8\n7 1\n5 4\n3 7\n1 4\n3 1\n3 2",
"output": "3\n1 3 4 "
},
{
"input": "4 4\n2 1\n3 1\n1 4\n3 2",
"output": "2\n1 3 "
},
{
"input": "2 1\n1 2",
"output": "2\n1 1 "
},
{
"input": "4 3\n1 3\n1 4\n2 3",
"output": "1\n4 "
},
{
"input": "3 1\n2 3",
"output": "1\n3 "
},
{
"input": "5 4\n1 4\n2 3\n4 3\n4 2",
"output": "1\n5 "
},
{
"input": "10 36\n7 8\n7 9\n2 3\n2 4\n2 5\n9 10\n2 7\n2 8\n2 9\n2 10\n4 5\n4 6\n4 7\n4 8\n4 10\n6 7\n6 9\n6 10\n1 2\n1 3\n1 4\n8 9\n1 5\n8 10\n1 7\n1 8\n1 9\n1 10\n3 4\n3 6\n3 7\n3 9\n5 6\n5 7\n5 9\n5 10",
"output": "2\n2 8 "
},
{
"input": "10 34\n7 10\n2 3\n2 4\n2 5\n9 10\n2 7\n2 8\n2 10\n4 5\n4 6\n4 7\n4 8\n4 9\n6 7\n6 8\n6 9\n6 10\n1 2\n1 3\n1 5\n8 9\n1 6\n1 7\n1 8\n1 9\n1 10\n3 4\n3 5\n3 6\n3 8\n3 10\n5 6\n5 9\n5 10",
"output": "1\n10 "
},
{
"input": "12 56\n9 5\n2 6\n9 8\n5 4\n1 11\n1 6\n4 1\n1 10\n10 3\n8 4\n5 1\n9 1\n5 10\n2 7\n11 5\n6 11\n5 8\n7 6\n3 2\n12 7\n8 6\n12 3\n1 2\n8 1\n2 11\n10 12\n4 6\n5 12\n2 4\n10 2\n7 3\n12 11\n7 10\n7 1\n9 2\n11 9\n9 10\n8 7\n11 3\n7 9\n5 7\n4 12\n3 5\n12 2\n4 10\n9 12\n5 2\n9 4\n11 8\n8 2\n3 6\n4 11\n8 10\n6 10\n3 9\n3 4",
"output": "3\n1 4 7 "
},
{
"input": "11 49\n10 3\n6 4\n11 3\n7 6\n10 6\n6 1\n4 3\n10 2\n4 5\n9 2\n10 1\n5 7\n1 5\n9 7\n2 11\n8 6\n3 9\n2 5\n9 5\n6 5\n1 4\n11 9\n1 7\n8 10\n3 6\n3 7\n11 5\n6 9\n4 10\n8 7\n4 9\n8 2\n4 2\n8 11\n7 4\n9 10\n8 1\n10 7\n3 2\n5 8\n8 9\n1 3\n2 7\n10 11\n5 3\n10 5\n4 11\n1 11\n8 3",
"output": "5\n1 1 1 2 6 "
}
] | 1,676,326,992 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 1 | 2,000 | 72,294,400 | from collections import defaultdict
n,m=map(int, input().split())
e=defaultdict(set)
for i in range(1, n+1):
for j in range(1,n+1):
if i==j: continue
e[i].add(j)
e[j].add(i)
for i in range(m):
a,b=map(int, input().split())
e[i].discard(j)
e[j].discard(i)
arr = [i for i in range(n+1)]
rank=[1 for i in arr]
def find(a):
if arr[a]==a: return a
return find(arr[a])
def union(a, b):
pa,pb=find(a), find(b)
if rank[pa]>rank[pb]:
rank[pa]+=rank[pb]
arr[pb]=pa
else:
rank[pb]+=rank[pa]
arr[pa]=pb
for i in range(1,n+1):
for j in e[i]:
union(i,j)
comps=set(arr[1:])
c={}
for i in arr[1:]:
if i not in c:
c[i]=0
c[i]+=1
print(len(comps))
out=list(map(str, sorted(list(c.values()))))
print(' '.join(out)) | Title: Connected Components?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an undirected graph consisting of *n* vertices and edges. Instead of giving you the edges that exist in the graph, we give you *m* unordered pairs (*x*,<=*y*) such that there is no edge between *x* and *y*, and if some pair of vertices is not listed in the input, then there is an edge between these vertices.
