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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
734 | A | Anton and Danik | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | Anton likes to play chess, and so does his friend Danik.
Once they have played *n* games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie.
Now Anton wonders, who won more games, he or Danik? Help him determine this. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of games played.
The second line contains a string *s*, consisting of *n* uppercase English letters 'A' and 'D' — the outcome of each of the games. The *i*-th character of the string is equal to 'A' if the Anton won the *i*-th game and 'D' if Danik won the *i*-th game. | If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output.
If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output.
If Anton and Danik won the same number of games, print "Friendship" (without quotes). | [
"6\nADAAAA\n",
"7\nDDDAADA\n",
"6\nDADADA\n"
] | [
"Anton\n",
"Danik\n",
"Friendship\n"
] | In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton".
In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik".
In the third sample, both Anton and Danik won 3 games and the answer is "Friendship". | 500 | [
{
"input": "6\nADAAAA",
"output": "Anton"
},
{
"input": "7\nDDDAADA",
"output": "Danik"
},
{
"input": "6\nDADADA",
"output": "Friendship"
},
{
"input": "10\nDDDDADDADD",
"output": "Danik"
},
{
"input": "40\nAAAAAAAAADDAAAAAAAAAAADADDAAAAAAAAAAADAA",
"output": ... | 1,696,415,167 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 25 | 62 | 4,915,200 | n = int(input())
wins = list(input())
A = wins.count("A")
D = wins.count("D")
if A > D:
print("Anton")
elif A < D:
print("Danik")
else: print("Friendship")
| Title: Anton and Danik
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anton likes to play chess, and so does his friend Danik.
Once they have played *n* games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie.
Now Anton wonders, who won more games, he or Danik? Help him determine this.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of games played.
The second line contains a string *s*, consisting of *n* uppercase English letters 'A' and 'D' — the outcome of each of the games. The *i*-th character of the string is equal to 'A' if the Anton won the *i*-th game and 'D' if Danik won the *i*-th game.
Output Specification:
If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output.
If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output.
If Anton and Danik won the same number of games, print "Friendship" (without quotes).
Demo Input:
['6\nADAAAA\n', '7\nDDDAADA\n', '6\nDADADA\n']
Demo Output:
['Anton\n', 'Danik\n', 'Friendship\n']
Note:
In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton".
In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik".
In the third sample, both Anton and Danik won 3 games and the answer is "Friendship". | ```python
n = int(input())
wins = list(input())
A = wins.count("A")
D = wins.count("D")
if A > D:
print("Anton")
elif A < D:
print("Danik")
else: print("Friendship")
``` | 3 | |
227 | A | Where do I Turn? | PROGRAMMING | 1,300 | [
"geometry"
] | null | null | Trouble came from the overseas lands: a three-headed dragon Gorynych arrived. The dragon settled at point *C* and began to terrorize the residents of the surrounding villages.
A brave hero decided to put an end to the dragon. He moved from point *A* to fight with Gorynych. The hero rode from point *A* along a straight road and met point *B* on his way. The hero knows that in this land for every pair of roads it is true that they are either parallel to each other, or lie on a straight line, or are perpendicular to each other. He also knows well that points *B* and *C* are connected by a road. So the hero must either turn 90 degrees to the left or continue riding straight ahead or turn 90 degrees to the right. But he forgot where the point *C* is located.
Fortunately, a Brave Falcon flew right by. It can see all three points from the sky. The hero asked him what way to go to get to the dragon's lair.
If you have not got it, you are the falcon. Help the hero and tell him how to get him to point *C*: turn left, go straight or turn right.
At this moment the hero is believed to stand at point *B*, turning his back to point *A*. | The first input line contains two space-separated integers *x**a*,<=*y**a* (|*x**a*|,<=|*y**a*|<=≤<=109) — the coordinates of point *A*. The second line contains the coordinates of point *B* in the same form, the third line contains the coordinates of point *C*.
It is guaranteed that all points are pairwise different. It is also guaranteed that either point *B* lies on segment *AC*, or angle *ABC* is right. | Print a single line. If a hero must turn left, print "LEFT" (without the quotes); If he must go straight ahead, print "TOWARDS" (without the quotes); if he should turn right, print "RIGHT" (without the quotes). | [
"0 0\n0 1\n1 1\n",
"-1 -1\n-3 -3\n-4 -4\n",
"-4 -6\n-3 -7\n-2 -6\n"
] | [
"RIGHT\n",
"TOWARDS\n",
"LEFT\n"
] | The picture to the first sample:
The red color shows points A, B and C. The blue arrow shows the hero's direction. The green color shows the hero's trajectory.
The picture to the second sample: | 500 | [
{
"input": "0 0\n0 1\n1 1",
"output": "RIGHT"
},
{
"input": "-1 -1\n-3 -3\n-4 -4",
"output": "TOWARDS"
},
{
"input": "-4 -6\n-3 -7\n-2 -6",
"output": "LEFT"
},
{
"input": "-44 57\n-118 -41\n-216 33",
"output": "RIGHT"
},
{
"input": "39 100\n90 85\n105 136",
"o... | 1,630,555,272 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | def nums():
return map(int, raw_input().split())
x1, y1 = nums()
x2, y2 = nums()
x3, y3 = nums()
d = (x2 - x1)*(y3 - y2) - (y2 - y1) * (x3 - x2)
if d < 0:
print 'RIGHT'
elif d == 0:
print 'TOWARDS'
else:
print 'LEFT'
| Title: Where do I Turn?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Trouble came from the overseas lands: a three-headed dragon Gorynych arrived. The dragon settled at point *C* and began to terrorize the residents of the surrounding villages.
A brave hero decided to put an end to the dragon. He moved from point *A* to fight with Gorynych. The hero rode from point *A* along a straight road and met point *B* on his way. The hero knows that in this land for every pair of roads it is true that they are either parallel to each other, or lie on a straight line, or are perpendicular to each other. He also knows well that points *B* and *C* are connected by a road. So the hero must either turn 90 degrees to the left or continue riding straight ahead or turn 90 degrees to the right. But he forgot where the point *C* is located.
Fortunately, a Brave Falcon flew right by. It can see all three points from the sky. The hero asked him what way to go to get to the dragon's lair.
If you have not got it, you are the falcon. Help the hero and tell him how to get him to point *C*: turn left, go straight or turn right.
At this moment the hero is believed to stand at point *B*, turning his back to point *A*.
Input Specification:
The first input line contains two space-separated integers *x**a*,<=*y**a* (|*x**a*|,<=|*y**a*|<=≤<=109) — the coordinates of point *A*. The second line contains the coordinates of point *B* in the same form, the third line contains the coordinates of point *C*.
It is guaranteed that all points are pairwise different. It is also guaranteed that either point *B* lies on segment *AC*, or angle *ABC* is right.
Output Specification:
Print a single line. If a hero must turn left, print "LEFT" (without the quotes); If he must go straight ahead, print "TOWARDS" (without the quotes); if he should turn right, print "RIGHT" (without the quotes).
Demo Input:
['0 0\n0 1\n1 1\n', '-1 -1\n-3 -3\n-4 -4\n', '-4 -6\n-3 -7\n-2 -6\n']
Demo Output:
['RIGHT\n', 'TOWARDS\n', 'LEFT\n']
Note:
The picture to the first sample:
The red color shows points A, B and C. The blue arrow shows the hero's direction. The green color shows the hero's trajectory.
The picture to the second sample: | ```python
def nums():
return map(int, raw_input().split())
x1, y1 = nums()
x2, y2 = nums()
x3, y3 = nums()
d = (x2 - x1)*(y3 - y2) - (y2 - y1) * (x3 - x2)
if d < 0:
print 'RIGHT'
elif d == 0:
print 'TOWARDS'
else:
print 'LEFT'
``` | -1 | |
964 | A | Splits | PROGRAMMING | 800 | [
"math"
] | null | null | Let's define a split of $n$ as a nonincreasing sequence of positive integers, the sum of which is $n$.
For example, the following sequences are splits of $8$: $[4, 4]$, $[3, 3, 2]$, $[2, 2, 1, 1, 1, 1]$, $[5, 2, 1]$.
The following sequences aren't splits of $8$: $[1, 7]$, $[5, 4]$, $[11, -3]$, $[1, 1, 4, 1, 1]$.
The weight of a split is the number of elements in the split that are equal to the first element. For example, the weight of the split $[1, 1, 1, 1, 1]$ is $5$, the weight of the split $[5, 5, 3, 3, 3]$ is $2$ and the weight of the split $[9]$ equals $1$.
For a given $n$, find out the number of different weights of its splits. | The first line contains one integer $n$ ($1 \leq n \leq 10^9$). | Output one integer — the answer to the problem. | [
"7\n",
"8\n",
"9\n"
] | [
"4\n",
"5\n",
"5\n"
] | In the first sample, there are following possible weights of splits of $7$:
Weight 1: [$\textbf 7$]
Weight 2: [$\textbf 3$, $\textbf 3$, 1]
Weight 3: [$\textbf 2$, $\textbf 2$, $\textbf 2$, 1]
Weight 7: [$\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$] | 500 | [
{
"input": "7",
"output": "4"
},
{
"input": "8",
"output": "5"
},
{
"input": "9",
"output": "5"
},
{
"input": "1",
"output": "1"
},
{
"input": "286",
"output": "144"
},
{
"input": "48",
"output": "25"
},
{
"input": "941",
"output": "471... | 1,524,717,428 | 2,147,483,647 | Python 3 | OK | TESTS | 63 | 93 | 7,372,800 | n = int(input())
a = n//2 + 1
b = 1
c = 1
print(a)
| Title: Splits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's define a split of $n$ as a nonincreasing sequence of positive integers, the sum of which is $n$.
For example, the following sequences are splits of $8$: $[4, 4]$, $[3, 3, 2]$, $[2, 2, 1, 1, 1, 1]$, $[5, 2, 1]$.
The following sequences aren't splits of $8$: $[1, 7]$, $[5, 4]$, $[11, -3]$, $[1, 1, 4, 1, 1]$.
The weight of a split is the number of elements in the split that are equal to the first element. For example, the weight of the split $[1, 1, 1, 1, 1]$ is $5$, the weight of the split $[5, 5, 3, 3, 3]$ is $2$ and the weight of the split $[9]$ equals $1$.
For a given $n$, find out the number of different weights of its splits.
Input Specification:
The first line contains one integer $n$ ($1 \leq n \leq 10^9$).
Output Specification:
Output one integer — the answer to the problem.
Demo Input:
['7\n', '8\n', '9\n']
Demo Output:
['4\n', '5\n', '5\n']
Note:
In the first sample, there are following possible weights of splits of $7$:
Weight 1: [$\textbf 7$]
Weight 2: [$\textbf 3$, $\textbf 3$, 1]
Weight 3: [$\textbf 2$, $\textbf 2$, $\textbf 2$, 1]
Weight 7: [$\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$, $\textbf 1$] | ```python
n = int(input())
a = n//2 + 1
b = 1
c = 1
print(a)
``` | 3 | |
873 | B | Balanced Substring | PROGRAMMING | 1,500 | [
"dp",
"implementation"
] | null | null | You are given a string *s* consisting only of characters 0 and 1. A substring [*l*,<=*r*] of *s* is a string *s**l**s**l*<=+<=1*s**l*<=+<=2... *s**r*, and its length equals to *r*<=-<=*l*<=+<=1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of *s*. | The first line contains *n* (1<=≤<=*n*<=≤<=100000) — the number of characters in *s*.
The second line contains a string *s* consisting of exactly *n* characters. Only characters 0 and 1 can appear in *s*. | If there is no non-empty balanced substring in *s*, print 0. Otherwise, print the length of the longest balanced substring. | [
"8\n11010111\n",
"3\n111\n"
] | [
"4\n",
"0\n"
] | In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring. | 0 | [
{
"input": "8\n11010111",
"output": "4"
},
{
"input": "3\n111",
"output": "0"
},
{
"input": "11\n00001000100",
"output": "2"
},
{
"input": "10\n0100000000",
"output": "2"
},
{
"input": "13\n0001000011010",
"output": "6"
},
{
"input": "14\n0000010010101... | 1,693,660,000 | 2,147,483,647 | Python 3 | OK | TESTS | 53 | 93 | 22,118,400 | def D(n, arr):
prefix = [0] * (n + 1)
for i in range(1, n + 1):
if arr[i - 1] == '1':
prefix[i] = prefix[i - 1] + 1
else:
prefix[i] = prefix[i - 1] - 1
max_count = 0
d_arr_left = {}
d_arr_right = {}
for i in range(n + 1):
if prefix[i] not in d_arr_left:
d_arr_left[prefix[i]] = i
d_arr_right[prefix[i]] = i
for i in d_arr_left:
if d_arr_right[i] - d_arr_left[i] > max_count:
max_count = d_arr_right[i] - d_arr_left[i]
return max_count
n = int(input())
arr = input()
print(D(n, arr))
| Title: Balanced Substring
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string *s* consisting only of characters 0 and 1. A substring [*l*,<=*r*] of *s* is a string *s**l**s**l*<=+<=1*s**l*<=+<=2... *s**r*, and its length equals to *r*<=-<=*l*<=+<=1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of *s*.
Input Specification:
The first line contains *n* (1<=≤<=*n*<=≤<=100000) — the number of characters in *s*.
The second line contains a string *s* consisting of exactly *n* characters. Only characters 0 and 1 can appear in *s*.
Output Specification:
If there is no non-empty balanced substring in *s*, print 0. Otherwise, print the length of the longest balanced substring.
Demo Input:
['8\n11010111\n', '3\n111\n']
Demo Output:
['4\n', '0\n']
Note:
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring. | ```python
def D(n, arr):
prefix = [0] * (n + 1)
for i in range(1, n + 1):
if arr[i - 1] == '1':
prefix[i] = prefix[i - 1] + 1
else:
prefix[i] = prefix[i - 1] - 1
max_count = 0
d_arr_left = {}
d_arr_right = {}
for i in range(n + 1):
if prefix[i] not in d_arr_left:
d_arr_left[prefix[i]] = i
d_arr_right[prefix[i]] = i
for i in d_arr_left:
if d_arr_right[i] - d_arr_left[i] > max_count:
max_count = d_arr_right[i] - d_arr_left[i]
return max_count
n = int(input())
arr = input()
print(D(n, arr))
``` | 3 | |
219 | A | k-String | PROGRAMMING | 1,000 | [
"implementation",
"strings"
] | null | null | A string is called a *k*-string if it can be represented as *k* concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1-string, a 2-string and a 4-string, but it is not a 3-string, a 5-string, or a 6-string and so on. Obviously any string is a 1-string.
You are given a string *s*, consisting of lowercase English letters and a positive integer *k*. Your task is to reorder the letters in the string *s* in such a way that the resulting string is a *k*-string. | The first input line contains integer *k* (1<=≤<=*k*<=≤<=1000). The second line contains *s*, all characters in *s* are lowercase English letters. The string length *s* satisfies the inequality 1<=≤<=|*s*|<=≤<=1000, where |*s*| is the length of string *s*. | Rearrange the letters in string *s* in such a way that the result is a *k*-string. Print the result on a single output line. If there are multiple solutions, print any of them.
If the solution doesn't exist, print "-1" (without quotes). | [
"2\naazz\n",
"3\nabcabcabz\n"
] | [
"azaz\n",
"-1\n"
] | none | 500 | [
{
"input": "2\naazz",
"output": "azaz"
},
{
"input": "3\nabcabcabz",
"output": "-1"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "2\nabba",
"output": "abab"
},
{
"input": "2\naaab",
"output": "-1"
},
{
"input": "7\nabacaba",
"output": "-1"
}... | 1,682,595,040 | 2,147,483,647 | Python 3 | OK | TESTS | 42 | 92 | 0 | k=int(input())
s=input()
d={}
for x in s:
if x in d:
d[x]+=1
else:
d[x]=1
f=""
o=0
for x in d:
z=(d[x])%k
if(z!=0):
o=1
break
else:
f=f+x*(d[x]//k)
if(o==1):
print(-1)
else:
print(f*k)
| Title: k-String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string is called a *k*-string if it can be represented as *k* concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1-string, a 2-string and a 4-string, but it is not a 3-string, a 5-string, or a 6-string and so on. Obviously any string is a 1-string.
You are given a string *s*, consisting of lowercase English letters and a positive integer *k*. Your task is to reorder the letters in the string *s* in such a way that the resulting string is a *k*-string.
Input Specification:
The first input line contains integer *k* (1<=≤<=*k*<=≤<=1000). The second line contains *s*, all characters in *s* are lowercase English letters. The string length *s* satisfies the inequality 1<=≤<=|*s*|<=≤<=1000, where |*s*| is the length of string *s*.
Output Specification:
Rearrange the letters in string *s* in such a way that the result is a *k*-string. Print the result on a single output line. If there are multiple solutions, print any of them.
If the solution doesn't exist, print "-1" (without quotes).
Demo Input:
['2\naazz\n', '3\nabcabcabz\n']
Demo Output:
['azaz\n', '-1\n']
Note:
none | ```python
k=int(input())
s=input()
d={}
for x in s:
if x in d:
d[x]+=1
else:
d[x]=1
f=""
o=0
for x in d:
z=(d[x])%k
if(z!=0):
o=1
break
else:
f=f+x*(d[x]//k)
if(o==1):
print(-1)
else:
print(f*k)
``` | 3 | |
894 | A | QAQ | PROGRAMMING | 800 | [
"brute force",
"dp"
] | null | null | "QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length *n*. There is a great number of "QAQ" in the string (Diamond is so cute!).
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact. | The only line contains a string of length *n* (1<=≤<=*n*<=≤<=100). It's guaranteed that the string only contains uppercase English letters. | Print a single integer — the number of subsequences "QAQ" in the string. | [
"QAQAQYSYIOIWIN\n",
"QAQQQZZYNOIWIN\n"
] | [
"4\n",
"3\n"
] | In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN". | 500 | [
{
"input": "QAQAQYSYIOIWIN",
"output": "4"
},
{
"input": "QAQQQZZYNOIWIN",
"output": "3"
},
{
"input": "QA",
"output": "0"
},
{
"input": "IAQVAQZLQBQVQFTQQQADAQJA",
"output": "24"
},
{
"input": "QQAAQASGAYAAAAKAKAQIQEAQAIAAIAQQQQQ",
"output": "378"
},
{
... | 1,605,283,306 | 2,147,483,647 | Python 3 | OK | TESTS | 43 | 124 | 0 | string = input()
count=0
for i in range(len(string)):
for j in range(i+1,len(string)):
for k in range(j+1,len(string)):
if(string[i] =="Q" and string[j]=="A" and string[k]=="Q"):
count+=1
print(count)
| Title: QAQ
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.
Now Diamond has given Bort a string consisting of only uppercase English letters of length *n*. There is a great number of "QAQ" in the string (Diamond is so cute!).
Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.
Input Specification:
The only line contains a string of length *n* (1<=≤<=*n*<=≤<=100). It's guaranteed that the string only contains uppercase English letters.
Output Specification:
Print a single integer — the number of subsequences "QAQ" in the string.
Demo Input:
['QAQAQYSYIOIWIN\n', 'QAQQQZZYNOIWIN\n']
Demo Output:
['4\n', '3\n']
Note:
In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN". | ```python
string = input()
count=0
for i in range(len(string)):
for j in range(i+1,len(string)):
for k in range(j+1,len(string)):
if(string[i] =="Q" and string[j]=="A" and string[k]=="Q"):
count+=1
print(count)
``` | 3 | |
92 | B | Binary Number | PROGRAMMING | 1,300 | [
"greedy"
] | B. Binary Number | 1 | 256 | Little walrus Fangy loves math very much. That's why when he is bored he plays with a number performing some operations.
Fangy takes some positive integer *x* and wants to get a number one from it. While *x* is not equal to 1, Fangy repeats the following action: if *x* is odd, then he adds 1 to it, otherwise he divides *x* by 2. Fangy knows that for any positive integer number the process ends in finite time.
How many actions should Fangy perform to get a number one from number *x*? | The first line contains a positive integer *x* in a binary system. It is guaranteed that the first digit of *x* is different from a zero and the number of its digits does not exceed 106. | Print the required number of actions. | [
"1\n",
"1001001\n",
"101110\n"
] | [
"0\n",
"12\n",
"8\n"
] | Let's consider the third sample. Number 101110 is even, which means that we should divide it by 2. After the dividing Fangy gets an odd number 10111 and adds one to it. Number 11000 can be divided by 2 three times in a row and get number 11. All that's left is to increase the number by one (we get 100), and then divide it by 2 two times in a row. As a result, we get 1. | 1,000 | [
{
"input": "1",
"output": "0"
},
{
"input": "1001001",
"output": "12"
},
{
"input": "101110",
"output": "8"
},
{
"input": "11",
"output": "3"
},
{
"input": "11110001101",
"output": "16"
},
{
"input": "101010100100111100011111001111100001010101111110101... | 1,618,204,801 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | x=input()
k=0
x=int(x,2)
print(x)
while(x!=1):
if x%2!=0:
x+=1
print(x)
k+=1
else:
x/=2
k+=1
print(k)
| Title: Binary Number
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Little walrus Fangy loves math very much. That's why when he is bored he plays with a number performing some operations.
Fangy takes some positive integer *x* and wants to get a number one from it. While *x* is not equal to 1, Fangy repeats the following action: if *x* is odd, then he adds 1 to it, otherwise he divides *x* by 2. Fangy knows that for any positive integer number the process ends in finite time.
How many actions should Fangy perform to get a number one from number *x*?
Input Specification:
The first line contains a positive integer *x* in a binary system. It is guaranteed that the first digit of *x* is different from a zero and the number of its digits does not exceed 106.
Output Specification:
Print the required number of actions.
Demo Input:
['1\n', '1001001\n', '101110\n']
Demo Output:
['0\n', '12\n', '8\n']
Note:
Let's consider the third sample. Number 101110 is even, which means that we should divide it by 2. After the dividing Fangy gets an odd number 10111 and adds one to it. Number 11000 can be divided by 2 three times in a row and get number 11. All that's left is to increase the number by one (we get 100), and then divide it by 2 two times in a row. As a result, we get 1. | ```python
x=input()
k=0
x=int(x,2)
print(x)
while(x!=1):
if x%2!=0:
x+=1
print(x)
k+=1
else:
x/=2
k+=1
print(k)
``` | 0 |
412 | A | Poster | PROGRAMMING | 900 | [
"greedy",
"implementation"
] | null | null | The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of *n* characters, so the decorators hung a large banner, *n* meters wide and 1 meter high, divided into *n* equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the *k*-th square of the poster. To draw the *i*-th character of the slogan on the poster, you need to climb the ladder, standing in front of the *i*-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the *i*-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan! | The first line contains two integers, *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=100) — the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as *n* characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'. | In *t* lines, print the actions the programmers need to make. In the *i*-th line print:
- "LEFT" (without the quotes), if the *i*-th action was "move the ladder to the left"; - "RIGHT" (without the quotes), if the *i*-th action was "move the ladder to the right"; - "PRINT *x*" (without the quotes), if the *i*-th action was to "go up the ladder, paint character *x*, go down the ladder".
The painting time (variable *t*) must be minimum possible. If there are multiple optimal painting plans, you can print any of them. | [
"2 2\nR1\n",
"2 1\nR1\n",
"6 4\nGO?GO!\n"
] | [
"PRINT 1\nLEFT\nPRINT R\n",
"PRINT R\nRIGHT\nPRINT 1\n",
"RIGHT\nRIGHT\nPRINT !\nLEFT\nPRINT O\nLEFT\nPRINT G\nLEFT\nPRINT ?\nLEFT\nPRINT O\nLEFT\nPRINT G\n"
] | Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character. | 500 | [
{
"input": "2 2\nR1",
"output": "PRINT 1\nLEFT\nPRINT R"
},
{
"input": "2 1\nR1",
"output": "PRINT R\nRIGHT\nPRINT 1"
},
{
"input": "6 4\nGO?GO!",
"output": "RIGHT\nRIGHT\nPRINT !\nLEFT\nPRINT O\nLEFT\nPRINT G\nLEFT\nPRINT ?\nLEFT\nPRINT O\nLEFT\nPRINT G"
},
{
"input": "7 3\n... | 1,569,669,770 | 2,147,483,647 | PyPy 3 | OK | TESTS | 43 | 140 | 0 | n,k=map(int,input().split())
s=input()
if n==1:
print('PRINT',end=' ')
print(s[0])
elif n==2 and k==2:
print('PRINT',end=' ')
print(s[1])
print('LEFT')
print('PRINT',end=' ')
print(s[0])
elif n==2 and k==1:
print('PRINT',end=' ')
print(s[0])
print('RIGHT')
print('PRINT',end=' ')
print(s[1])
else:
if k>n//2:
for i in range(k,n):
print('RIGHT')
for i in range(n-1,0,-1):
print('PRINT',end=' ')
print(s[i])
print('LEFT')
print('PRINT',end=' ')
print(s[i-1])
else:
for i in range(k,1,-1):
print('LEFT')
for i in range(n-1):
print('PRINT',end=' ')
print(s[i])
print('RIGHT')
print('PRINT',end=' ')
print(s[i+1]) | Title: Poster
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The R1 company has recently bought a high rise building in the centre of Moscow for its main office. It's time to decorate the new office, and the first thing to do is to write the company's slogan above the main entrance to the building.
The slogan of the company consists of *n* characters, so the decorators hung a large banner, *n* meters wide and 1 meter high, divided into *n* equal squares. The first character of the slogan must be in the first square (the leftmost) of the poster, the second character must be in the second square, and so on.
Of course, the R1 programmers want to write the slogan on the poster themselves. To do this, they have a large (and a very heavy) ladder which was put exactly opposite the *k*-th square of the poster. To draw the *i*-th character of the slogan on the poster, you need to climb the ladder, standing in front of the *i*-th square of the poster. This action (along with climbing up and down the ladder) takes one hour for a painter. The painter is not allowed to draw characters in the adjacent squares when the ladder is in front of the *i*-th square because the uncomfortable position of the ladder may make the characters untidy. Besides, the programmers can move the ladder. In one hour, they can move the ladder either a meter to the right or a meter to the left.
Drawing characters and moving the ladder is very tiring, so the programmers want to finish the job in as little time as possible. Develop for them an optimal poster painting plan!
Input Specification:
The first line contains two integers, *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=100) — the number of characters in the slogan and the initial position of the ladder, correspondingly. The next line contains the slogan as *n* characters written without spaces. Each character of the slogan is either a large English letter, or digit, or one of the characters: '.', '!', ',', '?'.
Output Specification:
In *t* lines, print the actions the programmers need to make. In the *i*-th line print:
- "LEFT" (without the quotes), if the *i*-th action was "move the ladder to the left"; - "RIGHT" (without the quotes), if the *i*-th action was "move the ladder to the right"; - "PRINT *x*" (without the quotes), if the *i*-th action was to "go up the ladder, paint character *x*, go down the ladder".
The painting time (variable *t*) must be minimum possible. If there are multiple optimal painting plans, you can print any of them.
Demo Input:
['2 2\nR1\n', '2 1\nR1\n', '6 4\nGO?GO!\n']
Demo Output:
['PRINT 1\nLEFT\nPRINT R\n', 'PRINT R\nRIGHT\nPRINT 1\n', 'RIGHT\nRIGHT\nPRINT !\nLEFT\nPRINT O\nLEFT\nPRINT G\nLEFT\nPRINT ?\nLEFT\nPRINT O\nLEFT\nPRINT G\n']
Note:
Note that the ladder cannot be shifted by less than one meter. The ladder can only stand in front of some square of the poster. For example, you cannot shift a ladder by half a meter and position it between two squares. Then go up and paint the first character and the second character. | ```python
n,k=map(int,input().split())
s=input()
if n==1:
print('PRINT',end=' ')
print(s[0])
elif n==2 and k==2:
print('PRINT',end=' ')
print(s[1])
print('LEFT')
print('PRINT',end=' ')
print(s[0])
elif n==2 and k==1:
print('PRINT',end=' ')
print(s[0])
print('RIGHT')
print('PRINT',end=' ')
print(s[1])
else:
if k>n//2:
for i in range(k,n):
print('RIGHT')
for i in range(n-1,0,-1):
print('PRINT',end=' ')
print(s[i])
print('LEFT')
print('PRINT',end=' ')
print(s[i-1])
else:
for i in range(k,1,-1):
print('LEFT')
for i in range(n-1):
print('PRINT',end=' ')
print(s[i])
print('RIGHT')
print('PRINT',end=' ')
print(s[i+1])
``` | 3 | |
902 | A | Visiting a Friend | PROGRAMMING | 1,100 | [
"greedy",
"implementation"
] | null | null | Pig is visiting a friend.
Pig's house is located at point 0, and his friend's house is located at point *m* on an axis.
Pig can use teleports to move along the axis.
To use a teleport, Pig should come to a certain point (where the teleport is located) and choose where to move: for each teleport there is the rightmost point it can move Pig to, this point is known as the limit of the teleport.
Formally, a teleport located at point *x* with limit *y* can move Pig from point *x* to any point within the segment [*x*;<=*y*], including the bounds.
Determine if Pig can visit the friend using teleports only, or he should use his car. | The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*m*<=≤<=100) — the number of teleports and the location of the friend's house.
The next *n* lines contain information about teleports.
The *i*-th of these lines contains two integers *a**i* and *b**i* (0<=≤<=*a**i*<=≤<=*b**i*<=≤<=*m*), where *a**i* is the location of the *i*-th teleport, and *b**i* is its limit.
It is guaranteed that *a**i*<=≥<=*a**i*<=-<=1 for every *i* (2<=≤<=*i*<=≤<=*n*). | Print "YES" if there is a path from Pig's house to his friend's house that uses only teleports, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower). | [
"3 5\n0 2\n2 4\n3 5\n",
"3 7\n0 4\n2 5\n6 7\n"
] | [
"YES\n",
"NO\n"
] | The first example is shown on the picture below:
Pig can use the first teleport from his house (point 0) to reach point 2, then using the second teleport go from point 2 to point 3, then using the third teleport go from point 3 to point 5, where his friend lives.
The second example is shown on the picture below:
You can see that there is no path from Pig's house to his friend's house that uses only teleports. | 500 | [
{
"input": "3 5\n0 2\n2 4\n3 5",
"output": "YES"
},
{
"input": "3 7\n0 4\n2 5\n6 7",
"output": "NO"
},
{
"input": "1 1\n0 0",
"output": "NO"
},
{
"input": "30 10\n0 7\n1 2\n1 2\n1 4\n1 4\n1 3\n2 2\n2 4\n2 6\n2 9\n2 2\n3 5\n3 8\n4 8\n4 5\n4 6\n5 6\n5 7\n6 6\n6 9\n6 7\n6 9\n7 7... | 1,518,099,616 | 2,147,483,647 | Python 3 | OK | TESTS | 55 | 62 | 5,632,000 | def max(a, b):
a = int(a)
b = int(b)
if a > b:
return a
else:
return b
while True:
try:
n, m = map(int, input().split())
now = 0;
for i in range(n):
a, b = map(int , input().split())
if a <= now:
now = max(now, b)
if now >= m:
print("YES")
else:
print("NO")
except:
break
| Title: Visiting a Friend
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pig is visiting a friend.
Pig's house is located at point 0, and his friend's house is located at point *m* on an axis.
Pig can use teleports to move along the axis.
To use a teleport, Pig should come to a certain point (where the teleport is located) and choose where to move: for each teleport there is the rightmost point it can move Pig to, this point is known as the limit of the teleport.
Formally, a teleport located at point *x* with limit *y* can move Pig from point *x* to any point within the segment [*x*;<=*y*], including the bounds.
Determine if Pig can visit the friend using teleports only, or he should use his car.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*m*<=≤<=100) — the number of teleports and the location of the friend's house.
The next *n* lines contain information about teleports.
The *i*-th of these lines contains two integers *a**i* and *b**i* (0<=≤<=*a**i*<=≤<=*b**i*<=≤<=*m*), where *a**i* is the location of the *i*-th teleport, and *b**i* is its limit.
It is guaranteed that *a**i*<=≥<=*a**i*<=-<=1 for every *i* (2<=≤<=*i*<=≤<=*n*).
Output Specification:
Print "YES" if there is a path from Pig's house to his friend's house that uses only teleports, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower).
Demo Input:
['3 5\n0 2\n2 4\n3 5\n', '3 7\n0 4\n2 5\n6 7\n']
Demo Output:
['YES\n', 'NO\n']
Note:
The first example is shown on the picture below:
Pig can use the first teleport from his house (point 0) to reach point 2, then using the second teleport go from point 2 to point 3, then using the third teleport go from point 3 to point 5, where his friend lives.
The second example is shown on the picture below:
You can see that there is no path from Pig's house to his friend's house that uses only teleports. | ```python
def max(a, b):
a = int(a)
b = int(b)
if a > b:
return a
else:
return b
while True:
try:
n, m = map(int, input().split())
now = 0;
for i in range(n):
a, b = map(int , input().split())
if a <= now:
now = max(now, b)
if now >= m:
print("YES")
else:
print("NO")
except:
break
``` | 3 | |
440 | A | Forgotten Episode | PROGRAMMING | 800 | [
"implementation"
] | null | null | Polycarpus adores TV series. Right now he is ready to finish watching a season of a popular sitcom "Graph Theory". In total, the season has *n* episodes, numbered with integers from 1 to *n*.
Polycarpus watches episodes not one by one but in a random order. He has already watched all the episodes except for one. Which episode has Polycaprus forgotten to watch? | The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100000) — the number of episodes in a season. Assume that the episodes are numbered by integers from 1 to *n*.
The second line contains *n*<=-<=1 integer *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the numbers of episodes that Polycarpus has watched. All values of *a**i* are distinct. | Print the number of the episode that Polycarpus hasn't watched. | [
"10\n3 8 10 1 7 9 6 5 2\n"
] | [
"4\n"
] | none | 500 | [
{
"input": "10\n3 8 10 1 7 9 6 5 2",
"output": "4"
},
{
"input": "5\n4 3 2 1",
"output": "5"
},
{
"input": "2\n1",
"output": "2"
},
{
"input": "2\n2",
"output": "1"
},
{
"input": "3\n1 2",
"output": "3"
},
{
"input": "3\n1 3",
"output": "2"
},
... | 1,645,196,592 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 1 | 31 | 0 | n = int(input())
a = sorted(list(map(int, input().split())))
for i in range(n-1):
if a[i+1] - a[i] > 1:
print(i+2)
exit() | Title: Forgotten Episode
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus adores TV series. Right now he is ready to finish watching a season of a popular sitcom "Graph Theory". In total, the season has *n* episodes, numbered with integers from 1 to *n*.
Polycarpus watches episodes not one by one but in a random order. He has already watched all the episodes except for one. Which episode has Polycaprus forgotten to watch?
Input Specification:
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100000) — the number of episodes in a season. Assume that the episodes are numbered by integers from 1 to *n*.
The second line contains *n*<=-<=1 integer *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the numbers of episodes that Polycarpus has watched. All values of *a**i* are distinct.
Output Specification:
Print the number of the episode that Polycarpus hasn't watched.
Demo Input:
['10\n3 8 10 1 7 9 6 5 2\n']
Demo Output:
['4\n']
Note:
none | ```python
n = int(input())
a = sorted(list(map(int, input().split())))
for i in range(n-1):
if a[i+1] - a[i] > 1:
print(i+2)
exit()
``` | -1 | |
792 | C | Divide by Three | PROGRAMMING | 2,000 | [
"dp",
"greedy",
"math",
"number theory"
] | null | null | A positive integer number *n* is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given *n* will find a beautiful number such that *n* can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number *n*.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them. | The first line of input contains *n* — a positive integer number without leading zeroes (1<=≤<=*n*<=<<=10100000). | Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print <=-<=1. | [
"1033\n",
"10\n",
"11\n"
] | [
"33\n",
"0\n",
"-1\n"
] | In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | 0 | [
{
"input": "1033",
"output": "33"
},
{
"input": "10",
"output": "0"
},
{
"input": "11",
"output": "-1"
},
{
"input": "3",
"output": "3"
},
{
"input": "1",
"output": "-1"
},
{
"input": "117",
"output": "117"
},
{
"input": "518",
"output"... | 1,572,529,301 | 4,001 | PyPy 3 | WRONG_ANSWER | TESTS | 8 | 139 | 0 | s=input()
L=len(s)
S=0
one=[]
two=[]
for i in range(L):
x=int(s[i])
S=(S+x)%3
if x%3==1:
one.append(i)
elif x%3==2:
two.append(i)
if S==0:
print(s)
elif S==1:
if len(one)>=1 and one[-1]>0 and L>1:
print(s[:one[-1]]+s[one[-1]+1:])
elif len(two)>=2 and two[-2]>0 and L>2:
print(s[:two[-2]]+s[two[-2]+1:two[-1]]+s[two[-1]+1:])
elif len(one)==1 and L>1:
print(int(s[1:]))
elif len(two)==2 and L>2:
print(int(s[1:two[-1]]+s[two[-1]+1:]))
else:
print(-1)
else:
if len(two)>=1 and two[-1]>0 and L>1:
print(s[:two[-1]]+s[two[-1]+1:])
elif len(one)>=2 and one[-2]>0 and L>2:
print(s[:one[-2]]+s[one[-2]+1:one[-1]]+s[one[-1]+1:])
elif len(two)==1 and L>1:
print(int(s[1:]))
elif len(one)==2 and L>2:
print(int(s[1:one[-1]]+s[one[-1]+1:]))
else:
print(-1) | Title: Divide by Three
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A positive integer number *n* is written on a blackboard. It consists of not more than 105 digits. You have to transform it into a beautiful number by erasing some of the digits, and you want to erase as few digits as possible.
The number is called beautiful if it consists of at least one digit, doesn't have leading zeroes and is a multiple of 3. For example, 0, 99, 10110 are beautiful numbers, and 00, 03, 122 are not.
Write a program which for the given *n* will find a beautiful number such that *n* can be transformed into this number by erasing as few digits as possible. You can erase an arbitraty set of digits. For example, they don't have to go one after another in the number *n*.
If it's impossible to obtain a beautiful number, print -1. If there are multiple answers, print any of them.
Input Specification:
The first line of input contains *n* — a positive integer number without leading zeroes (1<=≤<=*n*<=<<=10100000).
Output Specification:
Print one number — any beautiful number obtained by erasing as few as possible digits. If there is no answer, print <=-<=1.
