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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
255
|
A
|
Greg's Workout
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Greg is a beginner bodybuilder. Today the gym coach gave him the training plan. All it had was *n* integers *a*1,<=*a*2,<=...,<=*a**n*. These numbers mean that Greg needs to do exactly *n* exercises today. Besides, Greg should repeat the *i*-th in order exercise *a**i* times.
Greg now only does three types of exercises: "chest" exercises, "biceps" exercises and "back" exercises. Besides, his training is cyclic, that is, the first exercise he does is a "chest" one, the second one is "biceps", the third one is "back", the fourth one is "chest", the fifth one is "biceps", and so on to the *n*-th exercise.
Now Greg wonders, which muscle will get the most exercise during his training. We know that the exercise Greg repeats the maximum number of times, trains the corresponding muscle the most. Help Greg, determine which muscle will get the most training.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=20). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=25) — the number of times Greg repeats the exercises.
|
Print word "chest" (without the quotes), if the chest gets the most exercise, "biceps" (without the quotes), if the biceps gets the most exercise and print "back" (without the quotes) if the back gets the most exercise.
It is guaranteed that the input is such that the answer to the problem is unambiguous.
|
[
"2\n2 8\n",
"3\n5 1 10\n",
"7\n3 3 2 7 9 6 8\n"
] |
[
"biceps\n",
"back\n",
"chest\n"
] |
In the first sample Greg does 2 chest, 8 biceps and zero back exercises, so the biceps gets the most exercises.
In the second sample Greg does 5 chest, 1 biceps and 10 back exercises, so the back gets the most exercises.
In the third sample Greg does 18 chest, 12 biceps and 8 back exercises, so the chest gets the most exercise.
| 500
|
[
{
"input": "2\n2 8",
"output": "biceps"
},
{
"input": "3\n5 1 10",
"output": "back"
},
{
"input": "7\n3 3 2 7 9 6 8",
"output": "chest"
},
{
"input": "4\n5 6 6 2",
"output": "chest"
},
{
"input": "5\n8 2 2 6 3",
"output": "chest"
},
{
"input": "6\n8 7 2 5 3 4",
"output": "chest"
},
{
"input": "8\n7 2 9 10 3 8 10 6",
"output": "chest"
},
{
"input": "9\n5 4 2 3 4 4 5 2 2",
"output": "chest"
},
{
"input": "10\n4 9 8 5 3 8 8 10 4 2",
"output": "biceps"
},
{
"input": "11\n10 9 7 6 1 3 9 7 1 3 5",
"output": "chest"
},
{
"input": "12\n24 22 6 16 5 21 1 7 2 19 24 5",
"output": "chest"
},
{
"input": "13\n24 10 5 7 16 17 2 7 9 20 15 2 24",
"output": "chest"
},
{
"input": "14\n13 14 19 8 5 17 9 16 15 9 5 6 3 7",
"output": "back"
},
{
"input": "15\n24 12 22 21 25 23 21 5 3 24 23 13 12 16 12",
"output": "chest"
},
{
"input": "16\n12 6 18 6 25 7 3 1 1 17 25 17 6 8 17 8",
"output": "biceps"
},
{
"input": "17\n13 8 13 4 9 21 10 10 9 22 14 23 22 7 6 14 19",
"output": "chest"
},
{
"input": "18\n1 17 13 6 11 10 25 13 24 9 21 17 3 1 17 12 25 21",
"output": "back"
},
{
"input": "19\n22 22 24 25 19 10 7 10 4 25 19 14 1 14 3 18 4 19 24",
"output": "chest"
},
{
"input": "20\n9 8 22 11 18 14 15 10 17 11 2 1 25 20 7 24 4 25 9 20",
"output": "chest"
},
{
"input": "1\n10",
"output": "chest"
},
{
"input": "2\n15 3",
"output": "chest"
},
{
"input": "3\n21 11 19",
"output": "chest"
},
{
"input": "4\n19 24 13 15",
"output": "chest"
},
{
"input": "5\n4 24 1 9 19",
"output": "biceps"
},
{
"input": "6\n6 22 24 7 15 24",
"output": "back"
},
{
"input": "7\n10 8 23 23 14 18 14",
"output": "chest"
},
{
"input": "8\n5 16 8 9 17 16 14 7",
"output": "biceps"
},
{
"input": "9\n12 3 10 23 6 4 22 13 12",
"output": "chest"
},
{
"input": "10\n1 9 20 18 20 17 7 24 23 2",
"output": "back"
},
{
"input": "11\n22 25 8 2 18 15 1 13 1 11 4",
"output": "biceps"
},
{
"input": "12\n20 12 14 2 15 6 24 3 11 8 11 14",
"output": "chest"
},
{
"input": "13\n2 18 8 8 8 20 5 22 15 2 5 19 18",
"output": "back"
},
{
"input": "14\n1 6 10 25 17 13 21 11 19 4 15 24 5 22",
"output": "biceps"
},
{
"input": "15\n13 5 25 13 17 25 19 21 23 17 12 6 14 8 6",
"output": "back"
},
{
"input": "16\n10 15 2 17 22 12 14 14 6 11 4 13 9 8 21 14",
"output": "chest"
},
{
"input": "17\n7 22 9 22 8 7 20 22 23 5 12 11 1 24 17 20 10",
"output": "biceps"
},
{
"input": "18\n18 15 4 25 5 11 21 25 12 14 25 23 19 19 13 6 9 17",
"output": "chest"
},
{
"input": "19\n3 1 3 15 15 25 10 25 23 10 9 21 13 23 19 3 24 21 14",
"output": "back"
},
{
"input": "20\n19 18 11 3 6 14 3 3 25 3 1 19 25 24 23 12 7 4 8 6",
"output": "back"
},
{
"input": "1\n19",
"output": "chest"
},
{
"input": "2\n1 7",
"output": "biceps"
},
{
"input": "3\n18 18 23",
"output": "back"
},
{
"input": "4\n12 15 1 13",
"output": "chest"
},
{
"input": "5\n11 14 25 21 21",
"output": "biceps"
},
{
"input": "6\n11 9 12 11 22 18",
"output": "biceps"
},
{
"input": "7\n11 1 16 20 21 25 20",
"output": "chest"
},
{
"input": "8\n1 2 20 9 3 22 17 4",
"output": "back"
},
{
"input": "9\n19 2 10 19 15 20 3 1 13",
"output": "back"
},
{
"input": "10\n11 2 11 8 21 16 2 3 19 9",
"output": "back"
},
{
"input": "20\n25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 25 24",
"output": "chest"
},
{
"input": "12\n4 24 21 3 13 24 22 13 12 21 1 15",
"output": "back"
},
{
"input": "13\n14 14 16 2 13 5 1 14 9 4 16 8 3",
"output": "biceps"
},
{
"input": "14\n1 9 15 4 11 8 25 3 9 14 13 2 1 11",
"output": "biceps"
},
{
"input": "15\n4 19 10 6 16 12 5 11 7 23 1 24 11 7 17",
"output": "back"
},
{
"input": "16\n2 8 2 8 13 22 20 12 22 23 18 13 18 22 11 17",
"output": "chest"
},
{
"input": "17\n24 5 5 16 10 8 22 6 4 13 10 10 5 23 8 20 8",
"output": "chest"
},
{
"input": "18\n14 8 9 12 11 18 24 1 14 24 18 5 12 17 1 10 1 22",
"output": "chest"
},
{
"input": "19\n21 2 10 6 9 1 24 5 2 19 10 13 10 7 19 2 6 13 24",
"output": "chest"
},
{
"input": "20\n7 1 14 17 6 6 18 13 12 3 25 4 3 19 22 24 16 14 1 23",
"output": "biceps"
},
{
"input": "1\n19",
"output": "chest"
},
{
"input": "20\n2 1 2 2 1 2 2 1 2 1 1 1 1 1 1 1 1 1 1 22",
"output": "biceps"
}
] | 1,623,151,210
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 61
| 310
| 0
|
n=int(input())
l=list(map(int,input().split()))
c,b,ba=0,0,0
for i in range(0,n,3):
c+=l[i]
for i in range(1,n,3):
b+=l[i]
for i in range(2,n,3):
ba+=l[i]
x=max(c,b,ba)
if x==c:
print('chest')
elif x==b:
print('biceps')
else:
print('back')
|
Title: Greg's Workout
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greg is a beginner bodybuilder. Today the gym coach gave him the training plan. All it had was *n* integers *a*1,<=*a*2,<=...,<=*a**n*. These numbers mean that Greg needs to do exactly *n* exercises today. Besides, Greg should repeat the *i*-th in order exercise *a**i* times.
Greg now only does three types of exercises: "chest" exercises, "biceps" exercises and "back" exercises. Besides, his training is cyclic, that is, the first exercise he does is a "chest" one, the second one is "biceps", the third one is "back", the fourth one is "chest", the fifth one is "biceps", and so on to the *n*-th exercise.
Now Greg wonders, which muscle will get the most exercise during his training. We know that the exercise Greg repeats the maximum number of times, trains the corresponding muscle the most. Help Greg, determine which muscle will get the most training.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=20). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=25) — the number of times Greg repeats the exercises.
Output Specification:
Print word "chest" (without the quotes), if the chest gets the most exercise, "biceps" (without the quotes), if the biceps gets the most exercise and print "back" (without the quotes) if the back gets the most exercise.
It is guaranteed that the input is such that the answer to the problem is unambiguous.
Demo Input:
['2\n2 8\n', '3\n5 1 10\n', '7\n3 3 2 7 9 6 8\n']
Demo Output:
['biceps\n', 'back\n', 'chest\n']
Note:
In the first sample Greg does 2 chest, 8 biceps and zero back exercises, so the biceps gets the most exercises.
In the second sample Greg does 5 chest, 1 biceps and 10 back exercises, so the back gets the most exercises.
In the third sample Greg does 18 chest, 12 biceps and 8 back exercises, so the chest gets the most exercise.
|
```python
n=int(input())
l=list(map(int,input().split()))
c,b,ba=0,0,0
for i in range(0,n,3):
c+=l[i]
for i in range(1,n,3):
b+=l[i]
for i in range(2,n,3):
ba+=l[i]
x=max(c,b,ba)
if x==c:
print('chest')
elif x==b:
print('biceps')
else:
print('back')
```
| 3
|
|
688
|
B
|
Lovely Palindromes
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"math"
] | null | null |
Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not.
Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them.
Now Pari asks you to write a program that gets a huge integer *n* from the input and tells what is the *n*-th even-length positive palindrome number?
|
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10100<=000).
|
Print the *n*-th even-length palindrome number.
|
[
"1\n",
"10\n"
] |
[
"11\n",
"1001\n"
] |
The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001.
| 1,000
|
[
{
"input": "1",
"output": "11"
},
{
"input": "10",
"output": "1001"
},
{
"input": "11",
"output": "1111"
},
{
"input": "12",
"output": "1221"
},
{
"input": "100",
"output": "100001"
},
{
"input": "1321",
"output": "13211231"
},
{
"input": "2",
"output": "22"
},
{
"input": "3",
"output": "33"
},
{
"input": "4",
"output": "44"
},
{
"input": "5",
"output": "55"
},
{
"input": "6",
"output": "66"
},
{
"input": "7",
"output": "77"
},
{
"input": "8",
"output": "88"
},
{
"input": "9",
"output": "99"
},
{
"input": "13",
"output": "1331"
},
{
"input": "14",
"output": "1441"
},
{
"input": "15",
"output": "1551"
},
{
"input": "16",
"output": "1661"
},
{
"input": "17",
"output": "1771"
},
{
"input": "18",
"output": "1881"
},
{
"input": "19",
"output": "1991"
},
{
"input": "20",
"output": "2002"
},
{
"input": "26550",
"output": "2655005562"
},
{
"input": "16137",
"output": "1613773161"
},
{
"input": "91471",
"output": "9147117419"
},
{
"input": "41242",
"output": "4124224214"
},
{
"input": "30866",
"output": "3086666803"
},
{
"input": "4442231232741003009964183199713851566466079882929007253443848371978048610682180431295671905105980116",
"output": "44422312327410030099641831997138515664660798829290072534438483719780486106821804312956719051059801166110895015091765921340812860168408791738483443527009292889706646651583179913814699003001472321322444"
},
{
"input": "7806526084732010742850470902979267046151204705563201965818390803445852115032119297648697933316807252",
"output": "78065260847320107428504709029792670461512047055632019658183908034458521150321192976486979333168072522527086133397968467929112305112585443080938185691023655074021516407629792090740582470102374806256087"
},
{
"input": "2244399823612183124978103773977739948197709395321563148611877999503840818596183803323258317041875913",
"output": "22443998236121831249781037739777399481977093953215631486118779995038408185961838033232583170418759133195781407138523233083816958180483059997781168413651235939077918499377793773018794213812163289934422"
},
{
"input": "123451",
"output": "123451154321"
}
] | 1,609,856,486
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 109
| 614,400
|
import sys
try:
t=1
while(t):
t-=1
#n=int(input())
s=input()
print(s+''.join(reversed(s)))
except:
pass
|
Title: Lovely Palindromes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not.
Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them.
Now Pari asks you to write a program that gets a huge integer *n* from the input and tells what is the *n*-th even-length positive palindrome number?
Input Specification:
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10100<=000).
Output Specification:
Print the *n*-th even-length palindrome number.
Demo Input:
['1\n', '10\n']
Demo Output:
['11\n', '1001\n']
Note:
The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001.
|
```python
import sys
try:
t=1
while(t):
t-=1
#n=int(input())
s=input()
print(s+''.join(reversed(s)))
except:
pass
```
| 3
|
|
9
|
E
|
Interestring graph and Apples
|
PROGRAMMING
| 2,300
|
[
"dfs and similar",
"dsu",
"graphs"
] |
E. Interesting Graph and Apples
|
1
|
64
|
Hexadecimal likes drawing. She has drawn many graphs already, both directed and not. Recently she has started to work on a still-life «interesting graph and apples». An undirected graph is called interesting, if each of its vertices belongs to one cycle only — a funny ring — and does not belong to any other cycles. A funny ring is a cycle that goes through all the vertices just once. Moreover, loops are funny rings too.
She has already drawn the apples and some of the graph edges. But now it is not clear, how to connect the rest of the vertices to get an interesting graph as a result. The answer should contain the minimal amount of added edges. And furthermore, the answer should be the lexicographically smallest one. The set of edges (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*), where *x**i*<=≤<=*y**i*, is lexicographically smaller than the set (*u*1,<=*v*1),<=(*u*2,<=*v*2),<=...,<=(*u**n*,<=*v**n*), where *u**i*<=≤<=*v**i*, provided that the sequence of integers *x*1,<=*y*1,<=*x*2,<=*y*2,<=...,<=*x**n*,<=*y**n* is lexicographically smaller than the sequence *u*1,<=*v*1,<=*u*2,<=*v*2,<=...,<=*u**n*,<=*v**n*. If you do not cope, Hexadecimal will eat you. ...eat you alive.
|
The first line of the input data contains a pair of integers *n* and *m* (1<=≤<=*n*<=≤<=50, 0<=≤<=*m*<=≤<=2500) — the amount of vertices and edges respectively. The following lines contain pairs of numbers *x**i* and *y**i* (1<=≤<=*x**i*, *y**i*<=≤<=*n*) — the vertices that are already connected by edges. The initial graph may contain multiple edges and loops.
|
In the first line output «YES» or «NO»: if it is possible or not to construct an interesting graph. If the answer is «YES», in the second line output *k* — the amount of edges that should be added to the initial graph. Finally, output *k* lines: pairs of vertices *x**j* and *y**j*, between which edges should be drawn. The result may contain multiple edges and loops. *k* can be equal to zero.