You have to find the number of connected components in the graph and the size of each component. A connected component is a set of vertices *X* such that for every two vertices from this set there exists at least one path in the graph connecting these vertices, but adding any other vertex to *X* violates this rule.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=200000, ).
Then *m* lines follow, each containing a pair of integers *x* and *y* (1<=≤<=*x*,<=*y*<=≤<=*n*, *x*<=≠<=*y*) denoting that there is no edge between *x* and *y*. Each pair is listed at most once; (*x*,<=*y*) and (*y*,<=*x*) are considered the same (so they are never listed in the same test). If some pair of vertices is not listed in the input, then there exists an edge between those vertices.
Output Specification:
Firstly print *k* — the number of connected components in this graph.
Then print *k* integers — the sizes of components. You should output these integers in non-descending order.
Demo Input:
['5 5\n1 2\n3 4\n3 2\n4 2\n2 5\n']
Demo Output:
['2\n1 4 ']
Note:
none | ```python
from collections import defaultdict
n,m=map(int, input().split())
e=defaultdict(set)
for i in range(1, n+1):
for j in range(1,n+1):
if i==j: continue
e[i].add(j)
e[j].add(i)
for i in range(m):
a,b=map(int, input().split())
e[i].discard(j)
e[j].discard(i)
arr = [i for i in range(n+1)]
rank=[1 for i in arr]
def find(a):
if arr[a]==a: return a
return find(arr[a])
def union(a, b):
pa,pb=find(a), find(b)
if rank[pa]>rank[pb]:
rank[pa]+=rank[pb]
arr[pb]=pa
else:
rank[pb]+=rank[pa]
arr[pa]=pb
for i in range(1,n+1):
for j in e[i]:
union(i,j)
comps=set(arr[1:])
c={}
for i in arr[1:]:
if i not in c:
c[i]=0
c[i]+=1
print(len(comps))
out=list(map(str, sorted(list(c.values()))))
print(' '.join(out))
``` | 0 | |
52 | A | 123-sequence | PROGRAMMING | 900 | [
"implementation"
] | A. 123-sequence | 2 | 256 | There is a given sequence of integers *a*1,<=*a*2,<=...,<=*a**n*, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=106). The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3). | Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal. | [
"9\n1 3 2 2 2 1 1 2 3\n"
] | [
"5\n"
] | In the example all the numbers equal to 1 and 3 should be replaced by 2. | 500 | [
{
"input": "9\n1 3 2 2 2 1 1 2 3",
"output": "5"
},
{
"input": "6\n3 3 2 2 1 3",
"output": "3"
},
{
"input": "12\n3 1 3 1 2 1 3 2 2 1 2 1",
"output": "7"
},
{
"input": "15\n3 2 1 1 1 1 3 2 2 3 3 1 2 3 2",
"output": "10"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "2\n3 2",
"output": "1"
},
{
"input": "2\n3 1",
"output": "1"
},
{
"input": "18\n2 3 2 1 2 3 2 1 2 3 3 3 1 2 3 3 3 2",
"output": "10"
},
{
"input": "30\n2 1 3 2 3 2 2 2 2 3 2 2 3 2 1 1 3 1 3 2 1 2 3 1 1 3 3 1 3 1",
"output": "19"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n3",
"output": "0"
}
] | 1,568,066,645 | 2,147,483,647 | Python 3 | OK | TESTS | 23 | 342 | 11,161,600 | input()
s = input().split()
print(len(s)-max(s.count("1"),s.count("2"),s.count("3")))
| Title: 123-sequence
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There is a given sequence of integers *a*1,<=*a*2,<=...,<=*a**n*, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=106). The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3).
Output Specification:
Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal.
Demo Input:
['9\n1 3 2 2 2 1 1 2 3\n']
Demo Output:
['5\n']
Note:
In the example all the numbers equal to 1 and 3 should be replaced by 2. | ```python
input()
s = input().split()
print(len(s)-max(s.count("1"),s.count("2"),s.count("3")))
``` | 3.89371 |
253 | C | Text Editor | PROGRAMMING | 1,600 | [
"data structures",
"dfs and similar",
"graphs",
"greedy",
"shortest paths"
] | null | null | Vasya is pressing the keys on the keyboard reluctantly, squeezing out his ideas on the classical epos depicted in Homer's Odysseus... How can he explain to his literature teacher that he isn't going to become a writer? In fact, he is going to become a programmer. So, he would take great pleasure in writing a program, but none — in writing a composition.