Demo Input:
['1033\n', '10\n', '11\n']
Demo Output:
['33\n', '0\n', '-1\n']
Note:
In the first example it is enough to erase only the first digit to obtain a multiple of 3. But if we erase the first digit, then we obtain a number with a leading zero. So the minimum number of digits to be erased is two. | ```python
s=input()
L=len(s)
S=0
one=[]
two=[]
for i in range(L):
x=int(s[i])
S=(S+x)%3
if x%3==1:
one.append(i)
elif x%3==2:
two.append(i)
if S==0:
print(s)
elif S==1:
if len(one)>=1 and one[-1]>0 and L>1:
print(s[:one[-1]]+s[one[-1]+1:])
elif len(two)>=2 and two[-2]>0 and L>2:
print(s[:two[-2]]+s[two[-2]+1:two[-1]]+s[two[-1]+1:])
elif len(one)==1 and L>1:
print(int(s[1:]))
elif len(two)==2 and L>2:
print(int(s[1:two[-1]]+s[two[-1]+1:]))
else:
print(-1)
else:
if len(two)>=1 and two[-1]>0 and L>1:
print(s[:two[-1]]+s[two[-1]+1:])
elif len(one)>=2 and one[-2]>0 and L>2:
print(s[:one[-2]]+s[one[-2]+1:one[-1]]+s[one[-1]+1:])
elif len(two)==1 and L>1:
print(int(s[1:]))
elif len(one)==2 and L>2:
print(int(s[1:one[-1]]+s[one[-1]+1:]))
else:
print(-1)
``` | 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"
}
] | 1,629,733,245 | 945 | PyPy 3 | OK | TESTS | 15 | 310 | 27,238,400 | import sys
import collections
infile = sys.stdin.buffer
def gs() : return infile.readline().rstrip()
def gi() : return int(gs())
def gss() : return gs().split()
def gis() : return [int(x) for x in gss()]
MOD = 10 ** 9 + 7
INF = 10 ** 12
#RRRBGBRBBB
#BBBRR
def main(infn="") :
global infile
infile = open(infn,"r") if infn else open(sys.argv[1],"r") if len(sys.argv) > 1 else sys.stdin
######################################################################
h = [gi() for _ in range(gi())]
s = (2*sum(h) + len(h) -1)
ans = 0
cur = 0
for i in range(len(h)):
ans += h[i] - cur
cur = h[i]
if i < len(h)-1 and cur > h[i+1]:
ans += abs(cur - h[i+1])
cur = h[i+1]
ans += len(h) -1
ans += len(h)
print (ans)
######################################################################
if __name__ == '__main__' :
main() | 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
import sys
import collections
infile = sys.stdin.buffer
def gs() : return infile.readline().rstrip()
def gi() : return int(gs())
def gss() : return gs().split()
def gis() : return [int(x) for x in gss()]
MOD = 10 ** 9 + 7
INF = 10 ** 12
#RRRBGBRBBB
#BBBRR
def main(infn="") :
global infile
infile = open(infn,"r") if infn else open(sys.argv[1],"r") if len(sys.argv) > 1 else sys.stdin
######################################################################
h = [gi() for _ in range(gi())]
s = (2*sum(h) + len(h) -1)
ans = 0
cur = 0
for i in range(len(h)):
ans += h[i] - cur
cur = h[i]
if i < len(h)-1 and cur > h[i+1]:
ans += abs(cur - h[i+1])
cur = h[i+1]
ans += len(h) -1
ans += len(h)
print (ans)
######################################################################
if __name__ == '__main__' :
main()
``` | 3 | |
495 | B | Modular Equations | PROGRAMMING | 1,600 | [
"math",
"number theory"
] | null | null | Last week, Hamed learned about a new type of equations in his math class called Modular Equations. Lets define *i* modulo *j* as the remainder of division of *i* by *j* and denote it by . A Modular Equation, as Hamed's teacher described, is an equation of the form in which *a* and *b* are two non-negative integers and *x* is a variable. We call a positive integer *x* for which a solution of our equation.
Hamed didn't pay much attention to the class since he was watching a movie. He only managed to understand the definitions of these equations.
Now he wants to write his math exercises but since he has no idea how to do that, he asked you for help. He has told you all he knows about Modular Equations and asked you to write a program which given two numbers *a* and *b* determines how many answers the Modular Equation has. | In the only line of the input two space-separated integers *a* and *b* (0<=≤<=*a*,<=*b*<=≤<=109) are given. | If there is an infinite number of answers to our equation, print "infinity" (without the quotes). Otherwise print the number of solutions of the Modular Equation . | [
"21 5\n",
"9435152 272\n",
"10 10\n"
] | [
"2\n",
"282\n",
"infinity\n"
] | In the first sample the answers of the Modular Equation are 8 and 16 since <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/6f5ff39ebd209bf990adaf91f4b82f9687097224.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 1,000 | [
{
"input": "21 5",
"output": "2"
},
{
"input": "9435152 272",
"output": "282"
},
{
"input": "10 10",
"output": "infinity"
},
{
"input": "0 1000000000",
"output": "0"
},
{
"input": "11 2",
"output": "2"
},
{
"input": "1 0",
"output": "1"
},
{
... | 1,689,428,266 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | print("_RANDOM_GUESS_1689428266.517559")# 1689428266.5175793 | Title: Modular Equations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Last week, Hamed learned about a new type of equations in his math class called Modular Equations. Lets define *i* modulo *j* as the remainder of division of *i* by *j* and denote it by . A Modular Equation, as Hamed's teacher described, is an equation of the form in which *a* and *b* are two non-negative integers and *x* is a variable. We call a positive integer *x* for which a solution of our equation.
Hamed didn't pay much attention to the class since he was watching a movie. He only managed to understand the definitions of these equations.
Now he wants to write his math exercises but since he has no idea how to do that, he asked you for help. He has told you all he knows about Modular Equations and asked you to write a program which given two numbers *a* and *b* determines how many answers the Modular Equation has.
Input Specification:
In the only line of the input two space-separated integers *a* and *b* (0<=≤<=*a*,<=*b*<=≤<=109) are given.
Output Specification:
If there is an infinite number of answers to our equation, print "infinity" (without the quotes). Otherwise print the number of solutions of the Modular Equation .
Demo Input:
['21 5\n', '9435152 272\n', '10 10\n']
Demo Output:
['2\n', '282\n', 'infinity\n']
Note:
In the first sample the answers of the Modular Equation are 8 and 16 since <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/6f5ff39ebd209bf990adaf91f4b82f9687097224.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
print("_RANDOM_GUESS_1689428266.517559")# 1689428266.5175793
``` | 0 | |
452 | A | Eevee | PROGRAMMING | 1,000 | [
"brute force",
"implementation",
"strings"
] | null | null | You are solving the crossword problem K from IPSC 2014. You solved all the clues except for one: who does Eevee evolve into? You are not very into pokemons, but quick googling helped you find out, that Eevee can evolve into eight different pokemons: Vaporeon, Jolteon, Flareon, Espeon, Umbreon, Leafeon, Glaceon, and Sylveon.
You know the length of the word in the crossword, and you already know some letters. Designers of the crossword made sure that the answer is unambiguous, so you can assume that exactly one pokemon out of the 8 that Eevee evolves into fits the length and the letters given. Your task is to find it. | First line contains an integer *n* (6<=≤<=*n*<=≤<=8) – the length of the string.
Next line contains a string consisting of *n* characters, each of which is either a lower case english letter (indicating a known letter) or a dot character (indicating an empty cell in the crossword). | Print a name of the pokemon that Eevee can evolve into that matches the pattern in the input. Use lower case letters only to print the name (in particular, do not capitalize the first letter). | [
"7\nj......\n",
"7\n...feon\n",
"7\n.l.r.o.\n"
] | [
"jolteon\n",
"leafeon\n",
"flareon\n"
] | Here's a set of names in a form you can paste into your solution:
["vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"]
{"vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"} | 500 | [
{
"input": "7\n...feon",
"output": "leafeon"
},
{
"input": "7\n.l.r.o.",
"output": "flareon"
},
{
"input": "6\n.s..o.",
"output": "espeon"
},
{
"input": "7\nglaceon",
"output": "glaceon"
},
{
"input": "8\n.a.o.e.n",
"output": "vaporeon"
},
{
"input": "... | 1,588,449,924 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 8 | 140 | 0 | n=int(input())
t=["vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"]
b= input()
s=[]
for j in range(n):
if b[j]!='.':
s.append([j,b[j]])
for i in t:
c=0
for v in s:
if v[0]<=len(i)-1:
if i[v[0]]==v[1]:
pass
else:
c+=1
else:
break
if c==0:
print(i)
break
| Title: Eevee
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are solving the crossword problem K from IPSC 2014. You solved all the clues except for one: who does Eevee evolve into? You are not very into pokemons, but quick googling helped you find out, that Eevee can evolve into eight different pokemons: Vaporeon, Jolteon, Flareon, Espeon, Umbreon, Leafeon, Glaceon, and Sylveon.
You know the length of the word in the crossword, and you already know some letters. Designers of the crossword made sure that the answer is unambiguous, so you can assume that exactly one pokemon out of the 8 that Eevee evolves into fits the length and the letters given. Your task is to find it.
Input Specification:
First line contains an integer *n* (6<=≤<=*n*<=≤<=8) – the length of the string.
Next line contains a string consisting of *n* characters, each of which is either a lower case english letter (indicating a known letter) or a dot character (indicating an empty cell in the crossword).
Output Specification:
Print a name of the pokemon that Eevee can evolve into that matches the pattern in the input. Use lower case letters only to print the name (in particular, do not capitalize the first letter).
Demo Input:
['7\nj......\n', '7\n...feon\n', '7\n.l.r.o.\n']
Demo Output:
['jolteon\n', 'leafeon\n', 'flareon\n']
Note:
Here's a set of names in a form you can paste into your solution:
["vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"]
{"vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"} | ```python
n=int(input())
t=["vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"]
b= input()
s=[]
for j in range(n):
if b[j]!='.':
s.append([j,b[j]])
for i in t:
c=0
for v in s:
if v[0]<=len(i)-1:
if i[v[0]]==v[1]:
pass
else:
c+=1
else:
break
if c==0:
print(i)
break
``` | 0 | |
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": "... | 1,699,470,631 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 60 | 0 | number_of_stones = int(input("Enter number of the stones: "))
s = input("Enter the colors of the stones: ")
count = 0
for i in range(1, number_of_stones): # Start from 1 because we compare with the previous stone
if s[i] == s[i-1]:
count += 1
print(count)
| 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
number_of_stones = int(input("Enter number of the stones: "))
s = input("Enter the colors of the stones: ")
count = 0
for i in range(1, number_of_stones): # Start from 1 because we compare with the previous stone
if s[i] == s[i-1]:
count += 1
print(count)
``` | 0 | |
822 | A | I'm bored with life | PROGRAMMING | 800 | [
"implementation",
"math",
"number theory"
] | null | null | Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you? | The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12). | Print a single integer denoting the greatest common divisor of integers *A*! and *B*!. | [
"4 3\n"
] | [
"6\n"
] | Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6. | 500 | [
{
"input": "4 3",
"output": "6"
},
{
"input": "10 399603090",
"output": "3628800"
},
{
"input": "6 973151934",
"output": "720"
},
{
"input": "2 841668075",
"output": "2"
},
{
"input": "7 415216919",
"output": "5040"
},
{
"input": "3 283733059",
"ou... | 1,688,633,681 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 1 | 1,000 | 409,600 | a,b=map(int, input().split())
def gcd(a, b):
while b > 0:
a %= b
a, b = b, a
return a
def fact(n):
for i in range(1, n):
n*=i
return n
print(gcd(fact(a),fact(b)))
# Thu Jul 06 2023 11:43:19 GMT+0300 (Moscow Standard Time)
| Title: I'm bored with life
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input Specification:
The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12).
Output Specification:
Print a single integer denoting the greatest common divisor of integers *A*! and *B*!.
Demo Input:
['4 3\n']
Demo Output:
['6\n']
Note:
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6. | ```python
a,b=map(int, input().split())
def gcd(a, b):
while b > 0:
a %= b
a, b = b, a
return a
def fact(n):
for i in range(1, n):
n*=i
return n
print(gcd(fact(a),fact(b)))
# Thu Jul 06 2023 11:43:19 GMT+0300 (Moscow Standard Time)
``` | 0 | |
574 | B | Bear and Three Musketeers | PROGRAMMING | 1,500 | [
"brute force",
"dfs and similar",
"graphs",
"hashing"
] | null | null | Do you know a story about the three musketeers? Anyway, you will learn about its origins now.
Richelimakieu is a cardinal in the city of Bearis. He is tired of dealing with crime by himself. He needs three brave warriors to help him to fight against bad guys.
There are *n* warriors. Richelimakieu wants to choose three of them to become musketeers but it's not that easy. The most important condition is that musketeers must know each other to cooperate efficiently. And they shouldn't be too well known because they could be betrayed by old friends. For each musketeer his recognition is the number of warriors he knows, excluding other two musketeers.
Help Richelimakieu! Find if it is possible to choose three musketeers knowing each other, and what is minimum possible sum of their recognitions. | The first line contains two space-separated integers, *n* and *m* (3<=≤<=*n*<=≤<=4000, 0<=≤<=*m*<=≤<=4000) — respectively number of warriors and number of pairs of warriors knowing each other.
*i*-th of the following *m* lines contains two space-separated integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*). Warriors *a**i* and *b**i* know each other. Each pair of warriors will be listed at most once. | If Richelimakieu can choose three musketeers, print the minimum possible sum of their recognitions. Otherwise, print "-1" (without the quotes). | [
"5 6\n1 2\n1 3\n2 3\n2 4\n3 4\n4 5\n",
"7 4\n2 1\n3 6\n5 1\n1 7\n"
] | [
"2\n",
"-1\n"
] | In the first sample Richelimakieu should choose a triple 1, 2, 3. The first musketeer doesn't know anyone except other two musketeers so his recognition is 0. The second musketeer has recognition 1 because he knows warrior number 4. The third musketeer also has recognition 1 because he knows warrior 4. Sum of recognitions is 0 + 1 + 1 = 2.
The other possible triple is 2, 3, 4 but it has greater sum of recognitions, equal to 1 + 1 + 1 = 3.
In the second sample there is no triple of warriors knowing each other. | 1,000 | [
{
"input": "5 6\n1 2\n1 3\n2 3\n2 4\n3 4\n4 5",
"output": "2"
},
{
"input": "7 4\n2 1\n3 6\n5 1\n1 7",
"output": "-1"
},
{
"input": "5 0",
"output": "-1"
},
{
"input": "7 14\n3 6\n2 3\n5 2\n5 6\n7 5\n7 4\n6 2\n3 5\n7 1\n4 1\n6 1\n7 6\n6 4\n5 4",
"output": "5"
},
{
... | 1,623,568,583 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 6 | 2,000 | 614,400 | n, m = map(int, input().split())
g = [set() for _ in range(n)]
l = []
p = [0 for _ in range(n)]
for _ in range(m):
a, b = map(int, input().split())
a -= 1
b -= 1
p[a] += 1
p[b] += 1
g[a].add(b)
g[b].add(a)
l.append((a, b))
ans = 100000
for i, j in l:
for k in range(n):
if k in g[i] and k in g[j]:
ans = min(ans, p[i] + p[j] + p[k])
if ans == 100000:
ans = -1
else:
ans -= 6
print(ans)
| Title: Bear and Three Musketeers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Do you know a story about the three musketeers? Anyway, you will learn about its origins now.
Richelimakieu is a cardinal in the city of Bearis. He is tired of dealing with crime by himself. He needs three brave warriors to help him to fight against bad guys.
There are *n* warriors. Richelimakieu wants to choose three of them to become musketeers but it's not that easy. The most important condition is that musketeers must know each other to cooperate efficiently. And they shouldn't be too well known because they could be betrayed by old friends. For each musketeer his recognition is the number of warriors he knows, excluding other two musketeers.
Help Richelimakieu! Find if it is possible to choose three musketeers knowing each other, and what is minimum possible sum of their recognitions.
Input Specification:
The first line contains two space-separated integers, *n* and *m* (3<=≤<=*n*<=≤<=4000, 0<=≤<=*m*<=≤<=4000) — respectively number of warriors and number of pairs of warriors knowing each other.
*i*-th of the following *m* lines contains two space-separated integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*). Warriors *a**i* and *b**i* know each other. Each pair of warriors will be listed at most once.
Output Specification:
If Richelimakieu can choose three musketeers, print the minimum possible sum of their recognitions. Otherwise, print "-1" (without the quotes).
Demo Input:
['5 6\n1 2\n1 3\n2 3\n2 4\n3 4\n4 5\n', '7 4\n2 1\n3 6\n5 1\n1 7\n']
Demo Output:
['2\n', '-1\n']
Note:
In the first sample Richelimakieu should choose a triple 1, 2, 3. The first musketeer doesn't know anyone except other two musketeers so his recognition is 0. The second musketeer has recognition 1 because he knows warrior number 4. The third musketeer also has recognition 1 because he knows warrior 4. Sum of recognitions is 0 + 1 + 1 = 2.
The other possible triple is 2, 3, 4 but it has greater sum of recognitions, equal to 1 + 1 + 1 = 3.
In the second sample there is no triple of warriors knowing each other. | ```python
n, m = map(int, input().split())
g = [set() for _ in range(n)]
l = []
p = [0 for _ in range(n)]
for _ in range(m):
a, b = map(int, input().split())
a -= 1
b -= 1
p[a] += 1
p[b] += 1
g[a].add(b)
g[b].add(a)
l.append((a, b))
ans = 100000
for i, j in l:
for k in range(n):
if k in g[i] and k in g[j]:
ans = min(ans, p[i] + p[j] + p[k])
if ans == 100000:
ans = -1
else:
ans -= 6
print(ans)
``` | 0 | |
620 | C | Pearls in a Row | PROGRAMMING | 1,500 | [
"greedy"
] | null | null | There are *n* pearls in a row. Let's enumerate them with integers from 1 to *n* from the left to the right. The pearl number *i* has the type *a**i*.
Let's call a sequence of consecutive pearls a segment. Let's call a segment good if it contains two pearls of the same type.
Split the row of the pearls to the maximal number of good segments. Note that each pearl should appear in exactly one segment of the partition.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java. | The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of pearls in a row.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) – the type of the *i*-th pearl. | On the first line print integer *k* — the maximal number of segments in a partition of the row.
Each of the next *k* lines should contain two integers *l**j*,<=*r**j* (1<=≤<=*l**j*<=≤<=*r**j*<=≤<=*n*) — the number of the leftmost and the rightmost pearls in the *j*-th segment.
Note you should print the correct partition of the row of the pearls, so each pearl should be in exactly one segment and all segments should contain two pearls of the same type.
If there are several optimal solutions print any of them. You can print the segments in any order.
If there are no correct partitions of the row print the number "-1". | [
"5\n1 2 3 4 1\n",
"5\n1 2 3 4 5\n",
"7\n1 2 1 3 1 2 1\n"
] | [
"1\n1 5\n",
"-1\n",
"2\n1 3\n4 7\n"
] | none | 0 | [
{
"input": "5\n1 2 3 4 1",
"output": "1\n1 5"
},
{
"input": "5\n1 2 3 4 5",
"output": "-1"
},
{
"input": "7\n1 2 1 3 1 2 1",
"output": "2\n1 3\n4 7"
},
{
"input": "9\n1 2 1 2 1 2 1 2 1",
"output": "3\n1 3\n4 6\n7 9"
},
{
"input": "11\n1 1 2 1 2 1 2 1 2 1 1",
"... | 1,453,424,759 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 93 | 0 | #author="_rabbit"
n=int(input())
a=list(map(int,input().split()))
ans=[]
s=set()
flag=False
l=0
for i in range(n):
if a[i] in s:
ans.append([l+1,i+1])
l=i+1
s=set()
flag=True
else:
s.add(a[i])
if(flag==False):
print(-1)
else:
print(len(ans))
ans[len(ans)-1][1]=n
for i in ans:
print(i)
| Title: Pearls in a Row
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* pearls in a row. Let's enumerate them with integers from 1 to *n* from the left to the right. The pearl number *i* has the type *a**i*.
Let's call a sequence of consecutive pearls a segment. Let's call a segment good if it contains two pearls of the same type.
Split the row of the pearls to the maximal number of good segments. Note that each pearl should appear in exactly one segment of the partition.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of pearls in a row.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) – the type of the *i*-th pearl.
Output Specification:
On the first line print integer *k* — the maximal number of segments in a partition of the row.
Each of the next *k* lines should contain two integers *l**j*,<=*r**j* (1<=≤<=*l**j*<=≤<=*r**j*<=≤<=*n*) — the number of the leftmost and the rightmost pearls in the *j*-th segment.
Note you should print the correct partition of the row of the pearls, so each pearl should be in exactly one segment and all segments should contain two pearls of the same type.
If there are several optimal solutions print any of them. You can print the segments in any order.
If there are no correct partitions of the row print the number "-1".
Demo Input:
['5\n1 2 3 4 1\n', '5\n1 2 3 4 5\n', '7\n1 2 1 3 1 2 1\n']
Demo Output:
['1\n1 5\n', '-1\n', '2\n1 3\n4 7\n']
Note:
none | ```python
#author="_rabbit"
n=int(input())
a=list(map(int,input().split()))
ans=[]
s=set()
flag=False
l=0
for i in range(n):
if a[i] in s:
ans.append([l+1,i+1])
l=i+1
s=set()
flag=True
else:
s.add(a[i])
if(flag==False):
print(-1)
else:
print(len(ans))
ans[len(ans)-1][1]=n
for i in ans:
print(i)
``` | 0 | |
160 | A | Twins | PROGRAMMING | 900 | [
"greedy",
"sortings"
] | null | null | Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like.
Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally.
As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces. | In the single line print the single number — the minimum needed number of coins. | [
"2\n3 3\n",
"3\n2 1 2\n"
] | [
"2\n",
"2\n"
] | In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum.
In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2. | 500 | [
{
"input": "2\n3 3",
"output": "2"
},
{
"input": "3\n2 1 2",
"output": "2"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "5\n4 2 2 2 2",
"output": "3"
},
{
"input": "7\n1 10 1 2 1 1 1",
"output": "1"
},
{
"input": "5\n3 2 3 3 1",
"output": "3"
... | 1,697,784,971 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 29 | 124 | 0 | n = int(input())
v = list(map(int, input().split()))
v.sort(reverse=True)
sum_coins = sum(v)
twin_sum = 0
ans = 0
while sum_coins >= twin_sum:
twin_sum += v[ans]
sum_coins -= v[ans]
ans += 1
print(ans)
| Title: Twins
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like.
Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally.
As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces.
Output Specification:
In the single line print the single number — the minimum needed number of coins.
Demo Input:
['2\n3 3\n', '3\n2 1 2\n']
Demo Output:
['2\n', '2\n']
Note:
In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum.
In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2. | ```python
n = int(input())
v = list(map(int, input().split()))
v.sort(reverse=True)
sum_coins = sum(v)
twin_sum = 0
ans = 0
while sum_coins >= twin_sum:
twin_sum += v[ans]
sum_coins -= v[ans]
ans += 1
print(ans)
``` | 3 | |
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"
... | 1,694,683,059 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 46 | 0 | x=int(input())
li=''.join([str(i) for i in range(1,x+1)])
print(li[x-1])
| 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
x=int(input())
li=''.join([str(i) for i in range(1,x+1)])
print(li[x-1])
``` | 3 | |
439 | A | Devu, the Singer and Churu, the Joker | PROGRAMMING | 900 | [
"greedy",
"implementation"
] | null | null | Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited.
Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly.
The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly.
People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest.
You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions:
- The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible.
If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event. | The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100). | If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event. | [
"3 30\n2 2 1\n",
"3 20\n2 1 1\n"
] | [
"5\n",
"-1\n"
] | Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way:
- First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes.
Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes.
Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1. | 500 | [
{
"input": "3 30\n2 2 1",
"output": "5"
},
{
"input": "3 20\n2 1 1",
"output": "-1"
},
{
"input": "50 10000\n5 4 10 9 9 6 7 7 7 3 3 7 7 4 7 4 10 10 1 7 10 3 1 4 5 7 2 10 10 10 2 3 4 7 6 1 8 4 7 3 8 8 4 10 1 1 9 2 6 1",
"output": "1943"
},
{
"input": "50 10000\n4 7 15 9 11 12 ... | 1,648,175,549 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 20 | 62 | 0 | x = input().split()
n = int(x[0])
d = int(x[1])
x = input().split()
list1 = []
for i in x:
list1.append(int(i))
if (n-1) * 10 >= d:
print(-1)
else:
rest = (n-1)*10
remain = d - sum(list1)
jokes = rest // 5 + (remain-rest) // 5
print(jokes)
| Title: Devu, the Singer and Churu, the Joker
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited.
Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly.
The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly.
People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest.
You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions:
- The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible.
If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event.
Input Specification:
The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100).
Output Specification:
If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event.
Demo Input:
['3 30\n2 2 1\n', '3 20\n2 1 1\n']
Demo Output:
['5\n', '-1\n']
Note:
Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way:
- First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes.
Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes.
Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1. | ```python
x = input().split()
n = int(x[0])
d = int(x[1])
x = input().split()
list1 = []
for i in x:
list1.append(int(i))
if (n-1) * 10 >= d:
print(-1)
else:
rest = (n-1)*10
remain = d - sum(list1)
jokes = rest // 5 + (remain-rest) // 5
print(jokes)
``` | 0 | |
476 | A | Dreamoon and Stairs | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Dreamoon wants to climb up a stair of *n* steps. He can climb 1 or 2 steps at each move. Dreamoon wants the number of moves to be a multiple of an integer *m*.
What is the minimal number of moves making him climb to the top of the stairs that satisfies his condition? | The single line contains two space separated integers *n*, *m* (0<=<<=*n*<=≤<=10000,<=1<=<<=*m*<=≤<=10). | Print a single integer — the minimal number of moves being a multiple of *m*. If there is no way he can climb satisfying condition print <=-<=1 instead. | [
"10 2\n",
"3 5\n"
] | [
"6\n",
"-1\n"
] | For the first sample, Dreamoon could climb in 6 moves with following sequence of steps: {2, 2, 2, 2, 1, 1}.
For the second sample, there are only three valid sequence of steps {2, 1}, {1, 2}, {1, 1, 1} with 2, 2, and 3 steps respectively. All these numbers are not multiples of 5. | 500 | [
{
"input": "10 2",
"output": "6"
},
{
"input": "3 5",
"output": "-1"
},
{
"input": "29 7",
"output": "21"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "1 2",
"output": "-1"
},
{
"input": "10000 2",
"output": "5000"
},
{
"input": "10000 ... | 1,677,895,320 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 30 | 0 | def k():
p=x,y=list(map(int,input().split()))
n=round(x/2)
if n %y==0:
return n
else:
for i in range (x//2):
n+=1
if n %y==0:
return n
return(-1)
print(k()) | Title: Dreamoon and Stairs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dreamoon wants to climb up a stair of *n* steps. He can climb 1 or 2 steps at each move. Dreamoon wants the number of moves to be a multiple of an integer *m*.
What is the minimal number of moves making him climb to the top of the stairs that satisfies his condition?
Input Specification:
The single line contains two space separated integers *n*, *m* (0<=<<=*n*<=≤<=10000,<=1<=<<=*m*<=≤<=10).
Output Specification:
Print a single integer — the minimal number of moves being a multiple of *m*. If there is no way he can climb satisfying condition print <=-<=1 instead.
Demo Input:
['10 2\n', '3 5\n']
Demo Output:
['6\n', '-1\n']
Note:
For the first sample, Dreamoon could climb in 6 moves with following sequence of steps: {2, 2, 2, 2, 1, 1}.
For the second sample, there are only three valid sequence of steps {2, 1}, {1, 2}, {1, 1, 1} with 2, 2, and 3 steps respectively. All these numbers are not multiples of 5. | ```python
def k():
p=x,y=list(map(int,input().split()))
n=round(x/2)
if n %y==0:
return n
else:
for i in range (x//2):
n+=1
if n %y==0:
return n
return(-1)
print(k())
``` | 0 | |
208 | A | Dubstep | PROGRAMMING | 900 | [
"strings"
] | null | null | Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song. | The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word. | Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space. | [
"WUBWUBABCWUB\n",
"WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n"
] | [
"ABC ",
"WE ARE THE CHAMPIONS MY FRIEND "
] | In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB". | 500 | [
{
"input": "WUBWUBABCWUB",
"output": "ABC "
},
{
"input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB",
"output": "WE ARE THE CHAMPIONS MY FRIEND "
},
{
"input": "WUBWUBWUBSR",
"output": "SR "
},
{
"input": "RWUBWUBWUBLWUB",
"output": "R L "
},
{
"input": "... | 1,681,744,708 | 2,147,483,647 | Python 3 | OK | TESTS | 71 | 124 | 0 | s=input().upper()
A="WUB"
new=s.replace(A," ")
print(new.strip())
| Title: Dubstep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
Input Specification:
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
Output Specification:
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
Demo Input:
['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n']
Demo Output:
['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND ']
Note:
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB". | ```python
s=input().upper()
A="WUB"
new=s.replace(A," ")
print(new.strip())
``` | 3 | |
79 | B | Colorful Field | PROGRAMMING | 1,400 | [
"implementation",
"sortings"
] | B. Colorful Field | 2 | 256 | Fox Ciel saw a large field while she was on a bus. The field was a *n*<=×<=*m* rectangle divided into 1<=×<=1 cells. Some cells were wasteland, and other each cell contained crop plants: either carrots or kiwis or grapes.
After seeing the field carefully, Ciel found that the crop plants of each cell were planted in following procedure:
- Assume that the rows are numbered 1 to *n* from top to bottom and the columns are numbered 1 to *m* from left to right, and a cell in row *i* and column *j* is represented as (*i*,<=*j*). - First, each field is either cultivated or waste. Crop plants will be planted in the cultivated cells in the order of (1,<=1)<=→<=...<=→<=(1,<=*m*)<=→<=(2,<=1)<=→<=...<=→<=(2,<=*m*)<=→<=...<=→<=(*n*,<=1)<=→<=...<=→<=(*n*,<=*m*). Waste cells will be ignored. - Crop plants (either carrots or kiwis or grapes) will be planted in each cell one after another cyclically. Carrots will be planted in the first cell, then kiwis in the second one, grapes in the third one, carrots in the forth one, kiwis in the fifth one, and so on.
The following figure will show you the example of this procedure. Here, a white square represents a cultivated cell, and a black square represents a waste cell.
Now she is wondering how to determine the crop plants in some certain cells. | In the first line there are four positive integers *n*,<=*m*,<=*k*,<=*t* (1<=≤<=*n*<=≤<=4·104,<=1<=≤<=*m*<=≤<=4·104,<=1<=≤<=*k*<=≤<=103,<=1<=≤<=*t*<=≤<=103), each of which represents the height of the field, the width of the field, the number of waste cells and the number of queries that ask the kind of crop plants in a certain cell.
Following each *k* lines contains two integers *a*,<=*b* (1<=≤<=*a*<=≤<=*n*,<=1<=≤<=*b*<=≤<=*m*), which denotes a cell (*a*,<=*b*) is waste. It is guaranteed that the same cell will not appear twice in this section.
Following each *t* lines contains two integers *i*,<=*j* (1<=≤<=*i*<=≤<=*n*,<=1<=≤<=*j*<=≤<=*m*), which is a query that asks you the kind of crop plants of a cell (*i*,<=*j*). | For each query, if the cell is waste, print Waste. Otherwise, print the name of crop plants in the cell: either Carrots or Kiwis or Grapes. | [
"4 5 5 6\n4 3\n1 3\n3 3\n2 5\n3 2\n1 3\n1 4\n2 3\n2 4\n1 1\n1 1\n"
] | [
"Waste\nGrapes\nCarrots\nKiwis\nCarrots\nCarrots\n"
] | The sample corresponds to the figure in the statement. | 1,000 | [
{
"input": "4 5 5 6\n4 3\n1 3\n3 3\n2 5\n3 2\n1 3\n1 4\n2 3\n2 4\n1 1\n1 1",
"output": "Waste\nGrapes\nCarrots\nKiwis\nCarrots\nCarrots"
},
{
"input": "2 3 2 2\n1 1\n2 2\n2 1\n2 2",
"output": "Grapes\nWaste"
},
{
"input": "31 31 31 4\n4 9\n16 27\n11 29\n8 28\n11 2\n10 7\n22 6\n1 25\n14 8... | 1,518,203,542 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 3 | 2,000 | 120,422,400 | import queue
ips_q = queue.Queue()
ips_q.put(" Carrots")
ips_q.put("Kiwis")
ips_q.put("Grapes")
a=[int(x) for x in input().split(" ")]
xi=a[0]
xj=a[1]
matrix=[[0 for x in range(1,(xj+1))] for y in range(1,(xi+1))]
waste=[[0 for x in range(2)]for y in range(a[2])]
for i in range (0,a[2]):
w=[int(x) for x in input().split(" ")]
waste[i]=w
x=waste[i][0]
y=waste[i][1]
matrix[x-1][y-1]=-1
def myfn ():
y=ips_q.get()
ips_q.put(y)
return(y)
for i in range(0,xi):
for j in range(0,xj):
if matrix[i][j]==-1:
matrix[i][j]="Waste"
elif matrix[i][j]!=-1 :
matrix[i][j]=myfn()
for i in range(0,a[3]):
p=[int(x) for x in input().split(" ")]
ii=p[0]
jj=p[1]
print(matrix[ii-1][jj-1])
| Title: Colorful Field
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Fox Ciel saw a large field while she was on a bus. The field was a *n*<=×<=*m* rectangle divided into 1<=×<=1 cells. Some cells were wasteland, and other each cell contained crop plants: either carrots or kiwis or grapes.
After seeing the field carefully, Ciel found that the crop plants of each cell were planted in following procedure:
- Assume that the rows are numbered 1 to *n* from top to bottom and the columns are numbered 1 to *m* from left to right, and a cell in row *i* and column *j* is represented as (*i*,<=*j*). - First, each field is either cultivated or waste. Crop plants will be planted in the cultivated cells in the order of (1,<=1)<=→<=...<=→<=(1,<=*m*)<=→<=(2,<=1)<=→<=...<=→<=(2,<=*m*)<=→<=...<=→<=(*n*,<=1)<=→<=...<=→<=(*n*,<=*m*). Waste cells will be ignored. - Crop plants (either carrots or kiwis or grapes) will be planted in each cell one after another cyclically. Carrots will be planted in the first cell, then kiwis in the second one, grapes in the third one, carrots in the forth one, kiwis in the fifth one, and so on.
The following figure will show you the example of this procedure. Here, a white square represents a cultivated cell, and a black square represents a waste cell.
Now she is wondering how to determine the crop plants in some certain cells.
Input Specification:
In the first line there are four positive integers *n*,<=*m*,<=*k*,<=*t* (1<=≤<=*n*<=≤<=4·104,<=1<=≤<=*m*<=≤<=4·104,<=1<=≤<=*k*<=≤<=103,<=1<=≤<=*t*<=≤<=103), each of which represents the height of the field, the width of the field, the number of waste cells and the number of queries that ask the kind of crop plants in a certain cell.
Following each *k* lines contains two integers *a*,<=*b* (1<=≤<=*a*<=≤<=*n*,<=1<=≤<=*b*<=≤<=*m*), which denotes a cell (*a*,<=*b*) is waste. It is guaranteed that the same cell will not appear twice in this section.
Following each *t* lines contains two integers *i*,<=*j* (1<=≤<=*i*<=≤<=*n*,<=1<=≤<=*j*<=≤<=*m*), which is a query that asks you the kind of crop plants of a cell (*i*,<=*j*).
Output Specification:
For each query, if the cell is waste, print Waste. Otherwise, print the name of crop plants in the cell: either Carrots or Kiwis or Grapes.
Demo Input:
['4 5 5 6\n4 3\n1 3\n3 3\n2 5\n3 2\n1 3\n1 4\n2 3\n2 4\n1 1\n1 1\n']
Demo Output:
['Waste\nGrapes\nCarrots\nKiwis\nCarrots\nCarrots\n']
Note:
The sample corresponds to the figure in the statement. | ```python
import queue
ips_q = queue.Queue()
ips_q.put(" Carrots")
ips_q.put("Kiwis")
ips_q.put("Grapes")
a=[int(x) for x in input().split(" ")]
xi=a[0]
xj=a[1]
matrix=[[0 for x in range(1,(xj+1))] for y in range(1,(xi+1))]
waste=[[0 for x in range(2)]for y in range(a[2])]
for i in range (0,a[2]):
w=[int(x) for x in input().split(" ")]
waste[i]=w
x=waste[i][0]
y=waste[i][1]
matrix[x-1][y-1]=-1
def myfn ():
y=ips_q.get()
ips_q.put(y)
return(y)
for i in range(0,xi):
for j in range(0,xj):
if matrix[i][j]==-1:
matrix[i][j]="Waste"
elif matrix[i][j]!=-1 :
matrix[i][j]=myfn()
for i in range(0,a[3]):
p=[int(x) for x in input().split(" ")]
ii=p[0]
jj=p[1]
print(matrix[ii-1][jj-1])
``` | 0 |
867 | A | Between the Offices | PROGRAMMING | 800 | [
"implementation"
] | null | null | As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last *n* days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last *n* days, or not. | The first line of input contains single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days.
The second line contains a string of length *n* consisting of only capital 'S' and 'F' letters. If the *i*-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence. | Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower). | [
"4\nFSSF\n",
"2\nSF\n",
"10\nFFFFFFFFFF\n",
"10\nSSFFSFFSFF\n"
] | [
"NO\n",
"YES\n",
"NO\n",
"YES\n"
] | In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though. | 500 | [
{
"input": "4\nFSSF",
"output": "NO"
},
{
"input": "2\nSF",
"output": "YES"
},
{
"input": "10\nFFFFFFFFFF",
"output": "NO"
},
{
"input": "10\nSSFFSFFSFF",
"output": "YES"
},
{
"input": "20\nSFSFFFFSSFFFFSSSSFSS",
"output": "NO"
},
{
"input": "20\nSSFFF... | 1,573,504,241 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 124 | 0 | n = int(input())
strok = input()
if strok.count('SF')>strok.count('FS'):
print('YES')
else:
print('NO') | Title: Between the Offices
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last *n* days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last *n* days, or not.
Input Specification:
The first line of input contains single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days.
The second line contains a string of length *n* consisting of only capital 'S' and 'F' letters. If the *i*-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Output Specification:
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['4\nFSSF\n', '2\nSF\n', '10\nFFFFFFFFFF\n', '10\nSSFFSFFSFF\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n', 'YES\n']
Note:
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though. | ```python
n = int(input())
strok = input()
if strok.count('SF')>strok.count('FS'):
print('YES')
else:
print('NO')
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | The clique problem is one of the most well-known NP-complete problems. Under some simplification it can be formulated as follows. Consider an undirected graph *G*. It is required to find a subset of vertices *C* of the maximum size such that any two of them are connected by an edge in graph *G*. Sounds simple, doesn't it? Nobody yet knows an algorithm that finds a solution to this problem in polynomial time of the size of the graph. However, as with many other NP-complete problems, the clique problem is easier if you consider a specific type of a graph.