|
[
"3 2\n1 2\n2 3\n"
] |
[
"YES\n1\n1 3\n"
] |
none
| 0
|
[
{
"input": "3 2\n1 2\n2 3",
"output": "YES\n1\n1 3"
},
{
"input": "1 1\n1 1",
"output": "YES\n0"
},
{
"input": "1 2\n1 1\n1 1",
"output": "NO"
},
{
"input": "1 3\n1 1\n1 1\n1 1",
"output": "NO"
},
{
"input": "2 0",
"output": "YES\n2\n1 2\n1 2"
},
{
"input": "2 1\n1 1",
"output": "NO"
},
{
"input": "2 1\n2 2",
"output": "NO"
},
{
"input": "2 1\n2 1",
"output": "YES\n1\n1 2"
},
{
"input": "2 2\n1 1\n2 2",
"output": "NO"
},
{
"input": "2 2\n1 2\n2 1",
"output": "YES\n0"
},
{
"input": "2 3\n1 1\n1 2\n2 1",
"output": "NO"
},
{
"input": "3 2\n1 2\n1 2",
"output": "NO"
},
{
"input": "1 0",
"output": "YES\n1\n1 1"
},
{
"input": "4 3\n1 2\n1 3\n1 4",
"output": "NO"
},
{
"input": "6 6\n1 2\n2 3\n3 1\n4 5\n5 6\n6 1",
"output": "NO"
},
{
"input": "49 0",
"output": "YES\n49\n1 2\n1 3\n2 4\n3 5\n4 6\n5 7\n6 8\n7 9\n8 10\n9 11\n10 12\n11 13\n12 14\n13 15\n14 16\n15 17\n16 18\n17 19\n18 20\n19 21\n20 22\n21 23\n22 24\n23 25\n24 26\n25 27\n26 28\n27 29\n28 30\n29 31\n30 32\n31 33\n32 34\n33 35\n34 36\n35 37\n36 38\n37 39\n38 40\n39 41\n40 42\n41 43\n42 44\n43 45\n44 46\n45 47\n46 48\n47 49\n48 49"
},
{
"input": "50 0",
"output": "YES\n50\n1 2\n1 3\n2 4\n3 5\n4 6\n5 7\n6 8\n7 9\n8 10\n9 11\n10 12\n11 13\n12 14\n13 15\n14 16\n15 17\n16 18\n17 19\n18 20\n19 21\n20 22\n21 23\n22 24\n23 25\n24 26\n25 27\n26 28\n27 29\n28 30\n29 31\n30 32\n31 33\n32 34\n33 35\n34 36\n35 37\n36 38\n37 39\n38 40\n39 41\n40 42\n41 43\n42 44\n43 45\n44 46\n45 47\n46 48\n47 49\n48 50\n49 50"
},
{
"input": "50 1\n2 3",
"output": "YES\n49\n1 2\n1 4\n3 5\n4 6\n5 7\n6 8\n7 9\n8 10\n9 11\n10 12\n11 13\n12 14\n13 15\n14 16\n15 17\n16 18\n17 19\n18 20\n19 21\n20 22\n21 23\n22 24\n23 25\n24 26\n25 27\n26 28\n27 29\n28 30\n29 31\n30 32\n31 33\n32 34\n33 35\n34 36\n35 37\n36 38\n37 39\n38 40\n39 41\n40 42\n41 43\n42 44\n43 45\n44 46\n45 47\n46 48\n47 49\n48 50\n49 50"
},
{
"input": "3 2\n1 2\n2 3",
"output": "YES\n1\n1 3"
},
{
"input": "5 3\n3 5\n4 2\n5 1",
"output": "YES\n2\n1 2\n3 4"
},
{
"input": "6 6\n4 3\n3 5\n6 4\n1 6\n2 1\n5 2",
"output": "YES\n0"
},
{
"input": "7 4\n3 2\n2 6\n6 7\n1 5",
"output": "YES\n3\n1 3\n4 5\n4 7"
},
{
"input": "8 5\n4 7\n3 6\n8 3\n6 5\n1 2",
"output": "YES\n3\n1 4\n2 5\n7 8"
},
{
"input": "9 5\n5 2\n4 6\n8 4\n1 8\n2 1",
"output": "YES\n4\n3 5\n3 7\n6 9\n7 9"
},
{
"input": "3 3\n1 3\n2 1\n3 2",
"output": "YES\n0"
},
{
"input": "4 3\n1 2\n4 1\n2 3",
"output": "YES\n1\n3 4"
},
{
"input": "5 2\n1 5\n5 4",
"output": "YES\n3\n1 2\n2 3\n3 4"
},
{
"input": "6 3\n3 4\n1 3\n2 5",
"output": "YES\n3\n1 2\n4 6\n5 6"
},
{
"input": "7 6\n5 6\n2 7\n7 3\n4 1\n1 5\n3 4",
"output": "YES\n1\n2 6"
},
{
"input": "4 1\n3 1",
"output": "YES\n3\n1 2\n2 4\n3 4"
},
{
"input": "8 3\n3 8\n2 6\n1 7",
"output": "YES\n5\n1 2\n3 4\n4 5\n5 6\n7 8"
},
{
"input": "9 4\n7 6\n2 8\n3 5\n8 3",
"output": "YES\n5\n1 2\n1 4\n4 6\n5 9\n7 9"
},
{
"input": "45 22\n15 23\n14 30\n5 44\n43 21\n24 17\n37 38\n40 9\n41 43\n7 4\n38 22\n26 18\n44 41\n42 11\n4 33\n35 24\n36 15\n19 1\n1 37\n9 35\n12 40\n31 29\n18 25",
"output": "YES\n23\n2 3\n2 5\n3 6\n6 7\n8 10\n8 11\n10 12\n13 14\n13 16\n16 17\n19 20\n20 21\n22 23\n25 27\n26 28\n27 29\n28 30\n31 32\n32 33\n34 36\n34 39\n39 45\n42 45"
},
{
"input": "46 25\n44 40\n25 10\n28 44\n26 4\n38 7\n27 3\n46 8\n32 28\n22 20\n14 33\n30 14\n12 23\n13 30\n40 18\n37 35\n10 16\n23 22\n3 46\n36 24\n19 12\n18 42\n11 34\n34 36\n9 32\n24 19",
"output": "YES\n21\n1 2\n1 4\n2 5\n5 6\n6 7\n8 9\n11 13\n15 16\n15 17\n17 20\n21 25\n21 26\n27 29\n29 31\n31 33\n35 38\n37 39\n39 41\n41 43\n42 45\n43 45"
},
{
"input": "47 26\n24 2\n13 24\n25 14\n35 6\n4 10\n11 18\n29 41\n37 13\n38 3\n2 31\n30 29\n6 42\n33 25\n41 45\n40 8\n28 47\n43 39\n39 38\n1 5\n45 22\n19 21\n18 37\n36 17\n27 28\n16 11\n12 30",
"output": "YES\n21\n1 3\n4 5\n7 8\n7 9\n9 10\n12 14\n15 16\n15 17\n19 20\n20 22\n21 23\n23 26\n26 27\n31 32\n32 33\n34 35\n34 36\n40 42\n43 44\n44 46\n46 47"
},
{
"input": "48 26\n27 5\n13 21\n14 20\n41 31\n4 26\n21 39\n31 17\n18 4\n42 2\n28 43\n11 23\n35 22\n34 18\n23 15\n10 13\n7 48\n5 44\n19 25\n12 7\n15 27\n39 41\n33 10\n45 40\n20 42\n29 38\n17 28",
"output": "YES\n22\n1 2\n1 3\n3 6\n6 8\n8 9\n9 11\n12 14\n16 19\n16 22\n24 25\n24 26\n29 30\n30 32\n32 33\n34 36\n35 37\n36 38\n37 40\n43 44\n45 46\n46 47\n47 48"
},
{
"input": "49 26\n33 34\n43 21\n26 27\n46 33\n32 47\n6 3\n44 14\n34 42\n4 8\n27 29\n12 4\n42 7\n22 16\n5 31\n35 24\n39 40\n20 12\n17 44\n8 18\n38 26\n48 39\n31 17\n9 19\n10 23\n1 30\n49 38",
"output": "YES\n23\n1 2\n2 3\n5 6\n7 9\n10 11\n11 13\n13 14\n15 16\n15 18\n19 20\n21 22\n23 24\n25 28\n25 29\n28 30\n32 35\n36 37\n36 40\n37 41\n41 43\n45 46\n45 47\n48 49"
},
{
"input": "50 21\n27 16\n42 35\n15 28\n46 17\n30 39\n47 18\n35 25\n26 24\n24 30\n28 41\n40 38\n11 21\n33 20\n43 10\n37 14\n1 43\n32 49\n49 6\n10 45\n21 50\n39 3",
"output": "YES\n29\n1 2\n2 3\n4 5\n4 6\n5 7\n7 8\n8 9\n9 11\n12 13\n12 14\n13 15\n16 17\n18 19\n19 20\n22 23\n22 25\n23 26\n27 29\n29 31\n31 32\n33 34\n34 36\n36 37\n38 41\n40 42\n44 45\n44 46\n47 48\n48 50"
},
{
"input": "30 21\n6 14\n19 17\n25 20\n28 10\n10 3\n24 23\n22 13\n1 7\n11 26\n12 1\n16 8\n14 9\n30 15\n4 27\n13 21\n20 12\n24 14\n19 10\n7 10\n16 8\n26 11",
"output": "NO"
},
{
"input": "31 24\n6 25\n8 13\n29 20\n13 5\n26 8\n16 9\n31 2\n22 7\n24 21\n28 18\n9 12\n27 14\n20 24\n23 10\n10 27\n15 1\n21 28\n11 16\n12 29\n8 7\n10 28\n27 19\n17 3\n23 16",
"output": "NO"
},
{
"input": "5 2\n1 3\n4 1",
"output": "YES\n3\n2 3\n2 5\n4 5"
},
{
"input": "32 24\n9 15\n32 16\n26 7\n15 8\n30 21\n23 14\n22 17\n14 29\n19 1\n24 31\n3 22\n20 9\n5 23\n10 3\n27 24\n1 30\n8 18\n23 28\n14 4\n27 10\n11 9\n11 24\n11 18\n17 6",
"output": "NO"
},
{
"input": "33 19\n27 23\n17 16\n20 33\n3 11\n1 31\n26 24\n25 10\n21 15\n14 9\n12 4\n29 2\n7 21\n32 13\n33 6\n5 26\n13 28\n6 22\n3 24\n27 19",
"output": "YES\n14\n1 2\n4 5\n7 8\n8 9\n10 11\n12 14\n15 16\n17 18\n18 19\n20 23\n22 28\n25 29\n30 31\n30 32"
},
{
"input": "34 18\n9 14\n30 23\n19 3\n34 19\n26 2\n31 28\n7 21\n20 27\n16 15\n18 20\n5 34\n17 22\n10 12\n6 4\n8 32\n29 24\n24 10\n34 22",
"output": "NO"
},
{
"input": "35 28\n6 24\n35 10\n14 19\n30 34\n29 23\n21 16\n34 5\n22 6\n7 35\n13 29\n27 3\n8 27\n5 15\n26 11\n19 1\n31 28\n17 31\n18 20\n12 32\n4 17\n10 4\n32 8\n35 18\n9 5\n33 30\n24 25\n12 12\n34 3",
"output": "NO"
},
{
"input": "36 23\n27 31\n33 14\n17 24\n14 25\n3 8\n1 21\n24 27\n13 26\n23 6\n35 22\n34 33\n36 4\n19 16\n18 15\n32 36\n5 7\n20 30\n21 11\n11 27\n8 23\n6 10\n4 31\n15 31",
"output": "NO"
},
{
"input": "37 22\n2 15\n37 11\n14 29\n9 37\n15 23\n24 35\n18 3\n23 12\n34 33\n4 19\n22 14\n21 26\n28 27\n12 36\n8 6\n26 28\n31 1\n29 5\n27 25\n17 10\n33 18\n35 20",
"output": "YES\n15\n1 2\n3 4\n5 6\n7 8\n7 9\n10 11\n13 16\n13 17\n16 19\n20 21\n22 24\n25 30\n30 31\n32 34\n32 36"
},
{
"input": "38 30\n21 36\n20 21\n9 11\n27 10\n25 20\n33 16\n11 23\n31 4\n13 22\n36 27\n32 37\n12 6\n35 31\n5 34\n6 14\n7 38\n26 18\n4 24\n18 5\n23 17\n29 28\n38 13\n10 30\n18 3\n15 25\n1 24\n22 22\n17 22\n36 18\n23 13",
"output": "NO"
},
{
"input": "39 25\n8 23\n27 38\n6 32\n20 33\n7 34\n22 26\n32 12\n23 2\n28 20\n33 35\n18 10\n1 21\n11 18\n39 28\n17 9\n36 8\n15 17\n14 1\n19 24\n37 30\n21 39\n38 13\n28 5\n36 30\n33 13",
"output": "NO"
},
{
"input": "40 29\n23 2\n40 16\n35 31\n2 40\n39 35\n18 11\n21 7\n3 6\n15 5\n4 18\n17 19\n8 34\n16 17\n9 39\n37 21\n19 26\n26 36\n33 4\n10 9\n34 22\n13 20\n32 40\n35 11\n5 12\n14 5\n5 24\n40 6\n32 35\n21 21",
"output": "NO"
},
{
"input": "41 28\n6 28\n1 38\n11 7\n12 26\n10 36\n9 21\n8 3\n2 20\n33 32\n21 40\n34 10\n22 15\n30 22\n5 12\n19 35\n13 6\n31 37\n25 4\n15 23\n37 33\n19 19\n20 6\n14 8\n9 12\n27 33\n28 27\n37 11\n36 20",
"output": "NO"
},
{
"input": "6 1\n4 1",
"output": "YES\n5\n1 2\n2 3\n3 5\n4 6\n5 6"
},
{
"input": "42 28\n7 19\n15 24\n3 42\n18 5\n32 27\n26 20\n40 30\n35 2\n14 8\n22 10\n36 4\n16 14\n21 29\n37 40\n2 12\n30 21\n19 17\n39 34\n31 28\n20 3\n4 33\n11 42\n26 21\n9 10\n4 32\n6 1\n1 14\n14 12",
"output": "NO"
},
{
"input": "43 36\n3 24\n25 36\n36 11\n12 38\n11 32\n15 3\n8 9\n2 17\n5 40\n21 37\n39 20\n28 30\n16 22\n27 13\n31 6\n24 39\n34 19\n35 18\n43 21\n41 4\n7 31\n33 26\n6 5\n42 27\n29 2\n30 10\n40 1\n1 29\n20 14\n40 29\n29 6\n26 27\n37 21\n19 9\n31 4\n19 38",
"output": "NO"
},
{
"input": "44 31\n28 26\n5 36\n9 37\n36 29\n26 5\n25 42\n30 22\n29 3\n35 10\n44 28\n18 13\n16 6\n3 33\n22 9\n4 15\n27 19\n17 11\n19 41\n11 25\n10 30\n2 34\n12 7\n37 31\n16 40\n25 24\n28 44\n41 37\n21 21\n12 28\n20 23\n20 17",
"output": "NO"
},
{
"input": "45 20\n37 5\n41 6\n13 22\n28 24\n30 10\n39 35\n5 20\n38 32\n26 1\n23 37\n35 17\n21 12\n7 8\n1 7\n4 16\n8 40\n44 3\n27 23\n19 2\n33 27",
"output": "YES\n25\n2 3\n4 6\n9 10\n9 11\n11 12\n13 14\n14 15\n15 16\n17 18\n18 19\n20 21\n22 24\n25 26\n25 28\n29 30\n29 31\n31 32\n33 34\n34 36\n36 39\n38 40\n41 42\n42 43\n43 45\n44 45"
},
{
"input": "46 24\n24 43\n38 20\n8 38\n22 13\n25 24\n40 35\n21 10\n7 39\n18 5\n33 19\n26 7\n1 27\n43 26\n9 17\n3 44\n44 14\n20 11\n5 2\n15 32\n23 8\n10 37\n27 23\n43 23\n33 25",
"output": "NO"
},
{
"input": "47 36\n29 31\n25 45\n39 46\n12 19\n31 21\n4 41\n5 38\n33 3\n21 39\n40 1\n1 47\n35 12\n42 10\n2 4\n6 35\n17 16\n22 28\n14 22\n41 25\n10 14\n34 37\n27 20\n44 27\n20 2\n3 17\n45 13\n18 34\n47 15\n10 44\n25 15\n12 23\n27 17\n15 38\n17 32\n29 31\n3 39",
"output": "NO"
},
{
"input": "48 32\n45 23\n17 3\n2 48\n47 20\n27 18\n13 28\n18 26\n26 21\n48 31\n21 9\n43 19\n34 43\n10 36\n14 17\n6 12\n3 11\n15 1\n23 37\n37 13\n42 40\n35 5\n16 7\n40 44\n4 29\n24 25\n5 16\n31 45\n39 22\n46 34\n22 30\n28 33\n33 41",
"output": "YES\n16\n1 2\n4 6\n7 8\n8 9\n10 11\n12 14\n15 19\n20 24\n25 27\n29 30\n32 35\n32 36\n38 39\n38 41\n42 46\n44 47"
},
{
"input": "49 29\n43 18\n44 26\n49 31\n37 19\n20 16\n18 22\n30 5\n7 28\n12 2\n31 11\n27 43\n25 9\n19 4\n35 25\n4 30\n6 27\n46 41\n38 23\n17 37\n13 8\n11 38\n29 20\n40 10\n22 29\n36 7\n17 36\n35 48\n41 36\n39 27",
"output": "NO"
},
{
"input": "50 27\n10 7\n32 9\n17 33\n25 34\n47 28\n23 16\n15 46\n41 50\n18 24\n27 19\n35 36\n19 38\n50 31\n31 40\n4 14\n1 11\n6 48\n33 35\n36 30\n39 12\n28 45\n2 1\n22 13\n3 49\n29 36\n7 34\n36 8",
"output": "NO"
},
{
"input": "7 3\n7 4\n5 2\n1 3",
"output": "YES\n4\n1 2\n3 4\n5 6\n6 7"
},
{
"input": "8 4\n1 7\n2 4\n6 2\n5 8",
"output": "YES\n4\n1 3\n3 4\n5 6\n7 8"
},
{
"input": "9 2\n2 5\n1 6",
"output": "YES\n7\n1 2\n3 4\n3 5\n4 7\n6 8\n7 9\n8 9"
},
{
"input": "3 2\n3 2\n2 1",
"output": "YES\n1\n1 3"
},
{
"input": "4 2\n3 1\n4 2",
"output": "YES\n2\n1 2\n3 4"
}
] | 1,602,159,250
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 8
| 278
| 0
|
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
def NO():
print('NO')
exit()
def YES(edges):
edges.sort()
print('YES')
print(len(edges))
for e in edges:
print(e[0] + 1, e[1] + 1)
exit()
n, m = map(int, input().split())
adj = [[] for _ in range(n)]
deg = [0] * n
for u, v in (map(int, input().split()) for _ in range(m)):
adj[u - 1].append(v - 1)
adj[v - 1].append(u - 1)
deg[u - 1] += 1
deg[v - 1] += 1
if max(deg) > 2 or n < m:
NO()
if n == 1:
YES([] if m == 1 else [(0, 1), (0, 1)])
if min(deg) == 2:
v, par, used = 0, -1, [1] + [0] * (n - 1)
while True:
for dest in adj[v]:
if dest != par:
used[dest] = 1
v, par = dest, v
break
else:
break
if all(used):
YES([])
group = [0] * n
gi = 1
for i in range(n):
if group[i]:
continue
group[i] = gi
v = i
while True:
for dest in adj[v]:
if group[dest] == 0:
group[dest] = gi
v = dest
break
else:
break
gi += 1
ans = []
for i in range(n):
if deg[i] == 2:
continue
for j in range(i + 1, n):
if group[i] != group[j] and deg[j] < 2:
ans.append((i, j))
deg[i], deg[j] = deg[i] + 1, deg[j] + 1
for k in (k for itr in (range(j), range(j + 1, n), range(j, j + 1)) for k in itr):
if group[k] == group[j]:
group[k] = group[i]
break
if not (deg.count(1) + deg.count(2) == n and deg.count(1) in (0, 2)):
NO()
if deg.count(1) == 2:
i = deg.index(1)
j = deg.index(1, i + 1)
ans.append((i, j))
YES(ans)
|
Title: Interestring graph and Apples
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Hexadecimal likes drawing. She has drawn many graphs already, both directed and not. Recently she has started to work on a still-life «interesting graph and apples». An undirected graph is called interesting, if each of its vertices belongs to one cycle only — a funny ring — and does not belong to any other cycles. A funny ring is a cycle that goes through all the vertices just once. Moreover, loops are funny rings too.