As Vasya was fishing for a sentence in the dark pond of his imagination, he suddenly wondered: what is the least number of times he should push a key to shift the cursor from one position to another one?
Let's describe his question more formally: to type a text, Vasya is using the text editor. He has already written *n* lines, the *i*-th line contains *a**i* characters (including spaces). If some line contains *k* characters, then this line overall contains (*k*<=+<=1) positions where the cursor can stand: before some character or after all characters (at the end of the line). Thus, the cursor's position is determined by a pair of integers (*r*,<=*c*), where *r* is the number of the line and *c* is the cursor's position in the line (the positions are indexed starting from one from the beginning of the line).
Vasya doesn't use the mouse to move the cursor. He uses keys "Up", "Down", "Right" and "Left". When he pushes each of these keys, the cursor shifts in the needed direction. Let's assume that before the corresponding key is pressed, the cursor was located in the position (*r*,<=*c*), then Vasya pushed key:
- "Up": if the cursor was located in the first line (*r*<==<=1), then it does not move. Otherwise, it moves to the previous line (with number *r*<=-<=1), to the same position. At that, if the previous line was short, that is, the cursor couldn't occupy position *c* there, the cursor moves to the last position of the line with number *r*<=-<=1;- "Down": if the cursor was located in the last line (*r*<==<=*n*), then it does not move. Otherwise, it moves to the next line (with number *r*<=+<=1), to the same position. At that, if the next line was short, that is, the cursor couldn't occupy position *c* there, the cursor moves to the last position of the line with number *r*<=+<=1;- "Right": if the cursor can move to the right in this line (*c*<=<<=*a**r*<=+<=1), then it moves to the right (to position *c*<=+<=1). Otherwise, it is located at the end of the line and doesn't move anywhere when Vasya presses the "Right" key;- "Left": if the cursor can move to the left in this line (*c*<=><=1), then it moves to the left (to position *c*<=-<=1). Otherwise, it is located at the beginning of the line and doesn't move anywhere when Vasya presses the "Left" key.
You've got the number of lines in the text file and the number of characters, written in each line of this file. Find the least number of times Vasya should push the keys, described above, to shift the cursor from position (*r*1,<=*c*1) to position (*r*2,<=*c*2). | The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the file. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=105), separated by single spaces. The third line contains four integers *r*1,<=*c*1,<=*r*2,<=*c*2 (1<=≤<=*r*1,<=*r*2<=≤<=*n*,<=1<=≤<=*c*1<=≤<=*a**r*1<=+<=1,<=1<=≤<=*c*2<=≤<=*a**r*2<=+<=1). | Print a single integer — the minimum number of times Vasya should push a key to move the cursor from position (*r*1,<=*c*1) to position (*r*2,<=*c*2). | [
"4\n2 1 6 4\n3 4 4 2\n",
"4\n10 5 6 4\n1 11 4 2\n",
"3\n10 1 10\n1 10 1 1\n"
] | [
"3\n",
"6\n",
"3\n"
] | In the first sample the editor contains four lines. Let's represent the cursor's possible positions in the line as numbers. Letter *s* represents the cursor's initial position, letter *t* represents the last one. Then all possible positions of the cursor in the text editor are described by the following table.