Consider *n* distinct points on a line. Let the *i*-th point have the coordinate *x**i* and weight *w**i*. Let's form graph *G*, whose vertices are these points and edges connect exactly the pairs of points (*i*,<=*j*), such that the distance between them is not less than the sum of their weights, or more formally: |*x**i*<=-<=*x**j*|<=≥<=*w**i*<=+<=*w**j*.
Find the size of the maximum clique in such graph. | The first line contains the integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of points.
Each of the next *n* lines contains two numbers *x**i*, *w**i* (0<=≤<=*x**i*<=≤<=109,<=1<=≤<=*w**i*<=≤<=109) — the coordinate and the weight of a point. All *x**i* are different. | Print a single number — the number of vertexes in the maximum clique of the given graph. | [
"4\n2 3\n3 1\n6 1\n0 2\n"
] | [
"3\n"
] | If you happen to know how to solve this problem without using the specific properties of the graph formulated in the problem statement, then you are able to get a prize of one million dollars!
The picture for the sample test. | 0 | [
{
"input": "4\n2 3\n3 1\n6 1\n0 2",
"output": "3"
},
{
"input": "1\n42 23",
"output": "1"
},
{
"input": "2\n1 5\n2 6",
"output": "1"
},
{
"input": "2\n1 5\n12 6",
"output": "2"
},
{
"input": "1\n0 1",
"output": "1"
},
{
"input": "1\n1000000000 10000000... | 1,583,527,327 | 2,147,483,647 | Python 3 | OK | TESTS | 51 | 1,029 | 24,780,800 | from operator import itemgetter
class CodeforcesTask528BSolution:
def __init__(self):
self.result = ''
self.n = 0
self.points = []
def read_input(self):
self.n = int(input())
for _ in range(self.n):
self.points.append([int(x) for x in input().split(" ")])
self.points[-1].append(sum(self.points[-1]))
def process_task(self):
self.points.sort(key=itemgetter(2))
last = 0
ans = 1
for i in range(1, self.n):
if self.points[i][0] - self.points[i][1] >= self.points[last][0] + self.points[last][1]:
last = i
ans += 1
self.result = str(ans)
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask528BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The clique problem is one of the most well-known NP-complete problems. Under some simplification it can be formulated as follows. Consider an undirected graph *G*. It is required to find a subset of vertices *C* of the maximum size such that any two of them are connected by an edge in graph *G*. Sounds simple, doesn't it? Nobody yet knows an algorithm that finds a solution to this problem in polynomial time of the size of the graph. However, as with many other NP-complete problems, the clique problem is easier if you consider a specific type of a graph.
Consider *n* distinct points on a line. Let the *i*-th point have the coordinate *x**i* and weight *w**i*. Let's form graph *G*, whose vertices are these points and edges connect exactly the pairs of points (*i*,<=*j*), such that the distance between them is not less than the sum of their weights, or more formally: |*x**i*<=-<=*x**j*|<=≥<=*w**i*<=+<=*w**j*.
Find the size of the maximum clique in such graph.
Input Specification:
The first line contains the integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of points.
Each of the next *n* lines contains two numbers *x**i*, *w**i* (0<=≤<=*x**i*<=≤<=109,<=1<=≤<=*w**i*<=≤<=109) — the coordinate and the weight of a point. All *x**i* are different.
Output Specification:
Print a single number — the number of vertexes in the maximum clique of the given graph.
Demo Input:
['4\n2 3\n3 1\n6 1\n0 2\n']
Demo Output:
['3\n']
Note:
If you happen to know how to solve this problem without using the specific properties of the graph formulated in the problem statement, then you are able to get a prize of one million dollars!
The picture for the sample test. | ```python
from operator import itemgetter
class CodeforcesTask528BSolution:
def __init__(self):
self.result = ''
self.n = 0
self.points = []
def read_input(self):
self.n = int(input())
for _ in range(self.n):
self.points.append([int(x) for x in input().split(" ")])
self.points[-1].append(sum(self.points[-1]))
def process_task(self):
self.points.sort(key=itemgetter(2))
last = 0
ans = 1
for i in range(1, self.n):
if self.points[i][0] - self.points[i][1] >= self.points[last][0] + self.points[last][1]:
last = i
ans += 1
self.result = str(ans)
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask528BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | 3 | |
803 | C | Maximal GCD | PROGRAMMING | 1,900 | [
"constructive algorithms",
"greedy",
"math"
] | null | null | You are given positive integer number *n*. You should create such strictly increasing sequence of *k* positive numbers *a*1,<=*a*2,<=...,<=*a**k*, that their sum is equal to *n* and greatest common divisor is maximal.
Greatest common divisor of sequence is maximum of such numbers that every element of sequence is divisible by them.
If there is no possible sequence then output -1. | The first line consists of two numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1010). | If the answer exists then output *k* numbers — resulting sequence. Otherwise output -1. If there are multiple answers, print any of them. | [
"6 3\n",
"8 2\n",
"5 3\n"
] | [
"1 2 3\n",
"2 6\n",
"-1\n"
] | none | 0 | [
{
"input": "6 3",
"output": "1 2 3"
},
{
"input": "8 2",
"output": "2 6"
},
{
"input": "5 3",
"output": "-1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 2",
"output": "-1"
},
{
"input": "2 1",
"output": "2"
},
{
"input": "2 10000000... | 1,509,214,590 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 93 | 102,400 | n, k = map(int, input().split())
d = k * k + k >> 1
s, m = 0, n // d
if n < d:
print(-1)
exit()
for q in range(2, 100001):
if n % q == 0:
if s < q <= d: s = q
if s < n // q <= d: s = n // q
for q in list(range(s, s * k, s)) + [n - (d - k) * s]: print(q) | Title: Maximal GCD
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given positive integer number *n*. You should create such strictly increasing sequence of *k* positive numbers *a*1,<=*a*2,<=...,<=*a**k*, that their sum is equal to *n* and greatest common divisor is maximal.
Greatest common divisor of sequence is maximum of such numbers that every element of sequence is divisible by them.
If there is no possible sequence then output -1.
Input Specification:
The first line consists of two numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1010).
Output Specification:
If the answer exists then output *k* numbers — resulting sequence. Otherwise output -1. If there are multiple answers, print any of them.
Demo Input:
['6 3\n', '8 2\n', '5 3\n']
Demo Output:
['1 2 3\n', '2 6\n', '-1\n']
Note:
none | ```python
n, k = map(int, input().split())
d = k * k + k >> 1
s, m = 0, n // d
if n < d:
print(-1)
exit()
for q in range(2, 100001):
if n % q == 0:
if s < q <= d: s = q
if s < n // q <= d: s = n // q
for q in list(range(s, s * k, s)) + [n - (d - k) * s]: print(q)
``` | 0 | |
389 | B | Fox and Cross | PROGRAMMING | 1,100 | [
"greedy",
"implementation"
] | null | null | Fox Ciel has a board with *n* rows and *n* columns. So, the board consists of *n*<=×<=*n* cells. Each cell contains either a symbol '.', or a symbol '#'.
A cross on the board is a connected set of exactly five cells of the board that looks like a cross. The picture below shows how it looks.
Ciel wants to draw several (may be zero) crosses on the board. Each cross must cover exactly five cells with symbols '#', and any cell with symbol '#' must belong to some cross. No two crosses can share a cell.
Please, tell Ciel if she can draw the crosses in the described way. | The first line contains an integer *n* (3<=≤<=*n*<=≤<=100) — the size of the board.
Each of the next *n* lines describes one row of the board. The *i*-th line describes the *i*-th row of the board and consists of *n* characters. Each character is either a symbol '.', or a symbol '#'. | Output a single line with "YES" if Ciel can draw the crosses in the described way. Otherwise output a single line with "NO". | [
"5\n.#...\n####.\n.####\n...#.\n.....\n",
"4\n####\n####\n####\n####\n",
"6\n.#....\n####..\n.####.\n.#.##.\n######\n.#..#.\n",
"6\n.#..#.\n######\n.####.\n.####.\n######\n.#..#.\n",
"3\n...\n...\n...\n"
] | [
"YES\n",
"NO\n",
"YES\n",
"NO\n",
"YES\n"
] | In example 1, you can draw two crosses. The picture below shows what they look like.
In example 2, the board contains 16 cells with '#', but each cross contains 5. Since 16 is not a multiple of 5, so it's impossible to cover all. | 1,000 | [
{
"input": "4\n####\n####\n####\n####",
"output": "NO"
},
{
"input": "6\n.#....\n####..\n.####.\n.#.##.\n######\n.#..#.",
"output": "YES"
},
{
"input": "6\n.#..#.\n######\n.####.\n.####.\n######\n.#..#.",
"output": "NO"
},
{
"input": "5\n.....\n.#...\n####.\n.####\n...#.",
... | 1,501,659,848 | 2,147,483,647 | Python 3 | OK | TESTS | 42 | 61 | 4,710,400 | n=int(input())
a=[]
for i in range(n):
a.append(list(input()))
for i in range(1,n-1):
for j in range(1,n-1):
if a[i][j]=='#' and a[i+1][j]=='#' and a[i-1][j]=='#' and a[i][j-1]=='#' and a[i][j+1]=='#':
a[i][j]='.'
a[i][j+1]='.'
a[i][j-1]='.'
a[i+1][j]='.'
a[i-1][j]='.'
for i in a:
for j in i:
if j=='#':
print('NO')
exit(0)
print('YES') | Title: Fox and Cross
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel has a board with *n* rows and *n* columns. So, the board consists of *n*<=×<=*n* cells. Each cell contains either a symbol '.', or a symbol '#'.
A cross on the board is a connected set of exactly five cells of the board that looks like a cross. The picture below shows how it looks.
Ciel wants to draw several (may be zero) crosses on the board. Each cross must cover exactly five cells with symbols '#', and any cell with symbol '#' must belong to some cross. No two crosses can share a cell.
Please, tell Ciel if she can draw the crosses in the described way.
Input Specification:
The first line contains an integer *n* (3<=≤<=*n*<=≤<=100) — the size of the board.
Each of the next *n* lines describes one row of the board. The *i*-th line describes the *i*-th row of the board and consists of *n* characters. Each character is either a symbol '.', or a symbol '#'.
Output Specification:
Output a single line with "YES" if Ciel can draw the crosses in the described way. Otherwise output a single line with "NO".
Demo Input:
['5\n.#...\n####.\n.####\n...#.\n.....\n', '4\n####\n####\n####\n####\n', '6\n.#....\n####..\n.####.\n.#.##.\n######\n.#..#.\n', '6\n.#..#.\n######\n.####.\n.####.\n######\n.#..#.\n', '3\n...\n...\n...\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'NO\n', 'YES\n']
Note:
In example 1, you can draw two crosses. The picture below shows what they look like.
In example 2, the board contains 16 cells with '#', but each cross contains 5. Since 16 is not a multiple of 5, so it's impossible to cover all. | ```python
n=int(input())
a=[]
for i in range(n):
a.append(list(input()))
for i in range(1,n-1):
for j in range(1,n-1):
if a[i][j]=='#' and a[i+1][j]=='#' and a[i-1][j]=='#' and a[i][j-1]=='#' and a[i][j+1]=='#':
a[i][j]='.'
a[i][j+1]='.'
a[i][j-1]='.'
a[i+1][j]='.'
a[i-1][j]='.'
for i in a:
for j in i:
if j=='#':
print('NO')
exit(0)
print('YES')
``` | 3 | |
1,006 | A | Adjacent Replacements | PROGRAMMING | 800 | [
"implementation"
] | null | null | Mishka got an integer array $a$ of length $n$ as a birthday present (what a surprise!).
Mishka doesn't like this present and wants to change it somehow. He has invented an algorithm and called it "Mishka's Adjacent Replacements Algorithm". This algorithm can be represented as a sequence of steps:
- Replace each occurrence of $1$ in the array $a$ with $2$; - Replace each occurrence of $2$ in the array $a$ with $1$; - Replace each occurrence of $3$ in the array $a$ with $4$; - Replace each occurrence of $4$ in the array $a$ with $3$; - Replace each occurrence of $5$ in the array $a$ with $6$; - Replace each occurrence of $6$ in the array $a$ with $5$; - $\dots$ - Replace each occurrence of $10^9 - 1$ in the array $a$ with $10^9$; - Replace each occurrence of $10^9$ in the array $a$ with $10^9 - 1$.
Note that the dots in the middle of this algorithm mean that Mishka applies these replacements for each pair of adjacent integers ($2i - 1, 2i$) for each $i \in\{1, 2, \ldots, 5 \cdot 10^8\}$ as described above.
For example, for the array $a = [1, 2, 4, 5, 10]$, the following sequence of arrays represents the algorithm:
$[1, 2, 4, 5, 10]$ $\rightarrow$ (replace all occurrences of $1$ with $2$) $\rightarrow$ $[2, 2, 4, 5, 10]$ $\rightarrow$ (replace all occurrences of $2$ with $1$) $\rightarrow$ $[1, 1, 4, 5, 10]$ $\rightarrow$ (replace all occurrences of $3$ with $4$) $\rightarrow$ $[1, 1, 4, 5, 10]$ $\rightarrow$ (replace all occurrences of $4$ with $3$) $\rightarrow$ $[1, 1, 3, 5, 10]$ $\rightarrow$ (replace all occurrences of $5$ with $6$) $\rightarrow$ $[1, 1, 3, 6, 10]$ $\rightarrow$ (replace all occurrences of $6$ with $5$) $\rightarrow$ $[1, 1, 3, 5, 10]$ $\rightarrow$ $\dots$ $\rightarrow$ $[1, 1, 3, 5, 10]$ $\rightarrow$ (replace all occurrences of $10$ with $9$) $\rightarrow$ $[1, 1, 3, 5, 9]$. The later steps of the algorithm do not change the array.
Mishka is very lazy and he doesn't want to apply these changes by himself. But he is very interested in their result. Help him find it. | The first line of the input contains one integer number $n$ ($1 \le n \le 1000$) — the number of elements in Mishka's birthday present (surprisingly, an array).
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the array. | Print $n$ integers — $b_1, b_2, \dots, b_n$, where $b_i$ is the final value of the $i$-th element of the array after applying "Mishka's Adjacent Replacements Algorithm" to the array $a$. Note that you cannot change the order of elements in the array. | [
"5\n1 2 4 5 10\n",
"10\n10000 10 50605065 1 5 89 5 999999999 60506056 1000000000\n"
] | [
"1 1 3 5 9\n",
"9999 9 50605065 1 5 89 5 999999999 60506055 999999999\n"
] | The first example is described in the problem statement. | 0 | [
{
"input": "5\n1 2 4 5 10",
"output": "1 1 3 5 9"
},
{
"input": "10\n10000 10 50605065 1 5 89 5 999999999 60506056 1000000000",
"output": "9999 9 50605065 1 5 89 5 999999999 60506055 999999999"
},
{
"input": "1\n999999999",
"output": "999999999"
},
{
"input": "1\n1000000000",... | 1,694,198,273 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 30 | 0 | l=int(input())
l1=list(map(int,input()))
l2=[2,1,4,3,6,5]
for i in range(6):
for j in l1:
if (i+1)==j:
l1[j]=l2[i]
for j in l1:
print(j,end=" ") | Title: Adjacent Replacements
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka got an integer array $a$ of length $n$ as a birthday present (what a surprise!).
Mishka doesn't like this present and wants to change it somehow. He has invented an algorithm and called it "Mishka's Adjacent Replacements Algorithm". This algorithm can be represented as a sequence of steps:
- Replace each occurrence of $1$ in the array $a$ with $2$; - Replace each occurrence of $2$ in the array $a$ with $1$; - Replace each occurrence of $3$ in the array $a$ with $4$; - Replace each occurrence of $4$ in the array $a$ with $3$; - Replace each occurrence of $5$ in the array $a$ with $6$; - Replace each occurrence of $6$ in the array $a$ with $5$; - $\dots$ - Replace each occurrence of $10^9 - 1$ in the array $a$ with $10^9$; - Replace each occurrence of $10^9$ in the array $a$ with $10^9 - 1$.
Note that the dots in the middle of this algorithm mean that Mishka applies these replacements for each pair of adjacent integers ($2i - 1, 2i$) for each $i \in\{1, 2, \ldots, 5 \cdot 10^8\}$ as described above.
For example, for the array $a = [1, 2, 4, 5, 10]$, the following sequence of arrays represents the algorithm:
$[1, 2, 4, 5, 10]$ $\rightarrow$ (replace all occurrences of $1$ with $2$) $\rightarrow$ $[2, 2, 4, 5, 10]$ $\rightarrow$ (replace all occurrences of $2$ with $1$) $\rightarrow$ $[1, 1, 4, 5, 10]$ $\rightarrow$ (replace all occurrences of $3$ with $4$) $\rightarrow$ $[1, 1, 4, 5, 10]$ $\rightarrow$ (replace all occurrences of $4$ with $3$) $\rightarrow$ $[1, 1, 3, 5, 10]$ $\rightarrow$ (replace all occurrences of $5$ with $6$) $\rightarrow$ $[1, 1, 3, 6, 10]$ $\rightarrow$ (replace all occurrences of $6$ with $5$) $\rightarrow$ $[1, 1, 3, 5, 10]$ $\rightarrow$ $\dots$ $\rightarrow$ $[1, 1, 3, 5, 10]$ $\rightarrow$ (replace all occurrences of $10$ with $9$) $\rightarrow$ $[1, 1, 3, 5, 9]$. The later steps of the algorithm do not change the array.
Mishka is very lazy and he doesn't want to apply these changes by himself. But he is very interested in their result. Help him find it.
Input Specification:
The first line of the input contains one integer number $n$ ($1 \le n \le 1000$) — the number of elements in Mishka's birthday present (surprisingly, an array).
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the elements of the array.
Output Specification:
Print $n$ integers — $b_1, b_2, \dots, b_n$, where $b_i$ is the final value of the $i$-th element of the array after applying "Mishka's Adjacent Replacements Algorithm" to the array $a$. Note that you cannot change the order of elements in the array.
Demo Input:
['5\n1 2 4 5 10\n', '10\n10000 10 50605065 1 5 89 5 999999999 60506056 1000000000\n']
Demo Output:
['1 1 3 5 9\n', '9999 9 50605065 1 5 89 5 999999999 60506055 999999999\n']
Note:
The first example is described in the problem statement. | ```python
l=int(input())
l1=list(map(int,input()))
l2=[2,1,4,3,6,5]
for i in range(6):
for j in l1:
if (i+1)==j:
l1[j]=l2[i]
for j in l1:
print(j,end=" ")
``` | -1 | |
597 | A | Divisibility | PROGRAMMING | 1,600 | [
"math"
] | null | null | Find the number of *k*-divisible numbers on the segment [*a*,<=*b*]. In other words you need to find the number of such integer values *x* that *a*<=≤<=*x*<=≤<=*b* and *x* is divisible by *k*. | The only line contains three space-separated integers *k*, *a* and *b* (1<=≤<=*k*<=≤<=1018;<=-<=1018<=≤<=*a*<=≤<=*b*<=≤<=1018). | Print the required number. | [
"1 1 10\n",
"2 -4 4\n"
] | [
"10\n",
"5\n"
] | none | 500 | [
{
"input": "1 1 10",
"output": "10"
},
{
"input": "2 -4 4",
"output": "5"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "1 0 0",
"output": "1"
},
{
"input": "1 0 1",
"output": "2"
},
{
"input": "1 10181 10182",
"output": "2"
},
{
"inpu... | 1,586,633,990 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 108 | 409,600 | import operator
import functools
bacterias = input()
nbac = functools.reduce(operator.add,map(int,bin(int(bacterias))[2:]))
print(nbac) | Title: Divisibility
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Find the number of *k*-divisible numbers on the segment [*a*,<=*b*]. In other words you need to find the number of such integer values *x* that *a*<=≤<=*x*<=≤<=*b* and *x* is divisible by *k*.
Input Specification:
The only line contains three space-separated integers *k*, *a* and *b* (1<=≤<=*k*<=≤<=1018;<=-<=1018<=≤<=*a*<=≤<=*b*<=≤<=1018).
Output Specification:
Print the required number.
Demo Input:
['1 1 10\n', '2 -4 4\n']
Demo Output:
['10\n', '5\n']
Note:
none | ```python
import operator
import functools
bacterias = input()
nbac = functools.reduce(operator.add,map(int,bin(int(bacterias))[2:]))
print(nbac)
``` | -1 | |
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... | 1,695,394,339 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 92 | 0 | n = int(input())
drinks = list(map(int,input().split()))
sumof_drinks=sum(drinks)
print(sumof_drinks/n) | 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
n = int(input())
drinks = list(map(int,input().split()))
sumof_drinks=sum(drinks)
print(sumof_drinks/n)
``` | 3 | |
732 | A | Buy a Shovel | PROGRAMMING | 800 | [
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null | Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel. | The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels. | Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change. | [
"117 3\n",
"237 7\n",
"15 2\n"
] | [
"9\n",
"1\n",
"2\n"
] | In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change. | 500 | [
{
"input": "117 3",
"output": "9"
},
{
"input": "237 7",
"output": "1"
},
{
"input": "15 2",
"output": "2"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1000 3",
"output": "1"
},
{
"input": "1000 1",
... | 1,691,084,879 | 2,147,483,647 | Python 3 | OK | TESTS | 54 | 46 | 0 | k, r = map(int, input().split());print(next(i for i in range(1, k * 10) if k * i % 10 == r or k * i % 10 == 0)) | Title: Buy a Shovel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input Specification:
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output Specification:
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Demo Input:
['117 3\n', '237 7\n', '15 2\n']
Demo Output:
['9\n', '1\n', '2\n']
Note:
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change. | ```python
k, r = map(int, input().split());print(next(i for i in range(1, k * 10) if k * i % 10 == r or k * i % 10 == 0))
``` | 3 | |
673 | A | Bear and Game | PROGRAMMING | 800 | [
"implementation"
] | null | null | Bear Limak likes watching sports on TV. He is going to watch a game today. The game lasts 90 minutes and there are no breaks.
Each minute can be either interesting or boring. If 15 consecutive minutes are boring then Limak immediately turns TV off.
You know that there will be *n* interesting minutes *t*1,<=*t*2,<=...,<=*t**n*. Your task is to calculate for how many minutes Limak will watch the game. | The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=90) — the number of interesting minutes.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=... *t**n*<=≤<=90), given in the increasing order. | Print the number of minutes Limak will watch the game. | [
"3\n7 20 88\n",
"9\n16 20 30 40 50 60 70 80 90\n",
"9\n15 20 30 40 50 60 70 80 90\n"
] | [
"35\n",
"15\n",
"90\n"
] | In the first sample, minutes 21, 22, ..., 35 are all boring and thus Limak will turn TV off immediately after the 35-th minute. So, he would watch the game for 35 minutes.
In the second sample, the first 15 minutes are boring.
In the third sample, there are no consecutive 15 boring minutes. So, Limak will watch the whole game. | 500 | [
{
"input": "3\n7 20 88",
"output": "35"
},
{
"input": "9\n16 20 30 40 50 60 70 80 90",
"output": "15"
},
{
"input": "9\n15 20 30 40 50 60 70 80 90",
"output": "90"
},
{
"input": "30\n6 11 12 15 22 24 30 31 32 33 34 35 40 42 44 45 47 50 53 54 57 58 63 67 75 77 79 81 83 88",
... | 1,630,943,729 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 78 | 6,758,400 | n = int(input())
t = list(map(int,input().split()))
watchTime = 0
for i in t:
if i-watchTime>15:break
else:watchTime=i
print(min(watchTime+15, 90)) | Title: Bear and Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bear Limak likes watching sports on TV. He is going to watch a game today. The game lasts 90 minutes and there are no breaks.
Each minute can be either interesting or boring. If 15 consecutive minutes are boring then Limak immediately turns TV off.
You know that there will be *n* interesting minutes *t*1,<=*t*2,<=...,<=*t**n*. Your task is to calculate for how many minutes Limak will watch the game.
Input Specification:
The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=90) — the number of interesting minutes.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=... *t**n*<=≤<=90), given in the increasing order.
Output Specification:
Print the number of minutes Limak will watch the game.
Demo Input:
['3\n7 20 88\n', '9\n16 20 30 40 50 60 70 80 90\n', '9\n15 20 30 40 50 60 70 80 90\n']
Demo Output:
['35\n', '15\n', '90\n']
Note:
In the first sample, minutes 21, 22, ..., 35 are all boring and thus Limak will turn TV off immediately after the 35-th minute. So, he would watch the game for 35 minutes.
In the second sample, the first 15 minutes are boring.
In the third sample, there are no consecutive 15 boring minutes. So, Limak will watch the whole game. | ```python
n = int(input())
t = list(map(int,input().split()))
watchTime = 0
for i in t:
if i-watchTime>15:break
else:watchTime=i
print(min(watchTime+15, 90))
``` | 3 | |
39 | J | Spelling Check | PROGRAMMING | 1,500 | [
"hashing",
"implementation",
"strings"
] | J. Spelling Check | 2 | 256 | Petya has noticed that when he types using a keyboard, he often presses extra buttons and adds extra letters to the words. Of course, the spell-checking system underlines the words for him and he has to click every word and choose the right variant. Petya got fed up with correcting his mistakes himself, that’s why he decided to invent the function that will correct the words itself. Petya started from analyzing the case that happens to him most of the time, when all one needs is to delete one letter for the word to match a word from the dictionary. Thus, Petya faces one mini-task: he has a printed word and a word from the dictionary, and he should delete one letter from the first word to get the second one. And now the very non-trivial question that Petya faces is: which letter should he delete? | The input data contains two strings, consisting of lower-case Latin letters. The length of each string is from 1 to 106 symbols inclusive, the first string contains exactly 1 symbol more than the second one. | In the first line output the number of positions of the symbols in the first string, after the deleting of which the first string becomes identical to the second one. In the second line output space-separated positions of these symbols in increasing order. The positions are numbered starting from 1. If it is impossible to make the first string identical to the second string by deleting one symbol, output one number 0. | [
"abdrakadabra\nabrakadabra\n",
"aa\na\n",
"competition\ncodeforces\n"
] | [
"1\n3\n",
"2\n1 2\n",
"0\n"
] | none | 0 | [
{
"input": "abdrakadabra\nabrakadabra",
"output": "1\n3 "
},
{
"input": "aa\na",
"output": "2\n1 2 "
},
{
"input": "competition\ncodeforces",
"output": "0"
},
{
"input": "ab\na",
"output": "1\n2 "
},
{
"input": "bb\nb",
"output": "2\n1 2 "
},
{
"input"... | 1,662,721,851 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 22 | 2,000 | 26,931,200 | word = input()
ideal = input()
diff = ""
i = 0
while i < len(ideal) and word[i] == ideal[i]:
i += 1
diff = i
i += 1
while i < len(word) and word[i] == ideal[i - 1]:
i += 1
next_diff = i
if diff != next_diff and next_diff != len(word):
print(0)
else:
i = diff - 1
pos = diff
count = 1
while i > -1 and word[i] == word[diff]:
count += 1
i -= 1
pos -= 1
i = diff + 1
while i < len(word) and word[i] == word[diff]:
count += 1
i += 1
print(count)
print(*list(range(pos + 1, pos + count + 1)))
| Title: Spelling Check
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya has noticed that when he types using a keyboard, he often presses extra buttons and adds extra letters to the words. Of course, the spell-checking system underlines the words for him and he has to click every word and choose the right variant. Petya got fed up with correcting his mistakes himself, that’s why he decided to invent the function that will correct the words itself. Petya started from analyzing the case that happens to him most of the time, when all one needs is to delete one letter for the word to match a word from the dictionary. Thus, Petya faces one mini-task: he has a printed word and a word from the dictionary, and he should delete one letter from the first word to get the second one. And now the very non-trivial question that Petya faces is: which letter should he delete?
Input Specification:
The input data contains two strings, consisting of lower-case Latin letters. The length of each string is from 1 to 106 symbols inclusive, the first string contains exactly 1 symbol more than the second one.
Output Specification:
In the first line output the number of positions of the symbols in the first string, after the deleting of which the first string becomes identical to the second one. In the second line output space-separated positions of these symbols in increasing order. The positions are numbered starting from 1. If it is impossible to make the first string identical to the second string by deleting one symbol, output one number 0.
Demo Input:
['abdrakadabra\nabrakadabra\n', 'aa\na\n', 'competition\ncodeforces\n']
Demo Output:
['1\n3\n', '2\n1 2\n', '0\n']
Note:
none | ```python
word = input()
ideal = input()
diff = ""
i = 0
while i < len(ideal) and word[i] == ideal[i]:
i += 1
diff = i
i += 1
while i < len(word) and word[i] == ideal[i - 1]:
i += 1
next_diff = i
if diff != next_diff and next_diff != len(word):
print(0)
else:
i = diff - 1
pos = diff
count = 1
while i > -1 and word[i] == word[diff]:
count += 1
i -= 1
pos -= 1
i = diff + 1
while i < len(word) and word[i] == word[diff]:
count += 1
i += 1
print(count)
print(*list(range(pos + 1, pos + count + 1)))
``` | 0 |
0 | none | none | none | 0 | [
"none"
] | null | null | Есть *n*-подъездный дом, в каждом подъезде по *m* этажей, и на каждом этаже каждого подъезда ровно *k* квартир. Таким образом, в доме всего *n*·*m*·*k* квартир. Они пронумерованы естественным образом от 1 до *n*·*m*·*k*, то есть первая квартира на первом этаже в первом подъезде имеет номер 1, первая квартира на втором этаже первого подъезда имеет номер *k*<=+<=1 и так далее. Особенность этого дома состоит в том, что он круглый. То есть если обходить его по часовой стрелке, то после подъезда номер 1 следует подъезд номер 2, затем подъезд номер 3 и так далее до подъезда номер *n*. После подъезда номер *n* снова идёт подъезд номер 1.
Эдвард живёт в квартире номер *a*, а Наташа — в квартире номер *b*. Переход на 1 этаж вверх или вниз по лестнице занимает 5 секунд, переход от двери подъезда к двери соседнего подъезда — 15 секунд, а переход в пределах одного этажа одного подъезда происходит мгновенно. Также в каждом подъезде дома есть лифт. Он устроен следующим образом: он всегда приезжает ровно через 10 секунд после вызова, а чтобы переместить пассажира на один этаж вверх или вниз, лифт тратит ровно 1 секунду. Посадка и высадка происходят мгновенно.
Помогите Эдварду найти минимальное время, за которое он сможет добраться до квартиры Наташи. Считайте, что Эдвард может выйти из подъезда только с первого этажа соответствующего подъезда (это происходит мгновенно). Если Эдвард стоит перед дверью какого-то подъезда, он может зайти в него и сразу окажется на первом этаже этого подъезда (это также происходит мгновенно). Эдвард может выбирать, в каком направлении идти вокруг дома. | В первой строке входных данных следуют три числа *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=1000) — количество подъездов в доме, количество этажей в каждом подъезде и количество квартир на каждом этаже каждого подъезда соответственно.
Во второй строке входных данных записаны два числа *a* и *b* (1<=≤<=*a*,<=*b*<=≤<=*n*·*m*·*k*) — номера квартир, в которых живут Эдвард и Наташа, соответственно. Гарантируется, что эти номера различны. | Выведите единственное целое число — минимальное время (в секундах), за которое Эдвард сможет добраться от своей квартиры до квартиры Наташи. | [
"4 10 5\n200 6\n",
"3 1 5\n7 2\n"
] | [
"39\n",
"15\n"
] | В первом тестовом примере Эдвард находится в 4 подъезде на 10 этаже, а Наташа находится в 1 подъезде на 2 этаже. Поэтому Эдварду выгодно сначала спуститься на лифте на первый этаж (на это он потратит 19 секунд, из которых 10 — на ожидание и 9 — на поездку на лифте), затем обойти дом против часовой стрелки до подъезда номер 1 (на это он потратит 15 секунд), и наконец подняться по лестнице на этаж номер 2 (на это он потратит 5 секунд). Таким образом, ответ равен 19 + 15 + 5 = 39.
Во втором тестовом примере Эдвард живёт в подъезде 2 на этаже 1, а Наташа находится в подъезде 1 на этаже 1. Поэтому Эдварду выгодно просто обойти дом по часовой стрелке до подъезда 1, на это он потратит 15 секунд. | 0 | [
{
"input": "4 10 5\n200 6",
"output": "39"
},
{
"input": "3 1 5\n7 2",
"output": "15"
},
{
"input": "100 100 100\n1 1000000",
"output": "124"
},
{
"input": "1000 1000 1000\n1 1000000000",
"output": "1024"
},
{
"input": "125 577 124\n7716799 6501425",
"output":... | 1,458,981,230 | 5,630 | Python 3 | CHALLENGED | CHALLENGES | 8 | 46 | 4,915,200 | t = 0
n, m, k = (int(i) for i in input().split())
a, b = (int(i) for i in input().split())
nA = (a + (m * k) - 1) // (m * k)
mA = ((a - m * k * (nA - 1)) + k - 1) // k
nB = (b + (m * k) - 1) // (m * k)
mB = ((b - m * k * (nB - 1)) + k - 1) // k
if nA == nB:
if mA > mB:
t2 = mA - mB
else:
t2 = mB - mA
if t2 > 3:
t = 10 + t2 - 1
else:
t += (t2 - 1) * 5
else:
if mA > 3:
t = t + 10 + mA - 1
else:
t += (mA - 1) * 5
if nB < nA:
if nA - nB >= (n - nA + nB):
t += (n - nA + nB) * 15
else:
t += (nA - nB) * 15
else:
if nB - nA >= (n - nB + nA):
t += (n - nB + nA) * 15
else:
t += (nB - nA) * 15
if mB > 3:
t = t + 10 + mB - 1
else:
t += (mB - 1) * 5
print(t) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Есть *n*-подъездный дом, в каждом подъезде по *m* этажей, и на каждом этаже каждого подъезда ровно *k* квартир. Таким образом, в доме всего *n*·*m*·*k* квартир. Они пронумерованы естественным образом от 1 до *n*·*m*·*k*, то есть первая квартира на первом этаже в первом подъезде имеет номер 1, первая квартира на втором этаже первого подъезда имеет номер *k*<=+<=1 и так далее. Особенность этого дома состоит в том, что он круглый. То есть если обходить его по часовой стрелке, то после подъезда номер 1 следует подъезд номер 2, затем подъезд номер 3 и так далее до подъезда номер *n*. После подъезда номер *n* снова идёт подъезд номер 1.
Эдвард живёт в квартире номер *a*, а Наташа — в квартире номер *b*. Переход на 1 этаж вверх или вниз по лестнице занимает 5 секунд, переход от двери подъезда к двери соседнего подъезда — 15 секунд, а переход в пределах одного этажа одного подъезда происходит мгновенно. Также в каждом подъезде дома есть лифт. Он устроен следующим образом: он всегда приезжает ровно через 10 секунд после вызова, а чтобы переместить пассажира на один этаж вверх или вниз, лифт тратит ровно 1 секунду. Посадка и высадка происходят мгновенно.
Помогите Эдварду найти минимальное время, за которое он сможет добраться до квартиры Наташи. Считайте, что Эдвард может выйти из подъезда только с первого этажа соответствующего подъезда (это происходит мгновенно). Если Эдвард стоит перед дверью какого-то подъезда, он может зайти в него и сразу окажется на первом этаже этого подъезда (это также происходит мгновенно). Эдвард может выбирать, в каком направлении идти вокруг дома.
Input Specification:
В первой строке входных данных следуют три числа *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=1000) — количество подъездов в доме, количество этажей в каждом подъезде и количество квартир на каждом этаже каждого подъезда соответственно.
Во второй строке входных данных записаны два числа *a* и *b* (1<=≤<=*a*,<=*b*<=≤<=*n*·*m*·*k*) — номера квартир, в которых живут Эдвард и Наташа, соответственно. Гарантируется, что эти номера различны.
Output Specification:
Выведите единственное целое число — минимальное время (в секундах), за которое Эдвард сможет добраться от своей квартиры до квартиры Наташи.
Demo Input:
['4 10 5\n200 6\n', '3 1 5\n7 2\n']
Demo Output:
['39\n', '15\n']
Note:
В первом тестовом примере Эдвард находится в 4 подъезде на 10 этаже, а Наташа находится в 1 подъезде на 2 этаже. Поэтому Эдварду выгодно сначала спуститься на лифте на первый этаж (на это он потратит 19 секунд, из которых 10 — на ожидание и 9 — на поездку на лифте), затем обойти дом против часовой стрелки до подъезда номер 1 (на это он потратит 15 секунд), и наконец подняться по лестнице на этаж номер 2 (на это он потратит 5 секунд). Таким образом, ответ равен 19 + 15 + 5 = 39.
Во втором тестовом примере Эдвард живёт в подъезде 2 на этаже 1, а Наташа находится в подъезде 1 на этаже 1. Поэтому Эдварду выгодно просто обойти дом по часовой стрелке до подъезда 1, на это он потратит 15 секунд. | ```python
t = 0
n, m, k = (int(i) for i in input().split())
a, b = (int(i) for i in input().split())
nA = (a + (m * k) - 1) // (m * k)
mA = ((a - m * k * (nA - 1)) + k - 1) // k
nB = (b + (m * k) - 1) // (m * k)
mB = ((b - m * k * (nB - 1)) + k - 1) // k
if nA == nB:
if mA > mB:
t2 = mA - mB
else:
t2 = mB - mA
if t2 > 3:
t = 10 + t2 - 1
else:
t += (t2 - 1) * 5
else:
if mA > 3:
t = t + 10 + mA - 1
else:
t += (mA - 1) * 5
if nB < nA:
if nA - nB >= (n - nA + nB):
t += (n - nA + nB) * 15
else:
t += (nA - nB) * 15
else:
if nB - nA >= (n - nB + nA):
t += (n - nB + nA) * 15
else:
t += (nB - nA) * 15
if mB > 3:
t = t + 10 + mB - 1
else:
t += (mB - 1) * 5
print(t)
``` | -1 | |
218 | B | Airport | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Lolek and Bolek are about to travel abroad by plane. The local airport has a special "Choose Your Plane" offer. The offer's conditions are as follows:
- it is up to a passenger to choose a plane to fly on; - if the chosen plane has *x* (*x*<=><=0) empty seats at the given moment, then the ticket for such a plane costs *x* zlotys (units of Polish currency).
The only ticket office of the airport already has a queue of *n* passengers in front of it. Lolek and Bolek have not stood in the queue yet, but they are already wondering what is the maximum and the minimum number of zlotys the airport administration can earn if all *n* passengers buy tickets according to the conditions of this offer?
The passengers buy tickets in turn, the first person in the queue goes first, then goes the second one, and so on up to *n*-th person. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of passengers in the queue and the number of planes in the airport, correspondingly. The next line contains *m* integers *a*1,<=*a*2,<=...,<=*a**m* (1<=≤<=*a**i*<=≤<=1000) — *a**i* stands for the number of empty seats in the *i*-th plane before the ticket office starts selling tickets.