She has already drawn the apples and some of the graph edges. But now it is not clear, how to connect the rest of the vertices to get an interesting graph as a result. The answer should contain the minimal amount of added edges. And furthermore, the answer should be the lexicographically smallest one. The set of edges (*x*1,<=*y*1),<=(*x*2,<=*y*2),<=...,<=(*x**n*,<=*y**n*), where *x**i*<=≤<=*y**i*, is lexicographically smaller than the set (*u*1,<=*v*1),<=(*u*2,<=*v*2),<=...,<=(*u**n*,<=*v**n*), where *u**i*<=≤<=*v**i*, provided that the sequence of integers *x*1,<=*y*1,<=*x*2,<=*y*2,<=...,<=*x**n*,<=*y**n* is lexicographically smaller than the sequence *u*1,<=*v*1,<=*u*2,<=*v*2,<=...,<=*u**n*,<=*v**n*. If you do not cope, Hexadecimal will eat you. ...eat you alive.
Input Specification:
The first line of the input data contains a pair of integers *n* and *m* (1<=≤<=*n*<=≤<=50, 0<=≤<=*m*<=≤<=2500) — the amount of vertices and edges respectively. The following lines contain pairs of numbers *x**i* and *y**i* (1<=≤<=*x**i*, *y**i*<=≤<=*n*) — the vertices that are already connected by edges. The initial graph may contain multiple edges and loops.
Output Specification:
In the first line output «YES» or «NO»: if it is possible or not to construct an interesting graph. If the answer is «YES», in the second line output *k* — the amount of edges that should be added to the initial graph. Finally, output *k* lines: pairs of vertices *x**j* and *y**j*, between which edges should be drawn. The result may contain multiple edges and loops. *k* can be equal to zero.
Demo Input:
['3 2\n1 2\n2 3\n']
Demo Output:
['YES\n1\n1 3\n']
Note:
none
|
```python
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
def NO():
print('NO')
exit()
def YES(edges):
edges.sort()
print('YES')
print(len(edges))
for e in edges:
print(e[0] + 1, e[1] + 1)
exit()
n, m = map(int, input().split())
adj = [[] for _ in range(n)]
deg = [0] * n
for u, v in (map(int, input().split()) for _ in range(m)):
adj[u - 1].append(v - 1)
adj[v - 1].append(u - 1)
deg[u - 1] += 1
deg[v - 1] += 1
if max(deg) > 2 or n < m:
NO()
if n == 1:
YES([] if m == 1 else [(0, 1), (0, 1)])
if min(deg) == 2:
v, par, used = 0, -1, [1] + [0] * (n - 1)
while True:
for dest in adj[v]:
if dest != par:
used[dest] = 1
v, par = dest, v
break
else:
break
if all(used):
YES([])
group = [0] * n
gi = 1
for i in range(n):
if group[i]:
continue
group[i] = gi
v = i
while True:
for dest in adj[v]:
if group[dest] == 0:
group[dest] = gi
v = dest
break
else:
break
gi += 1
ans = []
for i in range(n):
if deg[i] == 2:
continue
for j in range(i + 1, n):
if group[i] != group[j] and deg[j] < 2:
ans.append((i, j))
deg[i], deg[j] = deg[i] + 1, deg[j] + 1
for k in (k for itr in (range(j), range(j + 1, n), range(j, j + 1)) for k in itr):
if group[k] == group[j]:
group[k] = group[i]
break
if not (deg.count(1) + deg.count(2) == n and deg.count(1) in (0, 2)):
NO()
if deg.count(1) == 2:
i = deg.index(1)
j = deg.index(1, i + 1)
ans.append((i, j))
YES(ans)
```
| 0
|
44
|
A
|
Indian Summer
|
PROGRAMMING
| 900
|
[
"implementation"
] |
A. Indian Summer
|
2
|
256
|
Indian summer is such a beautiful time of the year! A girl named Alyona is walking in the forest and picking a bouquet from fallen leaves. Alyona is very choosy — she doesn't take a leaf if it matches the color and the species of the tree of one of the leaves she already has. Find out how many leaves Alyona has picked.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of leaves Alyona has found. The next *n* lines contain the leaves' descriptions. Each leaf is characterized by the species of the tree it has fallen from and by the color. The species of the trees and colors are given in names, consisting of no more than 10 lowercase Latin letters. A name can not be an empty string. The species of a tree and the color are given in each line separated by a space.
|
Output the single number — the number of Alyona's leaves.
|
[
"5\nbirch yellow\nmaple red\nbirch yellow\nmaple yellow\nmaple green\n",
"3\noak yellow\noak yellow\noak yellow\n"
] |
[
"4\n",
"1\n"
] |
none
| 0
|
[
{
"input": "5\nbirch yellow\nmaple red\nbirch yellow\nmaple yellow\nmaple green",
"output": "4"
},
{
"input": "3\noak yellow\noak yellow\noak yellow",
"output": "1"
},
{
"input": "5\nxbnbkzn hp\nkaqkl vrgzbvqstu\nj aqidx\nhos gyul\nwefxmh tygpluae",
"output": "5"
},
{
"input": "1\nqvwli hz",
"output": "1"
},
{
"input": "4\nsrhk x\nsrhk x\nqfoe vnrjuab\nqfoe vnrjuab",
"output": "2"
},
{
"input": "4\nsddqllmmpk syded\nfprsq fnenjnaz\nn hdej\nsddqllmmpk syded",
"output": "3"
},
{
"input": "17\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw",
"output": "1"
},
{
"input": "18\nb d\nrfdb esp\nrfdb esp\nb d\nrfdb esp\nrfdb esp\nb d\nb d\nrfdb esp\nrfdb esp\nb d\nrfdb esp\nb d\nrfdb esp\nb d\nb d\nrfdb esp\nrfdb esp",
"output": "2"
},
{
"input": "13\nsvpzwtwn rykzfdce\nqweiyeck jkreouy\nhk nnli\ntwxrnbbdt vtuv\nxokqjtylly sz\nesdt dbfidjslq\ng ybqgomvw\nxcpfjmf zcqvz\nifajadhj invzueip\nvdemdnxifb hckwebmi\nsdpnhipam wvowzavh\nuqdlfskhgo vunbpghae\ne dtigwnb",
"output": "13"
},
{
"input": "20\nm vkfh\nvptikamead mvx\nitu mf\nklruxckw aqega\nekqkd enjllwol\ncc uybfdh\nimrfdngvo u\ne uh\ntwt jsslcfuogk\nbljwqsag tuqbdn\nqcv q\nasx gzhvwwmajj\nqcv q\nekqkd enjllwol\nasx gzhvwwmajj\nks vv\nkzyfi cn\ncc uybfdh\nitu mf\ncjbjhtbyvk vatwfmux",
"output": "15"
},
{
"input": "2\nab ab\na bab",
"output": "2"
}
] | 1,671,807,093
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 29
| 92
| 0
|
n = int(input())
b = []
for i in range(n):
s = ''
a = list(map(str,input().split()))
for i in range(len(a)):
s = s + ' ' + a[i]
b.append(s)
b = set(b)
print(len(b))
|
Title: Indian Summer
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Indian summer is such a beautiful time of the year! A girl named Alyona is walking in the forest and picking a bouquet from fallen leaves. Alyona is very choosy — she doesn't take a leaf if it matches the color and the species of the tree of one of the leaves she already has. Find out how many leaves Alyona has picked.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of leaves Alyona has found. The next *n* lines contain the leaves' descriptions. Each leaf is characterized by the species of the tree it has fallen from and by the color. The species of the trees and colors are given in names, consisting of no more than 10 lowercase Latin letters. A name can not be an empty string. The species of a tree and the color are given in each line separated by a space.
Output Specification:
Output the single number — the number of Alyona's leaves.
Demo Input:
['5\nbirch yellow\nmaple red\nbirch yellow\nmaple yellow\nmaple green\n', '3\noak yellow\noak yellow\noak yellow\n']
Demo Output:
['4\n', '1\n']
Note:
none
|
```python
n = int(input())
b = []
for i in range(n):
s = ''
a = list(map(str,input().split()))
for i in range(len(a)):
s = s + ' ' + a[i]
b.append(s)
b = set(b)
print(len(b))
```
| 3.977
|
808
|
B
|
Average Sleep Time
|
PROGRAMMING
| 1,300
|
[
"data structures",
"implementation",
"math"
] | null | null |
It's been almost a week since Polycarp couldn't get rid of insomnia. And as you may already know, one week in Berland lasts *k* days!
When Polycarp went to a doctor with his problem, the doctor asked him about his sleeping schedule (more specifically, the average amount of hours of sleep per week). Luckily, Polycarp kept records of sleep times for the last *n* days. So now he has a sequence *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* is the sleep time on the *i*-th day.
The number of records is so large that Polycarp is unable to calculate the average value by himself. Thus he is asking you to help him with the calculations. To get the average Polycarp is going to consider *k* consecutive days as a week. So there will be *n*<=-<=*k*<=+<=1 weeks to take into consideration. For example, if *k*<==<=2, *n*<==<=3 and *a*<==<=[3,<=4,<=7], then the result is .
You should write a program which will calculate average sleep times of Polycarp over all weeks.
|
The first line contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=2·105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105).
|
Output average sleeping time over all weeks.
The answer is considered to be correct if its absolute or relative error does not exceed 10<=-<=6. In particular, it is enough to output real number with at least 6 digits after the decimal point.
|
[
"3 2\n3 4 7\n",
"1 1\n10\n",
"8 2\n1 2 4 100000 123 456 789 1\n"
] |
[
"9.0000000000\n",
"10.0000000000\n",
"28964.2857142857\n"
] |
In the third example there are *n* - *k* + 1 = 7 weeks, so the answer is sums of all weeks divided by 7.
| 0
|
[
{
"input": "3 2\n3 4 7",
"output": "9.0000000000"
},
{
"input": "1 1\n10",
"output": "10.0000000000"
},
{
"input": "8 2\n1 2 4 100000 123 456 789 1",
"output": "28964.2857142857"
},
{
"input": "1 1\n1",
"output": "1.0000000000"
},
{
"input": "1 1\n100000",
"output": "100000.0000000000"
},
{
"input": "3 1\n1 2 3",
"output": "2.0000000000"
},
{
"input": "10 4\n11 3 5 20 12 7 9 2 2 20",
"output": "36.2857142857"
},
{
"input": "10 5\n15 9 3 2 17 10 9 18 4 19",
"output": "50.3333333333"
},
{
"input": "10 6\n19 3 20 16 14 10 1 13 7 3",
"output": "65.8000000000"
},
{
"input": "10 7\n8 16 2 13 15 9 5 13 9 2",
"output": "68.2500000000"
},
{
"input": "10 4\n127 1459 718 1183 880 1044 1857 1340 725 1496",
"output": "4574.4285714286"
},
{
"input": "10 5\n1384 1129 1780 1960 1567 1928 12 1523 1165 344",
"output": "6931.3333333333"
}
] | 1,495,393,670
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 8
| 62
| 0
|
import sys
#sys.stdin = open('input2','r')
n,k = map(int,input().split())
arr = list(map(int,input().split()))
d = n-k+1
s = 0
j = 1
for i in range(n):
s += arr[i] * j
if(i < k-1):
j += 1
elif(i >= n-k):
j -= 1
#print(s)
#print(round((s/d),6))
print ('%.6f' % (s/d) )
|
Title: Average Sleep Time
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's been almost a week since Polycarp couldn't get rid of insomnia. And as you may already know, one week in Berland lasts *k* days!
When Polycarp went to a doctor with his problem, the doctor asked him about his sleeping schedule (more specifically, the average amount of hours of sleep per week). Luckily, Polycarp kept records of sleep times for the last *n* days. So now he has a sequence *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* is the sleep time on the *i*-th day.
The number of records is so large that Polycarp is unable to calculate the average value by himself. Thus he is asking you to help him with the calculations. To get the average Polycarp is going to consider *k* consecutive days as a week. So there will be *n*<=-<=*k*<=+<=1 weeks to take into consideration. For example, if *k*<==<=2, *n*<==<=3 and *a*<==<=[3,<=4,<=7], then the result is .
You should write a program which will calculate average sleep times of Polycarp over all weeks.
Input Specification:
The first line contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=2·105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105).
Output Specification:
Output average sleeping time over all weeks.
The answer is considered to be correct if its absolute or relative error does not exceed 10<=-<=6. In particular, it is enough to output real number with at least 6 digits after the decimal point.
Demo Input:
['3 2\n3 4 7\n', '1 1\n10\n', '8 2\n1 2 4 100000 123 456 789 1\n']
Demo Output:
['9.0000000000\n', '10.0000000000\n', '28964.2857142857\n']
Note:
In the third example there are *n* - *k* + 1 = 7 weeks, so the answer is sums of all weeks divided by 7.
|
```python
import sys
#sys.stdin = open('input2','r')
n,k = map(int,input().split())
arr = list(map(int,input().split()))
d = n-k+1
s = 0
j = 1
for i in range(n):
s += arr[i] * j
if(i < k-1):
j += 1
elif(i >= n-k):
j -= 1
#print(s)
#print(round((s/d),6))
print ('%.6f' % (s/d) )
```
| 0
|
|
546
|
A
|
Soldier and Bananas
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation",
"math"
] | null | null |
A soldier wants to buy *w* bananas in the shop. He has to pay *k* dollars for the first banana, 2*k* dollars for the second one and so on (in other words, he has to pay *i*·*k* dollars for the *i*-th banana).
He has *n* dollars. How many dollars does he have to borrow from his friend soldier to buy *w* bananas?
|
The first line contains three positive integers *k*,<=*n*,<=*w* (1<=<=≤<=<=*k*,<=*w*<=<=≤<=<=1000, 0<=≤<=*n*<=≤<=109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants.
|
Output one integer — the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0.
|
[
"3 17 4\n"
] |
[
"13"
] |
none
| 500
|
[
{
"input": "3 17 4",
"output": "13"
},
{
"input": "1 2 1",
"output": "0"
},
{
"input": "1 1 1",
"output": "0"
},
{
"input": "1 5 6",
"output": "16"
},
{
"input": "1 1000000000 1",
"output": "0"
},
{
"input": "1000 0 1000",
"output": "500500000"
},
{
"input": "859 453892 543",
"output": "126416972"
},
{
"input": "1000 1000000000 1000",
"output": "0"
},
{
"input": "1000 500500000 1000",
"output": "0"
},
{
"input": "1000 500500001 1000",
"output": "0"
},
{
"input": "1000 500499999 1000",
"output": "1"
},
{
"input": "634 87973 214",
"output": "14497197"
},
{
"input": "432 10000 241",
"output": "12587552"
},
{
"input": "111 111111111 111",
"output": "0"
},
{
"input": "20 43 3",
"output": "77"
}
] | 1,694,597,831
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 46
| 0
|
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Wed Sep 13 17:22:41 2023
@author: huangxiaoyuan
"""
k,n,w=input().split()
k=int(k)
n=int(n)
w=int(w)
i=1
sum=0
if k>=1 and w<=1000 and 0<=n<=10**9:
k0=k
while i<=w:
sum+=k
k+=k0
i+=1
sum1=sum-n
print(sum1)
|
Title: Soldier and Bananas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A soldier wants to buy *w* bananas in the shop. He has to pay *k* dollars for the first banana, 2*k* dollars for the second one and so on (in other words, he has to pay *i*·*k* dollars for the *i*-th banana).
He has *n* dollars. How many dollars does he have to borrow from his friend soldier to buy *w* bananas?
Input Specification:
The first line contains three positive integers *k*,<=*n*,<=*w* (1<=<=≤<=<=*k*,<=*w*<=<=≤<=<=1000, 0<=≤<=*n*<=≤<=109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants.
Output Specification:
Output one integer — the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0.
Demo Input:
['3 17 4\n']
Demo Output:
['13']
Note:
none
|
```python
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Wed Sep 13 17:22:41 2023
@author: huangxiaoyuan
"""
k,n,w=input().split()
k=int(k)
n=int(n)
w=int(w)
i=1
sum=0
if k>=1 and w<=1000 and 0<=n<=10**9:
k0=k
while i<=w:
sum+=k
k+=k0
i+=1
sum1=sum-n
print(sum1)
```
| 0
|
|
637
|
B
|
Chat Order
|
PROGRAMMING
| 1,200
|
[
"*special",
"binary search",
"constructive algorithms",
"data structures",
"sortings"
] | null | null |
Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a message is sent to some friend, his friend's chat rises to the very top of the page. The relative order of the other chats doesn't change. If there was no chat with this friend before, then a new chat is simply inserted to the top of the list.