123
12
123s567
1t345
One of the possible answers in the given sample is: "Left", "Down", "Left". | 1,500 | [
{
"input": "4\n2 1 6 4\n3 4 4 2",
"output": "3"
},
{
"input": "4\n10 5 6 4\n1 11 4 2",
"output": "6"
},
{
"input": "3\n10 1 10\n1 10 1 1",
"output": "3"
},
{
"input": "4\n2 1 6 4\n4 2 3 5",
"output": "4"
},
{
"input": "3\n20 3 20\n1 20 1 1",
"output": "5"
},
{
"input": "2\n10 1\n1 3 2 1",
"output": "2"
},
{
"input": "20\n3 1 9 9 6 1 3 4 5 6 7 3 1 9 9 1 9 1 5 7\n17 7 19 5",
"output": "5"
},
{
"input": "20\n81 90 11 68 23 18 78 75 45 86 58 37 21 15 98 40 53 100 10 70\n11 55 8 19",
"output": "7"
},
{
"input": "25\n55 47 5 63 55 11 8 32 0 62 41 7 17 70 33 6 41 68 37 82 33 64 28 33 12\n6 11 14 12",
"output": "19"
},
{
"input": "30\n77 38 82 87 88 1 90 3 79 69 64 36 85 12 1 19 80 89 75 56 49 28 10 31 37 65 27 84 10 72\n26 65 19 3",
"output": "15"
},
{
"input": "100\n119 384 220 357 394 123 371 57 6 221 219 79 305 292 71 113 428 326 166 235 120 404 77 223 2 171 81 1 119 307 200 323 89 294 178 421 125 197 89 154 335 46 210 311 216 182 246 262 195 99 175 153 310 302 417 167 222 349 63 325 175 345 6 78 9 147 126 308 229 295 175 368 230 116 95 254 443 15 299 265 322 171 179 184 435 115 384 324 213 359 414 159 322 49 209 296 376 173 369 302\n8 47 23 65",
"output": "73"
},
{
"input": "100\n120 336 161 474 285 126 321 63 82 303 421 110 143 279 505 231 40 413 20 421 271 30 465 186 495 156 225 445 530 156 516 305 360 261 123 5 50 377 124 8 115 529 395 408 271 166 121 240 336 348 352 359 487 471 171 379 381 182 109 425 252 434 131 430 461 386 33 189 481 461 163 89 374 505 525 526 132 468 80 88 90 538 280 281 552 415 194 41 333 296 297 205 40 79 22 219 108 213 158 410\n58 119 82 196",
"output": "186"
},
{
"input": "100\n9 8 5 2 10 6 10 10 1 9 8 5 0 9 1 6 6 2 3 9 9 3 2 7 2 7 8 10 6 6 2 8 5 0 0 8 7 3 0 4 7 5 9 0 3 6 9 6 5 0 4 9 4 7 7 1 5 8 2 4 10 3 9 8 10 6 10 7 4 9 0 1 3 6 6 2 1 1 5 7 0 9 6 0 4 6 8 4 7 6 1 9 4 3 10 9 7 0 0 7\n72 2 87 2",
"output": "16"
},
{
"input": "100\n9 72 46 37 26 94 80 1 43 85 26 53 58 18 24 19 67 2 100 52 61 81 48 15 73 41 97 93 45 1 73 54 75 51 28 79 0 14 41 42 24 50 70 18 96 100 67 1 68 48 44 39 63 77 78 18 10 51 32 53 26 60 1 13 66 39 55 27 23 71 75 0 27 88 73 31 16 95 87 84 86 71 37 40 66 70 65 83 19 4 81 99 26 51 67 63 80 54 23 44\n6 76 89 15",
"output": "97"
},
{
"input": "100\n176 194 157 24 27 153 31 159 196 85 127 114 142 39 133 4 44 36 141 96 80 40 120 16 88 29 157 136 158 98 145 152 19 40 106 116 19 195 184 70 72 95 78 146 199 1 103 3 120 71 52 77 160 148 24 156 108 64 86 124 103 97 108 66 107 126 29 172 23 106 29 69 64 90 9 171 59 85 1 63 79 50 136 21 115 164 30 115 86 26 25 6 128 48 122 14 198 88 182 117\n71 4 85 80",
"output": "92"
},
{
"input": "100\n1622 320 1261 282 1604 57 1427 1382 904 911 1719 1682 984 1727 1301 1799 1110 1057 248 764 1642 1325 1172 1677 182 32 665 397 1146 73 412 554 973 874 774 1948 1676 1959 518 280 1467 568 613 760 594 252 224 1359 876 253 760 1566 929 1614 940 1079 288 245 1432 1647 1534 1768 1947 733 225 495 1239 644 124 522 1859 1856 1464 485 1962 131 1693 1622 242 1119 1290 538 998 1342 791 711 809 1407 1369 414 124 758 1104 1142 355 324 665 1155 551 1611\n36 1383 51 21",
"output": "47"
},
{
"input": "50\n966 151 777 841 507 884 487 813 29 230 966 819 390 482 137 365 391 693 56 756 327 500 895 22 361 619 8 516 21 770 572 53 497 682 162 32 308 309 110 470 699 318 947 658 720 679 435 645 481 42\n45 510 25 48",