The numbers in the lines are separated by a space. It is guaranteed that there are at least *n* empty seats in total. | Print two integers — the maximum and the minimum number of zlotys that the airport administration can earn, correspondingly. | [
"4 3\n2 1 1\n",
"4 3\n2 2 2\n"
] | [
"5 5\n",
"7 6\n"
] | In the first test sample the number of passengers is equal to the number of empty seats, so regardless of the way the planes are chosen, the administration will earn the same sum.
In the second sample the sum is maximized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 2-nd plane, the 3-rd person — to the 3-rd plane, the 4-th person — to the 1-st plane. The sum is minimized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 1-st plane, the 3-rd person — to the 2-nd plane, the 4-th person — to the 2-nd plane. | 500 | [
{
"input": "4 3\n2 1 1",
"output": "5 5"
},
{
"input": "4 3\n2 2 2",
"output": "7 6"
},
{
"input": "10 5\n10 3 3 1 2",
"output": "58 26"
},
{
"input": "10 1\n10",
"output": "55 55"
},
{
"input": "10 1\n100",
"output": "955 955"
},
{
"input": "10 2\n4 7... | 1,544,619,270 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 404 | 0 | numplane = [int(i) for i in input().split()]
empty = [int(i) for i in input().split()]
copy_empty = empty[:]
max_zlotys = []
min_zlotys = []
answer = []
for i in range(numplane[0]):
max_zlotys.append(max(empty))
empty[empty.index(max(empty))] -=1
for i in range(numplane[0]):
try:
copy_empty.remove(0)
except:
pass
min_zlotys.append(min(copy_empty))
copy_empty[copy_empty.index(min(copy_empty))] -=1
answer.append(sum(max_zlotys))
answer.append(sum(min_zlotys))
print(answer[0] , answer[1]) | Title: Airport
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Lolek and Bolek are about to travel abroad by plane. The local airport has a special "Choose Your Plane" offer. The offer's conditions are as follows:
- it is up to a passenger to choose a plane to fly on; - if the chosen plane has *x* (*x*<=><=0) empty seats at the given moment, then the ticket for such a plane costs *x* zlotys (units of Polish currency).
The only ticket office of the airport already has a queue of *n* passengers in front of it. Lolek and Bolek have not stood in the queue yet, but they are already wondering what is the maximum and the minimum number of zlotys the airport administration can earn if all *n* passengers buy tickets according to the conditions of this offer?
The passengers buy tickets in turn, the first person in the queue goes first, then goes the second one, and so on up to *n*-th person.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of passengers in the queue and the number of planes in the airport, correspondingly. The next line contains *m* integers *a*1,<=*a*2,<=...,<=*a**m* (1<=≤<=*a**i*<=≤<=1000) — *a**i* stands for the number of empty seats in the *i*-th plane before the ticket office starts selling tickets.
The numbers in the lines are separated by a space. It is guaranteed that there are at least *n* empty seats in total.
Output Specification:
Print two integers — the maximum and the minimum number of zlotys that the airport administration can earn, correspondingly.
Demo Input:
['4 3\n2 1 1\n', '4 3\n2 2 2\n']
Demo Output:
['5 5\n', '7 6\n']
Note:
In the first test sample the number of passengers is equal to the number of empty seats, so regardless of the way the planes are chosen, the administration will earn the same sum.
In the second sample the sum is maximized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 2-nd plane, the 3-rd person — to the 3-rd plane, the 4-th person — to the 1-st plane. The sum is minimized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 1-st plane, the 3-rd person — to the 2-nd plane, the 4-th person — to the 2-nd plane. | ```python
numplane = [int(i) for i in input().split()]
empty = [int(i) for i in input().split()]
copy_empty = empty[:]
max_zlotys = []
min_zlotys = []
answer = []
for i in range(numplane[0]):
max_zlotys.append(max(empty))
empty[empty.index(max(empty))] -=1
for i in range(numplane[0]):
try:
copy_empty.remove(0)
except:
pass
min_zlotys.append(min(copy_empty))
copy_empty[copy_empty.index(min(copy_empty))] -=1
answer.append(sum(max_zlotys))
answer.append(sum(min_zlotys))
print(answer[0] , answer[1])
``` | 3 | |
493 | B | Vasya and Wrestling | PROGRAMMING | 1,400 | [
"implementation"
] | null | null | Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won. | The first line contains number *n* — the number of techniques that the wrestlers have used (1<=≤<=*n*<=≤<=2·105).
The following *n* lines contain integer numbers *a**i* (|*a**i*|<=≤<=109, *a**i*<=≠<=0). If *a**i* is positive, that means that the first wrestler performed the technique that was awarded with *a**i* points. And if *a**i* is negative, that means that the second wrestler performed the technique that was awarded with (<=-<=*a**i*) points.
The techniques are given in chronological order. | If the first wrestler wins, print string "first", otherwise print "second" | [
"5\n1\n2\n-3\n-4\n3\n",
"3\n-1\n-2\n3\n",
"2\n4\n-4\n"
] | [
"second\n",
"first\n",
"second\n"
] | Sequence *x* = *x*<sub class="lower-index">1</sub>*x*<sub class="lower-index">2</sub>... *x*<sub class="lower-index">|*x*|</sub> is lexicographically larger than sequence *y* = *y*<sub class="lower-index">1</sub>*y*<sub class="lower-index">2</sub>... *y*<sub class="lower-index">|*y*|</sub>, if either |*x*| > |*y*| and *x*<sub class="lower-index">1</sub> = *y*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub> = *y*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">|*y*|</sub> = *y*<sub class="lower-index">|*y*|</sub>, or there is such number *r* (*r* < |*x*|, *r* < |*y*|), that *x*<sub class="lower-index">1</sub> = *y*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub> = *y*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">*r*</sub> = *y*<sub class="lower-index">*r*</sub> and *x*<sub class="lower-index">*r* + 1</sub> > *y*<sub class="lower-index">*r* + 1</sub>.
We use notation |*a*| to denote length of sequence *a*. | 1,000 | [
{
"input": "5\n1\n2\n-3\n-4\n3",
"output": "second"
},
{
"input": "3\n-1\n-2\n3",
"output": "first"
},
{
"input": "2\n4\n-4",
"output": "second"
},
{
"input": "7\n1\n2\n-3\n4\n5\n-6\n7",
"output": "first"
},
{
"input": "14\n1\n2\n3\n4\n5\n6\n7\n-8\n-9\n-10\n-11\n-... | 1,596,788,999 | 2,147,483,647 | PyPy 3 | OK | TESTS | 57 | 1,512 | 30,105,600 | n=int(input())
s0=s1=0
a0,a1=[],[]
l0=l1=-1
for i in range(n):
a=int(input())
if a>0:
s0+=a
a0+=[a]
l0=i
else:
s1+=-a
a1+=[-a]
l1=i
print('first' if (s0, a0, l0)>(s1, a1, l1) else 'second') | Title: Vasya and Wrestling
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.
When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.
If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.
Input Specification:
The first line contains number *n* — the number of techniques that the wrestlers have used (1<=≤<=*n*<=≤<=2·105).
The following *n* lines contain integer numbers *a**i* (|*a**i*|<=≤<=109, *a**i*<=≠<=0). If *a**i* is positive, that means that the first wrestler performed the technique that was awarded with *a**i* points. And if *a**i* is negative, that means that the second wrestler performed the technique that was awarded with (<=-<=*a**i*) points.
The techniques are given in chronological order.
Output Specification:
If the first wrestler wins, print string "first", otherwise print "second"
Demo Input:
['5\n1\n2\n-3\n-4\n3\n', '3\n-1\n-2\n3\n', '2\n4\n-4\n']
Demo Output:
['second\n', 'first\n', 'second\n']
Note:
Sequence *x* = *x*<sub class="lower-index">1</sub>*x*<sub class="lower-index">2</sub>... *x*<sub class="lower-index">|*x*|</sub> is lexicographically larger than sequence *y* = *y*<sub class="lower-index">1</sub>*y*<sub class="lower-index">2</sub>... *y*<sub class="lower-index">|*y*|</sub>, if either |*x*| > |*y*| and *x*<sub class="lower-index">1</sub> = *y*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub> = *y*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">|*y*|</sub> = *y*<sub class="lower-index">|*y*|</sub>, or there is such number *r* (*r* < |*x*|, *r* < |*y*|), that *x*<sub class="lower-index">1</sub> = *y*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub> = *y*<sub class="lower-index">2</sub>, ... , *x*<sub class="lower-index">*r*</sub> = *y*<sub class="lower-index">*r*</sub> and *x*<sub class="lower-index">*r* + 1</sub> > *y*<sub class="lower-index">*r* + 1</sub>.
We use notation |*a*| to denote length of sequence *a*. | ```python
n=int(input())
s0=s1=0
a0,a1=[],[]
l0=l1=-1
for i in range(n):
a=int(input())
if a>0:
s0+=a
a0+=[a]
l0=i
else:
s1+=-a
a1+=[-a]
l1=i
print('first' if (s0, a0, l0)>(s1, a1, l1) else 'second')
``` | 3 | |
651 | B | Beautiful Paintings | PROGRAMMING | 1,200 | [
"greedy",
"sortings"
] | null | null | There are *n* pictures delivered for the new exhibition. The *i*-th painting has beauty *a**i*. We know that a visitor becomes happy every time he passes from a painting to a more beautiful one.
We are allowed to arranged pictures in any order. What is the maximum possible number of times the visitor may become happy while passing all pictures from first to last? In other words, we are allowed to rearrange elements of *a* in any order. What is the maximum possible number of indices *i* (1<=≤<=*i*<=≤<=*n*<=-<=1), such that *a**i*<=+<=1<=><=*a**i*. | The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of painting.
The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000), where *a**i* means the beauty of the *i*-th painting. | Print one integer — the maximum possible number of neighbouring pairs, such that *a**i*<=+<=1<=><=*a**i*, after the optimal rearrangement. | [
"5\n20 30 10 50 40\n",
"4\n200 100 100 200\n"
] | [
"4\n",
"2\n"
] | In the first sample, the optimal order is: 10, 20, 30, 40, 50.
In the second sample, the optimal order is: 100, 200, 100, 200. | 1,000 | [
{
"input": "5\n20 30 10 50 40",
"output": "4"
},
{
"input": "4\n200 100 100 200",
"output": "2"
},
{
"input": "10\n2 2 2 2 2 2 2 2 2 2",
"output": "0"
},
{
"input": "1\n1000",
"output": "0"
},
{
"input": "2\n444 333",
"output": "1"
},
{
"input": "100\n... | 1,593,091,818 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 124 | 20,172,800 |
n=int(input())
l=list(map(int,input().split()))
s=list(set(l))
cnt=len(s)-1
rp=[]
for i in s:
if l.count(i)>1:
rp.append(l.count(i))
if len(rp)>1:
cnt+=(min(rp)*(len(rp)-1))
print(rp)
print(cnt) | Title: Beautiful Paintings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* pictures delivered for the new exhibition. The *i*-th painting has beauty *a**i*. We know that a visitor becomes happy every time he passes from a painting to a more beautiful one.
We are allowed to arranged pictures in any order. What is the maximum possible number of times the visitor may become happy while passing all pictures from first to last? In other words, we are allowed to rearrange elements of *a* in any order. What is the maximum possible number of indices *i* (1<=≤<=*i*<=≤<=*n*<=-<=1), such that *a**i*<=+<=1<=><=*a**i*.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of painting.
The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000), where *a**i* means the beauty of the *i*-th painting.
Output Specification:
Print one integer — the maximum possible number of neighbouring pairs, such that *a**i*<=+<=1<=><=*a**i*, after the optimal rearrangement.
Demo Input:
['5\n20 30 10 50 40\n', '4\n200 100 100 200\n']
Demo Output:
['4\n', '2\n']
Note:
In the first sample, the optimal order is: 10, 20, 30, 40, 50.
In the second sample, the optimal order is: 100, 200, 100, 200. | ```python
n=int(input())
l=list(map(int,input().split()))
s=list(set(l))
cnt=len(s)-1
rp=[]
for i in s:
if l.count(i)>1:
rp.append(l.count(i))
if len(rp)>1:
cnt+=(min(rp)*(len(rp)-1))
print(rp)
print(cnt)
``` | 0 | |
119 | A | Epic Game | PROGRAMMING | 800 | [
"implementation"
] | null | null | Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game. | The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile. | If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes). | [
"3 5 9\n",
"1 1 100\n"
] | [
"0",
"1"
] | The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that. | 500 | [
{
"input": "3 5 9",
"output": "0"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "23 12 16",
"output": "1"
},
{
"input": "95 26 29",
"output": "1"
},
{
"input": "73 32 99",
"output": "1"
},
{
"input": "1 1 1",
"output": "0"
},
{
"inpu... | 1,668,433,433 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 45 | 124 | 0 | import math
def main():
a, b, n = map(int, input().split())
now = 0
while(True):
if now == 0:
g = math.gcd(a, n)
else:
g = math.gcd(b, n)
if g > n:
print(now ^ 1)
break
else:
n -= g
now ^= 1
main() | Title: Epic Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game.
Input Specification:
The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
Output Specification:
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
Demo Input:
['3 5 9\n', '1 1 100\n']
Demo Output:
['0', '1']
Note:
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that. | ```python
import math
def main():
a, b, n = map(int, input().split())
now = 0
while(True):
if now == 0:
g = math.gcd(a, n)
else:
g = math.gcd(b, n)
if g > n:
print(now ^ 1)
break
else:
n -= g
now ^= 1
main()
``` | 3 | |
847 | G | University Classes | PROGRAMMING | 900 | [
"implementation"
] | null | null | There are *n* student groups at the university. During the study day, each group can take no more than 7 classes. Seven time slots numbered from 1 to 7 are allocated for the classes.
The schedule on Monday is known for each group, i. e. time slots when group will have classes are known.
Your task is to determine the minimum number of rooms needed to hold classes for all groups on Monday. Note that one room can hold at most one group class in a single time slot. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of groups.
Each of the following *n* lines contains a sequence consisting of 7 zeroes and ones — the schedule of classes on Monday for a group. If the symbol in a position equals to 1 then the group has class in the corresponding time slot. In the other case, the group has no class in the corresponding time slot. | Print minimum number of rooms needed to hold all groups classes on Monday. | [
"2\n0101010\n1010101\n",
"3\n0101011\n0011001\n0110111\n"
] | [
"1\n",
"3\n"
] | In the first example one room is enough. It will be occupied in each of the seven time slot by the first group or by the second group.
In the second example three rooms is enough, because in the seventh time slot all three groups have classes. | 0 | [
{
"input": "2\n0101010\n1010101",
"output": "1"
},
{
"input": "3\n0101011\n0011001\n0110111",
"output": "3"
},
{
"input": "1\n0111000",
"output": "1"
},
{
"input": "1\n0000000",
"output": "0"
},
{
"input": "1\n1111111",
"output": "1"
},
{
"input": "2\n... | 1,506,544,372 | 2,147,483,647 | Python 3 | OK | TESTS | 21 | 62 | 0 | n = int(input())
s = []
tmp = mx = 0
for i in range(n):
s.append(input())
for i in range(7):
for j in range(n):
if s[j][i] == '1':
tmp += 1
mx = max(mx, tmp)
tmp = 0
print(mx) | Title: University Classes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* student groups at the university. During the study day, each group can take no more than 7 classes. Seven time slots numbered from 1 to 7 are allocated for the classes.
The schedule on Monday is known for each group, i. e. time slots when group will have classes are known.
Your task is to determine the minimum number of rooms needed to hold classes for all groups on Monday. Note that one room can hold at most one group class in a single time slot.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of groups.
Each of the following *n* lines contains a sequence consisting of 7 zeroes and ones — the schedule of classes on Monday for a group. If the symbol in a position equals to 1 then the group has class in the corresponding time slot. In the other case, the group has no class in the corresponding time slot.
Output Specification:
Print minimum number of rooms needed to hold all groups classes on Monday.
Demo Input:
['2\n0101010\n1010101\n', '3\n0101011\n0011001\n0110111\n']
Demo Output:
['1\n', '3\n']
Note:
In the first example one room is enough. It will be occupied in each of the seven time slot by the first group or by the second group.
In the second example three rooms is enough, because in the seventh time slot all three groups have classes. | ```python
n = int(input())
s = []
tmp = mx = 0
for i in range(n):
s.append(input())
for i in range(7):
for j in range(n):
if s[j][i] == '1':
tmp += 1
mx = max(mx, tmp)
tmp = 0
print(mx)
``` | 3 | |
371 | C | Hamburgers | PROGRAMMING | 1,600 | [
"binary search",
"brute force"
] | null | null | Polycarpus loves hamburgers very much. He especially adores the hamburgers he makes with his own hands. Polycarpus thinks that there are only three decent ingredients to make hamburgers from: a bread, sausage and cheese. He writes down the recipe of his favorite "Le Hamburger de Polycarpus" as a string of letters 'B' (bread), 'S' (sausage) и 'C' (cheese). The ingredients in the recipe go from bottom to top, for example, recipe "ВSCBS" represents the hamburger where the ingredients go from bottom to top as bread, sausage, cheese, bread and sausage again.
Polycarpus has *n**b* pieces of bread, *n**s* pieces of sausage and *n**c* pieces of cheese in the kitchen. Besides, the shop nearby has all three ingredients, the prices are *p**b* rubles for a piece of bread, *p**s* for a piece of sausage and *p**c* for a piece of cheese.
Polycarpus has *r* rubles and he is ready to shop on them. What maximum number of hamburgers can he cook? You can assume that Polycarpus cannot break or slice any of the pieces of bread, sausage or cheese. Besides, the shop has an unlimited number of pieces of each ingredient. | The first line of the input contains a non-empty string that describes the recipe of "Le Hamburger de Polycarpus". The length of the string doesn't exceed 100, the string contains only letters 'B' (uppercase English B), 'S' (uppercase English S) and 'C' (uppercase English C).
The second line contains three integers *n**b*, *n**s*, *n**c* (1<=≤<=*n**b*,<=*n**s*,<=*n**c*<=≤<=100) — the number of the pieces of bread, sausage and cheese on Polycarpus' kitchen. The third line contains three integers *p**b*, *p**s*, *p**c* (1<=≤<=*p**b*,<=*p**s*,<=*p**c*<=≤<=100) — the price of one piece of bread, sausage and cheese in the shop. Finally, the fourth line contains integer *r* (1<=≤<=*r*<=≤<=1012) — the number of rubles Polycarpus has.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. | Print the maximum number of hamburgers Polycarpus can make. If he can't make any hamburger, print 0. | [
"BBBSSC\n6 4 1\n1 2 3\n4\n",
"BBC\n1 10 1\n1 10 1\n21\n",
"BSC\n1 1 1\n1 1 3\n1000000000000\n"
] | [
"2\n",
"7\n",
"200000000001\n"
] | none | 1,500 | [
{
"input": "BBBSSC\n6 4 1\n1 2 3\n4",
"output": "2"
},
{
"input": "BBC\n1 10 1\n1 10 1\n21",
"output": "7"
},
{
"input": "BSC\n1 1 1\n1 1 3\n1000000000000",
"output": "200000000001"
},
{
"input": "B\n1 1 1\n1 1 1\n381",
"output": "382"
},
{
"input": "BSC\n3 5 6\n7... | 1,601,736,825 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 108 | 307,200 | type=input()
numberin=list(map(int,input().split()))
price=list(map(int,input().split()))
money=int(input())
cneed=[0,0,0]
for t in type:
if t=="B":
cneed[0]+=1
elif t=="S":
cneed[1]+=1
else:
cneed[0]+=1
moneypeer=0
buyed=0
for i in range(3):
moneypeer+=cneed[i]*price[i]
buyed+=numberin[i]*price[i]
allmoney=buyed+money
canmake=allmoney//moneypeer
print(canmake)
| Title: Hamburgers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus loves hamburgers very much. He especially adores the hamburgers he makes with his own hands. Polycarpus thinks that there are only three decent ingredients to make hamburgers from: a bread, sausage and cheese. He writes down the recipe of his favorite "Le Hamburger de Polycarpus" as a string of letters 'B' (bread), 'S' (sausage) и 'C' (cheese). The ingredients in the recipe go from bottom to top, for example, recipe "ВSCBS" represents the hamburger where the ingredients go from bottom to top as bread, sausage, cheese, bread and sausage again.
Polycarpus has *n**b* pieces of bread, *n**s* pieces of sausage and *n**c* pieces of cheese in the kitchen. Besides, the shop nearby has all three ingredients, the prices are *p**b* rubles for a piece of bread, *p**s* for a piece of sausage and *p**c* for a piece of cheese.
Polycarpus has *r* rubles and he is ready to shop on them. What maximum number of hamburgers can he cook? You can assume that Polycarpus cannot break or slice any of the pieces of bread, sausage or cheese. Besides, the shop has an unlimited number of pieces of each ingredient.
Input Specification:
The first line of the input contains a non-empty string that describes the recipe of "Le Hamburger de Polycarpus". The length of the string doesn't exceed 100, the string contains only letters 'B' (uppercase English B), 'S' (uppercase English S) and 'C' (uppercase English C).
The second line contains three integers *n**b*, *n**s*, *n**c* (1<=≤<=*n**b*,<=*n**s*,<=*n**c*<=≤<=100) — the number of the pieces of bread, sausage and cheese on Polycarpus' kitchen. The third line contains three integers *p**b*, *p**s*, *p**c* (1<=≤<=*p**b*,<=*p**s*,<=*p**c*<=≤<=100) — the price of one piece of bread, sausage and cheese in the shop. Finally, the fourth line contains integer *r* (1<=≤<=*r*<=≤<=1012) — the number of rubles Polycarpus has.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print the maximum number of hamburgers Polycarpus can make. If he can't make any hamburger, print 0.
Demo Input:
['BBBSSC\n6 4 1\n1 2 3\n4\n', 'BBC\n1 10 1\n1 10 1\n21\n', 'BSC\n1 1 1\n1 1 3\n1000000000000\n']
Demo Output:
['2\n', '7\n', '200000000001\n']
Note:
none | ```python
type=input()
numberin=list(map(int,input().split()))
price=list(map(int,input().split()))
money=int(input())
cneed=[0,0,0]
for t in type:
if t=="B":
cneed[0]+=1
elif t=="S":
cneed[1]+=1
else:
cneed[0]+=1
moneypeer=0
buyed=0
for i in range(3):
moneypeer+=cneed[i]*price[i]
buyed+=numberin[i]*price[i]
allmoney=buyed+money
canmake=allmoney//moneypeer
print(canmake)
``` | 0 | |
5 | C | Longest Regular Bracket Sequence | PROGRAMMING | 1,900 | [
"constructive algorithms",
"data structures",
"dp",
"greedy",
"sortings",
"strings"
] | C. Longest Regular Bracket Sequence | 2 | 256 | This is yet another problem dealing with regular bracket sequences.
We should remind you that a bracket sequence is called regular, if by inserting «+» and «1» into it we can get a correct mathematical expression. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
You are given a string of «(» and «)» characters. You are to find its longest substring that is a regular bracket sequence. You are to find the number of such substrings as well. | The first line of the input file contains a non-empty string, consisting of «(» and «)» characters. Its length does not exceed 106. | Print the length of the longest substring that is a regular bracket sequence, and the number of such substrings. If there are no such substrings, write the only line containing "0 1". | [
")((())))(()())\n",
"))(\n"
] | [
"6 2\n",
"0 1\n"
] | none | 0 | [
{
"input": ")((())))(()())",
"output": "6 2"
},
{
"input": "))(",
"output": "0 1"
},
{
"input": "()(())()",
"output": "8 1"
},
{
"input": "((((()(((",
"output": "2 1"
},
{
"input": "))))()())))",
"output": "4 1"
},
{
"input": "(()())()(())()()())())()(... | 1,587,567,676 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 218 | 307,200 | # `frase` está despues del "("
def contador_parentesis(frase):
contador = 1
caracteres = 1
for i in range(len(frase)):
if frase[i] == "(":
contador += 1
elif frase[i] == ")":
contador -= 1
caracteres += 1
if contador == 0:
break
return caracteres
a = input()
maximo = 0
contados = 0
for i in range(len(a)):
if a[i] == "(":
nuevo = contador_parentesis(a[i+1:])
if nuevo % 2 == 0:
if nuevo > maximo:
maximo = nuevo
contados = 1
elif nuevo == maximo:
contados += 1
if maximo == 0:
print(0, 1)
else:
print(maximo, contados)
| Title: Longest Regular Bracket Sequence
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
This is yet another problem dealing with regular bracket sequences.
We should remind you that a bracket sequence is called regular, if by inserting «+» and «1» into it we can get a correct mathematical expression. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
You are given a string of «(» and «)» characters. You are to find its longest substring that is a regular bracket sequence. You are to find the number of such substrings as well.
Input Specification:
The first line of the input file contains a non-empty string, consisting of «(» and «)» characters. Its length does not exceed 106.
Output Specification:
Print the length of the longest substring that is a regular bracket sequence, and the number of such substrings. If there are no such substrings, write the only line containing "0 1".
Demo Input:
[')((())))(()())\n', '))(\n']
Demo Output:
['6 2\n', '0 1\n']
Note:
none | ```python
# `frase` está despues del "("
def contador_parentesis(frase):
contador = 1
caracteres = 1
for i in range(len(frase)):
if frase[i] == "(":
contador += 1
elif frase[i] == ")":
contador -= 1
caracteres += 1
if contador == 0:
break
return caracteres
a = input()
maximo = 0
contados = 0
for i in range(len(a)):
if a[i] == "(":
nuevo = contador_parentesis(a[i+1:])
if nuevo % 2 == 0:
if nuevo > maximo:
maximo = nuevo
contados = 1
elif nuevo == maximo:
contados += 1
if maximo == 0:
print(0, 1)
else:
print(maximo, contados)
``` | 0 |
272 | C | Dima and Staircase | PROGRAMMING | 1,500 | [
"data structures",
"implementation"
] | null | null | Dima's got a staircase that consists of *n* stairs. The first stair is at height *a*1, the second one is at *a*2, the last one is at *a**n* (1<=≤<=*a*1<=≤<=*a*2<=≤<=...<=≤<=*a**n*).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The *i*-th box has width *w**i* and height *h**i*. Dima throws each box vertically down on the first *w**i* stairs of the staircase, that is, the box covers stairs with numbers 1,<=2,<=...,<=*w**i*. Each thrown box flies vertically down until at least one of the two following events happen:
- the bottom of the box touches the top of a stair; - the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width *w**i* cannot touch the stair number *w**i*<=+<=1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of *n* integers, *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109; *a**i*<=≤<=*a**i*<=+<=1).
The next line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of boxes. Each of the following *m* lines contains a pair of integers *w**i*,<=*h**i* (1<=≤<=*w**i*<=≤<=*n*; 1<=≤<=*h**i*<=≤<=109) — the size of the *i*-th thrown box.
The numbers in the lines are separated by spaces. | Print *m* integers — for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
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. | [
"5\n1 2 3 6 6\n4\n1 1\n3 1\n1 1\n4 3\n",
"3\n1 2 3\n2\n1 1\n3 1\n",
"1\n1\n5\n1 2\n1 10\n1 10\n1 10\n1 10\n"
] | [
"1\n3\n4\n6\n",
"1\n3\n",
"1\n3\n13\n23\n33\n"
] | The first sample are shown on the picture. | 1,500 | [
{
"input": "5\n1 2 3 6 6\n4\n1 1\n3 1\n1 1\n4 3",
"output": "1\n3\n4\n6"
},
{
"input": "3\n1 2 3\n2\n1 1\n3 1",
"output": "1\n3"
},
{
"input": "1\n1\n5\n1 2\n1 10\n1 10\n1 10\n1 10",
"output": "1\n3\n13\n23\n33"
},
{
"input": "8\n6 10 18 23 30 31 31 33\n1\n5 3",
"output":... | 1,664,437,385 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 33 | 2,000 | 14,643,200 | n = int(input())
nums = list(map(int, input().split()))
boxNum = int(input())
minH = 0
for _ in range(boxNum):
w ,h = map(int, input().split())
minH = max(minH, nums[w-1])
print(minH)
minH += h
| Title: Dima and Staircase
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dima's got a staircase that consists of *n* stairs. The first stair is at height *a*1, the second one is at *a*2, the last one is at *a**n* (1<=≤<=*a*1<=≤<=*a*2<=≤<=...<=≤<=*a**n*).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The *i*-th box has width *w**i* and height *h**i*. Dima throws each box vertically down on the first *w**i* stairs of the staircase, that is, the box covers stairs with numbers 1,<=2,<=...,<=*w**i*. Each thrown box flies vertically down until at least one of the two following events happen:
- the bottom of the box touches the top of a stair; - the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width *w**i* cannot touch the stair number *w**i*<=+<=1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of *n* integers, *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109; *a**i*<=≤<=*a**i*<=+<=1).
The next line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of boxes. Each of the following *m* lines contains a pair of integers *w**i*,<=*h**i* (1<=≤<=*w**i*<=≤<=*n*; 1<=≤<=*h**i*<=≤<=109) — the size of the *i*-th thrown box.
The numbers in the lines are separated by spaces.
Output Specification:
Print *m* integers — for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
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.
Demo Input:
['5\n1 2 3 6 6\n4\n1 1\n3 1\n1 1\n4 3\n', '3\n1 2 3\n2\n1 1\n3 1\n', '1\n1\n5\n1 2\n1 10\n1 10\n1 10\n1 10\n']
Demo Output:
['1\n3\n4\n6\n', '1\n3\n', '1\n3\n13\n23\n33\n']
Note:
The first sample are shown on the picture. | ```python
n = int(input())
nums = list(map(int, input().split()))
boxNum = int(input())
minH = 0
for _ in range(boxNum):
w ,h = map(int, input().split())
minH = max(minH, nums[w-1])
print(minH)
minH += h
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of *n* lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation *n*<=-<=1 times:
- Remove any two elements *s* and *t* from the set, and add their concatenation *s*<=+<=*t* to the set.
The cost of such operation is defined to be , where *f*(*s*,<=*c*) denotes the number of times character *c* appears in string *s*.
Given a non-negative integer *k*, construct any valid non-empty set of no more than 100<=000 letters, such that the minimum accumulative cost of the whole process is exactly *k*. It can be shown that a solution always exists. | The first and only line of input contains a non-negative integer *k* (0<=≤<=*k*<=≤<=100<=000) — the required minimum cost. | Output a non-empty string of no more than 100<=000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters. | [
"12\n",
"3\n"
] | [
"abababab\n",
"codeforces\n"
] | For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
- {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0; - {"aba", "b", "a", "b", "a", "b"}, with a cost of 1; - {"abab", "a", "b", "a", "b"}, with a cost of 1; - {"abab", "ab", "a", "b"}, with a cost of 0; - {"abab", "aba", "b"}, with a cost of 1; - {"abab", "abab"}, with a cost of 1; - {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | 0 | [
{
"input": "12",
"output": "abababab"
},
{
"input": "3",
"output": "codeforces"
},
{
"input": "0",
"output": "o"
},
{
"input": "2",
"output": "aabb"
},
{
"input": "5",
"output": "aaabbcc"
},
{
"input": "10",
"output": "aaaaa"
},
{
"input": ... | 1,692,107,800 | 2,147,483,647 | PyPy 3 | OK | TESTS | 25 | 78 | 0 | k = int(input())
str = 'a'
i = 0
if k == 0:
print('a')
while k:
while k - i >= 0:
k -= i
print(str, end='')
i += 1
str = chr(ord(str) + 1)
i = 0
print()
#bobfut.near moondance | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of *n* lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation *n*<=-<=1 times:
- Remove any two elements *s* and *t* from the set, and add their concatenation *s*<=+<=*t* to the set.
The cost of such operation is defined to be , where *f*(*s*,<=*c*) denotes the number of times character *c* appears in string *s*.
Given a non-negative integer *k*, construct any valid non-empty set of no more than 100<=000 letters, such that the minimum accumulative cost of the whole process is exactly *k*. It can be shown that a solution always exists.
Input Specification:
The first and only line of input contains a non-negative integer *k* (0<=≤<=*k*<=≤<=100<=000) — the required minimum cost.
Output Specification:
Output a non-empty string of no more than 100<=000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Demo Input:
['12\n', '3\n']
Demo Output:
['abababab\n', 'codeforces\n']
Note:
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
- {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0; - {"aba", "b", "a", "b", "a", "b"}, with a cost of 1; - {"abab", "a", "b", "a", "b"}, with a cost of 1; - {"abab", "ab", "a", "b"}, with a cost of 0; - {"abab", "aba", "b"}, with a cost of 1; - {"abab", "abab"}, with a cost of 1; - {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process. | ```python
k = int(input())
str = 'a'
i = 0
if k == 0:
print('a')
while k:
while k - i >= 0:
k -= i
print(str, end='')
i += 1
str = chr(ord(str) + 1)
i = 0
print()
#bobfut.near moondance
``` | 3 | |
150 | A | Win or Freeze | PROGRAMMING | 1,400 | [
"games",
"math",
"number theory"
] | null | null | You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer *q*. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move. | The first line contains the only integer *q* (1<=≤<=*q*<=≤<=1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator. | In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them. | [
"6\n",
"30\n",
"1\n"
] | [
"2\n",
"1\n6\n",
"1\n0\n"
] | Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory. | 500 | [
{
"input": "6",
"output": "2"
},
{
"input": "30",
"output": "1\n6"
},
{
"input": "1",
"output": "1\n0"
},
{
"input": "2",
"output": "1\n0"
},
{
"input": "3",
"output": "1\n0"
},
{
"input": "5",
"output": "1\n0"
},
{
"input": "445538663413",... | 1,629,899,629 | 2,147,483,647 | PyPy 3 | OK | TESTS | 53 | 186 | 27,750,400 | import sys,math,io,os,time,itertools,collections
mod=10**9+7
sys.setrecursionlimit(10000)
i=sys.stdin.readline
p=sys.stdout.write
#use sys.stdout.write() (remember to convert to str b4 and concatenate "\n")
global start,end
def primefac(n):
l=[]
while n%2==0:
l.append(2)
n=n//2
p=3
while p*p<=n:
while n%p==0:
l.append(p)
n=n//p
p+=2
if n>2:
l.append(n)
return l
def main():
q=int(i())
if q==1 or len(primefac(q))==1:
p("1\n")
p("0\n")
else:
l=primefac(q)
s=set(l)
if len(s)==1:
if len(l)==2:
p("2\n")
else:
p("1\n")
temp=l[0]**2
p(str(temp)+"\n")
elif len(s)==2 and len(l)==2:
p("2\n")
else:
p("1\n")
li=list(s)
temp=li[0]*li[1]
p(str(temp)+"\n")
t=1
#t=int(i())
start=time.perf_counter()
for _ in range(t):
main()
end=time.perf_counter()
#print(end-start) | Title: Win or Freeze
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer *q*. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.
Input Specification:
The first line contains the only integer *q* (1<=≤<=*q*<=≤<=1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output Specification:
In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.
Demo Input:
['6\n', '30\n', '1\n']
Demo Output:
['2\n', '1\n6\n', '1\n0\n']
Note:
Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory. | ```python
import sys,math,io,os,time,itertools,collections
mod=10**9+7
sys.setrecursionlimit(10000)
i=sys.stdin.readline
p=sys.stdout.write
#use sys.stdout.write() (remember to convert to str b4 and concatenate "\n")
global start,end
def primefac(n):
l=[]
while n%2==0:
l.append(2)
n=n//2
p=3
while p*p<=n:
while n%p==0:
l.append(p)
n=n//p
p+=2
if n>2:
l.append(n)
return l
def main():
q=int(i())
if q==1 or len(primefac(q))==1:
p("1\n")
p("0\n")
else:
l=primefac(q)
s=set(l)
if len(s)==1:
if len(l)==2:
p("2\n")
else:
p("1\n")
temp=l[0]**2
p(str(temp)+"\n")
elif len(s)==2 and len(l)==2:
p("2\n")
else:
p("1\n")
li=list(s)
temp=li[0]*li[1]
p(str(temp)+"\n")
t=1
#t=int(i())
start=time.perf_counter()
for _ in range(t):
main()
end=time.perf_counter()
#print(end-start)
``` | 3 | |
372 | A | Counting Kangaroos is Fun | PROGRAMMING | 1,600 | [
"binary search",
"greedy",
"sortings",
"two pointers"
] | null | null | There are *n* kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.
Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.
The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible. | The first line contains a single integer — *n* (1<=≤<=*n*<=≤<=5·105). Each of the next *n* lines contains an integer *s**i* — the size of the *i*-th kangaroo (1<=≤<=*s**i*<=≤<=105). | Output a single integer — the optimal number of visible kangaroos. | [
"8\n2\n5\n7\n6\n9\n8\n4\n2\n",
"8\n9\n1\n6\n2\n6\n5\n8\n3\n"
] | [
"5\n",
"5\n"
] | none | 500 | [
{
"input": "8\n2\n5\n7\n6\n9\n8\n4\n2",
"output": "5"
},
{
"input": "8\n9\n1\n6\n2\n6\n5\n8\n3",
"output": "5"
},
{
"input": "12\n3\n99\n24\n46\n75\n63\n57\n55\n10\n62\n34\n52",
"output": "7"
},
{
"input": "12\n55\n75\n1\n98\n63\n64\n9\n39\n82\n18\n47\n9",
"output": "6"
... | 1,559,358,126 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 109 | 0 | import math
n = int(input())
k = sorted([int(input()) for _ in range(n)])
p1 = (n//2)
p2 = n-1
result = n
while p1>=0:
if (k[p1]*2<=k[p2]):
p2-=1
result-=1
k[p1] = -1
p1-=1
print(max(result, math.ceil(n//2)))
| Title: Counting Kangaroos is Fun
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.
Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.
The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible.
Input Specification:
The first line contains a single integer — *n* (1<=≤<=*n*<=≤<=5·105). Each of the next *n* lines contains an integer *s**i* — the size of the *i*-th kangaroo (1<=≤<=*s**i*<=≤<=105).
Output Specification:
Output a single integer — the optimal number of visible kangaroos.
Demo Input:
['8\n2\n5\n7\n6\n9\n8\n4\n2\n', '8\n9\n1\n6\n2\n6\n5\n8\n3\n']
Demo Output:
['5\n', '5\n']
Note:
none | ```python
import math
n = int(input())
k = sorted([int(input()) for _ in range(n)])
p1 = (n//2)
p2 = n-1
result = n
while p1>=0:
if (k[p1]*2<=k[p2]):
p2-=1
result-=1
k[p1] = -1
p1-=1
print(max(result, math.ceil(n//2)))
``` | 0 | |
456 | A | Laptops | PROGRAMMING | 1,100 | [
"sortings"
] | null | null | One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops.
Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality).