Assuming that the chat list is initially empty, given the sequence of Polycaprus' messages make a list of chats after all of his messages are processed. Assume that no friend wrote any message to Polycarpus.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of Polycarpus' messages. Next *n* lines enlist the message recipients in the order in which the messages were sent. The name of each participant is a non-empty sequence of lowercase English letters of length at most 10.
|
Print all the recipients to who Polycarp talked to in the order of chats with them, from top to bottom.
|
[
"4\nalex\nivan\nroman\nivan\n",
"8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina\n"
] |
[
"ivan\nroman\nalex\n",
"alina\nmaria\nekaterina\ndarya\n"
] |
In the first test case Polycarpus first writes to friend by name "alex", and the list looks as follows:
1. alex
Then Polycarpus writes to friend by name "ivan" and the list looks as follows:
1. ivan 1. alex
Polycarpus writes the third message to friend by name "roman" and the list looks as follows:
1. roman 1. ivan 1. alex
Polycarpus writes the fourth message to friend by name "ivan", to who he has already sent a message, so the list of chats changes as follows:
1. ivan 1. roman 1. alex
| 1,000
|
[
{
"input": "4\nalex\nivan\nroman\nivan",
"output": "ivan\nroman\nalex"
},
{
"input": "8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina",
"output": "alina\nmaria\nekaterina\ndarya"
},
{
"input": "1\nwdi",
"output": "wdi"
},
{
"input": "2\nypg\nypg",
"output": "ypg"
},
{
"input": "3\nexhll\nexhll\narruapexj",
"output": "arruapexj\nexhll"
},
{
"input": "3\nfv\nle\nle",
"output": "le\nfv"
},
{
"input": "8\nm\nm\nm\nm\nm\nm\nm\nm",
"output": "m"
},
{
"input": "10\nr\nr\ni\nw\nk\nr\nb\nu\nu\nr",
"output": "r\nu\nb\nk\nw\ni"
},
{
"input": "7\ne\nfau\ncmk\nnzs\nby\nwx\ntjmok",
"output": "tjmok\nwx\nby\nnzs\ncmk\nfau\ne"
},
{
"input": "6\nklrj\nwe\nklrj\nwe\nwe\nwe",
"output": "we\nklrj"
},
{
"input": "8\nzncybqmh\naeebef\nzncybqmh\nn\naeebef\nzncybqmh\nzncybqmh\nzncybqmh",
"output": "zncybqmh\naeebef\nn"
},
{
"input": "30\nkqqcbs\nvap\nkymomn\nj\nkqqcbs\nfuzlzoum\nkymomn\ndbh\nfuzlzoum\nkymomn\nvap\nvlgzs\ndbh\nvlgzs\nbvy\ndbh\nkymomn\nkymomn\neoqql\nkymomn\nkymomn\nkqqcbs\nvlgzs\nkqqcbs\nkqqcbs\nfuzlzoum\nvlgzs\nrylgdoo\nvlgzs\nrylgdoo",
"output": "rylgdoo\nvlgzs\nfuzlzoum\nkqqcbs\nkymomn\neoqql\ndbh\nbvy\nvap\nj"
},
{
"input": "40\nji\nv\nv\nns\nji\nn\nji\nv\nfvy\nvje\nns\nvje\nv\nhas\nv\nusm\nhas\nfvy\nvje\nkdb\nn\nv\nji\nji\nn\nhas\nv\nji\nkdb\nr\nvje\nns\nv\nusm\nn\nvje\nhas\nns\nhas\nn",
"output": "n\nhas\nns\nvje\nusm\nv\nr\nkdb\nji\nfvy"
},
{
"input": "50\njcg\nvle\njopb\nepdb\nnkef\nfv\nxj\nufe\nfuy\noqta\ngbc\nyuz\nec\nyji\nkuux\ncwm\ntq\nnno\nhp\nzry\nxxpp\ntjvo\ngyz\nkwo\nvwqz\nyaqc\njnj\nwoav\nqcv\ndcu\ngc\nhovn\nop\nevy\ndc\ntrpu\nyb\nuzfa\npca\noq\nnhxy\nsiqu\nde\nhphy\nc\nwovu\nf\nbvv\ndsik\nlwyg",
"output": "lwyg\ndsik\nbvv\nf\nwovu\nc\nhphy\nde\nsiqu\nnhxy\noq\npca\nuzfa\nyb\ntrpu\ndc\nevy\nop\nhovn\ngc\ndcu\nqcv\nwoav\njnj\nyaqc\nvwqz\nkwo\ngyz\ntjvo\nxxpp\nzry\nhp\nnno\ntq\ncwm\nkuux\nyji\nec\nyuz\ngbc\noqta\nfuy\nufe\nxj\nfv\nnkef\nepdb\njopb\nvle\njcg"
},
{
"input": "100\nvhh\nvhh\nvhh\nfa\nfa\nvhh\nvhh\nvhh\nfa\nfa\nfa\nvhh\nfa\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nfa\nfa\nfa\nfa\nfa\nfa\nvhh\nfa\nfa\nvhh\nvhh\nvhh\nfa\nfa\nfa\nvhh\nfa\nvhh\nfa\nvhh\nvhh\nfa\nvhh\nfa\nvhh\nvhh\nvhh\nfa\nvhh\nfa\nfa\nvhh\nfa\nvhh\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nvhh\nvhh\nvhh\nfa\nvhh\nvhh\nfa\nfa\nfa\nvhh\nfa\nfa\nvhh\nfa\nvhh\nfa\nfa\nfa\nfa\nfa\nfa\nvhh\nvhh\nfa\nvhh\nfa\nfa\nvhh\nfa\nfa\nvhh\nfa\nvhh\nvhh\nfa\nvhh",
"output": "vhh\nfa"
},
{
"input": "2\naa\nbb",
"output": "bb\naa"
},
{
"input": "2\naa\na",
"output": "a\naa"
},
{
"input": "3\naa\naa\naa",
"output": "aa"
},
{
"input": "5\naa\na\naa\na\naa",
"output": "aa\na"
},
{
"input": "7\naaaa\naaaa\naaa\na\naa\naaaaaaa\naaa",
"output": "aaa\naaaaaaa\naa\na\naaaa"
},
{
"input": "5\na\naa\naaa\naaaa\na",
"output": "a\naaaa\naaa\naa"
},
{
"input": "12\naaaaa\naaaaaa\naaaa\naaaaaa\naa\naaaa\naaaa\naaaaaa\na\naaa\naaaaaaaa\naa",
"output": "aa\naaaaaaaa\naaa\na\naaaaaa\naaaa\naaaaa"
},
{
"input": "3\na\naa\naaa",
"output": "aaa\naa\na"
},
{
"input": "9\nzzz\nzzzzz\nzzz\nzzzz\nzz\nzzzz\nzzzzz\nzzzz\nzzzzzzz",
"output": "zzzzzzz\nzzzz\nzzzzz\nzz\nzzz"
}
] | 1,656,514,192
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 95
| 639
| 15,974,400
|
n = int(input())
l = []
for i in range(n):
name = input()
l.append(name)
l.reverse()
output = []
final = set()
for i in l:
if i not in final:
final.add(i)
output.append(i)
for i in output:
print(i)
|
Title: Chat Order
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a message is sent to some friend, his friend's chat rises to the very top of the page. The relative order of the other chats doesn't change. If there was no chat with this friend before, then a new chat is simply inserted to the top of the list.
Assuming that the chat list is initially empty, given the sequence of Polycaprus' messages make a list of chats after all of his messages are processed. Assume that no friend wrote any message to Polycarpus.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of Polycarpus' messages. Next *n* lines enlist the message recipients in the order in which the messages were sent. The name of each participant is a non-empty sequence of lowercase English letters of length at most 10.
Output Specification:
Print all the recipients to who Polycarp talked to in the order of chats with them, from top to bottom.
Demo Input:
['4\nalex\nivan\nroman\nivan\n', '8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina\n']
Demo Output:
['ivan\nroman\nalex\n', 'alina\nmaria\nekaterina\ndarya\n']
Note:
In the first test case Polycarpus first writes to friend by name "alex", and the list looks as follows:
1. alex
Then Polycarpus writes to friend by name "ivan" and the list looks as follows:
1. ivan 1. alex
Polycarpus writes the third message to friend by name "roman" and the list looks as follows:
1. roman 1. ivan 1. alex
Polycarpus writes the fourth message to friend by name "ivan", to who he has already sent a message, so the list of chats changes as follows:
1. ivan 1. roman 1. alex
|
```python
n = int(input())
l = []
for i in range(n):
name = input()
l.append(name)
l.reverse()
output = []
final = set()
for i in l:
if i not in final:
final.add(i)
output.append(i)
for i in output:
print(i)
```
| 3
|
|
460
|
A
|
Vasya and Socks
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"math"
] | null | null |
Vasya has *n* pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every *m*-th day (at days with numbers *m*,<=2*m*,<=3*m*,<=...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks?
|
The single line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100; 2<=≤<=*m*<=≤<=100), separated by a space.
|
Print a single integer — the answer to the problem.
|
[
"2 2\n",
"9 3\n"
] |
[
"3\n",
"13\n"
] |
In the first sample Vasya spends the first two days wearing the socks that he had initially. Then on day three he puts on the socks that were bought on day two.
In the second sample Vasya spends the first nine days wearing the socks that he had initially. Then he spends three days wearing the socks that were bought on the third, sixth and ninth days. Than he spends another day wearing the socks that were bought on the twelfth day.
| 500
|
[
{
"input": "2 2",
"output": "3"
},
{
"input": "9 3",
"output": "13"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "1 99",
"output": "1"
},
{
"input": "4 4",
"output": "5"
},
{
"input": "10 2",
"output": "19"
},
{
"input": "10 9",
"output": "11"
},
{
"input": "100 100",
"output": "101"
},
{
"input": "2 27",
"output": "2"
},
{
"input": "99 100",
"output": "99"
},
{
"input": "99 2",
"output": "197"
},
{
"input": "100 3",
"output": "149"
},
{
"input": "98 3",
"output": "146"
},
{
"input": "4 4",
"output": "5"
},
{
"input": "100 2",
"output": "199"
},
{
"input": "62 4",
"output": "82"
},
{
"input": "99 10",
"output": "109"
},
{
"input": "100 5",
"output": "124"
},
{
"input": "80 80",
"output": "81"
},
{
"input": "95 16",
"output": "101"
},
{
"input": "75 16",
"output": "79"
},
{
"input": "99 74",
"output": "100"
},
{
"input": "20 21",
"output": "20"
},
{
"input": "52 96",
"output": "52"
},
{
"input": "24 5",
"output": "29"
}
] | 1,660,658,794
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 46
| 0
|
n,k = map(int,input().split())
c = 0
d = 0
c = n//k
d = c
e = 0
if c>=k:
e = c//k
d = d + e
print(n+d)
|
Title: Vasya and Socks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has *n* pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When he comes home in the evening, Vasya takes off the used socks and throws them away. Every *m*-th day (at days with numbers *m*,<=2*m*,<=3*m*,<=...) mom buys a pair of socks to Vasya. She does it late in the evening, so that Vasya cannot put on a new pair of socks before the next day. How many consecutive days pass until Vasya runs out of socks?
Input Specification:
The single line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=100; 2<=≤<=*m*<=≤<=100), separated by a space.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['2 2\n', '9 3\n']
Demo Output:
['3\n', '13\n']
Note:
In the first sample Vasya spends the first two days wearing the socks that he had initially. Then on day three he puts on the socks that were bought on day two.
In the second sample Vasya spends the first nine days wearing the socks that he had initially. Then he spends three days wearing the socks that were bought on the third, sixth and ninth days. Than he spends another day wearing the socks that were bought on the twelfth day.
|
```python
n,k = map(int,input().split())
c = 0
d = 0
c = n//k
d = c
e = 0
if c>=k:
e = c//k
d = d + e
print(n+d)
```
| 0
|
|
812
|
C
|
Sagheer and Nubian Market
|
PROGRAMMING
| 1,500
|
[
"binary search",
"sortings"
] | null | null |
On his trip to Luxor and Aswan, Sagheer went to a Nubian market to buy some souvenirs for his friends and relatives. The market has some strange rules. It contains *n* different items numbered from 1 to *n*. The *i*-th item has base cost *a**i* Egyptian pounds. If Sagheer buys *k* items with indices *x*1,<=*x*2,<=...,<=*x**k*, then the cost of item *x**j* is *a**x**j*<=+<=*x**j*·*k* for 1<=≤<=*j*<=≤<=*k*. In other words, the cost of an item is equal to its base cost in addition to its index multiplied by the factor *k*.
Sagheer wants to buy as many souvenirs as possible without paying more than *S* Egyptian pounds. Note that he cannot buy a souvenir more than once. If there are many ways to maximize the number of souvenirs, he will choose the way that will minimize the total cost. Can you help him with this task?
|
The first line contains two integers *n* and *S* (1<=≤<=*n*<=≤<=105 and 1<=≤<=*S*<=≤<=109) — the number of souvenirs in the market and Sagheer's budget.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the base costs of the souvenirs.
|
On a single line, print two integers *k*, *T* — the maximum number of souvenirs Sagheer can buy and the minimum total cost to buy these *k* souvenirs.
|
[
"3 11\n2 3 5\n",
"4 100\n1 2 5 6\n",
"1 7\n7\n"
] |
[
"2 11\n",
"4 54\n",
"0 0\n"
] |
In the first example, he cannot take the three items because they will cost him [5, 9, 14] with total cost 28. If he decides to take only two items, then the costs will be [4, 7, 11]. So he can afford the first and second items.
In the second example, he can buy all items as they will cost him [5, 10, 17, 22].
In the third example, there is only one souvenir in the market which will cost him 8 pounds, so he cannot buy it.
| 1,500
|
[
{
"input": "3 11\n2 3 5",
"output": "2 11"
},
{
"input": "4 100\n1 2 5 6",
"output": "4 54"
},
{
"input": "1 7\n7",
"output": "0 0"
},
{
"input": "1 7\n5",
"output": "1 6"
},
{
"input": "1 1\n1",
"output": "0 0"
},
{
"input": "4 33\n4 3 2 1",
"output": "3 27"
},
{
"input": "86 96\n89 48 14 55 5 35 7 79 49 70 74 18 64 63 35 93 63 97 90 77 33 11 100 75 60 99 54 38 3 6 55 1 7 64 56 90 21 76 35 16 61 78 38 78 93 21 89 1 58 53 34 77 56 37 46 59 30 5 85 1 52 87 84 99 97 9 15 66 29 60 17 16 59 23 88 93 32 2 98 89 63 42 9 86 70 80",
"output": "3 71"
},
{
"input": "9 2727\n73 41 68 90 51 7 20 48 69",
"output": "9 872"
},
{
"input": "35 792600\n61 11 82 29 3 50 65 60 62 86 83 78 15 82 7 77 38 87 100 12 93 86 96 79 14 58 60 47 94 39 36 23 69 93 18",
"output": "35 24043"
},
{
"input": "63 47677090\n53 4 59 68 6 12 47 63 28 93 9 53 61 63 53 70 77 63 49 76 70 23 4 40 4 34 24 70 42 83 84 95 11 46 38 83 26 85 34 29 67 96 3 62 97 7 42 65 49 45 50 54 81 74 83 59 10 87 95 87 89 27 3",
"output": "63 130272"
},
{
"input": "88 631662736\n93 75 25 7 6 55 92 23 22 32 4 48 61 29 91 79 16 18 18 9 66 9 57 62 3 81 48 16 21 90 93 58 30 8 31 47 44 70 34 85 52 71 58 42 99 53 43 54 96 26 6 13 38 4 13 60 1 48 32 100 52 8 27 99 66 34 98 45 19 50 37 59 31 56 58 70 61 14 100 66 74 85 64 57 92 89 7 92",
"output": "88 348883"
},
{
"input": "12 12\n1232 1848 2048 4694 5121 3735 9968 4687 2040 6033 5839 2507",
"output": "0 0"
},
{
"input": "37 5271\n368 6194 4856 8534 944 4953 2085 5350 788 7772 9786 1321 4310 4453 7078 9912 5799 4066 5471 5079 5161 9773 1300 5474 1202 1353 9499 9694 9020 6332 595 7619 1271 7430 1199 3127 8867",
"output": "5 4252"
},
{
"input": "65 958484\n9597 1867 5346 637 6115 5833 3318 6059 4430 9169 8155 7895 3534 7962 9900 9495 5694 3461 5370 1945 1724 9264 3475 618 3421 551 8359 6889 1843 6716 9216 2356 1592 6265 2945 6496 4947 2840 9057 6141 887 4823 4004 8027 1993 1391 796 7059 5500 4369 4012 4983 6495 8990 3633 5439 421 1129 6970 8796 7826 1200 8741 6555 5037",
"output": "65 468998"
},
{
"input": "90 61394040\n2480 6212 4506 829 8191 797 5336 6722 3178 1007 5849 3061 3588 6684 5983 5452 7654 5321 660 2569 2809 2179 679 4858 6887 2580 6880 6120 4159 5542 4999 8703 2386 8221 7046 1229 1662 4542 7089 3548 4298 1973 1854 2473 5507 241 359 5248 7907 5201 9624 4596 1723 2622 4800 4716 693 961 7402 9004 7994 8048 6590 5866 7502 3304 4331 5218 6906 1016 5342 6644 2205 5823 8525 4839 1914 2651 3940 7751 3489 4178 7234 6640 7602 9765 8559 7819 5827 163",
"output": "90 795634"
},
{
"input": "14 891190480\n1424 3077 9632 6506 4568 9650 5534 1085 6934 9340 2867 367 7075 618",
"output": "14 70147"
},
{
"input": "39 43\n22166 81842 15513 80979 39645 60168 96994 13493 12904 79871 49910 45356 93691 51829 18226 34288 11525 41944 40433 67295 30123 1081 55623 22279 75814 82316 2963 39329 38223 8445 43202 61912 15122 86367 37200 68113 57194 38541 49641",
"output": "0 0"
},
{
"input": "67 8824\n75515 67590 86373 34191 3446 27408 31581 24727 40005 23718 39738 30960 4786 51040 32590 80454 14335 47173 20079 41204 67289 58347 88969 88396 37681 43963 13886 85690 12259 14732 42036 62620 15011 41890 20150 59469 62104 30136 47163 19790 25699 27453 36151 52914 52684 20503 78622 81082 94500 55756 94030 54764 72763 37830 13210 64559 53600 87998 80472 19001 83769 79700 88794 10161 99980 95184 74439",
"output": "2 8268"
},
{
"input": "16 56532535\n84567 85265 99012 62115 58908 10120 27355 90226 17191 35588 43807 6327 89157 71156 16253 22387",
"output": "16 821610"
},
{
"input": "42 818723640\n57019 99450 43192 25131 80678 41786 52490 34454 51620 56132 41773 94622 23949 11597 96944 51420 44164 85727 72959 26622 1838 36302 99335 35572 92734 10651 22293 97601 38634 24066 81870 10904 5825 71968 2377 61198 3104 4191 97851 99920 65295 1385",
"output": "42 2094569"
}
] | 1,570,170,320
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 109
| 0
|
def BinarySearch(n, s, a):
N, S = int(n), int(s)
ans, total, _min, _max = 0, 0, 0, N
while total <= S:
K = N
b = sorted([int(a[i]) + (i + 1) * K for i in range(N)])
cost = sum([b[i] for i in range(K)])
if cost <= S:
ans = K
total = cost
break
else:
total = 0
N = N-1
print(str(ans) + " " + str(total))
N, S = str(input()).split(" ")
A = input().split(" ")
BinarySearch(N, S, A)
|
Title: Sagheer and Nubian Market
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On his trip to Luxor and Aswan, Sagheer went to a Nubian market to buy some souvenirs for his friends and relatives. The market has some strange rules. It contains *n* different items numbered from 1 to *n*. The *i*-th item has base cost *a**i* Egyptian pounds. If Sagheer buys *k* items with indices *x*1,<=*x*2,<=...,<=*x**k*, then the cost of item *x**j* is *a**x**j*<=+<=*x**j*·*k* for 1<=≤<=*j*<=≤<=*k*. In other words, the cost of an item is equal to its base cost in addition to its index multiplied by the factor *k*.