"output": "59"
},
{
"input": "50\n4143 2907 2028 539 3037 1198 6597 3658 972 9809 854 4931 642 3170 9777 2992 7121 8094 6634 684 5580 4684 3397 7909 3908 3822 2137 8299 8146 2105 7578 4338 7363 8237 530 301 4566 1153 4795 5342 3257 6953 4401 8311 9977 9260 7019 7705 5416 6754\n21 3413 23 218",
"output": "112"
},
{
"input": "50\n8974 13208 81051 72024 84908 49874 22875 64935 27340 38682 28512 43441 78752 83458 63344 5723 83425 54009 61980 7824 59956 43184 49274 3896 44079 67313 68565 9138 55087 68458 43009 3685 22879 85032 84273 93643 64957 73428 57016 33405 85961 47708 90325 1352 1551 20935 76821 75406 59309 40757\n14 45232 2 6810",
"output": "1102"
},
{
"input": "100\n34 80 42 99 7 49 109 61 20 7 92 2 62 96 65 77 70 5 16 83 99 39 88 66 106 1 80 68 71 74 28 75 19 97 38 100 30 1 55 86 3 13 61 82 72 50 68 18 77 89 96 27 26 35 46 13 83 77 40 31 85 108 15 5 40 80 1 108 44 18 66 26 46 7 36 80 34 76 17 9 23 57 109 90 88 1 54 66 71 94 6 89 50 22 93 82 32 74 41 74\n91 7 56 3",
"output": "36"
},
{
"input": "100\n156 150 75 72 205 133 139 99 212 82 58 104 133 88 46 157 49 179 32 72 159 188 42 47 36 58 127 215 125 115 209 118 109 11 62 159 110 151 92 202 203 25 44 209 153 8 199 168 126 34 21 106 31 40 48 212 106 0 131 166 2 126 13 126 103 44 2 66 33 25 194 41 37 198 199 6 22 1 161 16 95 11 198 198 166 145 214 159 143 2 181 130 159 118 176 165 192 178 42 168\n49 12 66 23",
"output": "39"
},
{
"input": "100\n289 16 321 129 0 121 61 86 93 5 63 276 259 144 275 236 309 257 244 138 107 18 158 14 295 162 7 113 58 101 142 196 181 329 115 109 62 237 110 87 19 205 68 257 252 0 166 45 310 244 140 251 262 315 213 206 290 128 287 230 198 83 135 40 8 273 319 295 288 274 34 260 288 252 172 129 201 110 294 111 95 180 34 98 16 188 170 40 274 153 11 159 245 51 328 290 112 11 105 182\n99 53 21 77",
"output": "154"
},
{
"input": "10\n11284 10942 14160 10062 1858 6457 1336 13842 5498 4236\n1 7123 5 664",
"output": "681"
},
{
"input": "53\n29496 9630 10781 25744 28508 15670 8252 14284 25995 20215 24251 14240 1370 15724 28268 30377 4839 16791 33515 23776 24252 1045 15245 12839 17531 28591 13091 27339 23361 10997 30438 26977 26789 18402 32938 2106 26599 10733 29549 9760 31507 33572 16934 7273 26477 15040 23704 19905 1941 3861 5950 1265 34\n11 6571 1 3145",
"output": "1788"
},
{
"input": "31\n14324 29226 58374 19956 61695 71586 13261 11436 58443 34879 12689 62786 68194 34303 99201 67616 51364 67539 56799 60130 22021 64546 28331 75746 45036 43950 2150 61718 33030 37781 34319\n24 57393 7 6152",
"output": "4024"
},
{
"input": "23\n5397 13279 11741 20182 18311 20961 16720 11864 2486 14081 15637 16216 3736 437 16346 12449 20205 10949 14237 2213 15281 15271 19138\n5 11479 13 68",
"output": "380"
},
{
"input": "40\n41997 20736 34699 73866 45509 41964 36050 16673 10454 21166 28306 69335 6172 65943 78569 16794 10439 68061 40392 52510 78248 63851 45294 49929 22580 5574 40993 18334 73897 59148 47727 76645 4280 23651 58772 64500 13704 60366 37099 20336\n14 29991 16 11904",
"output": "1468"
},
{
"input": "16\n922 7593 4748 4103 7672 6001 1573 3973 8524 8265 4747 3202 4796 2637 889 9359\n12 2165 12 1654",
"output": "90"
},
{
"input": "18\n22746 9084 3942 1120 25391 25307 7409 1189 23473 26175 10964 13584 5541 500 24338 12272 15824 27656\n3 1395 12 90",
"output": "424"
},
{