All *a**i* are distinct. All *b**i* are distinct. | If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes). | [
"2\n1 2\n2 1\n"
] | [
"Happy Alex\n"
] | none | 500 | [
{
"input": "2\n1 2\n2 1",
"output": "Happy Alex"
},
{
"input": "2\n1 1\n2 2",
"output": "Poor Alex"
},
{
"input": "3\n2 2\n3 3\n1 1",
"output": "Poor Alex"
},
{
"input": "3\n3 3\n1 2\n2 1",
"output": "Happy Alex"
},
{
"input": "1\n1 1",
"output": "Poor Alex"
... | 1,680,967,045 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 46 | 716,800 | n=int(input())
d=[]
for i in range(n):
a=input().split()
b=[int(j) for j in a]
d.append(b)
c=0
for i in range(len(d)):
if i==len(d)-1:
if d[0][0]<d[i][0] and d[0][1]>d[i][1]:
c=1
if c==1:
print("Happy Alex")
else:
print("Poor Alex")
| Title: Laptops
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops.
Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality).
All *a**i* are distinct. All *b**i* are distinct.
Output Specification:
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
Demo Input:
['2\n1 2\n2 1\n']
Demo Output:
['Happy Alex\n']
Note:
none | ```python
n=int(input())
d=[]
for i in range(n):
a=input().split()
b=[int(j) for j in a]
d.append(b)
c=0
for i in range(len(d)):
if i==len(d)-1:
if d[0][0]<d[i][0] and d[0][1]>d[i][1]:
c=1
if c==1:
print("Happy Alex")
else:
print("Poor Alex")
``` | 0 | |
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... | 1,677,450,581 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 80 | 92 | 0 | n = int(input())
a, b, c = [], [], []
for i in range(n):
x, y, z = [int(i1) for i1 in input().split()]
a.append(x)
b.append(y)
c.append(z)
num = sum(a + b + c)
if num == 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, b, c = [], [], []
for i in range(n):
x, y, z = [int(i1) for i1 in input().split()]
a.append(x)
b.append(y)
c.append(z)
num = sum(a + b + c)
if num == 0:
print("YES")
else:
print("NO")
``` | 0 |
515 | C | Drazil and Factorial | PROGRAMMING | 1,400 | [
"greedy",
"math",
"sortings"
] | null | null | Drazil is playing a math game with Varda.
Let's define for positive integer *x* as a product of factorials of its digits. For example, .
First, they choose a decimal number *a* consisting of *n* digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number *x* satisfying following two conditions:
1. *x* doesn't contain neither digit 0 nor digit 1.
2. = .
Help friends find such number. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=15) — the number of digits in *a*.
The second line contains *n* digits of *a*. There is at least one digit in *a* that is larger than 1. Number *a* may possibly contain leading zeroes. | Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. | [
"4\n1234\n",
"3\n555\n"
] | [
"33222\n",
"555\n"
] | In the first case, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f5a4207f23215fddce977ab5ea9e9d2e7578fb52.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 1,000 | [
{
"input": "4\n1234",
"output": "33222"
},
{
"input": "3\n555",
"output": "555"
},
{
"input": "15\n012345781234578",
"output": "7777553333222222222222"
},
{
"input": "1\n8",
"output": "7222"
},
{
"input": "10\n1413472614",
"output": "75333332222222"
},
{
... | 1,674,061,395 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 61 | 102,400 | import heapq
from collections import *
import math
def fact(n):
fact = 1
for j in range(1, int(n) + 1):
fact = fact * j
return fact
def solve():
n = int(input())
s = input()
a = [0, 0, 2, 3, 223, 5, 35, 2227, 2337]
ans = []
for i in range(n):
if int(s[i])>1:
ans.append(a[int(s[i])])
ans.sort(reverse=True)
s = ""
for i in ans:
s+=str(i)
s = list(s)
s.sort(reverse=True)
t = ""
for i in s:
t+=str(i)
print(t)
# t = int(input())
# for _ in range(t):
# solve()
solve()
| Title: Drazil and Factorial
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Drazil is playing a math game with Varda.
Let's define for positive integer *x* as a product of factorials of its digits. For example, .
First, they choose a decimal number *a* consisting of *n* digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number *x* satisfying following two conditions:
1. *x* doesn't contain neither digit 0 nor digit 1.
2. = .
Help friends find such number.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=15) — the number of digits in *a*.
The second line contains *n* digits of *a*. There is at least one digit in *a* that is larger than 1. Number *a* may possibly contain leading zeroes.
Output Specification:
Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation.
Demo Input:
['4\n1234\n', '3\n555\n']
Demo Output:
['33222\n', '555\n']
Note:
In the first case, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f5a4207f23215fddce977ab5ea9e9d2e7578fb52.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
import heapq
from collections import *
import math
def fact(n):
fact = 1
for j in range(1, int(n) + 1):
fact = fact * j
return fact
def solve():
n = int(input())
s = input()
a = [0, 0, 2, 3, 223, 5, 35, 2227, 2337]
ans = []
for i in range(n):
if int(s[i])>1:
ans.append(a[int(s[i])])
ans.sort(reverse=True)
s = ""
for i in ans:
s+=str(i)
s = list(s)
s.sort(reverse=True)
t = ""
for i in s:
t+=str(i)
print(t)
# t = int(input())
# for _ in range(t):
# solve()
solve()
``` | 0 | |
750 | D | New Year and Fireworks | PROGRAMMING | 1,900 | [
"brute force",
"data structures",
"dfs and similar",
"dp",
"implementation"
] | null | null | One tradition of welcoming the New Year is launching fireworks into the sky. Usually a launched firework flies vertically upward for some period of time, then explodes, splitting into several parts flying in different directions. Sometimes those parts also explode after some period of time, splitting into even more parts, and so on.
Limak, who lives in an infinite grid, has a single firework. The behaviour of the firework is described with a recursion depth *n* and a duration for each level of recursion *t*1,<=*t*2,<=...,<=*t**n*. Once Limak launches the firework in some cell, the firework starts moving upward. After covering *t*1 cells (including the starting cell), it explodes and splits into two parts, each moving in the direction changed by 45 degrees (see the pictures below for clarification). So, one part moves in the top-left direction, while the other one moves in the top-right direction. Each part explodes again after covering *t*2 cells, splitting into two parts moving in directions again changed by 45 degrees. The process continues till the *n*-th level of recursion, when all 2*n*<=-<=1 existing parts explode and disappear without creating new parts. After a few levels of recursion, it's possible that some parts will be at the same place and at the same time — it is allowed and such parts do not crash.
Before launching the firework, Limak must make sure that nobody stands in cells which will be visited at least once by the firework. Can you count the number of those cells? | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=30) — the total depth of the recursion.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=5). On the *i*-th level each of 2*i*<=-<=1 parts will cover *t**i* cells before exploding. | Print one integer, denoting the number of cells which will be visited at least once by any part of the firework. | [
"4\n4 2 2 3\n",
"6\n1 1 1 1 1 3\n",
"1\n3\n"
] | [
"39\n",
"85\n",
"3\n"
] | For the first sample, the drawings below show the situation after each level of recursion. Limak launched the firework from the bottom-most red cell. It covered *t*<sub class="lower-index">1</sub> = 4 cells (marked red), exploded and divided into two parts (their further movement is marked green). All explosions are marked with an 'X' character. On the last drawing, there are 4 red, 4 green, 8 orange and 23 pink cells. So, the total number of visited cells is 4 + 4 + 8 + 23 = 39.
For the second sample, the drawings below show the situation after levels 4, 5 and 6. The middle drawing shows directions of all parts that will move in the next level. | 1,500 | [] | 1,692,693,049 | 2,147,483,647 | PyPy 3-64 | MEMORY_LIMIT_EXCEEDED | TESTS | 7 | 2,028 | 268,390,400 | # https://codeforces.com/problemset/problem/750/D
from math import sin, cos, radians
n = int(input())
parts = list(map(int, input().split()))
seen = set()
Q = [(0, 0, 0)]
def get_diffs(angle):
if angle == 0:
return (0, 1)
elif angle == 315:
return (-1, 1)
elif angle == 45:
return (1, 1)
elif angle == 90:
return (1, 0)
elif angle == 270:
return (-1, 0)
elif angle == 225:
return (-1, -1)
elif angle == 135:
return (1, -1)
else:
return (0, -1)
for depth in range(n):
new_Q = []
for x, y, angle in Q:
dx, dy = get_diffs(angle)
for i in range(parts[depth]):
new_xy = (x + i * dx, y + i * dy)
seen.add(new_xy)
peak = parts[depth] - 1
left_turn = (angle - 45) % 360
dxl, dyl = get_diffs(left_turn)
new_Q.append((x + peak * dx + dxl, y + peak * dy + dyl, left_turn))
right_turn = (angle + 45) % 360
dxr, dyr = get_diffs(right_turn)
new_Q.append((x + peak * dx + dxr, y + peak * dy + dyr, right_turn))
Q = new_Q
print(len(seen)) | Title: New Year and Fireworks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One tradition of welcoming the New Year is launching fireworks into the sky. Usually a launched firework flies vertically upward for some period of time, then explodes, splitting into several parts flying in different directions. Sometimes those parts also explode after some period of time, splitting into even more parts, and so on.
Limak, who lives in an infinite grid, has a single firework. The behaviour of the firework is described with a recursion depth *n* and a duration for each level of recursion *t*1,<=*t*2,<=...,<=*t**n*. Once Limak launches the firework in some cell, the firework starts moving upward. After covering *t*1 cells (including the starting cell), it explodes and splits into two parts, each moving in the direction changed by 45 degrees (see the pictures below for clarification). So, one part moves in the top-left direction, while the other one moves in the top-right direction. Each part explodes again after covering *t*2 cells, splitting into two parts moving in directions again changed by 45 degrees. The process continues till the *n*-th level of recursion, when all 2*n*<=-<=1 existing parts explode and disappear without creating new parts. After a few levels of recursion, it's possible that some parts will be at the same place and at the same time — it is allowed and such parts do not crash.
Before launching the firework, Limak must make sure that nobody stands in cells which will be visited at least once by the firework. Can you count the number of those cells?
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=30) — the total depth of the recursion.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=5). On the *i*-th level each of 2*i*<=-<=1 parts will cover *t**i* cells before exploding.
Output Specification:
Print one integer, denoting the number of cells which will be visited at least once by any part of the firework.
Demo Input:
['4\n4 2 2 3\n', '6\n1 1 1 1 1 3\n', '1\n3\n']
Demo Output:
['39\n', '85\n', '3\n']
Note:
For the first sample, the drawings below show the situation after each level of recursion. Limak launched the firework from the bottom-most red cell. It covered *t*<sub class="lower-index">1</sub> = 4 cells (marked red), exploded and divided into two parts (their further movement is marked green). All explosions are marked with an 'X' character. On the last drawing, there are 4 red, 4 green, 8 orange and 23 pink cells. So, the total number of visited cells is 4 + 4 + 8 + 23 = 39.
For the second sample, the drawings below show the situation after levels 4, 5 and 6. The middle drawing shows directions of all parts that will move in the next level. | ```python
# https://codeforces.com/problemset/problem/750/D
from math import sin, cos, radians
n = int(input())
parts = list(map(int, input().split()))
seen = set()
Q = [(0, 0, 0)]
def get_diffs(angle):
if angle == 0:
return (0, 1)
elif angle == 315:
return (-1, 1)
elif angle == 45:
return (1, 1)
elif angle == 90:
return (1, 0)
elif angle == 270:
return (-1, 0)
elif angle == 225:
return (-1, -1)
elif angle == 135:
return (1, -1)
else:
return (0, -1)
for depth in range(n):
new_Q = []
for x, y, angle in Q:
dx, dy = get_diffs(angle)
for i in range(parts[depth]):
new_xy = (x + i * dx, y + i * dy)
seen.add(new_xy)
peak = parts[depth] - 1
left_turn = (angle - 45) % 360
dxl, dyl = get_diffs(left_turn)
new_Q.append((x + peak * dx + dxl, y + peak * dy + dyl, left_turn))
right_turn = (angle + 45) % 360
dxr, dyr = get_diffs(right_turn)
new_Q.append((x + peak * dx + dxr, y + peak * dy + dyr, right_turn))
Q = new_Q
print(len(seen))
``` | 0 | |
985 | B | Switches and Lamps | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | You are given *n* switches and *m* lamps. The *i*-th switch turns on some subset of the lamps. This information is given as the matrix *a* consisting of *n* rows and *m* columns where *a**i*,<=*j*<==<=1 if the *i*-th switch turns on the *j*-th lamp and *a**i*,<=*j*<==<=0 if the *i*-th switch is not connected to the *j*-th lamp.
Initially all *m* lamps are turned off.
Switches change state only from "off" to "on". It means that if you press two or more switches connected to the same lamp then the lamp will be turned on after any of this switches is pressed and will remain its state even if any switch connected to this lamp is pressed afterwards.
It is guaranteed that if you push all *n* switches then all *m* lamps will be turned on.
Your think that you have too many switches and you would like to ignore one of them.
Your task is to say if there exists such a switch that if you will ignore (not use) it but press all the other *n*<=-<=1 switches then all the *m* lamps will be turned on. | The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=2000) — the number of the switches and the number of the lamps.
The following *n* lines contain *m* characters each. The character *a**i*,<=*j* is equal to '1' if the *i*-th switch turns on the *j*-th lamp and '0' otherwise.
It is guaranteed that if you press all *n* switches all *m* lamps will be turned on. | Print "YES" if there is a switch that if you will ignore it and press all the other *n*<=-<=1 switches then all *m* lamps will be turned on. Print "NO" if there is no such switch. | [
"4 5\n10101\n01000\n00111\n10000\n",
"4 5\n10100\n01000\n00110\n00101\n"
] | [
"YES\n",
"NO\n"
] | none | 0 | [
{
"input": "4 5\n10101\n01000\n00111\n10000",
"output": "YES"
},
{
"input": "4 5\n10100\n01000\n00110\n00101",
"output": "NO"
},
{
"input": "1 5\n11111",
"output": "NO"
},
{
"input": "10 1\n1\n0\n0\n0\n0\n0\n0\n0\n0\n1",
"output": "YES"
},
{
"input": "1 1\n1",
... | 1,527,306,286 | 2,986 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 4 | 3,000 | 4,608,000 | n_m = list(map(int, input().split()))
n = n_m[0] # num of switches
m = n_m[1] # num of lamps
lamps = [0] * m # how many switches for this lamp
s = []
for i in range(n):
si = input()
s.append(si)
for j in range(m):
lamps[j] += int(si[j])
fine = [True]*n
for i in range(n):
si = s[i]
for j in range(m):
if si[j] == '1' and lamps[j] == 1:
fine[i] = False
ans = False
for f in fine:
ans = ans or f
if ans:
print("YES")
else:
print("NO")
| Title: Switches and Lamps
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given *n* switches and *m* lamps. The *i*-th switch turns on some subset of the lamps. This information is given as the matrix *a* consisting of *n* rows and *m* columns where *a**i*,<=*j*<==<=1 if the *i*-th switch turns on the *j*-th lamp and *a**i*,<=*j*<==<=0 if the *i*-th switch is not connected to the *j*-th lamp.
Initially all *m* lamps are turned off.
Switches change state only from "off" to "on". It means that if you press two or more switches connected to the same lamp then the lamp will be turned on after any of this switches is pressed and will remain its state even if any switch connected to this lamp is pressed afterwards.
It is guaranteed that if you push all *n* switches then all *m* lamps will be turned on.
Your think that you have too many switches and you would like to ignore one of them.
Your task is to say if there exists such a switch that if you will ignore (not use) it but press all the other *n*<=-<=1 switches then all the *m* lamps will be turned on.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=2000) — the number of the switches and the number of the lamps.
The following *n* lines contain *m* characters each. The character *a**i*,<=*j* is equal to '1' if the *i*-th switch turns on the *j*-th lamp and '0' otherwise.
It is guaranteed that if you press all *n* switches all *m* lamps will be turned on.
Output Specification:
Print "YES" if there is a switch that if you will ignore it and press all the other *n*<=-<=1 switches then all *m* lamps will be turned on. Print "NO" if there is no such switch.
Demo Input:
['4 5\n10101\n01000\n00111\n10000\n', '4 5\n10100\n01000\n00110\n00101\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
n_m = list(map(int, input().split()))
n = n_m[0] # num of switches
m = n_m[1] # num of lamps
lamps = [0] * m # how many switches for this lamp
s = []
for i in range(n):
si = input()
s.append(si)
for j in range(m):
lamps[j] += int(si[j])
fine = [True]*n
for i in range(n):
si = s[i]
for j in range(m):
if si[j] == '1' and lamps[j] == 1:
fine[i] = False
ans = False
for f in fine:
ans = ans or f
if ans:
print("YES")
else:
print("NO")
``` | 0 | |
166 | E | Tetrahedron | PROGRAMMING | 1,500 | [
"dp",
"math",
"matrices"
] | null | null | You are given a tetrahedron. Let's mark its vertices with letters *A*, *B*, *C* and *D* correspondingly.
An ant is standing in the vertex *D* of the tetrahedron. The ant is quite active and he wouldn't stay idle. At each moment of time he makes a step from one vertex to another one along some edge of the tetrahedron. The ant just can't stand on one place.
You do not have to do much to solve the problem: your task is to count the number of ways in which the ant can go from the initial vertex *D* to itself in exactly *n* steps. In other words, you are asked to find out the number of different cyclic paths with the length of *n* from vertex *D* to itself. As the number can be quite large, you should print it modulo 1000000007 (109<=+<=7). | The first line contains the only integer *n* (1<=≤<=*n*<=≤<=107) — the required length of the cyclic path. | Print the only integer — the required number of ways modulo 1000000007 (109<=+<=7). | [
"2\n",
"4\n"
] | [
"3\n",
"21\n"
] | The required paths in the first sample are:
- *D* - *A* - *D* - *D* - *B* - *D* - *D* - *C* - *D* | 1,000 | [
{
"input": "2",
"output": "3"
},
{
"input": "4",
"output": "21"
},
{
"input": "1",
"output": "0"
},
{
"input": "3",
"output": "6"
},
{
"input": "5",
"output": "60"
},
{
"input": "6",
"output": "183"
},
{
"input": "7",
"output": "546"
... | 1,633,737,797 | 2,147,483,647 | Python 3 | OK | TESTS | 36 | 154 | 6,758,400 |
n=int(input())
print((pow(3,n,4*10**9+28)+3*(-1)**n)//4)
| Title: Tetrahedron
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a tetrahedron. Let's mark its vertices with letters *A*, *B*, *C* and *D* correspondingly.
An ant is standing in the vertex *D* of the tetrahedron. The ant is quite active and he wouldn't stay idle. At each moment of time he makes a step from one vertex to another one along some edge of the tetrahedron. The ant just can't stand on one place.
You do not have to do much to solve the problem: your task is to count the number of ways in which the ant can go from the initial vertex *D* to itself in exactly *n* steps. In other words, you are asked to find out the number of different cyclic paths with the length of *n* from vertex *D* to itself. As the number can be quite large, you should print it modulo 1000000007 (109<=+<=7).
Input Specification:
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=107) — the required length of the cyclic path.
Output Specification:
Print the only integer — the required number of ways modulo 1000000007 (109<=+<=7).
Demo Input:
['2\n', '4\n']
Demo Output:
['3\n', '21\n']
Note:
The required paths in the first sample are:
- *D* - *A* - *D* - *D* - *B* - *D* - *D* - *C* - *D* | ```python
n=int(input())
print((pow(3,n,4*10**9+28)+3*(-1)**n)//4)
``` | 3 | |
676 | C | Vasya and String | PROGRAMMING | 1,500 | [
"binary search",
"dp",
"strings",
"two pointers"
] | null | null | High school student Vasya got a string of length *n* as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.
Vasya can change no more than *k* characters of the original string. What is the maximum beauty of the string he can achieve? | The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100<=000,<=0<=≤<=*k*<=≤<=*n*) — the length of the string and the maximum number of characters to change.
The second line contains the string, consisting of letters 'a' and 'b' only. | Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than *k* characters. | [
"4 2\nabba\n",
"8 1\naabaabaa\n"
] | [
"4\n",
"5\n"
] | In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".
In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa". | 1,500 | [
{
"input": "4 2\nabba",
"output": "4"
},
{
"input": "8 1\naabaabaa",
"output": "5"
},
{
"input": "1 0\na",
"output": "1"
},
{
"input": "1 1\nb",
"output": "1"
},
{
"input": "1 0\nb",
"output": "1"
},
{
"input": "1 1\na",
"output": "1"
},
{
... | 1,698,041,290 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 117 | 77 | 3,276,800 | n,k=map(int,input().split())
s=input()
def attempt(c):
j=0
u=k
best=0
for i in range(n):
while j<n and u-(s[j]!=c)>=0:
u-=s[j]!=c
j+=1
best=max(best,j-i)
u+=s[i]!=c
return best
ans=max(attempt('a'),attempt('b'))
print(ans) | Title: Vasya and String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
High school student Vasya got a string of length *n* as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.
Vasya can change no more than *k* characters of the original string. What is the maximum beauty of the string he can achieve?
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100<=000,<=0<=≤<=*k*<=≤<=*n*) — the length of the string and the maximum number of characters to change.
The second line contains the string, consisting of letters 'a' and 'b' only.
Output Specification:
Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than *k* characters.
Demo Input:
['4 2\nabba\n', '8 1\naabaabaa\n']
Demo Output:
['4\n', '5\n']
Note:
In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".
In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa". | ```python
n,k=map(int,input().split())
s=input()
def attempt(c):
j=0
u=k
best=0
for i in range(n):
while j<n and u-(s[j]!=c)>=0:
u-=s[j]!=c
j+=1
best=max(best,j-i)
u+=s[i]!=c
return best
ans=max(attempt('a'),attempt('b'))
print(ans)
``` | 3 | |
305 | C | Ivan and Powers of Two | PROGRAMMING | 1,600 | [
"greedy",
"implementation"
] | null | null | Ivan has got an array of *n* non-negative integers *a*1,<=*a*2,<=...,<=*a**n*. Ivan knows that the array is sorted in the non-decreasing order.
Ivan wrote out integers 2*a*1,<=2*a*2,<=...,<=2*a**n* on a piece of paper. Now he wonders, what minimum number of integers of form 2*b* (*b*<=≥<=0) need to be added to the piece of paper so that the sum of all integers written on the paper equalled 2*v*<=-<=1 for some integer *v* (*v*<=≥<=0).
Help Ivan, find the required quantity of numbers. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The second input line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=2·109). It is guaranteed that *a*1<=≤<=*a*2<=≤<=...<=≤<=*a**n*. | Print a single integer — the answer to the problem. | [
"4\n0 1 1 1\n",
"1\n3\n"
] | [
"0\n",
"3\n"
] | In the first sample you do not need to add anything, the sum of numbers already equals 2<sup class="upper-index">3</sup> - 1 = 7.
In the second sample you need to add numbers 2<sup class="upper-index">0</sup>, 2<sup class="upper-index">1</sup>, 2<sup class="upper-index">2</sup>. | 1,500 | [
{
"input": "4\n0 1 1 1",
"output": "0"
},
{
"input": "1\n3",
"output": "3"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n2000000000",
"output": "2000000000"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "26\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ... | 1,674,159,228 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 6 | 109 | 18,636,800 | from collections import defaultdict
n = int(input())
a = list(map(int, input().split()))
d = defaultdict(int)
for x in a:
d[x] += 1
while d[x] > 1:
d[x+1] += 1
d[x] -= 2
mx = max(a)
for x in range(mx, mx+30):
if d[x] == 0:
mx = x
break
while d[x] > 1:
d[x+1] += 1
d[x] -= 2
ct = 0
for x in d:
if d[x] == 0:
ct += 1
print(ct + mx - len(d)) | Title: Ivan and Powers of Two
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ivan has got an array of *n* non-negative integers *a*1,<=*a*2,<=...,<=*a**n*. Ivan knows that the array is sorted in the non-decreasing order.
Ivan wrote out integers 2*a*1,<=2*a*2,<=...,<=2*a**n* on a piece of paper. Now he wonders, what minimum number of integers of form 2*b* (*b*<=≥<=0) need to be added to the piece of paper so that the sum of all integers written on the paper equalled 2*v*<=-<=1 for some integer *v* (*v*<=≥<=0).
Help Ivan, find the required quantity of numbers.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The second input line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=2·109). It is guaranteed that *a*1<=≤<=*a*2<=≤<=...<=≤<=*a**n*.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['4\n0 1 1 1\n', '1\n3\n']
Demo Output:
['0\n', '3\n']
Note:
In the first sample you do not need to add anything, the sum of numbers already equals 2<sup class="upper-index">3</sup> - 1 = 7.
In the second sample you need to add numbers 2<sup class="upper-index">0</sup>, 2<sup class="upper-index">1</sup>, 2<sup class="upper-index">2</sup>. | ```python
from collections import defaultdict
n = int(input())
a = list(map(int, input().split()))
d = defaultdict(int)
for x in a:
d[x] += 1
while d[x] > 1:
d[x+1] += 1
d[x] -= 2
mx = max(a)
for x in range(mx, mx+30):
if d[x] == 0:
mx = x
break
while d[x] > 1:
d[x+1] += 1
d[x] -= 2
ct = 0
for x in d:
if d[x] == 0:
ct += 1
print(ct + mx - len(d))
``` | 0 | |
397 | A | On Segment's Own Points | PROGRAMMING | 0 | [
"implementation"
] | null | null | Our old friend Alexey has finally entered the University of City N — the Berland capital. Alexey expected his father to get him a place to live in but his father said it was high time for Alexey to practice some financial independence. So, Alexey is living in a dorm.
The dorm has exactly one straight dryer — a 100 centimeter long rope to hang clothes on. The dryer has got a coordinate system installed: the leftmost end of the dryer has coordinate 0, and the opposite end has coordinate 100. Overall, the university has *n* students. Dean's office allows *i*-th student to use the segment (*l**i*,<=*r**i*) of the dryer. However, the dean's office actions are contradictory and now one part of the dryer can belong to multiple students!
Alexey don't like when someone touch his clothes. That's why he want make it impossible to someone clothes touch his ones. So Alexey wonders: what is the total length of the parts of the dryer that he may use in a such way that clothes of the others (*n*<=-<=1) students aren't drying there. Help him! Note that Alexey, as the most respected student, has number 1. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100). The (*i*<=+<=1)-th line contains integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=100) — the endpoints of the corresponding segment for the *i*-th student. | On a single line print a single number *k*, equal to the sum of lengths of the parts of the dryer which are inside Alexey's segment and are outside all other segments. | [
"3\n0 5\n2 8\n1 6\n",
"3\n0 10\n1 5\n7 15\n"
] | [
"1\n",
"3\n"
] | Note that it's not important are clothes drying on the touching segments (e.g. (0, 1) and (1, 2)) considered to be touching or not because you need to find the length of segments.
In the first test sample Alexey may use the only segment (0, 1). In such case his clothes will not touch clothes on the segments (1, 6) and (2, 8). The length of segment (0, 1) is 1.
In the second test sample Alexey may dry his clothes on segments (0, 1) and (5, 7). Overall length of these segments is 3. | 500 | [
{
"input": "3\n0 5\n2 8\n1 6",
"output": "1"
},
{
"input": "3\n0 10\n1 5\n7 15",
"output": "3"
},
{
"input": "1\n0 100",
"output": "100"
},
{
"input": "2\n1 9\n1 9",
"output": "0"
},
{
"input": "2\n1 9\n5 10",
"output": "4"
},
{
"input": "2\n1 9\n3 5",... | 1,642,829,153 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 31 | 0 | n =int(input())
inp=[]
for i in range(0,n):
inp.append(list(map(int,input().split(' '))))
initiated=0
output=0
for i in range(inp[0][0],int(inp[0][1])+1):
for j in range(1,n):
if (inp[j][0]<i and inp[j][1]>=i):
pass
else:
output+=1
print(int(output/5)) | Title: On Segment's Own Points
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Our old friend Alexey has finally entered the University of City N — the Berland capital. Alexey expected his father to get him a place to live in but his father said it was high time for Alexey to practice some financial independence. So, Alexey is living in a dorm.
The dorm has exactly one straight dryer — a 100 centimeter long rope to hang clothes on. The dryer has got a coordinate system installed: the leftmost end of the dryer has coordinate 0, and the opposite end has coordinate 100. Overall, the university has *n* students. Dean's office allows *i*-th student to use the segment (*l**i*,<=*r**i*) of the dryer. However, the dean's office actions are contradictory and now one part of the dryer can belong to multiple students!
Alexey don't like when someone touch his clothes. That's why he want make it impossible to someone clothes touch his ones. So Alexey wonders: what is the total length of the parts of the dryer that he may use in a such way that clothes of the others (*n*<=-<=1) students aren't drying there. Help him! Note that Alexey, as the most respected student, has number 1.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100). The (*i*<=+<=1)-th line contains integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=100) — the endpoints of the corresponding segment for the *i*-th student.
Output Specification:
On a single line print a single number *k*, equal to the sum of lengths of the parts of the dryer which are inside Alexey's segment and are outside all other segments.
Demo Input:
['3\n0 5\n2 8\n1 6\n', '3\n0 10\n1 5\n7 15\n']
Demo Output:
['1\n', '3\n']
Note:
Note that it's not important are clothes drying on the touching segments (e.g. (0, 1) and (1, 2)) considered to be touching or not because you need to find the length of segments.
In the first test sample Alexey may use the only segment (0, 1). In such case his clothes will not touch clothes on the segments (1, 6) and (2, 8). The length of segment (0, 1) is 1.
In the second test sample Alexey may dry his clothes on segments (0, 1) and (5, 7). Overall length of these segments is 3. | ```python
n =int(input())
inp=[]
for i in range(0,n):
inp.append(list(map(int,input().split(' '))))
initiated=0
output=0
for i in range(inp[0][0],int(inp[0][1])+1):
for j in range(1,n):
if (inp[j][0]<i and inp[j][1]>=i):
pass
else:
output+=1
print(int(output/5))
``` | 0 | |
624 | B | Making a String | PROGRAMMING | 1,100 | [
"greedy",
"sortings"
] | null | null | You are given an alphabet consisting of *n* letters, your task is to make a string of the maximum possible length so that the following conditions are satisfied:
- the *i*-th letter occurs in the string no more than *a**i* times; - the number of occurrences of each letter in the string must be distinct for all the letters that occurred in the string at least once. | The first line of the input contains a single integer *n* (2<=<=≤<=<=*n*<=<=≤<=<=26) — the number of letters in the alphabet.
The next line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — *i*-th of these integers gives the limitation on the number of occurrences of the *i*-th character in the string. | Print a single integer — the maximum length of the string that meets all the requirements. | [
"3\n2 5 5\n",
"3\n1 1 2\n"
] | [
"11\n",
"3\n"
] | For convenience let's consider an alphabet consisting of three letters: "a", "b", "c". In the first sample, some of the optimal strings are: "cccaabbccbb", "aabcbcbcbcb". In the second sample some of the optimal strings are: "acc", "cbc". | 1,000 | [
{
"input": "3\n2 5 5",
"output": "11"
},
{
"input": "3\n1 1 2",
"output": "3"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "3\n1 1000000000 2",
"output": "1000000003"
},
{
"input": "26\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 10000... | 1,455,720,845 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | n = int(input())
a = [int(i) for i in input().split()]
a.sort(reverse=True)
b = set()
for i in range(n):
sz = len(b)
while len(b) == sz:
b.add(i)
i -= 1
print(sum(list(b))) | Title: Making a String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an alphabet consisting of *n* letters, your task is to make a string of the maximum possible length so that the following conditions are satisfied:
- the *i*-th letter occurs in the string no more than *a**i* times; - the number of occurrences of each letter in the string must be distinct for all the letters that occurred in the string at least once.
Input Specification:
The first line of the input contains a single integer *n* (2<=<=≤<=<=*n*<=<=≤<=<=26) — the number of letters in the alphabet.
The next line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — *i*-th of these integers gives the limitation on the number of occurrences of the *i*-th character in the string.
Output Specification:
Print a single integer — the maximum length of the string that meets all the requirements.
Demo Input:
['3\n2 5 5\n', '3\n1 1 2\n']
Demo Output:
['11\n', '3\n']
Note:
For convenience let's consider an alphabet consisting of three letters: "a", "b", "c". In the first sample, some of the optimal strings are: "cccaabbccbb", "aabcbcbcbcb". In the second sample some of the optimal strings are: "acc", "cbc". | ```python
n = int(input())
a = [int(i) for i in input().split()]
a.sort(reverse=True)
b = set()
for i in range(n):
sz = len(b)
while len(b) == sz:
b.add(i)
i -= 1
print(sum(list(b)))
``` | 0 | |
479 | A | Expression | PROGRAMMING | 1,000 | [
"brute force",
"math"
] | null | null | Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get. | The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10). | Print the maximum value of the expression that you can obtain. | [
"1\n2\n3\n",
"2\n10\n3\n"
] | [
"9\n",
"60\n"
] | none | 500 | [
{
"input": "1\n2\n3",
"output": "9"
},
{
"input": "2\n10\n3",
"output": "60"
},
{
"input": "1\n1\n1",
"output": "3"
},
{
"input": "1\n2\n1",
"output": "4"
},
{
"input": "10\n10\n10",
"output": "1000"
},
{
"input": "5\n1\n3",
"output": "20"
},
{... | 1,687,011,807 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 31 | 0 | l = []
for i in range(3) :
l.append(int(input()))
l.sort()
if l[0] == 0 or l[0] == 1 :
print((l[0]+l[1])*l[-1])
else :
print((l[0]*l[1])*l[-1])
| Title: Expression
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get.
Input Specification:
The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10).
Output Specification:
Print the maximum value of the expression that you can obtain.
Demo Input:
['1\n2\n3\n', '2\n10\n3\n']
Demo Output:
['9\n', '60\n']
Note:
none | ```python
l = []
for i in range(3) :
l.append(int(input()))
l.sort()
if l[0] == 0 or l[0] == 1 :
print((l[0]+l[1])*l[-1])
else :
print((l[0]*l[1])*l[-1])
``` | 0 | |
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":... | 1,688,059,235 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | n = int(input())
word_count = 0
for i in range(n):
word = input()
for j in word:
word_count += 1
if word_count >= 11:
print(word[0]+str(word_count-2)+word[-1])
else:
print(word) | 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())
word_count = 0
for i in range(n):
word = input()
for j in word:
word_count += 1
if word_count >= 11:
print(word[0]+str(word_count-2)+word[-1])
else:
print(word)
``` | 0 |
124 | A | The number of positions | PROGRAMMING | 1,000 | [
"math"
] | null | null | Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy. | The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100). | Print the single number — the number of the sought positions. | [
"3 1 1\n",
"5 2 3\n"
] | [
"2\n",
"3\n"
] | The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5. | 500 | [
{
"input": "3 1 1",
"output": "2"
},
{
"input": "5 2 3",
"output": "3"
},
{
"input": "5 4 0",
"output": "1"
},
{
"input": "6 5 5",
"output": "1"
},
{
"input": "9 4 3",
"output": "4"
},
{
"input": "11 4 6",
"output": "7"
},
{
"input": "13 8 ... | 1,650,334,888 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 124 | 0 | t = list(map(int,input().split()))
ans = t[0] - t[1]
print(ans)
| Title: The number of positions
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petr stands in line of *n* people, but he doesn't know exactly which position he occupies. He can say that there are no less than *a* people standing in front of him and no more than *b* people standing behind him. Find the number of different positions Petr can occupy.
Input Specification:
The only line contains three integers *n*, *a* and *b* (0<=≤<=*a*,<=*b*<=<<=*n*<=≤<=100).
Output Specification:
Print the single number — the number of the sought positions.
Demo Input:
['3 1 1\n', '5 2 3\n']
Demo Output:
['2\n', '3\n']
Note:
The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).
In the second sample they are 3, 4 and 5. | ```python
t = list(map(int,input().split()))
ans = t[0] - t[1]
print(ans)
``` | 0 | |
976 | B | Lara Croft and the New Game | PROGRAMMING | 1,300 | [
"implementation",
"math"
] | null | null | You might have heard about the next game in Lara Croft series coming out this year. You also might have watched its trailer. Though you definitely missed the main idea about its plot, so let me lift the veil of secrecy.
Lara is going to explore yet another dangerous dungeon. Game designers decided to use good old 2D environment. The dungeon can be represented as a rectangle matrix of *n* rows and *m* columns. Cell (*x*,<=*y*) is the cell in the *x*-th row in the *y*-th column. Lara can move between the neighbouring by side cells in all four directions.
Moreover, she has even chosen the path for herself to avoid all the traps. She enters the dungeon in cell (1,<=1), that is top left corner of the matrix. Then she goes down all the way to cell (*n*,<=1) — the bottom left corner. Then she starts moving in the snake fashion — all the way to the right, one cell up, then to the left to the cell in 2-nd column, one cell up. She moves until she runs out of non-visited cells. *n* and *m* given are such that she always end up in cell (1,<=2).
Lara has already moved to a neighbouring cell *k* times. Can you determine her current position? | The only line contains three integers *n*, *m* and *k* (2<=≤<=*n*,<=*m*<=≤<=109, *n* is always even, 0<=≤<=*k*<=<<=*n*·*m*). Note that *k* doesn't fit into 32-bit integer type! | Print the cell (the row and the column where the cell is situated) where Lara ends up after she moves *k* times. | [
"4 3 0\n",
"4 3 11\n",
"4 3 7\n"
] | [
"1 1\n",
"1 2\n",
"3 2\n"
] | Here is her path on matrix 4 by 3: | 0 | [
{
"input": "4 3 0",
"output": "1 1"
},
{
"input": "4 3 11",
"output": "1 2"
},
{
"input": "4 3 7",
"output": "3 2"
},
{
"input": "1000000000 2 1999999999",
"output": "1 2"
},
{
"input": "1000000000 1000000000 999999999999999999",
"output": "1 2"
},
{
"... | 1,615,817,605 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 77 | 0 | n,m,k = map(int,input().split())
k+=1
if k==n*m:
print(1,2)
elif k<=n:
print(1,k)
else:
k-=n
if (k//2)%2==1:
print(n-((m-1)//k),2+(m-1)%k)
else:
print(n-((m-1)//k),n-(m-1)%k)
| Title: Lara Croft and the New Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You might have heard about the next game in Lara Croft series coming out this year. You also might have watched its trailer. Though you definitely missed the main idea about its plot, so let me lift the veil of secrecy.
Lara is going to explore yet another dangerous dungeon. Game designers decided to use good old 2D environment. The dungeon can be represented as a rectangle matrix of *n* rows and *m* columns. Cell (*x*,<=*y*) is the cell in the *x*-th row in the *y*-th column. Lara can move between the neighbouring by side cells in all four directions.
Moreover, she has even chosen the path for herself to avoid all the traps. She enters the dungeon in cell (1,<=1), that is top left corner of the matrix. Then she goes down all the way to cell (*n*,<=1) — the bottom left corner. Then she starts moving in the snake fashion — all the way to the right, one cell up, then to the left to the cell in 2-nd column, one cell up. She moves until she runs out of non-visited cells. *n* and *m* given are such that she always end up in cell (1,<=2).
Lara has already moved to a neighbouring cell *k* times. Can you determine her current position?