Sagheer wants to buy as many souvenirs as possible without paying more than *S* Egyptian pounds. Note that he cannot buy a souvenir more than once. If there are many ways to maximize the number of souvenirs, he will choose the way that will minimize the total cost. Can you help him with this task?
Input Specification:
The first line contains two integers *n* and *S* (1<=≤<=*n*<=≤<=105 and 1<=≤<=*S*<=≤<=109) — the number of souvenirs in the market and Sagheer's budget.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the base costs of the souvenirs.
Output Specification:
On a single line, print two integers *k*, *T* — the maximum number of souvenirs Sagheer can buy and the minimum total cost to buy these *k* souvenirs.
Demo Input:
['3 11\n2 3 5\n', '4 100\n1 2 5 6\n', '1 7\n7\n']
Demo Output:
['2 11\n', '4 54\n', '0 0\n']
Note:
In the first example, he cannot take the three items because they will cost him [5, 9, 14] with total cost 28. If he decides to take only two items, then the costs will be [4, 7, 11]. So he can afford the first and second items.
In the second example, he can buy all items as they will cost him [5, 10, 17, 22].
In the third example, there is only one souvenir in the market which will cost him 8 pounds, so he cannot buy it.
|
```python
def BinarySearch(n, s, a):
N, S = int(n), int(s)
ans, total, _min, _max = 0, 0, 0, N
while total <= S:
K = N
b = sorted([int(a[i]) + (i + 1) * K for i in range(N)])
cost = sum([b[i] for i in range(K)])
if cost <= S:
ans = K
total = cost
break
else:
total = 0
N = N-1
print(str(ans) + " " + str(total))
N, S = str(input()).split(" ")
A = input().split(" ")
BinarySearch(N, S, A)
```
| 0
|
|
546
|
A
|
Soldier and Bananas
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation",
"math"
] | null | null |
A soldier wants to buy *w* bananas in the shop. He has to pay *k* dollars for the first banana, 2*k* dollars for the second one and so on (in other words, he has to pay *i*·*k* dollars for the *i*-th banana).
He has *n* dollars. How many dollars does he have to borrow from his friend soldier to buy *w* bananas?
|
The first line contains three positive integers *k*,<=*n*,<=*w* (1<=<=≤<=<=*k*,<=*w*<=<=≤<=<=1000, 0<=≤<=*n*<=≤<=109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants.
|
Output one integer — the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0.
|
[
"3 17 4\n"
] |
[
"13"
] |
none
| 500
|
[
{
"input": "3 17 4",
"output": "13"
},
{
"input": "1 2 1",
"output": "0"
},
{
"input": "1 1 1",
"output": "0"
},
{
"input": "1 5 6",
"output": "16"
},
{
"input": "1 1000000000 1",
"output": "0"
},
{
"input": "1000 0 1000",
"output": "500500000"
},
{
"input": "859 453892 543",
"output": "126416972"
},
{
"input": "1000 1000000000 1000",
"output": "0"
},
{
"input": "1000 500500000 1000",
"output": "0"
},
{
"input": "1000 500500001 1000",
"output": "0"
},
{
"input": "1000 500499999 1000",
"output": "1"
},
{
"input": "634 87973 214",
"output": "14497197"
},
{
"input": "432 10000 241",
"output": "12587552"
},
{
"input": "111 111111111 111",
"output": "0"
},
{
"input": "20 43 3",
"output": "77"
}
] | 1,695,258,403
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 31
| 0
|
a,b,c=map(int,input().split())
print(max(0,a*c*-~c//2-c))
|
Title: Soldier and Bananas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A soldier wants to buy *w* bananas in the shop. He has to pay *k* dollars for the first banana, 2*k* dollars for the second one and so on (in other words, he has to pay *i*·*k* dollars for the *i*-th banana).
He has *n* dollars. How many dollars does he have to borrow from his friend soldier to buy *w* bananas?
Input Specification:
The first line contains three positive integers *k*,<=*n*,<=*w* (1<=<=≤<=<=*k*,<=*w*<=<=≤<=<=1000, 0<=≤<=*n*<=≤<=109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants.
Output Specification:
Output one integer — the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0.
Demo Input:
['3 17 4\n']
Demo Output:
['13']
Note:
none
|
```python
a,b,c=map(int,input().split())
print(max(0,a*c*-~c//2-c))
```
| 0
|
|
450
|
A
|
Jzzhu and Children
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies.
Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:
1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty.
Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order?
|
The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
|
Output a single integer, representing the number of the last child.
|
[
"5 2\n1 3 1 4 2\n",
"6 4\n1 1 2 2 3 3\n"
] |
[
"4\n",
"6\n"
] |
Let's consider the first sample.
Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home.
Child 4 is the last one who goes home.
| 500
|
[
{
"input": "5 2\n1 3 1 4 2",
"output": "4"
},
{
"input": "6 4\n1 1 2 2 3 3",
"output": "6"
},
{
"input": "7 3\n6 1 5 4 2 3 1",
"output": "4"
},
{
"input": "10 5\n2 7 3 6 2 5 1 3 4 5",
"output": "4"
},
{
"input": "100 1\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "100"
},
{
"input": "9 3\n9 5 2 3 7 1 8 4 6",
"output": "7"
},
{
"input": "20 10\n58 4 32 10 73 7 30 39 47 6 59 21 24 66 79 79 46 13 29 58",
"output": "16"
},
{
"input": "50 5\n89 56 3 2 40 37 56 52 83 59 43 83 43 59 29 74 22 58 53 41 53 67 78 30 57 32 58 29 95 46 45 85 60 49 41 82 8 71 52 40 45 26 6 71 84 91 4 93 40 54",
"output": "48"
},
{
"input": "50 1\n4 3 9 7 6 8 3 7 10 9 8 8 10 2 9 3 2 4 4 10 4 6 8 10 9 9 4 2 8 9 4 4 9 5 1 5 2 4 4 9 10 2 5 10 7 2 8 6 8 1",
"output": "44"
},
{
"input": "50 5\n3 9 10 8 3 3 4 6 8 2 9 9 3 1 2 10 6 8 7 2 7 4 2 7 5 10 2 2 2 5 10 5 6 6 8 7 10 4 3 2 10 8 6 6 8 6 4 4 1 3",
"output": "46"
},
{
"input": "50 2\n56 69 72 15 95 92 51 1 74 87 100 29 46 54 18 81 84 72 84 83 20 63 71 27 45 74 50 89 48 8 21 15 47 3 39 73 80 84 6 99 17 25 56 3 74 64 71 39 89 78",
"output": "40"
},
{
"input": "50 3\n31 39 64 16 86 3 1 9 25 54 98 42 20 3 49 41 73 37 55 62 33 77 64 22 33 82 26 13 10 13 7 40 48 18 46 79 94 72 19 12 11 61 16 37 10 49 14 94 48 69",
"output": "11"
},
{
"input": "50 100\n67 67 61 68 42 29 70 77 12 61 71 27 4 73 87 52 59 38 93 90 31 27 87 47 26 57 76 6 28 72 81 68 50 84 69 79 39 93 52 6 88 12 46 13 90 68 71 38 90 95",
"output": "50"
},
{
"input": "100 3\n4 14 20 11 19 11 14 20 5 7 6 12 11 17 5 11 7 6 2 10 13 5 12 8 5 17 20 18 7 19 11 7 7 20 20 8 10 17 17 19 20 5 15 16 19 7 11 16 4 17 2 10 1 20 20 16 19 9 9 11 5 7 12 9 9 6 20 18 13 19 8 4 8 1 2 4 10 11 15 14 1 7 17 12 13 19 12 2 3 14 15 15 5 17 14 12 17 14 16 9",
"output": "86"
},
{
"input": "100 5\n16 8 14 16 12 11 17 19 19 2 8 9 5 6 19 9 11 18 6 9 14 16 14 18 17 17 17 5 15 20 19 7 7 10 10 5 14 20 5 19 11 16 16 19 17 9 7 12 14 10 2 11 14 5 20 8 10 11 19 2 14 14 19 17 5 10 8 8 4 2 1 10 20 12 14 11 7 6 6 15 1 5 9 15 3 17 16 17 5 14 11 9 16 15 1 11 10 6 15 7",
"output": "93"
},
{
"input": "100 1\n58 94 18 50 17 14 96 62 83 80 75 5 9 22 25 41 3 96 74 45 66 37 2 37 13 85 68 54 77 11 85 19 25 21 52 59 90 61 72 89 82 22 10 16 3 68 61 29 55 76 28 85 65 76 27 3 14 10 56 37 86 18 35 38 56 68 23 88 33 38 52 87 55 83 94 34 100 41 83 56 91 77 32 74 97 13 67 31 57 81 53 39 5 88 46 1 79 4 49 42",
"output": "77"
},
{
"input": "100 2\n1 51 76 62 34 93 90 43 57 59 52 78 3 48 11 60 57 48 5 54 28 81 87 23 44 77 67 61 14 73 29 53 21 89 67 41 47 9 63 37 1 71 40 85 4 14 77 40 78 75 89 74 4 70 32 65 81 95 49 90 72 41 76 55 69 83 73 84 85 93 46 6 74 90 62 37 97 7 7 37 83 30 37 88 34 16 11 59 85 19 57 63 85 20 63 97 97 65 61 48",
"output": "97"
},
{
"input": "100 3\n30 83 14 55 61 66 34 98 90 62 89 74 45 93 33 31 75 35 82 100 63 69 48 18 99 2 36 71 14 30 70 76 96 85 97 90 49 36 6 76 37 94 70 3 63 73 75 48 39 29 13 2 46 26 9 56 1 18 54 53 85 34 2 12 1 93 75 67 77 77 14 26 33 25 55 9 57 70 75 6 87 66 18 3 41 69 73 24 49 2 20 72 39 58 91 54 74 56 66 78",
"output": "20"
},
{
"input": "100 4\n69 92 76 3 32 50 15 38 21 22 14 3 67 41 95 12 10 62 83 52 78 1 18 58 94 35 62 71 58 75 13 73 60 34 50 97 50 70 19 96 53 10 100 26 20 39 62 59 88 26 24 83 70 68 66 8 6 38 16 93 2 91 81 89 78 74 21 8 31 56 28 53 77 5 81 5 94 42 77 75 92 15 59 36 61 18 55 45 69 68 81 51 12 42 85 74 98 31 17 41",
"output": "97"
},
{
"input": "100 5\n2 72 10 60 6 50 72 34 97 77 35 43 80 64 40 53 46 6 90 22 29 70 26 68 52 19 72 88 83 18 55 32 99 81 11 21 39 42 41 63 60 97 30 23 55 78 89 35 24 50 99 52 27 76 24 8 20 27 51 37 17 82 69 18 46 19 26 77 52 83 76 65 43 66 84 84 13 30 66 88 84 23 37 1 17 26 11 50 73 56 54 37 40 29 35 8 1 39 50 82",
"output": "51"
},
{
"input": "100 7\n6 73 7 54 92 33 66 65 80 47 2 53 28 59 61 16 54 89 37 48 77 40 49 59 27 52 17 22 78 80 81 80 8 93 50 7 87 57 29 16 89 55 20 7 51 54 30 98 44 96 27 70 1 1 32 61 22 92 84 98 31 89 91 90 28 56 49 25 86 49 55 16 19 1 18 8 88 47 16 18 73 86 2 96 16 91 74 49 38 98 94 25 34 85 29 27 99 31 31 58",
"output": "97"
},
{
"input": "100 9\n36 4 45 16 19 6 10 87 44 82 71 49 70 35 83 19 40 76 45 94 44 96 10 54 82 77 86 63 11 37 21 3 15 89 80 88 89 16 72 23 25 9 51 25 10 45 96 5 6 18 51 31 42 57 41 51 42 15 89 61 45 82 16 48 61 67 19 40 9 33 90 36 78 36 79 79 16 10 83 87 9 22 84 12 23 76 36 14 2 81 56 33 56 23 57 84 76 55 35 88",
"output": "47"
},
{
"input": "100 10\n75 81 39 64 90 58 92 28 75 9 96 78 92 83 77 68 76 71 14 46 58 60 80 25 78 11 13 63 22 82 65 68 47 6 33 63 90 50 85 43 73 94 80 48 67 11 83 17 22 15 94 80 66 99 66 4 46 35 52 1 62 39 96 57 37 47 97 49 64 12 36 63 90 16 4 75 85 82 85 56 13 4 92 45 44 93 17 35 22 46 18 44 29 7 52 4 100 98 87 51",
"output": "98"
},
{
"input": "100 20\n21 19 61 70 54 97 98 14 61 72 25 94 24 56 55 25 12 80 76 11 35 17 80 26 11 94 52 47 84 61 10 2 74 25 10 21 2 79 55 50 30 75 10 64 44 5 60 96 52 16 74 41 20 77 20 44 8 86 74 36 49 61 99 13 54 64 19 99 50 43 12 73 48 48 83 55 72 73 63 81 30 27 95 9 97 82 24 3 89 90 33 14 47 88 22 78 12 75 58 67",
"output": "94"
},
{
"input": "100 30\n56 79 59 23 11 23 67 82 81 80 99 79 8 58 93 36 98 81 46 39 34 67 3 50 4 68 70 71 2 21 52 30 75 23 33 21 16 100 56 43 8 27 40 8 56 24 17 40 94 10 67 49 61 36 95 87 17 41 7 94 33 19 17 50 26 11 94 54 38 46 77 9 53 35 98 42 50 20 43 6 78 6 38 24 100 45 43 16 1 50 16 46 14 91 95 88 10 1 50 19",
"output": "95"
},
{
"input": "100 40\n86 11 97 17 38 95 11 5 13 83 67 75 50 2 46 39 84 68 22 85 70 23 64 46 59 93 39 80 35 78 93 21 83 19 64 1 49 59 99 83 44 81 70 58 15 82 83 47 55 65 91 10 2 92 4 77 37 32 12 57 78 11 42 8 59 21 96 69 61 30 44 29 12 70 91 14 10 83 11 75 14 10 19 39 8 98 5 81 66 66 79 55 36 29 22 45 19 24 55 49",
"output": "88"
},
{
"input": "100 50\n22 39 95 69 94 53 80 73 33 90 40 60 2 4 84 50 70 38 92 12 36 74 87 70 51 36 57 5 54 6 35 81 52 17 55 100 95 81 32 76 21 1 100 1 95 1 40 91 98 59 84 19 11 51 79 19 47 86 45 15 62 2 59 77 31 68 71 92 17 33 10 33 85 57 5 2 88 97 91 99 63 20 63 54 79 93 24 62 46 27 30 87 3 64 95 88 16 50 79 1",
"output": "99"
},
{
"input": "100 70\n61 48 89 17 97 6 93 13 64 50 66 88 24 52 46 99 6 65 93 64 82 37 57 41 47 1 84 5 97 83 79 46 16 35 40 7 64 15 44 96 37 17 30 92 51 67 26 3 14 56 27 68 66 93 36 39 51 6 40 55 79 26 71 54 8 48 18 2 71 12 55 60 29 37 31 97 26 37 25 68 67 70 3 87 100 41 5 82 65 92 24 66 76 48 89 8 40 93 31 95",
"output": "100"
},
{
"input": "100 90\n87 32 30 15 10 52 93 63 84 1 82 41 27 51 75 32 42 94 39 53 70 13 4 22 99 35 44 38 5 23 18 100 61 80 9 12 42 93 9 77 3 7 60 95 66 78 95 42 69 8 1 88 93 66 96 20 76 63 15 36 92 52 2 72 36 57 48 63 29 20 74 88 49 47 81 61 94 74 70 93 47 3 19 52 59 41 5 40 22 3 76 97 91 37 95 88 91 99 76 15",
"output": "98"
},
{
"input": "100 100\n79 75 7 28 6 96 38 35 57 95 41 74 24 96 32 78 81 13 63 84 24 95 3 23 66 1 60 6 96 49 41 5 14 18 31 97 66 19 49 89 49 70 51 28 20 99 18 1 28 77 24 46 69 21 40 32 31 66 28 6 66 97 9 16 70 90 91 30 34 82 93 41 65 11 39 52 1 88 63 43 80 50 60 49 28 56 18 76 24 57 74 1 28 99 36 35 79 54 18 16",
"output": "100"
},
{
"input": "1 3\n5",
"output": "1"
},
{
"input": "1 1\n100",
"output": "1"
},
{
"input": "2 3\n4 2",
"output": "1"
},
{
"input": "2 5\n99 97",
"output": "2"
},
{
"input": "3 4\n7 5 2",
"output": "2"
},
{
"input": "3 50\n47 86 51",
"output": "3"
},
{
"input": "5 100\n82 100 85 1 37",
"output": "5"
},
{
"input": "5 20\n40 39 21 5 20",
"output": "3"
},
{
"input": "1 27\n81",
"output": "1"
},
{
"input": "20 13\n7 8 29 83 74 28 93 85 7 8 3 9 8 70 49 50 39 41 57 1",
"output": "7"
},
{
"input": "2 1\n100 2",
"output": "1"
},
{
"input": "2 2\n6 4",
"output": "1"
},
{
"input": "5 2\n6 4 4 1 1",
"output": "1"
},
{
"input": "1 4\n3",
"output": "1"
},
{
"input": "3 2\n1 5 3",
"output": "2"
},
{
"input": "3 1\n3 2 2",
"output": "1"
},
{
"input": "3 1\n2 3 2",
"output": "2"
},
{
"input": "5 1\n5 1 1 1 1",
"output": "1"
},
{
"input": "2 3\n7 4",
"output": "1"
}
] | 1,664,051,627
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 2
| 46
| 0
|
n,m=map(int,input().split())
A=list(map(int,input().split()))
ma=max(A)
ans=0
for i in range(n):
if A[i]>=ma:
ans=max(i+1,ans)
print(ans)
|
Title: Jzzhu and Children
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies.
Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:
1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty.
Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order?
Input Specification:
The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
Output Specification:
Output a single integer, representing the number of the last child.
Demo Input:
['5 2\n1 3 1 4 2\n', '6 4\n1 1 2 2 3 3\n']
Demo Output:
['4\n', '6\n']
Note:
Let's consider the first sample.
Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home.
Child 4 is the last one who goes home.
|
```python
n,m=map(int,input().split())
A=list(map(int,input().split()))
ma=max(A)
ans=0
for i in range(n):
if A[i]>=ma:
ans=max(i+1,ans)
print(ans)
```
| 0
|
|
559
|
B
|
Equivalent Strings
|
PROGRAMMING
| 1,700
|
[
"divide and conquer",
"hashing",
"sortings",
"strings"
] | null | null |
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
|
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
|
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
|
[
"aaba\nabaa\n",
"aabb\nabab\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
| 1,000
|
[
{
"input": "aaba\nabaa",
"output": "YES"
},
{
"input": "aabb\nabab",
"output": "NO"
},
{
"input": "a\na",
"output": "YES"
},
{
"input": "a\nb",
"output": "NO"
},
{
"input": "ab\nab",
"output": "YES"
},
{
"input": "ab\nba",
"output": "YES"
},
{
"input": "ab\nbb",
"output": "NO"
},
{
"input": "zzaa\naazz",
"output": "YES"
},
{
"input": "azza\nzaaz",
"output": "YES"
},
{
"input": "abc\nabc",
"output": "YES"
},
{
"input": "abc\nacb",
"output": "NO"
},
{
"input": "azzz\nzzaz",
"output": "YES"
},
{
"input": "abcd\ndcab",
"output": "YES"
},
{
"input": "abcd\ncdab",
"output": "YES"
},
{
"input": "abcd\ndcba",
"output": "YES"
},
{
"input": "abcd\nacbd",
"output": "NO"
},
{
"input": "oloaxgddgujq\noloaxgujqddg",
"output": "YES"
},
{
"input": "uwzwdxfmosmqatyv\ndxfmzwwusomqvyta",
"output": "YES"
},
{
"input": "hagnzomowtledfdotnll\nledfdotnllomowthagnz",
"output": "YES"
},
{
"input": "snyaydaeobufdg\nsnyaydaeobufdg",
"output": "YES"
},
{
"input": "baaaaa\nabaaaa",
"output": "NO"
},
{
"input": "hhiisug\nmzdjwju",
"output": "NO"
},
{
"input": "bbbabbabaaab\naaaabbabbbbb",
"output": "NO"
},
{
"input": "bbaaab\naababb",
"output": "NO"
},
{
"input": "aabbaaaa\naaaaabab",
"output": "NO"
},
{
"input": "aab\naba",
"output": "NO"
},
{
"input": "abcddd\nbacddd",
"output": "NO"
},
{
"input": "qgiufelsfhanx\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "aabaababaaba\naababaaababa",
"output": "NO"
},
{
"input": "nocdqzdriyyil\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "zdmctxl\nkojqhgw",
"output": "NO"
},
{
"input": "yhwepqwyhwepqwyhwepqweahnqtueahnqtueahnqtuyhwepqwyhwepqwyhwepqwyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtueahnqtueahnqtueahnqtueahnqtu\neahnqtueahnqtueahnqtuyhwepqweahnqtuyhwepqwyhwepqweahnqtuyhwepqweahnqtuyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtuyhwepqwyhwepqwyhwepqw",
"output": "NO"
},
{
"input": "abc\nbac",
"output": "NO"
},
{
"input": "ottceez\npcstdvz",
"output": "NO"
}
] | 1,569,050,315
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 2
| 233
| 5,939,200
|
def isEqual(a,b):
if a==b:return True
if len(a)==1 and len(b)==1:
return False
n=len(a)
a1=a[:n//2]
a2=a[n//2:]
b1=b[:n//2]
b2=b[n//2:]
if isEqual(a1,b1) and isEqual(a2,b2):
return True
if isEqual(a1,b2) and isEqual(a2,b1):
return True
return False
a=input()
b=input()
print("YES" if isEqual(a,b) else "NO")
|
Title: Equivalent Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input Specification:
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
Output Specification:
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Demo Input:
['aaba\nabaa\n', 'aabb\nabab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
|
```python
def isEqual(a,b):
if a==b:return True
if len(a)==1 and len(b)==1:
return False
n=len(a)
a1=a[:n//2]
a2=a[n//2:]
b1=b[:n//2]
b2=b[n//2:]
if isEqual(a1,b1) and isEqual(a2,b2):
return True
if isEqual(a1,b2) and isEqual(a2,b1):
return True
return False
a=input()
b=input()
print("YES" if isEqual(a,b) else "NO")
```
| -1
|
|
583
|
A
|
Asphalting Roads
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
City X consists of *n* vertical and *n* horizontal infinite roads, forming *n*<=×<=*n* intersections. Roads (both vertical and horizontal) are numbered from 1 to *n*, and the intersections are indicated by the numbers of the roads that form them.
Sand roads have long been recognized out of date, so the decision was made to asphalt them. To do this, a team of workers was hired and a schedule of work was made, according to which the intersections should be asphalted.
Road repairs are planned for *n*2 days. On the *i*-th day of the team arrives at the *i*-th intersection in the list and if none of the two roads that form the intersection were already asphalted they asphalt both roads. Otherwise, the team leaves the intersection, without doing anything with the roads.
According to the schedule of road works tell in which days at least one road will be asphalted.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of vertical and horizontal roads in the city.
Next *n*2 lines contain the order of intersections in the schedule. The *i*-th of them contains two numbers *h**i*,<=*v**i* (1<=≤<=*h**i*,<=*v**i*<=≤<=*n*), separated by a space, and meaning that the intersection that goes *i*-th in the timetable is at the intersection of the *h**i*-th horizontal and *v**i*-th vertical roads. It is guaranteed that all the intersections in the timetable are distinct.
|
In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1.
|
[
"2\n1 1\n1 2\n2 1\n2 2\n",
"1\n1 1\n"
] |
[
"1 4 \n",
"1 \n"
] |
In the sample the brigade acts like that:
1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road; 1. On the second day the brigade of the workers comes to the intersection of the 1-st horizontal and the 2-nd vertical road. The 2-nd vertical road hasn't been asphalted, but as the 1-st horizontal road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the third day the brigade of the workers come to the intersection of the 2-nd horizontal and the 1-st vertical road. The 2-nd horizontal road hasn't been asphalted but as the 1-st vertical road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the fourth day the brigade come to the intersection formed by the intersection of the 2-nd horizontal and 2-nd vertical road. As none of them has been asphalted, the workers asphalt the 2-nd vertical and the 2-nd horizontal road.
| 500
|
[
{
"input": "2\n1 1\n1 2\n2 1\n2 2",
"output": "1 4 "
},
{
"input": "1\n1 1",
"output": "1 "
},
{
"input": "2\n1 1\n2 2\n1 2\n2 1",
"output": "1 2 "
},
{
"input": "2\n1 2\n2 2\n2 1\n1 1",
"output": "1 3 "
},
{
"input": "3\n2 2\n1 2\n3 2\n3 3\n1 1\n2 3\n1 3\n3 1\n2 1",
"output": "1 4 5 "
},
{
"input": "3\n1 3\n3 1\n2 1\n1 1\n1 2\n2 2\n3 2\n3 3\n2 3",
"output": "1 2 6 "
},
{
"input": "4\n1 3\n2 3\n2 4\n4 4\n3 1\n1 1\n3 4\n2 1\n1 4\n4 3\n4 1\n3 2\n1 2\n4 2\n2 2\n3 3",
"output": "1 3 5 14 "
},
{
"input": "4\n3 3\n4 2\n2 3\n3 4\n4 4\n1 2\n3 2\n2 2\n1 4\n3 1\n4 1\n2 1\n1 3\n1 1\n4 3\n2 4",
"output": "1 2 9 12 "
},
{
"input": "9\n4 5\n2 3\n8 3\n5 6\n9 3\n4 4\n5 4\n4 7\n1 7\n8 4\n1 4\n1 5\n5 7\n7 8\n7 1\n9 9\n8 7\n7 5\n3 7\n6 6\n7 3\n5 2\n3 6\n7 4\n9 6\n5 8\n9 7\n6 3\n7 9\n1 2\n1 1\n6 2\n5 3\n7 2\n1 6\n4 1\n6 1\n8 9\n2 2\n3 9\n2 9\n7 7\n2 8\n9 4\n2 5\n8 6\n3 4\n2 1\n2 7\n6 5\n9 1\n3 3\n3 8\n5 5\n4 3\n3 1\n1 9\n6 4\n3 2\n6 8\n2 6\n5 9\n8 5\n8 8\n9 5\n6 9\n9 2\n3 5\n4 9\n4 8\n2 4\n5 1\n4 6\n7 6\n9 8\n1 3\n4 2\n8 1\n8 2\n6 7\n1 8",
"output": "1 2 4 9 10 14 16 32 56 "
},
{
"input": "8\n1 1\n1 2\n1 3\n1 4\n1 5\n8 6\n1 7\n1 8\n2 1\n8 5\n2 3\n2 4\n2 5\n2 6\n4 3\n2 2\n3 1\n3 2\n3 3\n3 4\n3 5\n3 6\n5 6\n3 8\n4 1\n4 2\n2 7\n4 4\n8 8\n4 6\n4 7\n4 8\n5 1\n5 2\n5 3\n6 5\n5 5\n3 7\n5 7\n5 8\n6 1\n6 2\n6 3\n6 4\n5 4\n6 6\n6 7\n6 8\n7 1\n7 2\n7 3\n7 4\n7 5\n7 6\n7 7\n7 8\n8 1\n8 2\n8 3\n8 4\n2 8\n1 6\n8 7\n4 5",
"output": "1 6 11 18 28 36 39 56 "
},
{
"input": "9\n9 9\n5 5\n8 8\n3 3\n2 2\n6 6\n4 4\n1 1\n7 7\n8 4\n1 4\n1 5\n5 7\n7 8\n7 1\n1 7\n8 7\n7 5\n3 7\n5 6\n7 3\n5 2\n3 6\n7 4\n9 6\n5 8\n9 7\n6 3\n7 9\n1 2\n4 5\n6 2\n5 3\n7 2\n1 6\n4 1\n6 1\n8 9\n2 3\n3 9\n2 9\n5 4\n2 8\n9 4\n2 5\n8 6\n3 4\n2 1\n2 7\n6 5\n9 1\n8 3\n3 8\n9 3\n4 3\n3 1\n1 9\n6 4\n3 2\n6 8\n2 6\n5 9\n8 5\n4 7\n9 5\n6 9\n9 2\n3 5\n4 9\n4 8\n2 4\n5 1\n4 6\n7 6\n9 8\n1 3\n4 2\n8 1\n8 2\n6 7\n1 8",
"output": "1 2 3 4 5 6 7 8 9 "
}
] | 1,443,891,150
| 450
|
Python 3
|
OK
|
TESTS
| 39
| 62
| 0
|
n = int(input())
res = []
x = set()
y = set()
for i in range(n*n):
a,b = map(int,input().split())
if not (a in x or b in y):
x.add(a)
y.add(b)
res.append(i+1)
print(*res)
|
Title: Asphalting Roads
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
City X consists of *n* vertical and *n* horizontal infinite roads, forming *n*<=×<=*n* intersections. Roads (both vertical and horizontal) are numbered from 1 to *n*, and the intersections are indicated by the numbers of the roads that form them.
Sand roads have long been recognized out of date, so the decision was made to asphalt them. To do this, a team of workers was hired and a schedule of work was made, according to which the intersections should be asphalted.
Road repairs are planned for *n*2 days. On the *i*-th day of the team arrives at the *i*-th intersection in the list and if none of the two roads that form the intersection were already asphalted they asphalt both roads. Otherwise, the team leaves the intersection, without doing anything with the roads.
According to the schedule of road works tell in which days at least one road will be asphalted.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of vertical and horizontal roads in the city.
Next *n*2 lines contain the order of intersections in the schedule. The *i*-th of them contains two numbers *h**i*,<=*v**i* (1<=≤<=*h**i*,<=*v**i*<=≤<=*n*), separated by a space, and meaning that the intersection that goes *i*-th in the timetable is at the intersection of the *h**i*-th horizontal and *v**i*-th vertical roads. It is guaranteed that all the intersections in the timetable are distinct.
Output Specification:
In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1.
Demo Input:
['2\n1 1\n1 2\n2 1\n2 2\n', '1\n1 1\n']
Demo Output:
['1 4 \n', '1 \n']
Note:
In the sample the brigade acts like that:
1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road; 1. On the second day the brigade of the workers comes to the intersection of the 1-st horizontal and the 2-nd vertical road. The 2-nd vertical road hasn't been asphalted, but as the 1-st horizontal road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the third day the brigade of the workers come to the intersection of the 2-nd horizontal and the 1-st vertical road. The 2-nd horizontal road hasn't been asphalted but as the 1-st vertical road has been asphalted on the first day, the workers leave and do not asphalt anything; 1. On the fourth day the brigade come to the intersection formed by the intersection of the 2-nd horizontal and 2-nd vertical road. As none of them has been asphalted, the workers asphalt the 2-nd vertical and the 2-nd horizontal road.
|
```python
n = int(input())
res = []
x = set()
y = set()
for i in range(n*n):
a,b = map(int,input().split())
if not (a in x or b in y):
x.add(a)
y.add(b)
res.append(i+1)
print(*res)
```
| 3
|
|
259
|
B
|
Little Elephant and Magic Square
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation"
] | null | null |
Little Elephant loves magic squares very much.
A magic square is a 3<=×<=3 table, each cell contains some positive integer. At that the sums of integers in all rows, columns and diagonals of the table are equal. The figure below shows the magic square, the sum of integers in all its rows, columns and diagonals equals 15.
The Little Elephant remembered one magic square. He started writing this square on a piece of paper, but as he wrote, he forgot all three elements of the main diagonal of the magic square. Fortunately, the Little Elephant clearly remembered that all elements of the magic square did not exceed 105.
Help the Little Elephant, restore the original magic square, given the Elephant's notes.
|
The first three lines of the input contain the Little Elephant's notes. The first line contains elements of the first row of the magic square. The second line contains the elements of the second row, the third line is for the third row. The main diagonal elements that have been forgotten by the Elephant are represented by zeroes.
It is guaranteed that the notes contain exactly three zeroes and they are all located on the main diagonal. It is guaranteed that all positive numbers in the table do not exceed 105.
|
Print three lines, in each line print three integers — the Little Elephant's magic square. If there are multiple magic squares, you are allowed to print any of them. Note that all numbers you print must be positive and not exceed 105.