"input": "45\n2286 4425 14666 34959 10792 3723 30132 34266 18100 22813 28627 23310 33911 27285 1211 993 15526 4751 13611 21400 25712 24437 27435 34808 33950 18373 33685 23487 5444 10249 21415 16368 35398 7889 30918 19940 1552 12164 34292 13922 10011 31377 24102 34539 11992\n20 21252 28 2058",
"output": "531"
},
{
"input": "29\n56328 80183 27682 79083 60680 12286 34299 8015 51808 50756 82133 45930 43695 65863 25178 70825 2288 15111 39667 39637 11453 62821 81484 84216 54524 53749 8396 67712 76146\n13 10739 9 3622",
"output": "1345"
},
{
"input": "46\n67864 68218 3593 30646 66413 65542 65322 26801 28984 61330 15247 16522 39142 14013 49272 41585 56739 6881 44227 7101 57657 21121 51857 39351 13500 71528 8488 66118 14756 43923 21284 20018 49049 60198 6181 62460 44141 55828 42636 14623 59758 68321 12192 29978 24745 16467\n27 5545 4 3766",
"output": "197"
},
{
"input": "70\n53691 15034 17444 13375 23285 29211 24567 21643 45514 10290 70111 24541 25072 5365 12162 34564 27535 48253 39581 13468 33718 35105 30468 50214 53365 74800 16749 33935 36346 54230 73796 26826 27866 41887 67566 40813 32267 58821 56828 26439 23708 32335 69515 33825 6092 20510 50174 11129 4592 74116 21498 77951 48056 28554 43904 21885 5967 40253 4990 70029 34374 41201 25399 6101 10354 61833 43646 20534 371 11111\n21 3911 45 1755",
"output": "1455"
},
{
"input": "10\n8121 10681 10179 10221 9410 5214 19040 17893 7862 4611\n7 7780 7 3369",
"output": "1249"
},
{
"input": "2\n1 2\n1 1 1 1",
"output": "0"
},
{
"input": "3\n1 10 20\n2 10 3 1",
"output": "4"
}
] | 1,655,022,628 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 0 | 154 | 3,379,200 | n = int(input())
X = [int(x) for x in input().split()]
Y = [int(y) for y in input().split()]
r1 = Y[0]
c1 = Y[1]
r2 = Y[2]
c2 = Y[3]
min_string = c1
min_answer = abs(r2-r1) + abs(c2-c1)
for i in range(n):
if (i+1-r1)*(i+1-r2) < 0:
min_string = min(X[i]+1, min_string)
min_answer = min(abs(r2-r1) + abs(c2-min_string), min_answer)
else:
if X[i]+1 < min_string:
min_string = min(X[i]+1, min_string)
min_answer = min(abs(min_string-c2) + abs(2*(r1-(i+1)) + (r2-r1)), min_answer)
print(min_answer)
# min_string = min(X[i]+1, min_string)
# answer = abs(r2-r1) + abs(c2-min_string)
# further_min = min_string
| Title: Text Editor
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya is pressing the keys on the keyboard reluctantly, squeezing out his ideas on the classical epos depicted in Homer's Odysseus... How can he explain to his literature teacher that he isn't going to become a writer? In fact, he is going to become a programmer. So, he would take great pleasure in writing a program, but none — in writing a composition.
As Vasya was fishing for a sentence in the dark pond of his imagination, he suddenly wondered: what is the least number of times he should push a key to shift the cursor from one position to another one?
Let's describe his question more formally: to type a text, Vasya is using the text editor. He has already written *n* lines, the *i*-th line contains *a**i* characters (including spaces). If some line contains *k* characters, then this line overall contains (*k*<=+<=1) positions where the cursor can stand: before some character or after all characters (at the end of the line). Thus, the cursor's position is determined by a pair of integers (*r*,<=*c*), where *r* is the number of the line and *c* is the cursor's position in the line (the positions are indexed starting from one from the beginning of the line).