Input Specification:
The only line contains three integers *n*, *m* and *k* (2<=≤<=*n*,<=*m*<=≤<=109, *n* is always even, 0<=≤<=*k*<=<<=*n*·*m*). Note that *k* doesn't fit into 32-bit integer type!
Output Specification:
Print the cell (the row and the column where the cell is situated) where Lara ends up after she moves *k* times.
Demo Input:
['4 3 0\n', '4 3 11\n', '4 3 7\n']
Demo Output:
['1 1\n', '1 2\n', '3 2\n']
Note:
Here is her path on matrix 4 by 3: | ```python
n,m,k = map(int,input().split())
k+=1
if k==n*m:
print(1,2)
elif k<=n:
print(1,k)
else:
k-=n
if (k//2)%2==1:
print(n-((m-1)//k),2+(m-1)%k)
else:
print(n-((m-1)//k),n-(m-1)%k)
``` | 0 | |
667 | B | Coat of Anticubism | PROGRAMMING | 1,100 | [
"constructive algorithms",
"geometry"
] | null | null | As some of you know, cubism is a trend in art, where the problem of constructing volumetrical shape on a plane with a combination of three-dimensional geometric shapes comes to the fore.
A famous sculptor Cicasso, whose self-portrait you can contemplate, hates cubism. He is more impressed by the idea to transmit two-dimensional objects through three-dimensional objects by using his magnificent sculptures. And his new project is connected with this. Cicasso wants to make a coat for the haters of anticubism. To do this, he wants to create a sculpture depicting a well-known geometric primitive — convex polygon.
Cicasso prepared for this a few blanks, which are rods with integer lengths, and now he wants to bring them together. The *i*-th rod is a segment of length *l**i*.
The sculptor plans to make a convex polygon with a nonzero area, using all rods he has as its sides. Each rod should be used as a side to its full length. It is forbidden to cut, break or bend rods. However, two sides may form a straight angle .
Cicasso knows that it is impossible to make a convex polygon with a nonzero area out of the rods with the lengths which he had chosen. Cicasso does not want to leave the unused rods, so the sculptor decides to make another rod-blank with an integer length so that his problem is solvable. Of course, he wants to make it as short as possible, because the materials are expensive, and it is improper deed to spend money for nothing.
Help sculptor! | The first line contains an integer *n* (3<=≤<=*n*<=≤<=105) — a number of rod-blanks.
The second line contains *n* integers *l**i* (1<=≤<=*l**i*<=≤<=109) — lengths of rods, which Cicasso already has. It is guaranteed that it is impossible to make a polygon with *n* vertices and nonzero area using the rods Cicasso already has. | Print the only integer *z* — the minimum length of the rod, so that after adding it it can be possible to construct convex polygon with (*n*<=+<=1) vertices and nonzero area from all of the rods. | [
"3\n1 2 1\n",
"5\n20 4 3 2 1\n"
] | [
"1\n",
"11\n"
] | In the first example triangle with sides {1 + 1 = 2, 2, 1} can be formed from a set of lengths {1, 1, 1, 2}.
In the second example you can make a triangle with lengths {20, 11, 4 + 3 + 2 + 1 = 10}. | 1,000 | [
{
"input": "3\n1 2 1",
"output": "1"
},
{
"input": "5\n20 4 3 2 1",
"output": "11"
},
{
"input": "7\n77486105 317474713 89523018 332007362 7897847 949616701 54820086",
"output": "70407571"
},
{
"input": "14\n245638694 2941428 4673577 12468 991349408 44735727 14046308 60637707... | 1,461,948,250 | 550 | PyPy 3 | OK | TESTS | 51 | 140 | 29,184,000 | #!/usr/bin/env python3
try:
while True:
n = int(input())
a = list(map(int, input().split()))
m = max(a)
print((m << 1) - sum(a) + 1)
except EOFError:
pass
| Title: Coat of Anticubism
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As some of you know, cubism is a trend in art, where the problem of constructing volumetrical shape on a plane with a combination of three-dimensional geometric shapes comes to the fore.
A famous sculptor Cicasso, whose self-portrait you can contemplate, hates cubism. He is more impressed by the idea to transmit two-dimensional objects through three-dimensional objects by using his magnificent sculptures. And his new project is connected with this. Cicasso wants to make a coat for the haters of anticubism. To do this, he wants to create a sculpture depicting a well-known geometric primitive — convex polygon.
Cicasso prepared for this a few blanks, which are rods with integer lengths, and now he wants to bring them together. The *i*-th rod is a segment of length *l**i*.
The sculptor plans to make a convex polygon with a nonzero area, using all rods he has as its sides. Each rod should be used as a side to its full length. It is forbidden to cut, break or bend rods. However, two sides may form a straight angle .
Cicasso knows that it is impossible to make a convex polygon with a nonzero area out of the rods with the lengths which he had chosen. Cicasso does not want to leave the unused rods, so the sculptor decides to make another rod-blank with an integer length so that his problem is solvable. Of course, he wants to make it as short as possible, because the materials are expensive, and it is improper deed to spend money for nothing.
Help sculptor!
Input Specification:
The first line contains an integer *n* (3<=≤<=*n*<=≤<=105) — a number of rod-blanks.
The second line contains *n* integers *l**i* (1<=≤<=*l**i*<=≤<=109) — lengths of rods, which Cicasso already has. It is guaranteed that it is impossible to make a polygon with *n* vertices and nonzero area using the rods Cicasso already has.
Output Specification:
Print the only integer *z* — the minimum length of the rod, so that after adding it it can be possible to construct convex polygon with (*n*<=+<=1) vertices and nonzero area from all of the rods.
Demo Input:
['3\n1 2 1\n', '5\n20 4 3 2 1\n']
Demo Output:
['1\n', '11\n']
Note:
In the first example triangle with sides {1 + 1 = 2, 2, 1} can be formed from a set of lengths {1, 1, 1, 2}.
In the second example you can make a triangle with lengths {20, 11, 4 + 3 + 2 + 1 = 10}. | ```python
#!/usr/bin/env python3
try:
while True:
n = int(input())
a = list(map(int, input().split()))
m = max(a)
print((m << 1) - sum(a) + 1)
except EOFError:
pass
``` | 3 | |
588 | A | Duff and Meat | PROGRAMMING | 900 | [
"greedy"
] | null | null | Duff is addicted to meat! Malek wants to keep her happy for *n* days. In order to be happy in *i*-th day, she needs to eat exactly *a**i* kilograms of meat.
There is a big shop uptown and Malek wants to buy meat for her from there. In *i*-th day, they sell meat for *p**i* dollars per kilogram. Malek knows all numbers *a*1,<=...,<=*a**n* and *p*1,<=...,<=*p**n*. In each day, he can buy arbitrary amount of meat, also he can keep some meat he has for the future.
Malek is a little tired from cooking meat, so he asked for your help. Help him to minimize the total money he spends to keep Duff happy for *n* days. | The first line of input contains integer *n* (1<=≤<=*n*<=≤<=105), the number of days.
In the next *n* lines, *i*-th line contains two integers *a**i* and *p**i* (1<=≤<=*a**i*,<=*p**i*<=≤<=100), the amount of meat Duff needs and the cost of meat in that day. | Print the minimum money needed to keep Duff happy for *n* days, in one line. | [
"3\n1 3\n2 2\n3 1\n",
"3\n1 3\n2 1\n3 2\n"
] | [
"10\n",
"8\n"
] | In the first sample case: An optimal way would be to buy 1 kg on the first day, 2 kg on the second day and 3 kg on the third day.
In the second sample case: An optimal way would be to buy 1 kg on the first day and 5 kg (needed meat for the second and third day) on the second day. | 750 | [
{
"input": "3\n1 3\n2 2\n3 1",
"output": "10"
},
{
"input": "3\n1 3\n2 1\n3 2",
"output": "8"
},
{
"input": "1\n39 52",
"output": "2028"
},
{
"input": "2\n25 56\n94 17",
"output": "2998"
},
{
"input": "5\n39 21\n95 89\n73 90\n9 55\n85 32",
"output": "6321"
}... | 1,580,905,154 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 327 | 0 | n = int(input())
p = 1000
s = 0
for i in range(n):
x, y = map(int, input().split())
p = min(p, y)
s += x * p
print(s)
| Title: Duff and Meat
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Duff is addicted to meat! Malek wants to keep her happy for *n* days. In order to be happy in *i*-th day, she needs to eat exactly *a**i* kilograms of meat.
There is a big shop uptown and Malek wants to buy meat for her from there. In *i*-th day, they sell meat for *p**i* dollars per kilogram. Malek knows all numbers *a*1,<=...,<=*a**n* and *p*1,<=...,<=*p**n*. In each day, he can buy arbitrary amount of meat, also he can keep some meat he has for the future.
Malek is a little tired from cooking meat, so he asked for your help. Help him to minimize the total money he spends to keep Duff happy for *n* days.
Input Specification:
The first line of input contains integer *n* (1<=≤<=*n*<=≤<=105), the number of days.
In the next *n* lines, *i*-th line contains two integers *a**i* and *p**i* (1<=≤<=*a**i*,<=*p**i*<=≤<=100), the amount of meat Duff needs and the cost of meat in that day.
Output Specification:
Print the minimum money needed to keep Duff happy for *n* days, in one line.
Demo Input:
['3\n1 3\n2 2\n3 1\n', '3\n1 3\n2 1\n3 2\n']
Demo Output:
['10\n', '8\n']
Note:
In the first sample case: An optimal way would be to buy 1 kg on the first day, 2 kg on the second day and 3 kg on the third day.
In the second sample case: An optimal way would be to buy 1 kg on the first day and 5 kg (needed meat for the second and third day) on the second day. | ```python
n = int(input())
p = 1000
s = 0
for i in range(n):
x, y = map(int, input().split())
p = min(p, y)
s += x * p
print(s)
``` | 3 | |
248 | A | Cupboards | PROGRAMMING | 800 | [
"implementation"
] | null | null | One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*. | The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces. | In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs. | [
"5\n0 1\n1 0\n0 1\n1 1\n0 1\n"
] | [
"3\n"
] | none | 500 | [
{
"input": "5\n0 1\n1 0\n0 1\n1 1\n0 1",
"output": "3"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1\n1 1\n1 1",
"output": "1"
},
{
"input": "8\n0 1\n1 0\n0 1\n1 1\n0 1\n1 0\n0 1\n1 0",
"output": "7"
},
{
"input": "8\n1 0\n1 0\n1 0\n0 1\n0 1\n1 ... | 1,665,232,843 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 30 | 0 | cupboards = int(input())
left, right = 0, 0
for _ in range(cupboards):
l, r = [int(x) for x in input().split()]
left += l
right += r
print(min(cupboards + left - right, cupboards - left + right))
| Title: Cupboards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*.
Input Specification:
The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces.
Output Specification:
In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs.
Demo Input:
['5\n0 1\n1 0\n0 1\n1 1\n0 1\n']
Demo Output:
['3\n']
Note:
none | ```python
cupboards = int(input())
left, right = 0, 0
for _ in range(cupboards):
l, r = [int(x) for x in input().split()]
left += l
right += r
print(min(cupboards + left - right, cupboards - left + right))
``` | 0 | |
978 | A | Remove Duplicates | PROGRAMMING | 800 | [
"implementation"
] | null | null | Petya has an array $a$ consisting of $n$ integers. He wants to remove duplicate (equal) elements.
Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed. | The first line contains a single integer $n$ ($1 \le n \le 50$) — the number of elements in Petya's array.
The following line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1\,000$) — the Petya's array. | In the first line print integer $x$ — the number of elements which will be left in Petya's array after he removed the duplicates.
In the second line print $x$ integers separated with a space — Petya's array after he removed the duplicates. For each unique element only the rightmost entry should be left. | [
"6\n1 5 5 1 6 1\n",
"5\n2 4 2 4 4\n",
"5\n6 6 6 6 6\n"
] | [
"3\n5 6 1 \n",
"2\n2 4 \n",
"1\n6 \n"
] | In the first example you should remove two integers $1$, which are in the positions $1$ and $4$. Also you should remove the integer $5$, which is in the position $2$.
In the second example you should remove integer $2$, which is in the position $1$, and two integers $4$, which are in the positions $2$ and $4$.
In the third example you should remove four integers $6$, which are in the positions $1$, $2$, $3$ and $4$. | 0 | [
{
"input": "6\n1 5 5 1 6 1",
"output": "3\n5 6 1 "
},
{
"input": "5\n2 4 2 4 4",
"output": "2\n2 4 "
},
{
"input": "5\n6 6 6 6 6",
"output": "1\n6 "
},
{
"input": "7\n1 2 3 4 2 2 3",
"output": "4\n1 4 2 3 "
},
{
"input": "9\n100 100 100 99 99 99 100 100 100",
... | 1,677,331,973 | 2,147,483,647 | Python 3 | OK | TESTS | 28 | 46 | 0 | n = int(input())
el = input()
arr = list(map(int, el.split()))
arr.reverse()
arr1 = []
for i in arr:
if not i in arr1:
arr1.append(i)
arr1.reverse()
print(len(arr1))
for i in arr1:
print(i,end=' ') | Title: Remove Duplicates
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya has an array $a$ consisting of $n$ integers. He wants to remove duplicate (equal) elements.
Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 50$) — the number of elements in Petya's array.
The following line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1\,000$) — the Petya's array.
Output Specification:
In the first line print integer $x$ — the number of elements which will be left in Petya's array after he removed the duplicates.
In the second line print $x$ integers separated with a space — Petya's array after he removed the duplicates. For each unique element only the rightmost entry should be left.
Demo Input:
['6\n1 5 5 1 6 1\n', '5\n2 4 2 4 4\n', '5\n6 6 6 6 6\n']
Demo Output:
['3\n5 6 1 \n', '2\n2 4 \n', '1\n6 \n']
Note:
In the first example you should remove two integers $1$, which are in the positions $1$ and $4$. Also you should remove the integer $5$, which is in the position $2$.
In the second example you should remove integer $2$, which is in the position $1$, and two integers $4$, which are in the positions $2$ and $4$.
In the third example you should remove four integers $6$, which are in the positions $1$, $2$, $3$ and $4$. | ```python
n = int(input())
el = input()
arr = list(map(int, el.split()))
arr.reverse()
arr1 = []
for i in arr:
if not i in arr1:
arr1.append(i)
arr1.reverse()
print(len(arr1))
for i in arr1:
print(i,end=' ')
``` | 3 | |
870 | C | Maximum splitting | PROGRAMMING | 1,300 | [
"dp",
"greedy",
"math",
"number theory"
] | null | null | You are given several queries. In the *i*-th query you are given a single positive integer *n**i*. You are to represent *n**i* as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings.
An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself. | The first line contains single integer *q* (1<=≤<=*q*<=≤<=105) — the number of queries.
*q* lines follow. The (*i*<=+<=1)-th line contains single integer *n**i* (1<=≤<=*n**i*<=≤<=109) — the *i*-th query. | For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings. | [
"1\n12\n",
"2\n6\n8\n",
"3\n1\n2\n3\n"
] | [
"3\n",
"1\n2\n",
"-1\n-1\n-1\n"
] | 12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands.
8 = 4 + 4, 6 can't be split into several composite summands.
1, 2, 3 are less than any composite number, so they do not have valid splittings. | 1,500 | [
{
"input": "1\n12",
"output": "3"
},
{
"input": "2\n6\n8",
"output": "1\n2"
},
{
"input": "3\n1\n2\n3",
"output": "-1\n-1\n-1"
},
{
"input": "6\n1\n2\n3\n5\n7\n11",
"output": "-1\n-1\n-1\n-1\n-1\n-1"
},
{
"input": "3\n4\n6\n9",
"output": "1\n1\n1"
},
{
... | 1,604,021,036 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 1,028 | 0 | maximum = [-1, -1, -1, -1, 1, -1, 1, -1, 2, 1, 2, -1, 3, 2, 3, 2, 4]
q = int(input())
for _ in range(q):
n = int(input())
base = 12 + n % 4
if n < len(maximum):
print(maximum[n])
continue
print(maximum[base] + (n - base) // 4)
| Title: Maximum splitting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given several queries. In the *i*-th query you are given a single positive integer *n**i*. You are to represent *n**i* as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings.
An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself.
Input Specification:
The first line contains single integer *q* (1<=≤<=*q*<=≤<=105) — the number of queries.
*q* lines follow. The (*i*<=+<=1)-th line contains single integer *n**i* (1<=≤<=*n**i*<=≤<=109) — the *i*-th query.
Output Specification:
For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings.
Demo Input:
['1\n12\n', '2\n6\n8\n', '3\n1\n2\n3\n']
Demo Output:
['3\n', '1\n2\n', '-1\n-1\n-1\n']
Note:
12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands.
8 = 4 + 4, 6 can't be split into several composite summands.
1, 2, 3 are less than any composite number, so they do not have valid splittings. | ```python
maximum = [-1, -1, -1, -1, 1, -1, 1, -1, 2, 1, 2, -1, 3, 2, 3, 2, 4]
q = int(input())
for _ in range(q):
n = int(input())
base = 12 + n % 4
if n < len(maximum):
print(maximum[n])
continue
print(maximum[base] + (n - base) // 4)
``` | 3 | |
230 | A | Dragons | PROGRAMMING | 1,000 | [
"greedy",
"sortings"
] | null | null | Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all *n* dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals *s*.
If Kirito starts duelling with the *i*-th (1<=≤<=*i*<=≤<=*n*) dragon and Kirito's strength is not greater than the dragon's strength *x**i*, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by *y**i*.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss. | The first line contains two space-separated integers *s* and *n* (1<=≤<=*s*<=≤<=104, 1<=≤<=*n*<=≤<=103). Then *n* lines follow: the *i*-th line contains space-separated integers *x**i* and *y**i* (1<=≤<=*x**i*<=≤<=104, 0<=≤<=*y**i*<=≤<=104) — the *i*-th dragon's strength and the bonus for defeating it. | On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't. | [
"2 2\n1 99\n100 0\n",
"10 1\n100 100\n"
] | [
"YES\n",
"NO\n"
] | In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win. | 500 | [
{
"input": "2 2\n1 99\n100 0",
"output": "YES"
},
{
"input": "10 1\n100 100",
"output": "NO"
},
{
"input": "123 2\n78 10\n130 0",
"output": "YES"
},
{
"input": "999 2\n1010 10\n67 89",
"output": "YES"
},
{
"input": "2 5\n5 1\n2 1\n3 1\n1 1\n4 1",
"output": "YE... | 1,698,067,696 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 92 | 0 | s,n = map(int, input().split())
x = []
y = []
for i in range(n):
temp = list(map(int,input().split()))
x.append(temp[0])
y.append(temp[1])
for i in range(len(x)):
if s >= x[i]:
s -= x[i]
s += y[i]
x[i] = 0
if s >= sum(x):
print("YES")
else:
print("NO")
| Title: Dragons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat all *n* dragons that live on this level. Kirito and the dragons have strength, which is represented by an integer. In the duel between two opponents the duel's outcome is determined by their strength. Initially, Kirito's strength equals *s*.
If Kirito starts duelling with the *i*-th (1<=≤<=*i*<=≤<=*n*) dragon and Kirito's strength is not greater than the dragon's strength *x**i*, then Kirito loses the duel and dies. But if Kirito's strength is greater than the dragon's strength, then he defeats the dragon and gets a bonus strength increase by *y**i*.
Kirito can fight the dragons in any order. Determine whether he can move on to the next level of the game, that is, defeat all dragons without a single loss.
Input Specification:
The first line contains two space-separated integers *s* and *n* (1<=≤<=*s*<=≤<=104, 1<=≤<=*n*<=≤<=103). Then *n* lines follow: the *i*-th line contains space-separated integers *x**i* and *y**i* (1<=≤<=*x**i*<=≤<=104, 0<=≤<=*y**i*<=≤<=104) — the *i*-th dragon's strength and the bonus for defeating it.
Output Specification:
On a single line print "YES" (without the quotes), if Kirito can move on to the next level and print "NO" (without the quotes), if he can't.
Demo Input:
['2 2\n1 99\n100 0\n', '10 1\n100 100\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample Kirito's strength initially equals 2. As the first dragon's strength is less than 2, Kirito can fight it and defeat it. After that he gets the bonus and his strength increases to 2 + 99 = 101. Now he can defeat the second dragon and move on to the next level.
In the second sample Kirito's strength is too small to defeat the only dragon and win. | ```python
s,n = map(int, input().split())
x = []
y = []
for i in range(n):
temp = list(map(int,input().split()))
x.append(temp[0])
y.append(temp[1])
for i in range(len(x)):
if s >= x[i]:
s -= x[i]
s += y[i]
x[i] = 0
if s >= sum(x):
print("YES")
else:
print("NO")
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string *s* consisting of *n* lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking! | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the interview.
The second line contains the string *s* of length *n*, consisting of lowercase English letters. | Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences. | [
"7\naogogob\n",
"13\nogogmgogogogo\n",
"9\nogoogoogo\n"
] | [
"a***b\n",
"***gmg***\n",
"*********\n"
] | The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***". | 0 | [
{
"input": "7\naogogob",
"output": "a***b"
},
{
"input": "13\nogogmgogogogo",
"output": "***gmg***"
},
{
"input": "9\nogoogoogo",
"output": "*********"
},
{
"input": "32\nabcdefogoghijklmnogoopqrstuvwxyz",
"output": "abcdef***ghijklmn***opqrstuvwxyz"
},
{
"input":... | 1,482,919,725 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 46 | 4,608,000 | N = input()
str = input()
state = 0
pr = ""
for s in str:
# print (state, s , end=" ", sep="")
if state == 0:
if s == 'o':
state = 1
else:
pr += s
state = 5
elif state == 1:
if s == 'g':
state = 2
else:
pr += ('o'+s)
state = 5
elif state == 2:
if s == 'o':
state = 3
else:
pr += ("og"+s)
state = 5
elif state == 3:
if s == 'g':
state = 4
elif s == 'o':
pr += "***"
state = 1
else:
pr += ("***"+s)
state = 5
elif state == 4:
if s == 'o':
state = 3
else:
pr += ("***g"+s)
state = 5
elif state == 5:
if s == 'o':
state = 1
else:
pr += s
state = 5
if state == 1:
pr += "o"
elif state == 2:
pr += "og"
elif state == 3:
pr += "***"
elif state == 4:
pr += "***g"
print (pr)
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string *s* consisting of *n* lowercase English letters.
There is a filler word ogo in Oleg's speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.
The fillers have maximal size, for example, for ogogoo speech we can't consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.
To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.
Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking!
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the interview.
The second line contains the string *s* of length *n*, consisting of lowercase English letters.
Output Specification:
Print the interview text after the replacement of each of the fillers with "***". It is allowed for the substring "***" to have several consecutive occurences.
Demo Input:
['7\naogogob\n', '13\nogogmgogogogo\n', '9\nogoogoogo\n']
Demo Output:
['a***b\n', '***gmg***\n', '*********\n']
Note:
The first sample contains one filler word ogogo, so the interview for printing is "a***b".
The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to "***gmg***". | ```python
N = input()
str = input()
state = 0
pr = ""
for s in str:
# print (state, s , end=" ", sep="")
if state == 0:
if s == 'o':
state = 1
else:
pr += s
state = 5
elif state == 1:
if s == 'g':
state = 2
else:
pr += ('o'+s)
state = 5
elif state == 2:
if s == 'o':
state = 3
else:
pr += ("og"+s)
state = 5
elif state == 3:
if s == 'g':
state = 4
elif s == 'o':
pr += "***"
state = 1
else:
pr += ("***"+s)
state = 5
elif state == 4:
if s == 'o':
state = 3
else:
pr += ("***g"+s)
state = 5
elif state == 5:
if s == 'o':
state = 1
else:
pr += s
state = 5
if state == 1:
pr += "o"
elif state == 2:
pr += "og"
elif state == 3:
pr += "***"
elif state == 4:
pr += "***g"
print (pr)
``` | 0 | |
478 | B | Random Teams | PROGRAMMING | 1,300 | [
"combinatorics",
"constructive algorithms",
"greedy",
"math"
] | null | null | *n* participants of the competition were split into *m* teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition. | The only line of input contains two integers *n* and *m*, separated by a single space (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the number of participants and the number of teams respectively. | The only line of the output should contain two integers *k**min* and *k**max* — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively. | [
"5 1\n",
"3 2\n",
"6 3\n"
] | [
"10 10\n",
"1 1\n",
"3 6\n"
] | In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people. | 1,000 | [
{
"input": "5 1",
"output": "10 10"
},
{
"input": "3 2",
"output": "1 1"
},
{
"input": "6 3",
"output": "3 6"
},
{
"input": "5 3",
"output": "2 3"
},
{
"input": "10 2",
"output": "20 36"
},
{
"input": "10 6",
"output": "4 10"
},
{
"input": ... | 1,695,443,867 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 46 | 0 | import math
n,x = map(int,input().split())
if x==1:
k = math.comb(n,2)
print(k,k)
else:
k = n-x
if k<x:
m = k
elif k ==x:
m = k
else:
m = x-1
k = k-(x-1)
m+=math.comb(k+1,2)
z = math.comb(k+1,2)
print(m,z) | Title: Random Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
*n* participants of the competition were split into *m* teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input Specification:
The only line of input contains two integers *n* and *m*, separated by a single space (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the number of participants and the number of teams respectively.
Output Specification:
The only line of the output should contain two integers *k**min* and *k**max* — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Demo Input:
['5 1\n', '3 2\n', '6 3\n']
Demo Output:
['10 10\n', '1 1\n', '3 6\n']
Note:
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people. | ```python
import math
n,x = map(int,input().split())
if x==1:
k = math.comb(n,2)
print(k,k)
else:
k = n-x
if k<x:
m = k
elif k ==x:
m = k
else:
m = x-1
k = k-(x-1)
m+=math.comb(k+1,2)
z = math.comb(k+1,2)
print(m,z)
``` | 0 | |
842 | B | Gleb And Pizza | PROGRAMMING | 1,100 | [
"geometry"
] | null | null | Gleb ordered pizza home. When the courier delivered the pizza, he was very upset, because several pieces of sausage lay on the crust, and he does not really like the crust.
The pizza is a circle of radius *r* and center at the origin. Pizza consists of the main part — circle of radius *r*<=-<=*d* with center at the origin, and crust around the main part of the width *d*. Pieces of sausage are also circles. The radius of the *i* -th piece of the sausage is *r**i*, and the center is given as a pair (*x**i*, *y**i*).
Gleb asks you to help determine the number of pieces of sausage caught on the crust. A piece of sausage got on the crust, if it completely lies on the crust. | First string contains two integer numbers *r* and *d* (0<=≤<=*d*<=<<=*r*<=≤<=500) — the radius of pizza and the width of crust.
Next line contains one integer number *n* — the number of pieces of sausage (1<=≤<=*n*<=≤<=105).
Each of next *n* lines contains three integer numbers *x**i*, *y**i* and *r**i* (<=-<=500<=≤<=*x**i*,<=*y**i*<=≤<=500, 0<=≤<=*r**i*<=≤<=500), where *x**i* and *y**i* are coordinates of the center of *i*-th peace of sausage, *r**i* — radius of *i*-th peace of sausage. | Output the number of pieces of sausage that lay on the crust. | [
"8 4\n7\n7 8 1\n-7 3 2\n0 2 1\n0 -2 2\n-3 -3 1\n0 6 2\n5 3 1\n",
"10 8\n4\n0 0 9\n0 0 10\n1 0 1\n1 0 2\n"
] | [
"2\n",
"0\n"
] | Below is a picture explaining the first example. Circles of green color denote pieces of sausage lying on the crust. | 1,000 | [
{
"input": "8 4\n7\n7 8 1\n-7 3 2\n0 2 1\n0 -2 2\n-3 -3 1\n0 6 2\n5 3 1",
"output": "2"
},
{
"input": "10 8\n4\n0 0 9\n0 0 10\n1 0 1\n1 0 2",
"output": "0"
},
{
"input": "1 0\n1\n1 1 0",
"output": "0"
},
{
"input": "3 0\n5\n3 0 0\n0 3 0\n-3 0 0\n0 -3 0\n3 0 1",
"output": ... | 1,648,103,712 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 31 | 0 | # index = list(input())
# rowCount = int(index[0])
# columnCount = int(index[2])
index = list(input())
n = int(input())
for i in range(n):
x = input()
print(2)
# arr = []
# for i in range(rowCount):
# arr.append(input())
# print(arr)
| Title: Gleb And Pizza
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Gleb ordered pizza home. When the courier delivered the pizza, he was very upset, because several pieces of sausage lay on the crust, and he does not really like the crust.
The pizza is a circle of radius *r* and center at the origin. Pizza consists of the main part — circle of radius *r*<=-<=*d* with center at the origin, and crust around the main part of the width *d*. Pieces of sausage are also circles. The radius of the *i* -th piece of the sausage is *r**i*, and the center is given as a pair (*x**i*, *y**i*).
Gleb asks you to help determine the number of pieces of sausage caught on the crust. A piece of sausage got on the crust, if it completely lies on the crust.
Input Specification:
First string contains two integer numbers *r* and *d* (0<=≤<=*d*<=<<=*r*<=≤<=500) — the radius of pizza and the width of crust.
Next line contains one integer number *n* — the number of pieces of sausage (1<=≤<=*n*<=≤<=105).
Each of next *n* lines contains three integer numbers *x**i*, *y**i* and *r**i* (<=-<=500<=≤<=*x**i*,<=*y**i*<=≤<=500, 0<=≤<=*r**i*<=≤<=500), where *x**i* and *y**i* are coordinates of the center of *i*-th peace of sausage, *r**i* — radius of *i*-th peace of sausage.
Output Specification:
Output the number of pieces of sausage that lay on the crust.
Demo Input:
['8 4\n7\n7 8 1\n-7 3 2\n0 2 1\n0 -2 2\n-3 -3 1\n0 6 2\n5 3 1\n', '10 8\n4\n0 0 9\n0 0 10\n1 0 1\n1 0 2\n']
Demo Output:
['2\n', '0\n']
Note:
Below is a picture explaining the first example. Circles of green color denote pieces of sausage lying on the crust. | ```python
# index = list(input())
# rowCount = int(index[0])
# columnCount = int(index[2])
index = list(input())
n = int(input())
for i in range(n):
x = input()
print(2)
# arr = []
# for i in range(rowCount):
# arr.append(input())
# print(arr)
``` | 0 | |
897 | A | Scarborough Fair | PROGRAMMING | 800 | [
"implementation"
] | null | null | Parsley, sage, rosemary and thyme.
Remember me to one who lives there.
He once was the true love of mine.
Willem is taking the girl to the highest building in island No.28, however, neither of them knows how to get there.
Willem asks his friend, Grick for directions, Grick helped them, and gave them a task.
Although the girl wants to help, Willem insists on doing it by himself.
Grick gave Willem a string of length *n*.
Willem needs to do *m* operations, each operation has four parameters *l*,<=*r*,<=*c*1,<=*c*2, which means that all symbols *c*1 in range [*l*,<=*r*] (from *l*-th to *r*-th, including *l* and *r*) are changed into *c*2. String is 1-indexed.
Grick wants to know the final string after all the *m* operations. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
The second line contains a string *s* of length *n*, consisting of lowercase English letters.
Each of the next *m* lines contains four parameters *l*,<=*r*,<=*c*1,<=*c*2 (1<=≤<=*l*<=≤<=*r*<=≤<=*n*, *c*1,<=*c*2 are lowercase English letters), separated by space. | Output string *s* after performing *m* operations described above. | [
"3 1\nioi\n1 1 i n\n",
"5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g\n"
] | [
"noi",
"gaaak"
] | For the second example:
After the first operation, the string is wxxak.
After the second operation, the string is waaak.
After the third operation, the string is gaaak. | 500 | [
{
"input": "3 1\nioi\n1 1 i n",
"output": "noi"
},
{
"input": "5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g",
"output": "gaaak"
},
{
"input": "9 51\nbhfbdcgff\n2 3 b b\n2 8 e f\n3 8 g f\n5 7 d a\n1 5 e b\n3 4 g b\n6 7 c d\n3 6 e g\n3 6 e h\n5 6 a e\n7 9 a c\n4 9 a h\n3 7 c b\n6 9 b g\n1 7 h b\n... | 1,666,012,279 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 15 | 0 | n,m = map(int,input().split())
string = input()
l,r,c1,c2 = map(int,input().split())
string[l:r].replace(c1,c2)
print(string) | Title: Scarborough Fair
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Parsley, sage, rosemary and thyme.
Remember me to one who lives there.
He once was the true love of mine.
Willem is taking the girl to the highest building in island No.28, however, neither of them knows how to get there.
Willem asks his friend, Grick for directions, Grick helped them, and gave them a task.
Although the girl wants to help, Willem insists on doing it by himself.
Grick gave Willem a string of length *n*.
Willem needs to do *m* operations, each operation has four parameters *l*,<=*r*,<=*c*1,<=*c*2, which means that all symbols *c*1 in range [*l*,<=*r*] (from *l*-th to *r*-th, including *l* and *r*) are changed into *c*2. String is 1-indexed.
Grick wants to know the final string after all the *m* operations.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
The second line contains a string *s* of length *n*, consisting of lowercase English letters.
Each of the next *m* lines contains four parameters *l*,<=*r*,<=*c*1,<=*c*2 (1<=≤<=*l*<=≤<=*r*<=≤<=*n*, *c*1,<=*c*2 are lowercase English letters), separated by space.
Output Specification:
Output string *s* after performing *m* operations described above.
Demo Input:
['3 1\nioi\n1 1 i n\n', '5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g\n']
Demo Output:
['noi', 'gaaak']
Note:
For the second example:
After the first operation, the string is wxxak.
After the second operation, the string is waaak.
After the third operation, the string is gaaak. | ```python
n,m = map(int,input().split())
string = input()
l,r,c1,c2 = map(int,input().split())
string[l:r].replace(c1,c2)
print(string)
``` | -1 | |
910 | A | The Way to Home | PROGRAMMING | 800 | [
"dfs and similar",
"dp",
"greedy",
"implementation"
] | null | null | A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*.
For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*.
Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1. | The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump.
The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one. | If the frog can not reach the home, print -1.
In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1. | [
"8 4\n10010101\n",
"4 2\n1001\n",
"8 4\n11100101\n",
"12 3\n101111100101\n"
] | [
"2\n",
"-1\n",
"3\n",
"4\n"
] | In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four).
In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two. | 500 | [
{
"input": "8 4\n10010101",
"output": "2"
},
{
"input": "4 2\n1001",
"output": "-1"
},
{
"input": "8 4\n11100101",
"output": "3"
},
{
"input": "12 3\n101111100101",
"output": "4"
},
{
"input": "5 4\n11011",
"output": "1"
},
{
"input": "5 4\n10001",
... | 1,514,863,307 | 2,147,483,647 | Python 3 | OK | TESTS | 63 | 62 | 5,529,600 | length, jump = input().strip().split()
plot = input()
length, jump = int(length), int(jump)
n_jumps = 0
pos_index = 0
while pos_index != length - 1:
land_index = plot[pos_index: pos_index + jump + 1].rfind('1') + pos_index
if land_index == pos_index:
n_jumps = -1
break
pos_index = land_index
n_jumps += 1
print (n_jumps) | Title: The Way to Home
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*.
For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*.
Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1.
Input Specification:
The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump.
The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one.
Output Specification:
If the frog can not reach the home, print -1.
In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1.
Demo Input:
['8 4\n10010101\n', '4 2\n1001\n', '8 4\n11100101\n', '12 3\n101111100101\n']
Demo Output:
['2\n', '-1\n', '3\n', '4\n']
Note:
In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four).
In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two. | ```python
length, jump = input().strip().split()
plot = input()
length, jump = int(length), int(jump)
n_jumps = 0
pos_index = 0
while pos_index != length - 1:
land_index = plot[pos_index: pos_index + jump + 1].rfind('1') + pos_index
if land_index == pos_index:
n_jumps = -1
break
pos_index = land_index
n_jumps += 1
print (n_jumps)
``` | 3 | |
551 | A | GukiZ and Contest | PROGRAMMING | 800 | [
"brute force",
"implementation",
"sortings"
] | null | null | Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.
In total, *n* students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to *n*. Let's denote the rating of *i*-th student as *a**i*. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.
He thinks that each student will take place equal to . In particular, if student *A* has rating strictly lower then student *B*, *A* will get the strictly better position than *B*, and if two students have equal ratings, they will share the same position.
GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected. | The first line contains integer *n* (1<=≤<=*n*<=≤<=2000), number of GukiZ's students.
The second line contains *n* numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=2000) where *a**i* is the rating of *i*-th student (1<=≤<=*i*<=≤<=*n*). | In a single line, print the position after the end of the contest for each of *n* students in the same order as they appear in the input. | [
"3\n1 3 3\n",
"1\n1\n",
"5\n3 5 3 4 5\n"
] | [
"3 1 1\n",
"1\n",
"4 1 4 3 1\n"
] | In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.
In the second sample, first student is the only one on the contest.
In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3 are the last sharing fourth position. | 500 | [
{
"input": "3\n1 3 3",
"output": "3 1 1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "5\n3 5 3 4 5",
"output": "4 1 4 3 1"
},
{
"input": "7\n1 3 5 4 2 2 1",
"output": "6 3 1 2 4 4 6"
},
{
"input": "11\n5 6 4 2 9 7 6 6 6 6 7",
"output": "9 4 10 11 1 2 4 4... | 1,582,726,771 | 2,147,483,647 | PyPy 3 | OK | TESTS | 36 | 233 | 3,379,200 | def solve(n, a):
s = []
for i in a:
k = sum(1 for j in a if j > i)
s.append(1 + k)
return ' '.join(map(str, s))
def main():
n = int(input())
a = list(map(int, input().split()))
print(solve(n, a))
main()
| Title: GukiZ and Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.
In total, *n* students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to *n*. Let's denote the rating of *i*-th student as *a**i*. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.
He thinks that each student will take place equal to . In particular, if student *A* has rating strictly lower then student *B*, *A* will get the strictly better position than *B*, and if two students have equal ratings, they will share the same position.
GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000), number of GukiZ's students.
The second line contains *n* numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=2000) where *a**i* is the rating of *i*-th student (1<=≤<=*i*<=≤<=*n*).
Output Specification:
In a single line, print the position after the end of the contest for each of *n* students in the same order as they appear in the input.
Demo Input:
['3\n1 3 3\n', '1\n1\n', '5\n3 5 3 4 5\n']
Demo Output:
['3 1 1\n', '1\n', '4 1 4 3 1\n']
Note:
In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.
In the second sample, first student is the only one on the contest.
In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3 are the last sharing fourth position. | ```python
def solve(n, a):
s = []
for i in a:
k = sum(1 for j in a if j > i)
s.append(1 + k)
return ' '.join(map(str, s))
def main():
n = int(input())
a = list(map(int, input().split()))
print(solve(n, a))
main()
``` | 3 | |
465 | B | Inbox (100500) | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Over time, Alexey's mail box got littered with too many letters. Some of them are read, while others are unread.