It is guaranteed that there exists at least one magic square that meets the conditions.
|
[
"0 1 1\n1 0 1\n1 1 0\n",
"0 3 6\n5 0 5\n4 7 0\n"
] |
[
"1 1 1\n1 1 1\n1 1 1\n",
"6 3 6\n5 5 5\n4 7 4\n"
] |
none
| 1,000
|
[
{
"input": "0 1 1\n1 0 1\n1 1 0",
"output": "1 1 1\n1 1 1\n1 1 1"
},
{
"input": "0 3 6\n5 0 5\n4 7 0",
"output": "6 3 6\n5 5 5\n4 7 4"
},
{
"input": "0 4 4\n4 0 4\n4 4 0",
"output": "4 4 4\n4 4 4\n4 4 4"
},
{
"input": "0 54 48\n36 0 78\n66 60 0",
"output": "69 54 48\n36 57 78\n66 60 45"
},
{
"input": "0 17 14\n15 0 15\n16 13 0",
"output": "14 17 14\n15 15 15\n16 13 16"
},
{
"input": "0 97 56\n69 0 71\n84 43 0",
"output": "57 97 56\n69 70 71\n84 43 83"
},
{
"input": "0 1099 1002\n1027 0 1049\n1074 977 0",
"output": "1013 1099 1002\n1027 1038 1049\n1074 977 1063"
},
{
"input": "0 98721 99776\n99575 0 99123\n98922 99977 0",
"output": "99550 98721 99776\n99575 99349 99123\n98922 99977 99148"
},
{
"input": "0 6361 2304\n1433 0 8103\n7232 3175 0",
"output": "5639 6361 2304\n1433 4768 8103\n7232 3175 3897"
},
{
"input": "0 99626 99582\n99766 0 99258\n99442 99398 0",
"output": "99328 99626 99582\n99766 99512 99258\n99442 99398 99696"
},
{
"input": "0 99978 99920\n99950 0 99918\n99948 99890 0",
"output": "99904 99978 99920\n99950 99934 99918\n99948 99890 99964"
},
{
"input": "0 840 666\n612 0 948\n894 720 0",
"output": "834 840 666\n612 780 948\n894 720 726"
},
{
"input": "0 28 10\n12 0 24\n26 8 0",
"output": "16 28 10\n12 18 24\n26 8 20"
},
{
"input": "0 120 83\n98 0 90\n105 68 0",
"output": "79 120 83\n98 94 90\n105 68 109"
},
{
"input": "0 86900 85807\n85836 0 86842\n86871 85778 0",
"output": "86310 86900 85807\n85836 86339 86842\n86871 85778 86368"
},
{
"input": "0 74 78\n78 0 74\n74 78 0",
"output": "76 74 78\n78 76 74\n74 78 76"
},
{
"input": "0 505 681\n605 0 657\n581 757 0",
"output": "707 505 681\n605 631 657\n581 757 555"
},
{
"input": "0 662 918\n822 0 854\n758 1014 0",
"output": "934 662 918\n822 838 854\n758 1014 742"
},
{
"input": "0 93 95\n93 0 97\n95 97 0",
"output": "97 93 95\n93 95 97\n95 97 93"
},
{
"input": "0 709 712\n719 0 695\n702 705 0",
"output": "700 709 712\n719 707 695\n702 705 714"
},
{
"input": "0 7 6\n9 0 1\n4 3 0",
"output": "2 7 6\n9 5 1\n4 3 8"
},
{
"input": "0 9 2\n3 0 7\n8 1 0",
"output": "4 9 2\n3 5 7\n8 1 6"
},
{
"input": "0 1 43\n13 0 61\n31 73 0",
"output": "67 1 43\n13 37 61\n31 73 7"
},
{
"input": "0 100000 100000\n100000 0 100000\n100000 100000 0",
"output": "100000 100000 100000\n100000 100000 100000\n100000 100000 100000"
}
] | 1,686,636,493
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 24
| 62
| 0
|
x11,x12,x13 = map(int, input().split())
x21,x22,x23 = map(int, input().split())
x31, x32, x33 = map(int, input().split())
su = (x11+x12+x13+x21+x22+x23+x31+x32+x33)//2
x11 = su - x12 - x13
x22 = su - x21 - x23
x33 = su - x31 - x32
print(x11,x12,x13)
print(x21,x22,x23)
print(x31,x32,x33)
|
Title: Little Elephant and Magic Square
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Elephant loves magic squares very much.
A magic square is a 3<=×<=3 table, each cell contains some positive integer. At that the sums of integers in all rows, columns and diagonals of the table are equal. The figure below shows the magic square, the sum of integers in all its rows, columns and diagonals equals 15.
The Little Elephant remembered one magic square. He started writing this square on a piece of paper, but as he wrote, he forgot all three elements of the main diagonal of the magic square. Fortunately, the Little Elephant clearly remembered that all elements of the magic square did not exceed 105.
Help the Little Elephant, restore the original magic square, given the Elephant's notes.
Input Specification:
The first three lines of the input contain the Little Elephant's notes. The first line contains elements of the first row of the magic square. The second line contains the elements of the second row, the third line is for the third row. The main diagonal elements that have been forgotten by the Elephant are represented by zeroes.
It is guaranteed that the notes contain exactly three zeroes and they are all located on the main diagonal. It is guaranteed that all positive numbers in the table do not exceed 105.
Output Specification:
Print three lines, in each line print three integers — the Little Elephant's magic square. If there are multiple magic squares, you are allowed to print any of them. Note that all numbers you print must be positive and not exceed 105.
It is guaranteed that there exists at least one magic square that meets the conditions.
Demo Input:
['0 1 1\n1 0 1\n1 1 0\n', '0 3 6\n5 0 5\n4 7 0\n']
Demo Output:
['1 1 1\n1 1 1\n1 1 1\n', '6 3 6\n5 5 5\n4 7 4\n']
Note:
none
|
```python
x11,x12,x13 = map(int, input().split())
x21,x22,x23 = map(int, input().split())
x31, x32, x33 = map(int, input().split())
su = (x11+x12+x13+x21+x22+x23+x31+x32+x33)//2
x11 = su - x12 - x13
x22 = su - x21 - x23
x33 = su - x31 - x32
print(x11,x12,x13)
print(x21,x22,x23)
print(x31,x32,x33)
```
| 3
|
|
432
|
A
|
Choosing Teams
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times?
|
The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship.
|
Print a single number — the answer to the problem.
|
[
"5 2\n0 4 5 1 0\n",
"6 4\n0 1 2 3 4 5\n",
"6 5\n0 0 0 0 0 0\n"
] |
[
"1\n",
"0\n",
"2\n"
] |
In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits.
| 500
|
[
{
"input": "5 2\n0 4 5 1 0",
"output": "1"
},
{
"input": "6 4\n0 1 2 3 4 5",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0 0",
"output": "2"
},
{
"input": "3 4\n0 1 0",
"output": "1"
},
{
"input": "3 4\n0 2 0",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0 0",
"output": "2"
},
{
"input": "12 2\n0 1 2 3 4 0 1 2 3 4 0 1",
"output": "3"
},
{
"input": "15 2\n0 1 2 3 4 0 1 2 3 4 0 1 2 3 4",
"output": "4"
},
{
"input": "13 1\n5 0 5 0 1 5 0 4 1 1 1 4 1",
"output": "3"
},
{
"input": "20 1\n5 0 4 2 2 3 2 1 2 4 1 3 5 5 5 4 4 1 3 0",
"output": "5"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "20 2\n1 4 5 1 3 4 0 5 0 3 3 0 4 0 0 1 4 2 0 4",
"output": "4"
},
{
"input": "6 3\n4 4 4 4 4 4",
"output": "0"
},
{
"input": "1 4\n1",
"output": "0"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "2 3\n0 0",
"output": "0"
}
] | 1,654,109,689
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 46
| 0
|
n,k = map(int,input().split(" "))
y = list(map(int,input().split(" ")))
y.sort()
teams = 0
i = 0
while i < n :
if y[i] + k <= 5:
teams += 1
i += 1
print(teams//3)
|
Title: Choosing Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times?
Input Specification:
The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship.
Output Specification:
Print a single number — the answer to the problem.
Demo Input:
['5 2\n0 4 5 1 0\n', '6 4\n0 1 2 3 4 5\n', '6 5\n0 0 0 0 0 0\n']
Demo Output:
['1\n', '0\n', '2\n']
Note:
In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits.
|
```python
n,k = map(int,input().split(" "))
y = list(map(int,input().split(" ")))
y.sort()
teams = 0
i = 0
while i < n :
if y[i] + k <= 5:
teams += 1
i += 1
print(teams//3)
```
| 3
|
|
200
|
B
|
Drinks
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink.
|
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space.
|
Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4.
|
[
"3\n50 50 100\n",
"4\n0 25 50 75\n"
] |
[
"66.666666666667\n",
"37.500000000000\n"
] |
Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent.
| 500
|
[
{
"input": "3\n50 50 100",
"output": "66.666666666667"
},
{
"input": "4\n0 25 50 75",
"output": "37.500000000000"
},
{
"input": "3\n0 1 8",
"output": "3.000000000000"
},
{
"input": "5\n96 89 93 95 70",
"output": "88.600000000000"
},
{
"input": "7\n62 41 78 4 38 39 75",
"output": "48.142857142857"
},
{
"input": "13\n2 22 7 0 1 17 3 17 11 2 21 26 22",
"output": "11.615384615385"
},
{
"input": "21\n5 4 11 7 0 5 45 21 0 14 51 6 0 16 10 19 8 9 7 12 18",
"output": "12.761904761905"
},
{
"input": "26\n95 70 93 74 94 70 91 70 39 79 80 57 87 75 37 93 48 67 51 90 85 26 23 64 66 84",
"output": "69.538461538462"
},
{
"input": "29\n84 99 72 96 83 92 95 98 97 93 76 84 99 93 81 76 93 99 99 100 95 100 96 95 97 100 71 98 94",
"output": "91.551724137931"
},
{
"input": "33\n100 99 100 100 99 99 99 100 100 100 99 99 99 100 100 100 100 99 100 99 100 100 97 100 100 100 100 100 100 100 98 98 100",
"output": "99.515151515152"
},
{
"input": "34\n14 9 10 5 4 26 18 23 0 1 0 20 18 15 2 2 3 5 14 1 9 4 2 15 7 1 7 19 10 0 0 11 0 2",
"output": "8.147058823529"
},
{
"input": "38\n99 98 100 100 99 92 99 99 98 84 88 94 86 99 93 100 98 99 65 98 85 84 64 97 96 89 79 96 91 84 99 93 72 96 94 97 96 93",
"output": "91.921052631579"
},
{
"input": "52\n100 94 99 98 99 99 99 95 97 97 98 100 100 98 97 100 98 90 100 99 97 94 90 98 100 100 90 99 100 95 98 95 94 85 97 94 96 94 99 99 99 98 100 100 94 99 99 100 98 87 100 100",
"output": "97.019230769231"
},
{
"input": "58\n10 70 12 89 1 82 100 53 40 100 21 69 92 91 67 66 99 77 25 48 8 63 93 39 46 79 82 14 44 42 1 79 0 69 56 73 67 17 59 4 65 80 20 60 77 52 3 61 16 76 33 18 46 100 28 59 9 6",
"output": "50.965517241379"
},
{
"input": "85\n7 8 1 16 0 15 1 7 0 11 15 6 2 12 2 8 9 8 2 0 3 7 15 7 1 8 5 7 2 26 0 3 11 1 8 10 31 0 7 6 1 8 1 0 9 14 4 8 7 16 9 1 0 16 10 9 6 1 1 4 2 7 4 5 4 1 20 6 16 16 1 1 10 17 8 12 14 19 3 8 1 7 10 23 10",
"output": "7.505882352941"
},
{
"input": "74\n5 3 0 7 13 10 12 10 18 5 0 18 2 13 7 17 2 7 5 2 40 19 0 2 2 3 0 45 4 20 0 4 2 8 1 19 3 9 17 1 15 0 16 1 9 4 0 9 32 2 6 18 11 18 1 15 16 12 7 19 5 3 9 28 26 8 3 10 33 29 4 13 28 6",
"output": "10.418918918919"
},
{
"input": "98\n42 9 21 11 9 11 22 12 52 20 10 6 56 9 26 27 1 29 29 14 38 17 41 21 7 45 15 5 29 4 51 20 6 8 34 17 13 53 30 45 0 10 16 41 4 5 6 4 14 2 31 6 0 11 13 3 3 43 13 36 51 0 7 16 28 23 8 36 30 22 8 54 21 45 39 4 50 15 1 30 17 8 18 10 2 20 16 50 6 68 15 6 38 7 28 8 29 41",
"output": "20.928571428571"
},
{
"input": "99\n60 65 40 63 57 44 30 84 3 10 39 53 40 45 72 20 76 11 61 32 4 26 97 55 14 57 86 96 34 69 52 22 26 79 31 4 21 35 82 47 81 28 72 70 93 84 40 4 69 39 83 58 30 7 32 73 74 12 92 23 61 88 9 58 70 32 75 40 63 71 46 55 39 36 14 97 32 16 95 41 28 20 85 40 5 50 50 50 75 6 10 64 38 19 77 91 50 72 96",
"output": "49.191919191919"
},
{
"input": "99\n100 88 40 30 81 80 91 98 69 73 88 96 79 58 14 100 87 84 52 91 83 88 72 83 99 35 54 80 46 79 52 72 85 32 99 39 79 79 45 83 88 50 75 75 50 59 65 75 97 63 92 58 89 46 93 80 89 33 69 86 99 99 66 85 72 74 79 98 85 95 46 63 77 97 49 81 89 39 70 76 68 91 90 56 31 93 51 87 73 95 74 69 87 95 57 68 49 95 92",
"output": "73.484848484848"
},
{
"input": "100\n18 15 17 0 3 3 0 4 1 8 2 22 7 21 5 0 0 8 3 16 1 0 2 9 9 3 10 8 17 20 5 4 8 12 2 3 1 1 3 2 23 0 1 0 5 7 4 0 1 3 3 4 25 2 2 14 8 4 9 3 0 11 0 3 12 3 14 16 7 7 14 1 17 9 0 35 42 12 3 1 25 9 3 8 5 3 2 8 22 14 11 6 3 9 6 8 7 7 4 6",
"output": "7.640000000000"
},
{
"input": "100\n88 77 65 87 100 63 91 96 92 89 77 95 76 80 84 83 100 71 85 98 26 54 74 78 69 59 96 86 88 91 95 26 52 88 64 70 84 81 76 84 94 82 100 66 97 98 43 94 59 94 100 80 98 73 69 83 94 70 74 79 91 31 62 88 69 55 62 97 40 64 62 83 87 85 50 90 69 72 67 49 100 51 69 96 81 90 83 91 86 34 79 69 100 66 97 98 47 97 74 100",
"output": "77.660000000000"
},
{
"input": "100\n91 92 90 91 98 84 85 96 83 98 99 87 94 70 87 75 86 90 89 88 82 83 91 94 88 86 90 99 100 98 97 75 95 99 95 100 91 92 76 93 95 97 88 93 95 81 96 89 88 100 98 87 90 96 100 99 58 90 96 77 92 82 100 100 93 93 98 99 79 88 97 95 98 66 96 83 96 100 99 92 98 98 92 93 100 97 98 100 98 97 100 100 94 90 99 100 98 79 80 81",
"output": "91.480000000000"
},
{
"input": "1\n0",
"output": "0.000000000000"
},
{
"input": "1\n100",
"output": "100.000000000000"
},
{
"input": "1\n78",
"output": "78.000000000000"
},
{
"input": "2\n0 100",
"output": "50.000000000000"
},
{
"input": "2\n100 100",
"output": "100.000000000000"
},
{
"input": "5\n0 0 0 0 1",
"output": "0.200000000000"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99",
"output": "99.990000000000"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1",
"output": "0.010000000000"
},
{
"input": "5\n100 100 100 100 100",
"output": "100.000000000000"
}
] | 1,694,332,859
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 92
| 0
|
a = int(input())
b = list(map(int,input().split()))
print((sum(b)*100)/(a*100))
|
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
a = int(input())
b = list(map(int,input().split()))
print((sum(b)*100)/(a*100))
```
| 3
|
|
121
|
A
|
Lucky Sum
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let *next*(*x*) be the minimum lucky number which is larger than or equals *x*. Petya is interested what is the value of the expression *next*(*l*)<=+<=*next*(*l*<=+<=1)<=+<=...<=+<=*next*(*r*<=-<=1)<=+<=*next*(*r*). Help him solve this problem.
|
The single line contains two integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=109) — the left and right interval limits.
|
In the single line print the only number — the sum *next*(*l*)<=+<=*next*(*l*<=+<=1)<=+<=...<=+<=*next*(*r*<=-<=1)<=+<=*next*(*r*).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
|
[
"2 7\n",
"7 7\n"
] |
[
"33\n",
"7\n"
] |
In the first sample: *next*(2) + *next*(3) + *next*(4) + *next*(5) + *next*(6) + *next*(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: *next*(7) = 7
| 500
|
[
{
"input": "2 7",
"output": "33"
},
{
"input": "7 7",
"output": "7"
},
{
"input": "1 9",
"output": "125"
},
{
"input": "4 7",
"output": "25"
},
{
"input": "12 47",
"output": "1593"
},
{
"input": "6 77",
"output": "4012"
},
{
"input": "1 100",
"output": "14247"
},
{
"input": "1000000000 1000000000",
"output": "4444444444"
},
{
"input": "77 77",
"output": "77"
},
{
"input": "69 788",
"output": "452195"
},
{
"input": "474 747",
"output": "202794"
},
{
"input": "4 77777",
"output": "4070145675"
},
{
"input": "1 1000000",
"output": "1394675359387"
},
{
"input": "47 744447",
"output": "381286992761"
},
{
"input": "47444 1000000000",
"output": "1394672348253941136"
},
{
"input": "48 854888",
"output": "749733199853"
},
{
"input": "854444 985555",
"output": "582719941728"
},
{
"input": "774744 774747",
"output": "3098985"
},
{
"input": "654 987654",
"output": "1339803940266"
},
{
"input": "477777 1000000000",
"output": "1394672167300009765"
},
{
"input": "77777 777777777",
"output": "407018021649898097"
},
{
"input": "963 85555574",
"output": "7526978888069560"
},
{
"input": "47 7444",
"output": "38125896"
},
{
"input": "1 1000000000",
"output": "1394672350065645019"
},
{
"input": "474 999888555",
"output": "1394177038954402791"
},
{
"input": "777 1000000000",
"output": "1394672350065239125"
},
{
"input": "7 777777774",
"output": "407018023386632646"
},
{
"input": "369 852",
"output": "606732"
},
{
"input": "47 47",
"output": "47"
},
{
"input": "747 748",
"output": "1521"
},
{
"input": "77777440 77777444",
"output": "388887220"
},
{
"input": "987545885 999875584",
"output": "54798666661186800"
},
{
"input": "2 777777776",
"output": "407018024942188226"
},
{
"input": "1 1",
"output": "4"
},
{
"input": "1 2",
"output": "8"
},
{
"input": "999999999 1000000000",
"output": "8888888888"
},
{
"input": "777777777 1000000000",
"output": "987654325123456789"
},
{
"input": "777777778 1000000000",
"output": "987654324345679012"
},
{
"input": "5 6",
"output": "14"
},
{
"input": "49 49",
"output": "74"
},
{
"input": "6 6",
"output": "7"
},
{
"input": "3 4",
"output": "8"
}
] | 1,629,070,392
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 14
| 2,000
| 22,220,800
|
from itertools import product
def main():
a, b = map(int, input().split())
chars = ['4', '7']
res = []
for i in range(1, 11):
for j in product(chars, repeat = i):
res.append(int(''.join(j)))
ans = 0
start = 0
while a <= b:
if res[start] < a:
start += 1
elif res[start] >= a:
ans += res[start]
a += 1
print(ans)
main()
|
Title: Lucky Sum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let *next*(*x*) be the minimum lucky number which is larger than or equals *x*. Petya is interested what is the value of the expression *next*(*l*)<=+<=*next*(*l*<=+<=1)<=+<=...<=+<=*next*(*r*<=-<=1)<=+<=*next*(*r*). Help him solve this problem.