Vasya doesn't use the mouse to move the cursor. He uses keys "Up", "Down", "Right" and "Left". When he pushes each of these keys, the cursor shifts in the needed direction. Let's assume that before the corresponding key is pressed, the cursor was located in the position (*r*,<=*c*), then Vasya pushed key:
- "Up": if the cursor was located in the first line (*r*<==<=1), then it does not move. Otherwise, it moves to the previous line (with number *r*<=-<=1), to the same position. At that, if the previous line was short, that is, the cursor couldn't occupy position *c* there, the cursor moves to the last position of the line with number *r*<=-<=1;- "Down": if the cursor was located in the last line (*r*<==<=*n*), then it does not move. Otherwise, it moves to the next line (with number *r*<=+<=1), to the same position. At that, if the next line was short, that is, the cursor couldn't occupy position *c* there, the cursor moves to the last position of the line with number *r*<=+<=1;- "Right": if the cursor can move to the right in this line (*c*<=<<=*a**r*<=+<=1), then it moves to the right (to position *c*<=+<=1). Otherwise, it is located at the end of the line and doesn't move anywhere when Vasya presses the "Right" key;- "Left": if the cursor can move to the left in this line (*c*<=><=1), then it moves to the left (to position *c*<=-<=1). Otherwise, it is located at the beginning of the line and doesn't move anywhere when Vasya presses the "Left" key.
You've got the number of lines in the text file and the number of characters, written in each line of this file. Find the least number of times Vasya should push the keys, described above, to shift the cursor from position (*r*1,<=*c*1) to position (*r*2,<=*c*2).
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the file. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=105), separated by single spaces. The third line contains four integers *r*1,<=*c*1,<=*r*2,<=*c*2 (1<=≤<=*r*1,<=*r*2<=≤<=*n*,<=1<=≤<=*c*1<=≤<=*a**r*1<=+<=1,<=1<=≤<=*c*2<=≤<=*a**r*2<=+<=1).
Output Specification:
Print a single integer — the minimum number of times Vasya should push a key to move the cursor from position (*r*1,<=*c*1) to position (*r*2,<=*c*2).
Demo Input:
['4\n2 1 6 4\n3 4 4 2\n', '4\n10 5 6 4\n1 11 4 2\n', '3\n10 1 10\n1 10 1 1\n']
Demo Output:
['3\n', '6\n', '3\n']
Note:
In the first sample the editor contains four lines. Let's represent the cursor's possible positions in the line as numbers. Letter *s* represents the cursor's initial position, letter *t* represents the last one. Then all possible positions of the cursor in the text editor are described by the following table.
123
12
123s567
1t345
One of the possible answers in the given sample is: "Left", "Down", "Left". | ```python
n = int(input())
X = [int(x) for x in input().split()]
Y = [int(y) for y in input().split()]
r1 = Y[0]
c1 = Y[1]
r2 = Y[2]
c2 = Y[3]
min_string = c1
min_answer = abs(r2-r1) + abs(c2-c1)
for i in range(n):
if (i+1-r1)*(i+1-r2) < 0:
min_string = min(X[i]+1, min_string)
min_answer = min(abs(r2-r1) + abs(c2-min_string), min_answer)
else:
if X[i]+1 < min_string:
min_string = min(X[i]+1, min_string)
min_answer = min(abs(min_string-c2) + abs(2*(r1-(i+1)) + (r2-r1)), min_answer)
print(min_answer)
# min_string = min(X[i]+1, min_string)
# answer = abs(r2-r1) + abs(c2-min_string)
# further_min = min_string
``` | -1 | |
791 | A | Bear and Big Brother | PROGRAMMING | 800 | [
"implementation"
] | null | null | Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob.
Right now, Limak and Bob weigh *a* and *b* respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight.
Limak eats a lot and his weight is tripled after every year, while Bob's weight is doubled after every year.
After how many full years will Limak become strictly larger (strictly heavier) than Bob? | The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10) — the weight of Limak and the weight of Bob respectively. | Print one integer, denoting the integer number of years after which Limak will become strictly larger than Bob. | [
"4 7\n",
"4 9\n",
"1 1\n"
] | [
"2\n",
"3\n",
"1\n"
] | In the first sample, Limak weighs 4 and Bob weighs 7 initially. After one year their weights are 4·3 = 12 and 7·2 = 14 respectively (one weight is tripled while the other one is doubled). Limak isn't larger than Bob yet. After the second year weights are 36 and 28, so the first weight is greater than the second one. Limak became larger than Bob after two years so you should print 2.
In the second sample, Limak's and Bob's weights in next years are: 12 and 18, then 36 and 36, and finally 108 and 72 (after three years). The answer is 3. Remember that Limak wants to be larger than Bob and he won't be satisfied with equal weights.