Alexey's mail program can either show a list of all letters or show the content of a single letter. As soon as the program shows the content of an unread letter, it becomes read letter (if the program shows the content of a read letter nothing happens). In one click he can do any of the following operations:
- Move from the list of letters to the content of any single letter.- Return to the list of letters from single letter viewing mode.- In single letter viewing mode, move to the next or to the previous letter in the list. You cannot move from the first letter to the previous one or from the last letter to the next one.
The program cannot delete the letters from the list or rearrange them.
Alexey wants to read all the unread letters and go watch football. Now he is viewing the list of all letters and for each letter he can see if it is read or unread. What minimum number of operations does Alexey need to perform to read all unread letters? | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of letters in the mailbox.
The second line contains *n* space-separated integers (zeros and ones) — the state of the letter list. The *i*-th number equals either 1, if the *i*-th number is unread, or 0, if the *i*-th letter is read. | Print a single number — the minimum number of operations needed to make all the letters read. | [
"5\n0 1 0 1 0\n",
"5\n1 1 0 0 1\n",
"2\n0 0\n"
] | [
"3\n",
"4\n",
"0\n"
] | In the first sample Alexey needs three operations to cope with the task: open the second letter, move to the third one, move to the fourth one.
In the second sample the action plan: open the first letter, move to the second letter, return to the list, open the fifth letter.
In the third sample all letters are already read. | 1,000 | [
{
"input": "5\n0 1 0 1 0",
"output": "3"
},
{
"input": "5\n1 1 0 0 1",
"output": "4"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "9\n1 0 1 0 1 0 1 0 1",
"output": "9"
},
{
"input": "5\n1 1 1 1 1",
"output": "5"
},
{
"input": "14\n0 0 1 1 1 0 1 ... | 1,520,301,441 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 5,632,000 | # Q3
n = int(input())
a = list(map(int, input().split(" ")))
flag = True
for i in range(n):
if a[i] == 0:
flag = False
continue
if not flag:
flag = True
a[i] += 1
else:
pass
print(sum(a)) | Title: Inbox (100500)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Over time, Alexey's mail box got littered with too many letters. Some of them are read, while others are unread.
Alexey's mail program can either show a list of all letters or show the content of a single letter. As soon as the program shows the content of an unread letter, it becomes read letter (if the program shows the content of a read letter nothing happens). In one click he can do any of the following operations:
- Move from the list of letters to the content of any single letter.- Return to the list of letters from single letter viewing mode.- In single letter viewing mode, move to the next or to the previous letter in the list. You cannot move from the first letter to the previous one or from the last letter to the next one.
The program cannot delete the letters from the list or rearrange them.
Alexey wants to read all the unread letters and go watch football. Now he is viewing the list of all letters and for each letter he can see if it is read or unread. What minimum number of operations does Alexey need to perform to read all unread letters?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of letters in the mailbox.
The second line contains *n* space-separated integers (zeros and ones) — the state of the letter list. The *i*-th number equals either 1, if the *i*-th number is unread, or 0, if the *i*-th letter is read.
Output Specification:
Print a single number — the minimum number of operations needed to make all the letters read.
Demo Input:
['5\n0 1 0 1 0\n', '5\n1 1 0 0 1\n', '2\n0 0\n']
Demo Output:
['3\n', '4\n', '0\n']
Note:
In the first sample Alexey needs three operations to cope with the task: open the second letter, move to the third one, move to the fourth one.
In the second sample the action plan: open the first letter, move to the second letter, return to the list, open the fifth letter.
In the third sample all letters are already read. | ```python
# Q3
n = int(input())
a = list(map(int, input().split(" ")))
flag = True
for i in range(n):
if a[i] == 0:
flag = False
continue
if not flag:
flag = True
a[i] += 1
else:
pass
print(sum(a))
``` | 0 | |
117 | C | Cycle | PROGRAMMING | 2,000 | [
"dfs and similar",
"graphs"
] | null | null | A tournament is a directed graph without self-loops in which every pair of vertexes is connected by exactly one directed edge. That is, for any two vertexes *u* and *v* (*u*<=≠<=*v*) exists either an edge going from *u* to *v*, or an edge from *v* to *u*.
You are given a tournament consisting of *n* vertexes. Your task is to find there a cycle of length three. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=5000). Next *n* lines contain the adjacency matrix *A* of the graph (without spaces). *A**i*,<=*j*<==<=1 if the graph has an edge going from vertex *i* to vertex *j*, otherwise *A**i*,<=*j*<==<=0. *A**i*,<=*j* stands for the *j*-th character in the *i*-th line.
It is guaranteed that the given graph is a tournament, that is, *A**i*,<=*i*<==<=0,<=*A**i*,<=*j*<=≠<=*A**j*,<=*i* (1<=≤<=*i*,<=*j*<=≤<=*n*,<=*i*<=≠<=*j*). | Print three distinct vertexes of the graph *a*1, *a*2, *a*3 (1<=≤<=*a**i*<=≤<=*n*), such that *A**a*1,<=*a*2<==<=*A**a*2,<=*a*3<==<=*A**a*3,<=*a*1<==<=1, or "-1", if a cycle whose length equals three does not exist.
If there are several solutions, print any of them. | [
"5\n00100\n10000\n01001\n11101\n11000\n",
"5\n01111\n00000\n01000\n01100\n01110\n"
] | [
"1 3 2 ",
"-1\n"
] | none | 1,500 | [
{
"input": "5\n00100\n10000\n01001\n11101\n11000",
"output": "1 3 2 "
},
{
"input": "5\n01111\n00000\n01000\n01100\n01110",
"output": "-1"
},
{
"input": "5\n01000\n00101\n10010\n11001\n10100",
"output": "1 2 3 "
},
{
"input": "5\n00110\n10110\n00011\n00000\n11010",
"outpu... | 1,698,404,916 | 2,147,483,647 | Python 3 | OK | TESTS | 71 | 1,682 | 2,252,800 | def f():
p, q, n = [0], [0], int(input())
input()
for i in range(1, n):
t = input()[:i]
if '0' in t:
if '1' in t:
for l, j in enumerate(p):
if t[j] == '1':
for r, j in enumerate(q):
if t[j] == '0':
if l + r == i:
break
return str(p[l] + 1) + ' ' + str(q[r] + 1) + ' ' + str(i + 1)
break
p.insert(l, i)
q.insert(i - l, i)
else:
p.append(i)
q = [i] + q
else:
p = [i] + p
q.append(i)
return -1
print(f())
| Title: Cycle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A tournament is a directed graph without self-loops in which every pair of vertexes is connected by exactly one directed edge. That is, for any two vertexes *u* and *v* (*u*<=≠<=*v*) exists either an edge going from *u* to *v*, or an edge from *v* to *u*.
You are given a tournament consisting of *n* vertexes. Your task is to find there a cycle of length three.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=5000). Next *n* lines contain the adjacency matrix *A* of the graph (without spaces). *A**i*,<=*j*<==<=1 if the graph has an edge going from vertex *i* to vertex *j*, otherwise *A**i*,<=*j*<==<=0. *A**i*,<=*j* stands for the *j*-th character in the *i*-th line.
It is guaranteed that the given graph is a tournament, that is, *A**i*,<=*i*<==<=0,<=*A**i*,<=*j*<=≠<=*A**j*,<=*i* (1<=≤<=*i*,<=*j*<=≤<=*n*,<=*i*<=≠<=*j*).
Output Specification:
Print three distinct vertexes of the graph *a*1, *a*2, *a*3 (1<=≤<=*a**i*<=≤<=*n*), such that *A**a*1,<=*a*2<==<=*A**a*2,<=*a*3<==<=*A**a*3,<=*a*1<==<=1, or "-1", if a cycle whose length equals three does not exist.
If there are several solutions, print any of them.
Demo Input:
['5\n00100\n10000\n01001\n11101\n11000\n', '5\n01111\n00000\n01000\n01100\n01110\n']
Demo Output:
['1 3 2 ', '-1\n']
Note:
none | ```python
def f():
p, q, n = [0], [0], int(input())
input()
for i in range(1, n):
t = input()[:i]
if '0' in t:
if '1' in t:
for l, j in enumerate(p):
if t[j] == '1':
for r, j in enumerate(q):
if t[j] == '0':
if l + r == i:
break
return str(p[l] + 1) + ' ' + str(q[r] + 1) + ' ' + str(i + 1)
break
p.insert(l, i)
q.insert(i - l, i)
else:
p.append(i)
q = [i] + q
else:
p = [i] + p
q.append(i)
return -1
print(f())
``` | 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"
},... | 1,654,785,272 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<string>
#include<cmath>
#include<stack>
#include<queue>
#include<set>
#include<map>
using namespace std;
int main()
{
#ifdef _DEBUG
freopen("input.txt","rt",stdin);
freopen("output.txt","wt",stdout);
#endif
int n;
cin>>n;
vector<string> A(n);
for(int i=0;i<n;i++)
{
cin>>A[i];
}
long long otv=0;
long long u=0;
for(int i=0;i<n;i++)
{
u=0;
for(int j=0;j<n;j++)
{
if(A[i][j]=='C')
u++;
}
otv+=(u*(u-1))/2;
}
for(int j=0;j<n;j++)
{
u=0;
for(int i=0;i<n;i++)
{
if(A[i][j]=='C')
u++;
}
otv+=(u*(u-1))/2;
}
cout<<otv;
} | 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
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<string>
#include<cmath>
#include<stack>
#include<queue>
#include<set>
#include<map>
using namespace std;
int main()
{
#ifdef _DEBUG
freopen("input.txt","rt",stdin);
freopen("output.txt","wt",stdout);
#endif
int n;
cin>>n;
vector<string> A(n);
for(int i=0;i<n;i++)
{
cin>>A[i];
}
long long otv=0;
long long u=0;
for(int i=0;i<n;i++)
{
u=0;
for(int j=0;j<n;j++)
{
if(A[i][j]=='C')
u++;
}
otv+=(u*(u-1))/2;
}
for(int j=0;j<n;j++)
{
u=0;
for(int i=0;i<n;i++)
{
if(A[i][j]=='C')
u++;
}
otv+=(u*(u-1))/2;
}
cout<<otv;
}
``` | -1 | |
614 | A | Link/Cut Tree | PROGRAMMING | 1,500 | [
"brute force",
"implementation"
] | null | null | Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, which is based on Splay trees. Specifically, he is now studying the *expose* procedure.
Unfortunately, Rostislav is unable to understand the definition of this procedure, so he decided to ask programmer Serezha to help him. Serezha agreed to help if Rostislav solves a simple task (and if he doesn't, then why would he need Splay trees anyway?)
Given integers *l*, *r* and *k*, you need to print all powers of number *k* within range from *l* to *r* inclusive. However, Rostislav doesn't want to spent time doing this, as he got interested in playing a network game called Agar with Gleb. Help him! | The first line of the input contains three space-separated integers *l*, *r* and *k* (1<=≤<=*l*<=≤<=*r*<=≤<=1018, 2<=≤<=*k*<=≤<=109). | Print all powers of number *k*, that lie within range from *l* to *r* in the increasing order. If there are no such numbers, print "-1" (without the quotes). | [
"1 10 2\n",
"2 4 5\n"
] | [
"1 2 4 8 ",
"-1"
] | Note to the first sample: numbers 2<sup class="upper-index">0</sup> = 1, 2<sup class="upper-index">1</sup> = 2, 2<sup class="upper-index">2</sup> = 4, 2<sup class="upper-index">3</sup> = 8 lie within the specified range. The number 2<sup class="upper-index">4</sup> = 16 is greater then 10, thus it shouldn't be printed. | 500 | [
{
"input": "1 10 2",
"output": "1 2 4 8 "
},
{
"input": "2 4 5",
"output": "-1"
},
{
"input": "18102 43332383920 28554",
"output": "28554 815330916 "
},
{
"input": "19562 31702689720 17701",
"output": "313325401 "
},
{
"input": "11729 55221128400 313",
"output... | 1,626,839,566 | 2,147,483,647 | Python 3 | OK | TESTS | 44 | 77 | 6,758,400 | (l,r,k)=(int(i) for i in input().split())
num=1
cnt=0
while num<l:
num*=k
while num<=r:
print(num,end=' ')
num*=k
cnt=1
if cnt==0:
print(-1)
| Title: Link/Cut Tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, which is based on Splay trees. Specifically, he is now studying the *expose* procedure.
Unfortunately, Rostislav is unable to understand the definition of this procedure, so he decided to ask programmer Serezha to help him. Serezha agreed to help if Rostislav solves a simple task (and if he doesn't, then why would he need Splay trees anyway?)
Given integers *l*, *r* and *k*, you need to print all powers of number *k* within range from *l* to *r* inclusive. However, Rostislav doesn't want to spent time doing this, as he got interested in playing a network game called Agar with Gleb. Help him!
Input Specification:
The first line of the input contains three space-separated integers *l*, *r* and *k* (1<=≤<=*l*<=≤<=*r*<=≤<=1018, 2<=≤<=*k*<=≤<=109).
Output Specification:
Print all powers of number *k*, that lie within range from *l* to *r* in the increasing order. If there are no such numbers, print "-1" (without the quotes).
Demo Input:
['1 10 2\n', '2 4 5\n']
Demo Output:
['1 2 4 8 ', '-1']
Note:
Note to the first sample: numbers 2<sup class="upper-index">0</sup> = 1, 2<sup class="upper-index">1</sup> = 2, 2<sup class="upper-index">2</sup> = 4, 2<sup class="upper-index">3</sup> = 8 lie within the specified range. The number 2<sup class="upper-index">4</sup> = 16 is greater then 10, thus it shouldn't be printed. | ```python
(l,r,k)=(int(i) for i in input().split())
num=1
cnt=0
while num<l:
num*=k
while num<=r:
print(num,end=' ')
num*=k
cnt=1
if cnt==0:
print(-1)
``` | 3 | |
897 | B | Chtholly's request | PROGRAMMING | 1,300 | [
"brute force"
] | null | null | — I experienced so many great things.
— You gave me memories like dreams... But I have to leave now...
— One last request, can you...
— Help me solve a Codeforces problem?
— ......
— What?
Chtholly has been thinking about a problem for days:
If a number is palindrome and length of its decimal representation without leading zeros is even, we call it a zcy number. A number is palindrome means when written in decimal representation, it contains no leading zeros and reads the same forwards and backwards. For example 12321 and 1221 are palindromes and 123 and 12451 are not. Moreover, 1221 is zcy number and 12321 is not.
Given integers *k* and *p*, calculate the sum of the *k* smallest zcy numbers and output this sum modulo *p*.
Unfortunately, Willem isn't good at solving this kind of problems, so he asks you for help! | The first line contains two integers *k* and *p* (1<=≤<=*k*<=≤<=105,<=1<=≤<=*p*<=≤<=109). | Output single integer — answer to the problem. | [
"2 100\n",
"5 30\n"
] | [
"33\n",
"15\n"
] | In the first example, the smallest zcy number is 11, and the second smallest zcy number is 22.
In the second example, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/68fffad54395f7d920ad0384e07c6215ddc64141.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 1,000 | [
{
"input": "2 100",
"output": "33"
},
{
"input": "5 30",
"output": "15"
},
{
"input": "42147 412393322",
"output": "251637727"
},
{
"input": "77809 868097296",
"output": "440411873"
},
{
"input": "5105 443422097",
"output": "363192634"
},
{
"input": "7... | 1,512,271,570 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 77 | 23,040,000 | import math
k,p = map(int,input().strip().split(' '))
sm = 0
for i in range(1,k+1):
ds = str(i)
vl = i
for j in range(len(ds)-1,-1,-1):
vl =vl*10 + int(ds[j])
print(vl)
sm = (sm + vl)%p
print(sm)
| Title: Chtholly's request
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
— I experienced so many great things.
— You gave me memories like dreams... But I have to leave now...
— One last request, can you...
— Help me solve a Codeforces problem?
— ......
— What?
Chtholly has been thinking about a problem for days:
If a number is palindrome and length of its decimal representation without leading zeros is even, we call it a zcy number. A number is palindrome means when written in decimal representation, it contains no leading zeros and reads the same forwards and backwards. For example 12321 and 1221 are palindromes and 123 and 12451 are not. Moreover, 1221 is zcy number and 12321 is not.
Given integers *k* and *p*, calculate the sum of the *k* smallest zcy numbers and output this sum modulo *p*.
Unfortunately, Willem isn't good at solving this kind of problems, so he asks you for help!
Input Specification:
The first line contains two integers *k* and *p* (1<=≤<=*k*<=≤<=105,<=1<=≤<=*p*<=≤<=109).
Output Specification:
Output single integer — answer to the problem.
Demo Input:
['2 100\n', '5 30\n']
Demo Output:
['33\n', '15\n']
Note:
In the first example, the smallest zcy number is 11, and the second smallest zcy number is 22.
In the second example, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/68fffad54395f7d920ad0384e07c6215ddc64141.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
import math
k,p = map(int,input().strip().split(' '))
sm = 0
for i in range(1,k+1):
ds = str(i)
vl = i
for j in range(len(ds)-1,-1,-1):
vl =vl*10 + int(ds[j])
print(vl)
sm = (sm + vl)%p
print(sm)
``` | 0 | |
519 | B | A and B and Compilation Errors | PROGRAMMING | 1,100 | [
"data structures",
"implementation",
"sortings"
] | null | null | A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected? | The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one. | Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively. | [
"5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n",
"6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n"
] | [
"8\n123\n",
"1\n3\n"
] | In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | 1,000 | [
{
"input": "5\n1 5 8 123 7\n123 7 5 1\n5 1 7",
"output": "8\n123"
},
{
"input": "6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5",
"output": "1\n3"
},
{
"input": "3\n1 2 3\n3 2\n2",
"output": "1\n3"
},
{
"input": "10\n460626451 802090732 277246428 661369649 388684428 784303821 376287098 6... | 1,660,025,569 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 33 | 233 | 39,321,600 | from collections import Counter
n = int(input())
a1, a2, a3 = [int(_) for _ in input().split()], [int(_) for _ in input().split()], [int(_) for _ in input().split()]
print(*list((Counter(a1) - Counter(a2)).elements()))
print(*list((Counter(a2) - Counter(a3)).elements())) | Title: A and B and Compilation Errors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output Specification:
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Demo Input:
['5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n', '6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n']
Demo Output:
['8\n123\n', '1\n3\n']
Note:
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | ```python
from collections import Counter
n = int(input())
a1, a2, a3 = [int(_) for _ in input().split()], [int(_) for _ in input().split()], [int(_) for _ in input().split()]
print(*list((Counter(a1) - Counter(a2)).elements()))
print(*list((Counter(a2) - Counter(a3)).elements()))
``` | 3 | |
552 | B | Vanya and Books | PROGRAMMING | 1,200 | [
"implementation",
"math"
] | null | null | Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the *n* books should be assigned with a number from 1 to *n*. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books. | The first line contains integer *n* (1<=≤<=*n*<=≤<=109) — the number of books in the library. | Print the number of digits needed to number all the books. | [
"13\n",
"4\n"
] | [
"17\n",
"4\n"
] | Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits. | 1,000 | [
{
"input": "13",
"output": "17"
},
{
"input": "4",
"output": "4"
},
{
"input": "100",
"output": "192"
},
{
"input": "99",
"output": "189"
},
{
"input": "1000000000",
"output": "8888888899"
},
{
"input": "1000000",
"output": "5888896"
},
{
"... | 1,647,939,794 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 46 | 0 | # coding=utf-8
n=int(input())
t=n//10+1
if t==1:
print(n)
elif t>1:
print(9+(n-9)*2)
| Title: Vanya and Books
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya got an important task — he should enumerate books in the library and label each book with its number. Each of the *n* books should be assigned with a number from 1 to *n*. Naturally, distinct books should be assigned distinct numbers.
Vanya wants to know how many digits he will have to write down as he labels the books.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=109) — the number of books in the library.
Output Specification:
Print the number of digits needed to number all the books.
Demo Input:
['13\n', '4\n']
Demo Output:
['17\n', '4\n']
Note:
Note to the first test. The books get numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, which totals to 17 digits.
Note to the second sample. The books get numbers 1, 2, 3, 4, which totals to 4 digits. | ```python
# coding=utf-8
n=int(input())
t=n//10+1
if t==1:
print(n)
elif t>1:
print(9+(n-9)*2)
``` | 0 | |
6 | A | Triangle | PROGRAMMING | 900 | [
"brute force",
"geometry"
] | A. Triangle | 2 | 64 | Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him. | The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks. | Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length. | [
"4 2 1 3\n",
"7 2 2 4\n",
"3 5 9 1\n"
] | [
"TRIANGLE\n",
"SEGMENT\n",
"IMPOSSIBLE\n"
] | none | 0 | [
{
"input": "4 2 1 3",
"output": "TRIANGLE"
},
{
"input": "7 2 2 4",
"output": "SEGMENT"
},
{
"input": "3 5 9 1",
"output": "IMPOSSIBLE"
},
{
"input": "3 1 5 1",
"output": "IMPOSSIBLE"
},
{
"input": "10 10 10 10",
"output": "TRIANGLE"
},
{
"input": "11 ... | 1,591,206,089 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 1 | 156 | 0 | n = [int(a) for a in input().split()]
n.sort()
seg=0
tri=0
for i in range(2):
if(n[i]+n[i+1]>n[i+2]):
print("TRIANGLE")
t=1
break
elif(n[i]+n[i+1]==n[i+2]):
seg=1
if t!=1:
if seg==1:
print("SEGMENT")
else:
print("IMPOSSIBLE") | Title: Triangle
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input Specification:
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output Specification:
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Demo Input:
['4 2 1 3\n', '7 2 2 4\n', '3 5 9 1\n']
Demo Output:
['TRIANGLE\n', 'SEGMENT\n', 'IMPOSSIBLE\n']
Note:
none | ```python
n = [int(a) for a in input().split()]
n.sort()
seg=0
tri=0
for i in range(2):
if(n[i]+n[i+1]>n[i+2]):
print("TRIANGLE")
t=1
break
elif(n[i]+n[i+1]==n[i+2]):
seg=1
if t!=1:
if seg==1:
print("SEGMENT")
else:
print("IMPOSSIBLE")
``` | -1 |
305 | B | Continued Fractions | PROGRAMMING | 1,700 | [
"brute force",
"implementation",
"math"
] | null | null | A continued fraction of height *n* is a fraction of form . You are given two rational numbers, one is represented as and the other one is represented as a finite fraction of height *n*. Check if they are equal. | The first line contains two space-separated integers *p*,<=*q* (1<=≤<=*q*<=≤<=*p*<=≤<=1018) — the numerator and the denominator of the first fraction.
The second line contains integer *n* (1<=≤<=*n*<=≤<=90) — the height of the second fraction. The third line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1018) — the continued fraction.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. | Print "YES" if these fractions are equal and "NO" otherwise. | [
"9 4\n2\n2 4\n",
"9 4\n3\n2 3 1\n",
"9 4\n3\n1 2 4\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | In the first sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5ff92f27aebea2560d99ad61202d20bab5ee5390.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/221368c79c05fc0ecad4e5f7a64f30b832fd99f5.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4fb4b411afc0fbad27a1c8fdd08ba88ec3830ef5.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 1,000 | [
{
"input": "9 4\n2\n2 4",
"output": "YES"
},
{
"input": "9 4\n3\n2 3 1",
"output": "YES"
},
{
"input": "9 4\n3\n1 2 4",
"output": "NO"
},
{
"input": "39088169 24157817\n36\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 2",
"output": "YES"
},
{
... | 1,596,204,599 | 2,147,483,647 | PyPy 3 | OK | TESTS | 49 | 310 | 20,172,800 | import math
p,q=map(int,input().split())
n=int(input())
arr=list(map(int,input().split()))
arr=arr[::-1]
while arr:
if len(arr)==1:
if q*arr[-1]==p:
print('YES')
exit(0)
p,q=q,p-arr[-1]*q
g=math.gcd(p,q)
p//=g
q//=g
arr.pop()
print('NO')
| Title: Continued Fractions
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A continued fraction of height *n* is a fraction of form . You are given two rational numbers, one is represented as and the other one is represented as a finite fraction of height *n*. Check if they are equal.
Input Specification:
The first line contains two space-separated integers *p*,<=*q* (1<=≤<=*q*<=≤<=*p*<=≤<=1018) — the numerator and the denominator of the first fraction.
The second line contains integer *n* (1<=≤<=*n*<=≤<=90) — the height of the second fraction. The third line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1018) — the continued fraction.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print "YES" if these fractions are equal and "NO" otherwise.
Demo Input:
['9 4\n2\n2 4\n', '9 4\n3\n2 3 1\n', '9 4\n3\n1 2 4\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5ff92f27aebea2560d99ad61202d20bab5ee5390.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/221368c79c05fc0ecad4e5f7a64f30b832fd99f5.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4fb4b411afc0fbad27a1c8fdd08ba88ec3830ef5.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
import math
p,q=map(int,input().split())
n=int(input())
arr=list(map(int,input().split()))
arr=arr[::-1]
while arr:
if len(arr)==1:
if q*arr[-1]==p:
print('YES')
exit(0)
p,q=q,p-arr[-1]*q
g=math.gcd(p,q)
p//=g
q//=g
arr.pop()
print('NO')
``` | 3 | |
295 | B | Greg and Graph | PROGRAMMING | 1,700 | [
"dp",
"graphs",
"shortest paths"
] | null | null | Greg has a weighed directed graph, consisting of *n* vertices. In this graph any pair of distinct vertices has an edge between them in both directions. Greg loves playing with the graph and now he has invented a new game:
- The game consists of *n* steps. - On the *i*-th step Greg removes vertex number *x**i* from the graph. As Greg removes a vertex, he also removes all the edges that go in and out of this vertex. - Before executing each step, Greg wants to know the sum of lengths of the shortest paths between all pairs of the remaining vertices. The shortest path can go through any remaining vertex. In other words, if we assume that *d*(*i*,<=*v*,<=*u*) is the shortest path between vertices *v* and *u* in the graph that formed before deleting vertex *x**i*, then Greg wants to know the value of the following sum: .
Help Greg, print the value of the required sum before each step. | The first line contains integer *n* (1<=≤<=*n*<=≤<=500) — the number of vertices in the graph.
Next *n* lines contain *n* integers each — the graph adjacency matrix: the *j*-th number in the *i*-th line *a**ij* (1<=≤<=*a**ij*<=≤<=105,<=*a**ii*<==<=0) represents the weight of the edge that goes from vertex *i* to vertex *j*.
The next line contains *n* distinct integers: *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=*n*) — the vertices that Greg deletes. | Print *n* integers — the *i*-th number equals the required sum before the *i*-th step.
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 of the %I64d specifier. | [
"1\n0\n1\n",
"2\n0 5\n4 0\n1 2\n",
"4\n0 3 1 1\n6 0 400 1\n2 4 0 1\n1 1 1 0\n4 1 2 3\n"
] | [
"0 ",
"9 0 ",
"17 23 404 0 "
] | none | 1,000 | [
{
"input": "1\n0\n1",
"output": "0 "
},
{
"input": "2\n0 5\n4 0\n1 2",
"output": "9 0 "
},
{
"input": "4\n0 3 1 1\n6 0 400 1\n2 4 0 1\n1 1 1 0\n4 1 2 3",
"output": "17 23 404 0 "
},
{
"input": "4\n0 57148 51001 13357\n71125 0 98369 67226\n49388 90852 0 66291\n39573 38165 9700... | 1,639,604,287 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 12 | 3,000 | 7,065,600 | from typing import List
def pathes(W, n):
A = [[W[i][j] for j in range(n)] for i in range(n)]
for k in range(n):
for i in range(n):
for j in range(n):
A[i][j] = min(A[i][j], A[i][k] + A[k][j])
return A
def get_graph(N, edges):
graph = dict()
for i in range(1, N + 1):
graph[i] = []
for path in edges:
graph[path[0]].append((path[1], path[2]))
graph[path[1]].append(((path[0], path[2])))
return graph
def slices(g, to_del):
g1 = []
deleted = []
for i in range(len(to_del)):
if i not in deleted:
yield []
deleted.append(i)
def task2(vert_count: int, g: List[List[int]], to_del: List[int]) -> List[int]:
res = []
for d in to_del:
res.append(sum(map(sum, pathes(g, vert_count))))
g[d-1] = [10000000000000]*vert_count
for r in g:
r[d-1] = 10000000000000
return list(map(lambda x: x%10000000000000, res))
c = int(input())
m = []
for i in range(c):
m.append(list(map(int, input().split())))
td = list(map(int, input().split()))
print(*task2(c, m, td))
| Title: Greg and Graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greg has a weighed directed graph, consisting of *n* vertices. In this graph any pair of distinct vertices has an edge between them in both directions. Greg loves playing with the graph and now he has invented a new game:
- The game consists of *n* steps. - On the *i*-th step Greg removes vertex number *x**i* from the graph. As Greg removes a vertex, he also removes all the edges that go in and out of this vertex. - Before executing each step, Greg wants to know the sum of lengths of the shortest paths between all pairs of the remaining vertices. The shortest path can go through any remaining vertex. In other words, if we assume that *d*(*i*,<=*v*,<=*u*) is the shortest path between vertices *v* and *u* in the graph that formed before deleting vertex *x**i*, then Greg wants to know the value of the following sum: .
Help Greg, print the value of the required sum before each step.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=500) — the number of vertices in the graph.
Next *n* lines contain *n* integers each — the graph adjacency matrix: the *j*-th number in the *i*-th line *a**ij* (1<=≤<=*a**ij*<=≤<=105,<=*a**ii*<==<=0) represents the weight of the edge that goes from vertex *i* to vertex *j*.
The next line contains *n* distinct integers: *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=*n*) — the vertices that Greg deletes.
Output Specification:
Print *n* integers — the *i*-th number equals the required sum before the *i*-th step.
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 of the %I64d specifier.
Demo Input:
['1\n0\n1\n', '2\n0 5\n4 0\n1 2\n', '4\n0 3 1 1\n6 0 400 1\n2 4 0 1\n1 1 1 0\n4 1 2 3\n']
Demo Output:
['0 ', '9 0 ', '17 23 404 0 ']
Note:
none | ```python
from typing import List
def pathes(W, n):
A = [[W[i][j] for j in range(n)] for i in range(n)]
for k in range(n):
for i in range(n):
for j in range(n):
A[i][j] = min(A[i][j], A[i][k] + A[k][j])
return A
def get_graph(N, edges):
graph = dict()
for i in range(1, N + 1):
graph[i] = []
for path in edges:
graph[path[0]].append((path[1], path[2]))
graph[path[1]].append(((path[0], path[2])))
return graph
def slices(g, to_del):
g1 = []
deleted = []
for i in range(len(to_del)):
if i not in deleted:
yield []
deleted.append(i)
def task2(vert_count: int, g: List[List[int]], to_del: List[int]) -> List[int]:
res = []
for d in to_del:
res.append(sum(map(sum, pathes(g, vert_count))))
g[d-1] = [10000000000000]*vert_count
for r in g:
r[d-1] = 10000000000000
return list(map(lambda x: x%10000000000000, res))
c = int(input())
m = []
for i in range(c):
m.append(list(map(int, input().split())))
td = list(map(int, input().split()))
print(*task2(c, m, td))
``` | 0 | |
724 | D | Dense Subsequence | PROGRAMMING | 1,900 | [
"data structures",
"greedy",
"strings"
] | null | null | You are given a string *s*, consisting of lowercase English letters, and the integer *m*.
One should choose some symbols from the given string so that any contiguous subsegment of length *m* has at least one selected symbol. Note that here we choose positions of symbols, not the symbols themselves.
Then one uses the chosen symbols to form a new string. All symbols from the chosen position should be used, but we are allowed to rearrange them in any order.
Formally, we choose a subsequence of indices 1<=≤<=*i*1<=<<=*i*2<=<<=...<=<<=*i**t*<=≤<=|*s*|. The selected sequence must meet the following condition: for every *j* such that 1<=≤<=*j*<=≤<=|*s*|<=-<=*m*<=+<=1, there must be at least one selected index that belongs to the segment [*j*,<= *j*<=+<=*m*<=-<=1], i.e. there should exist a *k* from 1 to *t*, such that *j*<=≤<=*i**k*<=≤<=*j*<=+<=*m*<=-<=1.
Then we take any permutation *p* of the selected indices and form a new string *s**i**p*1*s**i**p*2... *s**i**p**t*.
Find the lexicographically smallest string, that can be obtained using this procedure. | The first line of the input contains a single integer *m* (1<=≤<=*m*<=≤<=100<=000).
The second line contains the string *s* consisting of lowercase English letters. It is guaranteed that this string is non-empty and its length doesn't exceed 100<=000. It is also guaranteed that the number *m* doesn't exceed the length of the string *s*. | Print the single line containing the lexicographically smallest string, that can be obtained using the procedure described above. | [
"3\ncbabc\n",
"2\nabcab\n",
"3\nbcabcbaccba\n"
] | [
"a\n",
"aab\n",
"aaabb\n"
] | In the first sample, one can choose the subsequence {3} and form a string "a".
In the second sample, one can choose the subsequence {1, 2, 4} (symbols on this positions are 'a', 'b' and 'a') and rearrange the chosen symbols to form a string "aab". | 1,500 | [
{
"input": "3\ncbabc",
"output": "a"
},
{
"input": "2\nabcab",
"output": "aab"
},
{
"input": "3\nbcabcbaccba",
"output": "aaabb"
},
{
"input": "5\nimmaydobun",
"output": "ab"
},
{
"input": "5\nwjjdqawypvtgrncmqvcsergermprauyevcegjtcrrblkwiugrcjfpjyxngyryxntauxlouv... | 1,511,212,889 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | m = int(input()) - 1
k = input()
r = ''
i = 97
t = [k]
while 1:
q = chr(i)
p = []
for d in t:
for s in d.split(q):
if len(s) > m: p += [s]
if not p: break
r += q * k.count(q)
i += 1
t = p
y = chr(i)
for d in t:
i = 0
for x in d:
if x == y: j = i
if i == m:
r += y
i -= j
i += 1
print(r) | Title: Dense Subsequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string *s*, consisting of lowercase English letters, and the integer *m*.
One should choose some symbols from the given string so that any contiguous subsegment of length *m* has at least one selected symbol. Note that here we choose positions of symbols, not the symbols themselves.
Then one uses the chosen symbols to form a new string. All symbols from the chosen position should be used, but we are allowed to rearrange them in any order.
Formally, we choose a subsequence of indices 1<=≤<=*i*1<=<<=*i*2<=<<=...<=<<=*i**t*<=≤<=|*s*|. The selected sequence must meet the following condition: for every *j* such that 1<=≤<=*j*<=≤<=|*s*|<=-<=*m*<=+<=1, there must be at least one selected index that belongs to the segment [*j*,<= *j*<=+<=*m*<=-<=1], i.e. there should exist a *k* from 1 to *t*, such that *j*<=≤<=*i**k*<=≤<=*j*<=+<=*m*<=-<=1.
Then we take any permutation *p* of the selected indices and form a new string *s**i**p*1*s**i**p*2... *s**i**p**t*.
Find the lexicographically smallest string, that can be obtained using this procedure.
Input Specification:
The first line of the input contains a single integer *m* (1<=≤<=*m*<=≤<=100<=000).
The second line contains the string *s* consisting of lowercase English letters. It is guaranteed that this string is non-empty and its length doesn't exceed 100<=000. It is also guaranteed that the number *m* doesn't exceed the length of the string *s*.
Output Specification:
Print the single line containing the lexicographically smallest string, that can be obtained using the procedure described above.
Demo Input:
['3\ncbabc\n', '2\nabcab\n', '3\nbcabcbaccba\n']
Demo Output:
['a\n', 'aab\n', 'aaabb\n']
Note:
In the first sample, one can choose the subsequence {3} and form a string "a".
In the second sample, one can choose the subsequence {1, 2, 4} (symbols on this positions are 'a', 'b' and 'a') and rearrange the chosen symbols to form a string "aab". | ```python
m = int(input()) - 1
k = input()
r = ''
i = 97
t = [k]
while 1:
q = chr(i)
p = []
for d in t:
for s in d.split(q):
if len(s) > m: p += [s]
if not p: break
r += q * k.count(q)
i += 1
t = p
y = chr(i)
for d in t:
i = 0
for x in d:
if x == y: j = i
if i == m:
r += y
i -= j
i += 1
print(r)
``` | 0 | |
58 | A | Chat room | PROGRAMMING | 1,000 | [
"greedy",
"strings"
] | A. Chat room | 1 | 256 | Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. | The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. | If Vasya managed to say hello, print "YES", otherwise print "NO". | [
"ahhellllloou\n",
"hlelo\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymeda... | 1,491,038,438 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 46 | 5,529,600 | s=input()
n=len(s)
com='hello'
j=0
if n<5:
print('NO')
else:
for i in s:
if com[j]==i:
j+=1
if j>=4:
break;
if j>4:
print('YES')
else:
print('NO')
| Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
s=input()
n=len(s)
com='hello'
j=0
if n<5:
print('NO')
else:
for i in s:
if com[j]==i:
j+=1
if j>=4:
break;
if j>4:
print('YES')
else:
print('NO')
``` | 0 |
845 | A | Chess Tourney | PROGRAMMING | 1,100 | [
"implementation",
"sortings"
] | null | null | Berland annual chess tournament is coming!
Organizers have gathered 2·*n* chess players who should be divided into two teams with *n* people each. The first team is sponsored by BerOil and the second team is sponsored by BerMobile. Obviously, organizers should guarantee the win for the team of BerOil.
Thus, organizers should divide all 2·*n* players into two teams with *n* people each in such a way that the first team always wins.
Every chess player has its rating *r**i*. It is known that chess player with the greater rating always wins the player with the lower rating. If their ratings are equal then any of the players can win.
After teams assignment there will come a drawing to form *n* pairs of opponents: in each pair there is a player from the first team and a player from the second team. Every chess player should be in exactly one pair. Every pair plays once. The drawing is totally random.
Is it possible to divide all 2·*n* players into two teams with *n* people each so that the player from the first team in every pair wins regardless of the results of the drawing? | The first line contains one integer *n* (1<=≤<=*n*<=≤<=100).
The second line contains 2·*n* integers *a*1,<=*a*2,<=... *a*2*n* (1<=≤<=*a**i*<=≤<=1000). | If it's possible to divide all 2·*n* players into two teams with *n* people each so that the player from the first team in every pair wins regardless of the results of the drawing, then print "YES". Otherwise print "NO". | [
"2\n1 3 2 4\n",
"1\n3 3\n"
] | [
"YES\n",
"NO\n"
] | none | 0 | [
{
"input": "2\n1 3 2 4",
"output": "YES"
},
{
"input": "1\n3 3",
"output": "NO"
},
{
"input": "5\n1 1 1 1 2 2 3 3 3 3",
"output": "NO"
},
{
"input": "5\n1 1 1 1 1 2 2 2 2 2",
"output": "YES"
},
{
"input": "10\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000... | 1,564,172,786 | 2,147,483,647 | PyPy 3 | OK | TESTS | 88 | 140 | 0 | n, a = int(input()), list(sorted(map(int, input().split())))
print(('NO', 'YES')[a[n - 1] != a[n]])
| Title: Chess Tourney
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Berland annual chess tournament is coming!
Organizers have gathered 2·*n* chess players who should be divided into two teams with *n* people each. The first team is sponsored by BerOil and the second team is sponsored by BerMobile. Obviously, organizers should guarantee the win for the team of BerOil.
Thus, organizers should divide all 2·*n* players into two teams with *n* people each in such a way that the first team always wins.
Every chess player has its rating *r**i*. It is known that chess player with the greater rating always wins the player with the lower rating. If their ratings are equal then any of the players can win.
After teams assignment there will come a drawing to form *n* pairs of opponents: in each pair there is a player from the first team and a player from the second team. Every chess player should be in exactly one pair. Every pair plays once. The drawing is totally random.
Is it possible to divide all 2·*n* players into two teams with *n* people each so that the player from the first team in every pair wins regardless of the results of the drawing?
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100).
The second line contains 2·*n* integers *a*1,<=*a*2,<=... *a*2*n* (1<=≤<=*a**i*<=≤<=1000).
Output Specification:
If it's possible to divide all 2·*n* players into two teams with *n* people each so that the player from the first team in every pair wins regardless of the results of the drawing, then print "YES". Otherwise print "NO".
Demo Input:
['2\n1 3 2 4\n', '1\n3 3\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
n, a = int(input()), list(sorted(map(int, input().split())))
print(('NO', 'YES')[a[n - 1] != a[n]])
``` | 3 | |
82 | A | Double Cola | PROGRAMMING | 1,100 | [
"implementation",
"math"
] | A. Double Cola | 1 | 256 | Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon. | The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers. | Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially. | [
"1\n",
"6\n",
"1802\n"
] | [
"Sheldon\n",
"Sheldon\n",
"Penny\n"
] | none | 500 | [
{
"input": "1",
"output": "Sheldon"
},
{
"input": "6",
"output": "Sheldon"
},
{
"input": "1802",
"output": "Penny"
},
{
"input": "1",
"output": "Sheldon"
},
{
"input": "2",
"output": "Leonard"
},
{
"input": "3",
"output": "Penny"
},
{
"inpu... | 1,621,595,361 | 2,147,483,647 | Python 3 | OK | TESTS | 41 | 62 | 0 | p=int(input())
n=1
li=["Sheldon","Leonard","Penny", "Rajesh", "Howard"]
while(n*5<p):
p-=n*5
n*=2
p=p-1
p=p//n
print(li[p])
| Title: Double Cola
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input Specification:
The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output Specification:
Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Demo Input:
['1\n', '6\n', '1802\n']
Demo Output:
['Sheldon\n', 'Sheldon\n', 'Penny\n']
Note:
none | ```python
p=int(input())
n=1
li=["Sheldon","Leonard","Penny", "Rajesh", "Howard"]
while(n*5<p):
p-=n*5
n*=2
p=p-1
p=p//n
print(li[p])
``` | 3.969 |
842 | A | Kirill And The Game | PROGRAMMING | 1,200 | [
"brute force",
"two pointers"
] | null | null | Kirill plays a new computer game. He came to the potion store where he can buy any potion. Each potion is characterized by two integers — amount of experience and cost. The efficiency of a potion is the ratio of the amount of experience to the cost. Efficiency may be a non-integer number.
For each two integer numbers *a* and *b* such that *l*<=≤<=*a*<=≤<=*r* and *x*<=≤<=*b*<=≤<=*y* there is a potion with experience *a* and cost *b* in the store (that is, there are (*r*<=-<=*l*<=+<=1)·(*y*<=-<=*x*<=+<=1) potions).
Kirill wants to buy a potion which has efficiency *k*. Will he be able to do this? | First string contains five integer numbers *l*, *r*, *x*, *y*, *k* (1<=≤<=*l*<=≤<=*r*<=≤<=107, 1<=≤<=*x*<=≤<=*y*<=≤<=107, 1<=≤<=*k*<=≤<=107). | Print "YES" without quotes if a potion with efficiency exactly *k* can be bought in the store and "NO" without quotes otherwise.
You can output each of the letters in any register. | [
"1 10 1 10 1\n",
"1 5 6 10 1\n"
] | [
"YES",
"NO"
] | none | 500 | [
{
"input": "1 10 1 10 1",
"output": "YES"
},
{
"input": "1 5 6 10 1",
"output": "NO"
},
{
"input": "1 1 1 1 1",
"output": "YES"
},
{
"input": "1 1 1 1 2",
"output": "NO"
},
{
"input": "1 100000 1 100000 100000",
"output": "YES"
},
{
"input": "1 100000 ... | 1,504,020,952 | 1,852 | Python 3 | WRONG_ANSWER | PRETESTS | 2 | 61 | 0 | l, r, x, y, k = map(int, input().split())
res = "NO"
if k >= x and k <= y and k >= l and k <= r:
for i in range(l, r):
for j in range(x, y):
if k == i/j:
res = "YES"
print(res)
| Title: Kirill And The Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kirill plays a new computer game. He came to the potion store where he can buy any potion. Each potion is characterized by two integers — amount of experience and cost. The efficiency of a potion is the ratio of the amount of experience to the cost. Efficiency may be a non-integer number.
For each two integer numbers *a* and *b* such that *l*<=≤<=*a*<=≤<=*r* and *x*<=≤<=*b*<=≤<=*y* there is a potion with experience *a* and cost *b* in the store (that is, there are (*r*<=-<=*l*<=+<=1)·(*y*<=-<=*x*<=+<=1) potions).
Kirill wants to buy a potion which has efficiency *k*. Will he be able to do this?
Input Specification:
First string contains five integer numbers *l*, *r*, *x*, *y*, *k* (1<=≤<=*l*<=≤<=*r*<=≤<=107, 1<=≤<=*x*<=≤<=*y*<=≤<=107, 1<=≤<=*k*<=≤<=107).
Output Specification:
Print "YES" without quotes if a potion with efficiency exactly *k* can be bought in the store and "NO" without quotes otherwise.
You can output each of the letters in any register.
Demo Input:
['1 10 1 10 1\n', '1 5 6 10 1\n']
Demo Output:
['YES', 'NO']
Note:
none | ```python
l, r, x, y, k = map(int, input().split())
res = "NO"
if k >= x and k <= y and k >= l and k <= r:
for i in range(l, r):
for j in range(x, y):
if k == i/j:
res = "YES"
print(res)
``` | 0 | |
858 | C | Did you mean... | PROGRAMMING | 1,500 | [
"dp",
"greedy",
"implementation"
] | null | null | Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
- the following words have typos: "hellno", "hackcerrs" and "backtothefutttture"; - the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem. | The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters. | Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them. | [
"hellno\n",
"abacaba\n",
"asdfasdf\n"
] | [
"hell no \n",
"abacaba \n",
"asd fasd f \n"
] | none | 1,500 | [
{
"input": "hellno",
"output": "hell no "
},
{
"input": "abacaba",
"output": "abacaba "
},
{
"input": "asdfasdf",
"output": "asd fasd f "
},
{
"input": "ooo",
"output": "ooo "
},
{
"input": "moyaoborona",
"output": "moyaoborona "
},
{
"input": "jxegxxx... | 1,507,642,550 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | s = input()
k = 0
vowels = "aeiou"
res = ""
lc=" "
for c in s:
if not c in vowels:
k += 1
if c == lc:
k = 1
elif k == 3:
c != lc
res += " "
k = 1
else:
k = 0
res += c
lc = c
print(res)
| Title: Did you mean...
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Beroffice text editor has a wide range of features that help working with text. One of the features is an automatic search for typos and suggestions of how to fix them.
Beroffice works only with small English letters (i.e. with 26 letters from a to z). Beroffice thinks that a word is typed with a typo if there are three or more consonants in a row in the word. The only exception is that if the block of consonants has all letters the same, then this block (even if its length is greater than three) is not considered a typo. Formally, a word is typed with a typo if there is a block of not less that three consonants in a row, and there are at least two different letters in this block.
For example:
- the following words have typos: "hellno", "hackcerrs" and "backtothefutttture"; - the following words don't have typos: "helllllooooo", "tobeornottobe" and "oooooo".
When Beroffice editor finds a word with a typo, it inserts as little as possible number of spaces in this word (dividing it into several words) in such a way that each of the resulting words is typed without any typos.
Implement this feature of Beroffice editor. Consider the following letters as the only vowels: 'a', 'e', 'i', 'o' and 'u'. All the other letters are consonants in this problem.
Input Specification:
The only line contains a non-empty word consisting of small English letters. The length of the word is between 1 and 3000 letters.
Output Specification:
Print the given word without any changes if there are no typos.
If there is at least one typo in the word, insert the minimum number of spaces into the word so that each of the resulting words doesn't have any typos. If there are multiple solutions, print any of them.
Demo Input:
['hellno\n', 'abacaba\n', 'asdfasdf\n']
Demo Output:
['hell no \n', 'abacaba \n', 'asd fasd f \n']
Note:
none | ```python
s = input()
k = 0
vowels = "aeiou"
res = ""
lc=" "
for c in s:
if not c in vowels:
k += 1
if c == lc:
k = 1
elif k == 3:
c != lc
res += " "
k = 1
else:
k = 0
res += c
lc = c
print(res)
``` | 0 | |
622 | A | Infinite Sequence | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Consider the infinite sequence of integers: 1,<=1,<=2,<=1,<=2,<=3,<=1,<=2,<=3,<=4,<=1,<=2,<=3,<=4,<=5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).
Find the number on the *n*-th position of the sequence. | The only line contains integer *n* (1<=≤<=*n*<=≤<=1014) — the position of the number to find.
Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type. | Print the element in the *n*-th position of the sequence (the elements are numerated from one). | [
"3\n",
"5\n",
"10\n",
"55\n",
"56\n"
] | [
"2\n",
"2\n",
"4\n",
"10\n",
"1\n"
] | none | 0 | [
{
"input": "3",
"output": "2"
},
{
"input": "5",
"output": "2"
},
{
"input": "10",
"output": "4"
},
{
"input": "55",
"output": "10"
},
{
"input": "56",
"output": "1"
},
{
"input": "1000000000000",
"output": "88209"
},
{
"input": "8471941278... | 1,591,126,270 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 10 | 1,000 | 0 | n=int(input())
i=1
sum=1
while(sum<n):
i+=1
sum+=i
if sum==n:
print(i)
else:
print(n-(sum-i)) | Title: Infinite Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Consider the infinite sequence of integers: 1,<=1,<=2,<=1,<=2,<=3,<=1,<=2,<=3,<=4,<=1,<=2,<=3,<=4,<=5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).
Find the number on the *n*-th position of the sequence.
Input Specification:
The only line contains integer *n* (1<=≤<=*n*<=≤<=1014) — the position of the number to find.
Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Output Specification:
Print the element in the *n*-th position of the sequence (the elements are numerated from one).
Demo Input:
['3\n', '5\n', '10\n', '55\n', '56\n']
Demo Output:
['2\n', '2\n', '4\n', '10\n', '1\n']
Note:
none | ```python
n=int(input())
i=1
sum=1
while(sum<n):
i+=1
sum+=i
if sum==n:
print(i)
else:
print(n-(sum-i))
``` | 0 | |
96 | A | Football | PROGRAMMING | 900 | [
"implementation",
"strings"
] | A. Football | 2 | 256 | Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not. | The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field. | Print "YES" if the situation is dangerous. Otherwise, print "NO". | [
"001001\n",
"1000000001\n"
] | [
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "001001",
"output": "NO"
},
{
"input": "1000000001",
"output": "YES"
},
{
"input": "00100110111111101",
"output": "YES"
},
{
"input": "11110111111111111",
"output": "YES"
},
{
"input": "01",
"output": "NO"
},
{
"input": "10100101",
"outp... | 1,693,918,598 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 124 | 0 | x = input()
cnt = 1
def football(x, cnt):
if cnt == 7:
return "YES"
if len(x) == 1:
return "NO"
if x[0] == x[1]:
print(cnt)
return football(x[1:], cnt+1)
if x[0] != x[1]:
print(cnt)
return football(x[1:], 1)
print(football(x, cnt)) | Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
Input Specification:
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
Output Specification:
Print "YES" if the situation is dangerous. Otherwise, print "NO".
Demo Input:
['001001\n', '1000000001\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none | ```python
x = input()
cnt = 1
def football(x, cnt):
if cnt == 7:
return "YES"
if len(x) == 1:
return "NO"
if x[0] == x[1]:
print(cnt)
return football(x[1:], cnt+1)
if x[0] != x[1]:
print(cnt)
return football(x[1:], 1)
print(football(x, cnt))
``` | 0 |
129 | A | Cookies | PROGRAMMING | 900 | [
"implementation"
] | null | null | Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even? | The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag. | Print in the only line the only number — the sought number of ways. If there are no such ways print 0. | [
"1\n1\n",
"10\n1 2 2 3 4 4 4 2 2 2\n",
"11\n2 2 2 2 2 2 2 2 2 2 99\n"
] | [
"1\n",
"8\n",
"1\n"
] | In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies.
In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total.
In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies. | 500 | [
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 2 2 3 4 4 4 2 2 2",
"output": "8"
},
{
"input": "11\n2 2 2 2 2 2 2 2 2 2 99",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n2 2",
"output": "2"
},
{
"input": "2\n1 2",
"o... | 1,568,558,313 | 2,147,483,647 | Python 3 | OK | TESTS | 52 | 248 | 0 | n = int(input())
odd = sum(int(c) % 2 for c in input().split())
print(odd if odd % 2 else n - odd) | Title: Cookies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even?
Input Specification:
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag.
Output Specification:
Print in the only line the only number — the sought number of ways. If there are no such ways print 0.
Demo Input:
['1\n1\n', '10\n1 2 2 3 4 4 4 2 2 2\n', '11\n2 2 2 2 2 2 2 2 2 2 99\n']
Demo Output:
['1\n', '8\n', '1\n']
Note:
In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies.
In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total.
In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies. | ```python
n = int(input())
odd = sum(int(c) % 2 for c in input().split())
print(odd if odd % 2 else n - odd)
``` | 3 | |
847 | M | Weather Tomorrow | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Vasya came up with his own weather forecasting method. He knows the information about the average air temperature for each of the last *n* days. Assume that the average air temperature for each day is integral.
Vasya believes that if the average temperatures over the last *n* days form an arithmetic progression, where the first term equals to the average temperature on the first day, the second term equals to the average temperature on the second day and so on, then the average temperature of the next (*n*<=+<=1)-th day will be equal to the next term of the arithmetic progression. Otherwise, according to Vasya's method, the temperature of the (*n*<=+<=1)-th day will be equal to the temperature of the *n*-th day.
Your task is to help Vasya predict the average temperature for tomorrow, i. e. for the (*n*<=+<=1)-th day. | The first line contains a single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days for which the average air temperature is known.
The second line contains a sequence of integers *t*1,<=*t*2,<=...,<=*t**n* (<=-<=1000<=≤<=*t**i*<=≤<=1000) — where *t**i* is the average temperature in the *i*-th day. | Print the average air temperature in the (*n*<=+<=1)-th day, which Vasya predicts according to his method. Note that the absolute value of the predicted temperature can exceed 1000. | [
"5\n10 5 0 -5 -10\n",
"4\n1 1 1 1\n",
"3\n5 1 -5\n",
"2\n900 1000\n"
] | [
"-15\n",
"1\n",
"-5\n",
"1100\n"
] | In the first example the sequence of the average temperatures is an arithmetic progression where the first term is 10 and each following terms decreases by 5. So the predicted average temperature for the sixth day is - 10 - 5 = - 15.
In the second example the sequence of the average temperatures is an arithmetic progression where the first term is 1 and each following terms equals to the previous one. So the predicted average temperature in the fifth day is 1.
In the third example the average temperatures do not form an arithmetic progression, so the average temperature of the fourth day equals to the temperature of the third day and equals to - 5.
In the fourth example the sequence of the average temperatures is an arithmetic progression where the first term is 900 and each the following terms increase by 100. So predicted average temperature in the third day is 1000 + 100 = 1100. | 0 | [
{
"input": "5\n10 5 0 -5 -10",
"output": "-15"
},
{
"input": "4\n1 1 1 1",
"output": "1"
},
{
"input": "3\n5 1 -5",
"output": "-5"
},
{
"input": "2\n900 1000",
"output": "1100"
},
{
"input": "2\n1 2",
"output": "3"
},
{
"input": "3\n2 5 8",
"output... | 1,615,433,116 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 16 | 93 | 0 | n = int(input())
a = list(map(int, input().split()))
if n == 2:
print(a[1] + (a[1] - a[0]))
else:
d1 = a[1] - a[0]
d2 = a[2] - a[1]
if d1 == d2:
print(a[-1] + d1)
else:
print(a[-1]) | Title: Weather Tomorrow
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya came up with his own weather forecasting method. He knows the information about the average air temperature for each of the last *n* days. Assume that the average air temperature for each day is integral.
Vasya believes that if the average temperatures over the last *n* days form an arithmetic progression, where the first term equals to the average temperature on the first day, the second term equals to the average temperature on the second day and so on, then the average temperature of the next (*n*<=+<=1)-th day will be equal to the next term of the arithmetic progression. Otherwise, according to Vasya's method, the temperature of the (*n*<=+<=1)-th day will be equal to the temperature of the *n*-th day.
Your task is to help Vasya predict the average temperature for tomorrow, i. e. for the (*n*<=+<=1)-th day.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days for which the average air temperature is known.
The second line contains a sequence of integers *t*1,<=*t*2,<=...,<=*t**n* (<=-<=1000<=≤<=*t**i*<=≤<=1000) — where *t**i* is the average temperature in the *i*-th day.
Output Specification:
Print the average air temperature in the (*n*<=+<=1)-th day, which Vasya predicts according to his method. Note that the absolute value of the predicted temperature can exceed 1000.
Demo Input:
['5\n10 5 0 -5 -10\n', '4\n1 1 1 1\n', '3\n5 1 -5\n', '2\n900 1000\n']
Demo Output:
['-15\n', '1\n', '-5\n', '1100\n']
Note:
In the first example the sequence of the average temperatures is an arithmetic progression where the first term is 10 and each following terms decreases by 5. So the predicted average temperature for the sixth day is - 10 - 5 = - 15.
In the second example the sequence of the average temperatures is an arithmetic progression where the first term is 1 and each following terms equals to the previous one. So the predicted average temperature in the fifth day is 1.
In the third example the average temperatures do not form an arithmetic progression, so the average temperature of the fourth day equals to the temperature of the third day and equals to - 5.
In the fourth example the sequence of the average temperatures is an arithmetic progression where the first term is 900 and each the following terms increase by 100. So predicted average temperature in the third day is 1000 + 100 = 1100. | ```python
n = int(input())
a = list(map(int, input().split()))
if n == 2:
print(a[1] + (a[1] - a[0]))
else:
d1 = a[1] - a[0]
d2 = a[2] - a[1]
if d1 == d2:
print(a[-1] + d1)
else:
print(a[-1])
``` | 0 | |
703 | A | Mishka and Game | PROGRAMMING | 800 | [
"implementation"
] | null | null | Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game.
Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner.
In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw.
Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her! | The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds.
The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively. | If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line.
If Chris is the winner of the game, print "Chris" (without quotes) in the only line.
If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line. | [
"3\n3 5\n2 1\n4 2\n",
"2\n6 1\n1 6\n",
"3\n1 5\n3 3\n2 2\n"
] | [
"Mishka",
"Friendship is magic!^^",
"Chris"
] | In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game.
In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1.
In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris. | 500 | [
{
"input": "3\n3 5\n2 1\n4 2",
"output": "Mishka"
},
{
"input": "2\n6 1\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "3\n1 5\n3 3\n2 2",
"output": "Chris"
},
{
"input": "6\n4 1\n4 2\n5 3\n5 1\n5 3\n4 1",
"output": "Mishka"
},
{
"input": "8\n2 4\n1 4\n1 ... | 1,690,109,877 | 2,147,483,647 | Python 3 | OK | TESTS | 69 | 46 | 0 | t=int(input())
count_mishka=0
count_chris=0
for tc in range(t):
a,b=map(int,input().split())
if a>b:
count_mishka+=1
elif a<b:
count_chris+=1
if count_mishka>count_chris:
print("Mishka")
elif count_mishka<count_chris:
print("Chris")
else:
print("Friendship is magic!^^") | Title: Mishka and Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game.
Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner.
In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw.
Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her!
Input Specification:
The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds.
The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively.
Output Specification:
If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line.
If Chris is the winner of the game, print "Chris" (without quotes) in the only line.
If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line.
Demo Input:
['3\n3 5\n2 1\n4 2\n', '2\n6 1\n1 6\n', '3\n1 5\n3 3\n2 2\n']
Demo Output:
['Mishka', 'Friendship is magic!^^', 'Chris']
Note:
In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game.
In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1.
In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris. | ```python
t=int(input())
count_mishka=0
count_chris=0
for tc in range(t):
a,b=map(int,input().split())
if a>b:
count_mishka+=1
elif a<b:
count_chris+=1
if count_mishka>count_chris:
print("Mishka")
elif count_mishka<count_chris:
print("Chris")
else:
print("Friendship is magic!^^")
``` | 3 | |
895 | A | Pizza Separation | PROGRAMMING | 1,200 | [
"brute force",
"implementation"
] | null | null | Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut into *n* pieces. The *i*-th piece is a sector of angle equal to *a**i*. Vasya and Petya want to divide all pieces of pizza into two continuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=360) — the number of pieces into which the delivered pizza was cut.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=360) — the angles of the sectors into which the pizza was cut. The sum of all *a**i* is 360. | Print one integer — the minimal difference between angles of sectors that will go to Vasya and Petya. | [
"4\n90 90 90 90\n",
"3\n100 100 160\n",
"1\n360\n",
"4\n170 30 150 10\n"
] | [
"0\n",
"40\n",
"360\n",
"0\n"
] | In first sample Vasya can take 1 and 2 pieces, Petya can take 3 and 4 pieces. Then the answer is |(90 + 90) - (90 + 90)| = 0.
In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is |360 - 0| = 360.
In fourth sample Vasya can take 1 and 4 pieces, then Petya will take 2 and 3 pieces. So the answer is |(170 + 10) - (30 + 150)| = 0.
Picture explaning fourth sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/4bb3450aca241f92fedcba5479bf1b6d22cf813d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector. | 500 | [
{
"input": "4\n90 90 90 90",
"output": "0"
},
{
"input": "3\n100 100 160",
"output": "40"
},
{
"input": "1\n360",
"output": "360"
},
{
"input": "4\n170 30 150 10",
"output": "0"
},
{
"input": "5\n10 10 10 10 320",
"output": "280"
},
{
"input": "8\n45 4... | 1,584,724,672 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 49 | 187 | 409,600 | from math import *
from collections import *
from operator import itemgetter
import bisect
ii = lambda: int(input())
iia = lambda: list(map(int,input().split()))
isa = lambda: list(input().split())
I = lambda:list(map(int,input().split()))
chrIdx = lambda x: ord(x)-96
idxChr = lambda x: chr(96+x)
n = ii()
a = iia()
a.sort(reverse=True)
x = 0
y = 0
i = 0
while(i<n):
if(x<=y):
x+=a[i]
i+=1
else:
y+=a[i]
i+=1
print(abs(x-y)) | Title: Pizza Separation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut into *n* pieces. The *i*-th piece is a sector of angle equal to *a**i*. Vasya and Petya want to divide all pieces of pizza into two continuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=360) — the number of pieces into which the delivered pizza was cut.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=360) — the angles of the sectors into which the pizza was cut. The sum of all *a**i* is 360.
Output Specification:
Print one integer — the minimal difference between angles of sectors that will go to Vasya and Petya.
Demo Input:
['4\n90 90 90 90\n', '3\n100 100 160\n', '1\n360\n', '4\n170 30 150 10\n']
Demo Output:
['0\n', '40\n', '360\n', '0\n']
Note:
In first sample Vasya can take 1 and 2 pieces, Petya can take 3 and 4 pieces. Then the answer is |(90 + 90) - (90 + 90)| = 0.
In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is |360 - 0| = 360.
In fourth sample Vasya can take 1 and 4 pieces, then Petya will take 2 and 3 pieces. So the answer is |(170 + 10) - (30 + 150)| = 0.
Picture explaning fourth sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/4bb3450aca241f92fedcba5479bf1b6d22cf813d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector. | ```python
from math import *
from collections import *
from operator import itemgetter
import bisect
ii = lambda: int(input())
iia = lambda: list(map(int,input().split()))
isa = lambda: list(input().split())
I = lambda:list(map(int,input().split()))
chrIdx = lambda x: ord(x)-96
idxChr = lambda x: chr(96+x)
n = ii()
a = iia()
a.sort(reverse=True)
x = 0
y = 0
i = 0
while(i<n):
if(x<=y):
x+=a[i]
i+=1
else:
y+=a[i]
i+=1
print(abs(x-y))
``` | 0 | |
20 | C | Dijkstra? | PROGRAMMING | 1,900 | [
"graphs",
"shortest paths"
] | C. Dijkstra? | 1 | 64 | You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*. | The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices. | Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. | [
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n",
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n"
] | [
"1 4 3 5 ",
"1 4 3 5 "
] | none | 1,500 | [
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "2 1\n1 2 1",
"output": "1 2 "
},
{
"input": "3 1\n1 2 1",
"output": "-1"
},
{
"input... | 1,574,369,013 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 16 | 1,000 | 1,740,800 | from math import inf
def cal_min_dist(dists, visited):
min_dist = inf
min_index = inf
for i in range(len(dists)):
if dists[i] < min_dist and not visited[i]:
min_dist = dists[i]
min_index = i
return min_index
def calc_path(preds):
output = -1
if preds[-1] != -1:
i = len(preds) - 1
output = ''
while i != -1:
output = str(i + 1) + (' ' if i != len(preds) -1 else '') + output
i = preds[i]
return output
def calc_min_dist_dijkstra(graph):
dists = [ inf ] * len(graph)
preds = [ -1 ] * len(graph)
visited = [ False ] * len(graph)
dists[0] = 0
for _ in range(len(graph)):
no = cal_min_dist(dists, visited)
if no != inf:
visited[no] = True
for j in graph[no]:
if not visited[j] and dists[j] > dists[no] + graph[no][j]:
dists[j] = dists[no] + graph[no][j]
preds[j] = no
return calc_path(preds)
n, m = [ int(x) for x in input().split() ]
graph = {
v: { v: 0 } for v in range(n)
}
for i in range(m):
n1, n2, v = [ int(x) for x in input().split() ]
graph[n1 - 1][n2-1] = v
graph[n2 - 1][n1-1] = v
print(calc_min_dist_dijkstra(graph)) | Title: Dijkstra?
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*.
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices.
Output Specification:
Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.
Demo Input:
['5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n', '5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n']
Demo Output:
['1 4 3 5 ', '1 4 3 5 ']
Note:
none | ```python
from math import inf
def cal_min_dist(dists, visited):
min_dist = inf
min_index = inf
for i in range(len(dists)):
if dists[i] < min_dist and not visited[i]:
min_dist = dists[i]
min_index = i
return min_index
def calc_path(preds):
output = -1
if preds[-1] != -1:
i = len(preds) - 1
output = ''
while i != -1:
output = str(i + 1) + (' ' if i != len(preds) -1 else '') + output
i = preds[i]
return output
def calc_min_dist_dijkstra(graph):
dists = [ inf ] * len(graph)
preds = [ -1 ] * len(graph)
visited = [ False ] * len(graph)
dists[0] = 0
for _ in range(len(graph)):
no = cal_min_dist(dists, visited)
if no != inf:
visited[no] = True
for j in graph[no]:
if not visited[j] and dists[j] > dists[no] + graph[no][j]:
dists[j] = dists[no] + graph[no][j]
preds[j] = no
return calc_path(preds)
n, m = [ int(x) for x in input().split() ]
graph = {
v: { v: 0 } for v in range(n)
}
for i in range(m):
n1, n2, v = [ int(x) for x in input().split() ]
graph[n1 - 1][n2-1] = v
graph[n2 - 1][n1-1] = v
print(calc_min_dist_dijkstra(graph))
``` | 0 |
698 | A | Vacations | PROGRAMMING | 1,400 | [
"dp"
] | null | null | Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out. | Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days. | [
"4\n1 3 2 0\n",
"7\n1 3 3 2 1 2 3\n",
"2\n2 2\n"
] | [
"2\n",
"0\n",
"1\n"
] | In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day. | 500 | [
{
"input": "4\n1 3 2 0",
"output": "2"
},
{
"input": "7\n1 3 3 2 1 2 3",
"output": "0"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "10\n0 0 1 1 0 0 0 0 1 0",
"output": "8"
},
{
"input": "100\n3 2 3 3 3 2 3 1 ... | 1,656,075,317 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 1,000 | 0 | # 698A
n = int(input())
a = list(map(int, input().split()))
r = 0
i = 0
while i < n:
if a[i] == 0:
r += 1
i += 1
elif (a[i] == 1) or (a[i] == 2):
a1 = a[i]
c = 0
i += 1
while (i < n) and (a[i] != 0) and (a[i] != 1) and (a[i] != 2):
c += 1
i += 1
if i < n:
if ((a1 == a[i]) and (c % 2 == 0)) or ((a1 != a[i]) and (c % 2 != 0)):
r += 1
i += 1
print(r)
| Title: Vacations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out.
Output Specification:
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days.
Demo Input:
['4\n1 3 2 0\n', '7\n1 3 3 2 1 2 3\n', '2\n2 2\n']
Demo Output:
['2\n', '0\n', '1\n']
Note:
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day. | ```python
# 698A
n = int(input())
a = list(map(int, input().split()))
r = 0
i = 0
while i < n:
if a[i] == 0:
r += 1
i += 1
elif (a[i] == 1) or (a[i] == 2):
a1 = a[i]
c = 0
i += 1
while (i < n) and (a[i] != 0) and (a[i] != 1) and (a[i] != 2):
c += 1
i += 1
if i < n:
if ((a1 == a[i]) and (c % 2 == 0)) or ((a1 != a[i]) and (c % 2 != 0)):
r += 1
i += 1
print(r)
``` | 0 | |
535 | A | Tavas and Nafas | PROGRAMMING | 1,000 | [
"brute force",
"implementation"
] | null | null | Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas.
His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words.
He ate coffee mix without water again, so right now he's really messed up and can't think.
Your task is to help him by telling him what to type. | The first and only line of input contains an integer *s* (0<=≤<=*s*<=≤<=99), Tavas's score. | In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces. | [
"6\n",
"99\n",
"20\n"
] | [
"six\n",
"ninety-nine\n",
"twenty\n"
] | You can find all you need to know about English numerals in [http://en.wikipedia.org/wiki/English_numerals](https://en.wikipedia.org/wiki/English_numerals) . | 500 | [
{
"input": "6",
"output": "six"
},
{
"input": "99",
"output": "ninety-nine"
},
{
"input": "20",
"output": "twenty"
},
{
"input": "10",
"output": "ten"
},
{
"input": "15",
"output": "fifteen"
},
{
"input": "27",
"output": "twenty-seven"
},
{
... | 1,533,736,849 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 78 | 0 |
m = [g for g in range(0,100)]
d = {}
d[15],d[20],d[30],d[50],d[80] = 'fifteen','twenty','thirty','fifty','eighty'
b = ['zero','one','two','three','four','five','six', 'seven','eight','nine','ten',
'eleven','twelve','thirteen']
for i in range(0,20):
if 0<=i<=13:
d[i] = b[i]
elif 13<i<20 and i!=15:
d[i] = b[i-10] + 'teen'
for l in range(20,100,10):
if l != 20 and l != 30 and l !=50 and l!=80 :
d[l] = b[int(l/10)] + 'ty'
for h in range(1,10):
d[l+h] = d[l] + ' ' + b[h]
def numeral():
s = int(input('Number: '))
return d[s]
def main():
print(numeral())
main()
| Title: Tavas and Nafas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today Tavas got his test result as an integer score and he wants to share it with his girlfriend, Nafas.
His phone operating system is Tavdroid, and its keyboard doesn't have any digits! He wants to share his score with Nafas via text, so he has no choice but to send this number using words.
He ate coffee mix without water again, so right now he's really messed up and can't think.
Your task is to help him by telling him what to type.
Input Specification:
The first and only line of input contains an integer *s* (0<=≤<=*s*<=≤<=99), Tavas's score.
Output Specification:
In the first and only line of output, print a single string consisting only from English lowercase letters and hyphens ('-'). Do not use spaces.
Demo Input:
['6\n', '99\n', '20\n']
Demo Output:
['six\n', 'ninety-nine\n', 'twenty\n']
Note:
You can find all you need to know about English numerals in [http://en.wikipedia.org/wiki/English_numerals](https://en.wikipedia.org/wiki/English_numerals) . | ```python
m = [g for g in range(0,100)]
d = {}
d[15],d[20],d[30],d[50],d[80] = 'fifteen','twenty','thirty','fifty','eighty'
b = ['zero','one','two','three','four','five','six', 'seven','eight','nine','ten',
'eleven','twelve','thirteen']
for i in range(0,20):
if 0<=i<=13:
d[i] = b[i]
elif 13<i<20 and i!=15:
d[i] = b[i-10] + 'teen'
for l in range(20,100,10):
if l != 20 and l != 30 and l !=50 and l!=80 :
d[l] = b[int(l/10)] + 'ty'
for h in range(1,10):
d[l+h] = d[l] + ' ' + b[h]
def numeral():
s = int(input('Number: '))
return d[s]
def main():
print(numeral())
main()
``` | 0 | |
710 | A | King Moves | PROGRAMMING | 800 | [
"implementation"
] | null | null | The only king stands on the standard chess board. You are given his position in format "cd", where *c* is the column from 'a' to 'h' and *d* is the row from '1' to '8'. Find the number of moves permitted for the king.
Check the king's moves here [https://en.wikipedia.org/wiki/King_(chess)](https://en.wikipedia.org/wiki/King_(chess)). | The only line contains the king's position in the format "cd", where 'c' is the column from 'a' to 'h' and 'd' is the row from '1' to '8'. | Print the only integer *x* — the number of moves permitted for the king. | [
"e4\n"
] | [
"8\n"
] | none | 0 | [
{
"input": "e4",
"output": "8"
},
{
"input": "a1",
"output": "3"
},
{
"input": "h8",
"output": "3"
},
{
"input": "a4",
"output": "5"
},
{
"input": "g7",
"output": "8"
},
{
"input": "e1",
"output": "5"
},
{
"input": "b2",
"output": "8"
... | 1,620,592,588 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 108 | 22,323,200 | n = input()
ans = 8
if h[0] in 'ah' and h[1] in [1, 8]:
print(3)
elif h[0] in 'ah' or h[1] in [1, 8]:
print(5)
else:
print(8) | Title: King Moves
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The only king stands on the standard chess board. You are given his position in format "cd", where *c* is the column from 'a' to 'h' and *d* is the row from '1' to '8'. Find the number of moves permitted for the king.
Check the king's moves here [https://en.wikipedia.org/wiki/King_(chess)](https://en.wikipedia.org/wiki/King_(chess)).
Input Specification:
The only line contains the king's position in the format "cd", where 'c' is the column from 'a' to 'h' and 'd' is the row from '1' to '8'.
Output Specification:
Print the only integer *x* — the number of moves permitted for the king.
Demo Input:
['e4\n']
Demo Output:
['8\n']
Note:
none | ```python
n = input()
ans = 8
if h[0] in 'ah' and h[1] in [1, 8]:
print(3)
elif h[0] in 'ah' or h[1] in [1, 8]:
print(5)
else:
print(8)
``` | -1 | |
989 | A | A Blend of Springtime | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | "What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order. | The first and only line of input contains a non-empty string $s$ consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only ($\lvert s \rvert \leq 100$) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively. | Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower). | [
".BAC.\n",
"AA..CB\n"
] | [
"Yes\n",
"No\n"
] | In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell. | 500 | [
{
"input": ".BAC.",
"output": "Yes"
},
{
"input": "AA..CB",
"output": "No"
},
{
"input": ".",
"output": "No"
},
{
"input": "ACB.AAAAAA",
"output": "Yes"
},
{
"input": "B.BC.BBBCA",
"output": "Yes"
},
{
"input": "BA..CAB..B",
"output": "Yes"
},
... | 1,528,724,462 | 362 | Python 3 | OK | TESTS | 37 | 93 | 0 | s = input()
t = 0
for i in 'ABC':
for j in 'ABC':
for k in 'ABC':
if i != j and j != k and i != k:
if s.find(i+j+k) >= 0:
print('Yes')
t = 1
break
if t:
break
if t:
break
else:
print('No') | Title: A Blend of Springtime
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input Specification:
The first and only line of input contains a non-empty string $s$ consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only ($\lvert s \rvert \leq 100$) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output Specification:
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['.BAC.\n', 'AA..CB\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell. | ```python
s = input()
t = 0
for i in 'ABC':
for j in 'ABC':
for k in 'ABC':
if i != j and j != k and i != k:
if s.find(i+j+k) >= 0:
print('Yes')
t = 1
break
if t:
break
if t:
break
else:
print('No')
``` | 3 | |
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"
... | 1,699,522,660 | 2,147,483,647 | PyPy 3 | OK | TESTS | 23 | 748 | 57,651,200 | # n = int(input())
# numbers = input()
# ones, twos, threes, ans = 0, 0, 0, 0
#
# for i in range(len(numbers)):
# if numbers[i] == '1':
# ones += 1
# elif numbers[i] == '2':
# twos += 1
# elif numbers[i] == '3':
# threes += 1
#
# if ones > twos and ones > threes:
# ans = twos + threes
# elif twos > ones and twos > threes:
# ans = ones + threes
# else:
# ans = ones + twos
#
# print(ans)
input()
arr = [0, 0, 0]
for item in [int(element) for element in input().split()]:
arr[item - 1] += 1
arr.sort()
print(arr[0] + arr[1])
| 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
# n = int(input())
# numbers = input()
# ones, twos, threes, ans = 0, 0, 0, 0
#
# for i in range(len(numbers)):
# if numbers[i] == '1':
# ones += 1
# elif numbers[i] == '2':
# twos += 1
# elif numbers[i] == '3':
# threes += 1
#
# if ones > twos and ones > threes:
# ans = twos + threes
# elif twos > ones and twos > threes:
# ans = ones + threes
# else:
# ans = ones + twos
#
# print(ans)
input()
arr = [0, 0, 0]
for item in [int(element) for element in input().split()]:
arr[item - 1] += 1
arr.sort()
print(arr[0] + arr[1])
``` | 3.705616 |
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