In the third sample, Limak becomes larger than Bob after the first year. Their weights will be 3 and 2 then. | 500 | [
{
"input": "4 7",
"output": "2"
},
{
"input": "4 9",
"output": "3"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "4 6",
"output": "2"
},
{
"input": "1 10",
"output": "6"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "1 5",
"output": "4"
},
{
"input": "1 6",
"output": "5"
},
{
"input": "1 7",
"output": "5"
},
{
"input": "1 8",
"output": "6"
},
{
"input": "1 9",
"output": "6"
},
{
"input": "1 10",
"output": "6"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "2 4",
"output": "2"
},
{
"input": "2 5",
"output": "3"
},
{
"input": "2 6",
"output": "3"
},
{
"input": "2 7",
"output": "4"
},
{
"input": "2 8",
"output": "4"
},
{
"input": "2 9",
"output": "4"
},
{
"input": "2 10",
"output": "4"
},
{
"input": "3 3",
"output": "1"
},
{
"input": "3 4",
"output": "1"
},
{
"input": "3 5",
"output": "2"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "3 7",
"output": "3"
},
{
"input": "3 8",
"output": "3"
},
{
"input": "3 9",
"output": "3"
},
{
"input": "3 10",
"output": "3"
},
{
"input": "4 4",
"output": "1"
},
{
"input": "4 5",
"output": "1"
},
{
"input": "4 6",
"output": "2"
},
{
"input": "4 7",
"output": "2"
},
{
"input": "4 8",
"output": "2"
},
{
"input": "4 9",
"output": "3"
},
{
"input": "4 10",
"output": "3"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "5 6",
"output": "1"
},
{
"input": "5 7",
"output": "1"
},
{
"input": "5 8",
"output": "2"
},
{
"input": "5 9",
"output": "2"
},
{
"input": "5 10",
"output": "2"
},
{
"input": "6 6",
"output": "1"
},
{
"input": "6 7",
"output": "1"
},
{
"input": "6 8",
"output": "1"
},
{
"input": "6 9",
"output": "2"
},
{
"input": "6 10",
"output": "2"
},
{
"input": "7 7",
"output": "1"
},
{
"input": "7 8",
"output": "1"
},
{
"input": "7 9",
"output": "1"
},
{
"input": "7 10",
"output": "1"
},
{
"input": "8 8",
"output": "1"
},
{
"input": "8 9",
"output": "1"
},
{
"input": "8 10",
"output": "1"
},
{
"input": "9 9",
"output": "1"
},
{
"input": "9 10",
"output": "1"
},
{
"input": "10 10",
"output": "1"
},
{
"input": "10 10",
"output": "1"
},
{
"input": "1 2",
"output": "2"
}
] | 1,694,011,625 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | x,y = map(int, input().split())
y=0
while x<=y:
x*= 3
y*= 2
y+= 1
print(y)
| Title: Bear and Big Brother
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob.
Right now, Limak and Bob weigh *a* and *b* respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight.
Limak eats a lot and his weight is tripled after every year, while Bob's weight is doubled after every year.
After how many full years will Limak become strictly larger (strictly heavier) than Bob?
Input Specification:
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10) — the weight of Limak and the weight of Bob respectively.
Output Specification:
Print one integer, denoting the integer number of years after which Limak will become strictly larger than Bob.
Demo Input:
['4 7\n', '4 9\n', '1 1\n']
Demo Output:
['2\n', '3\n', '1\n']
Note:
In the first sample, Limak weighs 4 and Bob weighs 7 initially. After one year their weights are 4·3 = 12 and 7·2 = 14 respectively (one weight is tripled while the other one is doubled). Limak isn't larger than Bob yet. After the second year weights are 36 and 28, so the first weight is greater than the second one. Limak became larger than Bob after two years so you should print 2.
In the second sample, Limak's and Bob's weights in next years are: 12 and 18, then 36 and 36, and finally 108 and 72 (after three years). The answer is 3. Remember that Limak wants to be larger than Bob and he won't be satisfied with equal weights.
In the third sample, Limak becomes larger than Bob after the first year. Their weights will be 3 and 2 then. | ```python
x,y = map(int, input().split())
y=0
while x<=y:
x*= 3
y*= 2
y+= 1
print(y)
``` | 0 |
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