Input Specification:
The single line contains two integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=109) — the left and right interval limits.
Output Specification:
In the single line print the only number — the sum *next*(*l*)<=+<=*next*(*l*<=+<=1)<=+<=...<=+<=*next*(*r*<=-<=1)<=+<=*next*(*r*).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
Demo Input:
['2 7\n', '7 7\n']
Demo Output:
['33\n', '7\n']
Note:
In the first sample: *next*(2) + *next*(3) + *next*(4) + *next*(5) + *next*(6) + *next*(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: *next*(7) = 7
|
```python
from itertools import product
def main():
a, b = map(int, input().split())
chars = ['4', '7']
res = []
for i in range(1, 11):
for j in product(chars, repeat = i):
res.append(int(''.join(j)))
ans = 0
start = 0
while a <= b:
if res[start] < a:
start += 1
elif res[start] >= a:
ans += res[start]
a += 1
print(ans)
main()
```
| 0
|
|
888
|
C
|
K-Dominant Character
|
PROGRAMMING
| 1,400
|
[
"binary search",
"implementation",
"two pointers"
] | null | null |
You are given a string *s* consisting of lowercase Latin letters. Character *c* is called *k*-dominant iff each substring of *s* with length at least *k* contains this character *c*.
You have to find minimum *k* such that there exists at least one *k*-dominant character.
|
The first line contains string *s* consisting of lowercase Latin letters (1<=≤<=|*s*|<=≤<=100000).
|
Print one number — the minimum value of *k* such that there exists at least one *k*-dominant character.
|
[
"abacaba\n",
"zzzzz\n",
"abcde\n"
] |
[
"2\n",
"1\n",
"3\n"
] |
none
| 0
|
[
{
"input": "abacaba",
"output": "2"
},
{
"input": "zzzzz",
"output": "1"
},
{
"input": "abcde",
"output": "3"
},
{
"input": "bcaccacaaabaacaabaaabcbbcbcaacacbcbaaaacccacbbcbaabcbacaacbabacacacaccbbccbcbacbbbbccccabcabaaab",
"output": "8"
},
{
"input": "daabcdabbabbacacbaacabacbcaabaacac",
"output": "4"
},
{
"input": "abghim",
"output": "4"
},
{
"input": "gfliflgfhhdkceacdljgkegmdlhcgkcmlelmbbbmdddgdeeljjhgbbffmemmmkhebgkhadkdajabcjkcgbkgbaeacdedlkklfech",
"output": "17"
},
{
"input": "a",
"output": "1"
},
{
"input": "aaabbb",
"output": "4"
},
{
"input": "ab",
"output": "2"
},
{
"input": "abbb",
"output": "2"
},
{
"input": "ob",
"output": "2"
},
{
"input": "abccba",
"output": "3"
},
{
"input": "saaaaaaaas",
"output": "2"
},
{
"input": "axxx",
"output": "2"
},
{
"input": "abcba",
"output": "2"
},
{
"input": "abb",
"output": "2"
},
{
"input": "abcdea",
"output": "4"
},
{
"input": "axcbb",
"output": "3"
},
{
"input": "tmivvdcbbfrfogjviirrximhttoskopwrcmkcborcxvr",
"output": "13"
},
{
"input": "dlchmmuateksgldkckljrovmeuniobjrelqjpnjljlvhpqrjsfklliqpufplgaelevmlcnodbllquubobecem",
"output": "15"
},
{
"input": "zx",
"output": "2"
},
{
"input": "zxz",
"output": "2"
},
{
"input": "bbhdgaefbhf",
"output": "6"
},
{
"input": "xx",
"output": "1"
},
{
"input": "zxx",
"output": "2"
},
{
"input": "cbcada",
"output": "4"
},
{
"input": "z",
"output": "1"
},
{
"input": "aab",
"output": "2"
},
{
"input": "jficc",
"output": "3"
},
{
"input": "cceeaabfba",
"output": "5"
},
{
"input": "jk",
"output": "2"
},
{
"input": "lzeznbwu",
"output": "5"
},
{
"input": "rr",
"output": "1"
}
] | 1,513,930,589
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 5,529,600
|
i=input()
s=set(i)
def F(c):
l=[i+1 for i,x in enumerate(i) if x==c]
d=(y-x for x,y in zip(l,l[1:]))
return max(l[0],len(i)-l[-1]+1,*d)
print(max(map(F,s)))
|
Title: K-Dominant Character
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string *s* consisting of lowercase Latin letters. Character *c* is called *k*-dominant iff each substring of *s* with length at least *k* contains this character *c*.
You have to find minimum *k* such that there exists at least one *k*-dominant character.
Input Specification:
The first line contains string *s* consisting of lowercase Latin letters (1<=≤<=|*s*|<=≤<=100000).
Output Specification:
Print one number — the minimum value of *k* such that there exists at least one *k*-dominant character.
Demo Input:
['abacaba\n', 'zzzzz\n', 'abcde\n']
Demo Output:
['2\n', '1\n', '3\n']
Note:
none
|
```python
i=input()
s=set(i)
def F(c):
l=[i+1 for i,x in enumerate(i) if x==c]
d=(y-x for x,y in zip(l,l[1:]))
return max(l[0],len(i)-l[-1]+1,*d)
print(max(map(F,s)))
```
| 0
|
|
356
|
A
|
Knight Tournament
|
PROGRAMMING
| 1,500
|
[
"data structures",
"dsu"
] | null | null |
Hooray! Berl II, the king of Berland is making a knight tournament. The king has already sent the message to all knights in the kingdom and they in turn agreed to participate in this grand event.
As for you, you're just a simple peasant. There's no surprise that you slept in this morning and were late for the tournament (it was a weekend, after all). Now you are really curious about the results of the tournament. This time the tournament in Berland went as follows:
- There are *n* knights participating in the tournament. Each knight was assigned his unique number — an integer from 1 to *n*. - The tournament consisted of *m* fights, in the *i*-th fight the knights that were still in the game with numbers at least *l**i* and at most *r**i* have fought for the right to continue taking part in the tournament. - After the *i*-th fight among all participants of the fight only one knight won — the knight number *x**i*, he continued participating in the tournament. Other knights left the tournament. - The winner of the last (the *m*-th) fight (the knight number *x**m*) became the winner of the tournament.
You fished out all the information about the fights from your friends. Now for each knight you want to know the name of the knight he was conquered by. We think that the knight number *b* was conquered by the knight number *a*, if there was a fight with both of these knights present and the winner was the knight number *a*.
Write the code that calculates for each knight, the name of the knight that beat him.
|
The first line contains two integers *n*, *m* (2<=≤<=*n*<=≤<=3·105; 1<=≤<=*m*<=≤<=3·105) — the number of knights and the number of fights. Each of the following *m* lines contains three integers *l**i*,<=*r**i*,<=*x**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=*n*; *l**i*<=≤<=*x**i*<=≤<=*r**i*) — the description of the *i*-th fight.
It is guaranteed that the input is correct and matches the problem statement. It is guaranteed that at least two knights took part in each battle.
|
Print *n* integers. If the *i*-th knight lost, then the *i*-th number should equal the number of the knight that beat the knight number *i*. If the *i*-th knight is the winner, then the *i*-th number must equal 0.
|
[
"4 3\n1 2 1\n1 3 3\n1 4 4\n",
"8 4\n3 5 4\n3 7 6\n2 8 8\n1 8 1\n"
] |
[
"3 1 4 0 ",
"0 8 4 6 4 8 6 1 "
] |
Consider the first test case. Knights 1 and 2 fought the first fight and knight 1 won. Knights 1 and 3 fought the second fight and knight 3 won. The last fight was between knights 3 and 4, knight 4 won.
| 500
|
[
{
"input": "4 3\n1 2 1\n1 3 3\n1 4 4",
"output": "3 1 4 0 "
},
{
"input": "8 4\n3 5 4\n3 7 6\n2 8 8\n1 8 1",
"output": "0 8 4 6 4 8 6 1 "
},
{
"input": "2 1\n1 2 1",
"output": "0 1 "
},
{
"input": "2 1\n1 2 2",
"output": "2 0 "
},
{
"input": "3 1\n1 3 1",
"output": "0 1 1 "
},
{
"input": "3 1\n1 3 2",
"output": "2 0 2 "
},
{
"input": "3 1\n1 3 3",
"output": "3 3 0 "
},
{
"input": "3 2\n1 2 1\n1 3 3",
"output": "3 1 0 "
},
{
"input": "3 2\n1 2 2\n1 3 2",
"output": "2 0 2 "
},
{
"input": "3 2\n2 3 3\n1 3 3",
"output": "3 3 0 "
},
{
"input": "11 6\n1 2 2\n7 8 7\n3 4 4\n6 9 6\n5 10 10\n2 11 11",
"output": "2 11 4 11 10 10 6 7 6 11 0 "
},
{
"input": "10 6\n9 10 10\n6 7 7\n2 4 2\n2 5 5\n1 7 5\n4 10 8",
"output": "5 5 2 2 8 7 5 0 10 8 "
},
{
"input": "11 8\n3 5 5\n8 9 9\n4 6 6\n8 10 10\n5 7 7\n2 7 2\n10 11 11\n1 11 1",
"output": "0 1 5 5 6 7 2 9 10 11 1 "
},
{
"input": "10 7\n7 8 7\n7 9 9\n5 9 5\n5 10 10\n1 2 2\n3 4 4\n2 10 4",
"output": "2 4 4 0 10 5 9 7 5 4 "
},
{
"input": "11 5\n8 10 9\n6 10 7\n6 11 11\n3 5 5\n1 11 1",
"output": "0 1 5 5 1 7 11 9 7 9 1 "
},
{
"input": "10 6\n6 7 6\n5 7 5\n3 7 4\n2 8 2\n2 10 10\n1 10 10",
"output": "10 10 4 2 4 5 6 2 10 0 "
},
{
"input": "11 7\n7 8 8\n5 6 5\n1 3 3\n7 9 9\n5 10 10\n10 11 11\n1 11 4",
"output": "3 3 4 0 10 5 8 9 10 11 4 "
},
{
"input": "10 7\n8 9 9\n3 4 3\n2 3 2\n1 5 2\n6 7 6\n6 10 10\n1 10 10",
"output": "2 10 2 3 2 10 6 9 10 0 "
},
{
"input": "11 6\n1 2 1\n8 9 9\n3 5 5\n3 6 6\n9 10 10\n1 11 10",
"output": "10 1 5 5 6 10 10 9 10 0 10 "
},
{
"input": "10 5\n1 2 1\n8 10 8\n3 6 4\n4 7 7\n1 8 7",
"output": "7 1 4 7 4 4 0 7 8 8 "
},
{
"input": "4 3\n1 2 2\n1 3 3\n1 4 4",
"output": "2 3 4 0 "
}
] | 1,647,113,130
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 10
| 3,000
| 2,867,200
|
n, m=map(int,input().split())
record=[0 for i in range(n)]
for i in range(m):
l, r, w=map(int,input().split())
for j in range(l,r+1):
if record[j-1]==0 and j!=w:
record[j-1]=w
print(' '.join(map(str,record)))
|
Title: Knight Tournament
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Hooray! Berl II, the king of Berland is making a knight tournament. The king has already sent the message to all knights in the kingdom and they in turn agreed to participate in this grand event.
As for you, you're just a simple peasant. There's no surprise that you slept in this morning and were late for the tournament (it was a weekend, after all). Now you are really curious about the results of the tournament. This time the tournament in Berland went as follows:
- There are *n* knights participating in the tournament. Each knight was assigned his unique number — an integer from 1 to *n*. - The tournament consisted of *m* fights, in the *i*-th fight the knights that were still in the game with numbers at least *l**i* and at most *r**i* have fought for the right to continue taking part in the tournament. - After the *i*-th fight among all participants of the fight only one knight won — the knight number *x**i*, he continued participating in the tournament. Other knights left the tournament. - The winner of the last (the *m*-th) fight (the knight number *x**m*) became the winner of the tournament.
You fished out all the information about the fights from your friends. Now for each knight you want to know the name of the knight he was conquered by. We think that the knight number *b* was conquered by the knight number *a*, if there was a fight with both of these knights present and the winner was the knight number *a*.
Write the code that calculates for each knight, the name of the knight that beat him.
Input Specification:
The first line contains two integers *n*, *m* (2<=≤<=*n*<=≤<=3·105; 1<=≤<=*m*<=≤<=3·105) — the number of knights and the number of fights. Each of the following *m* lines contains three integers *l**i*,<=*r**i*,<=*x**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=*n*; *l**i*<=≤<=*x**i*<=≤<=*r**i*) — the description of the *i*-th fight.
It is guaranteed that the input is correct and matches the problem statement. It is guaranteed that at least two knights took part in each battle.
Output Specification:
Print *n* integers. If the *i*-th knight lost, then the *i*-th number should equal the number of the knight that beat the knight number *i*. If the *i*-th knight is the winner, then the *i*-th number must equal 0.
Demo Input:
['4 3\n1 2 1\n1 3 3\n1 4 4\n', '8 4\n3 5 4\n3 7 6\n2 8 8\n1 8 1\n']
Demo Output:
['3 1 4 0 ', '0 8 4 6 4 8 6 1 ']
Note:
Consider the first test case. Knights 1 and 2 fought the first fight and knight 1 won. Knights 1 and 3 fought the second fight and knight 3 won. The last fight was between knights 3 and 4, knight 4 won.
|
```python
n, m=map(int,input().split())
record=[0 for i in range(n)]
for i in range(m):
l, r, w=map(int,input().split())
for j in range(l,r+1):
if record[j-1]==0 and j!=w:
record[j-1]=w
print(' '.join(map(str,record)))
```
| 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": "4"
},
{
"input": "9 0\n5\n8 1 0\n8 2 0\n8 3 0\n-8 3 0\n-8 2 0",
"output": "0"
},
{
"input": "10 2\n11\n1 1 0\n2 2 3\n3 3 0\n4 4 0\n5 5 0\n6 6 0\n7 7 4\n8 8 7\n9 9 3\n10 10 100\n9 0 1",
"output": "2"
},
{
"input": "5 3\n1\n500 500 10",
"output": "0"
}
] | 1,650,183,936
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 0
|
r,d = map(int,input().split())
ans = 0
r -= d
for _ in range(int(input())):
x,y,u = map(int,input().split())
if (x-u>=r and x+u<=r+d) and (y+u<=r+d or -y-u>=-(r+d)):
ans += 1
elif (x+u<=-r and x-u>=-r-d) and (y+u<=r+d or -y-u>=-(r+d)):
ans += 1
elif y-u>=r and y+u<=r+d and (x+u<=r+d or -x-u>=-(r+d)):
ans += 1
elif y+u<=-r and y-u>=-r-d and (x+u<=r+d or -x-u>=-(r+d)):
ans += 1
print(ans)
|
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
r,d = map(int,input().split())
ans = 0
r -= d
for _ in range(int(input())):
x,y,u = map(int,input().split())
if (x-u>=r and x+u<=r+d) and (y+u<=r+d or -y-u>=-(r+d)):
ans += 1
elif (x+u<=-r and x-u>=-r-d) and (y+u<=r+d or -y-u>=-(r+d)):
ans += 1
elif y-u>=r and y+u<=r+d and (x+u<=r+d or -x-u>=-(r+d)):
ans += 1
elif y+u<=-r and y-u>=-r-d and (x+u<=r+d or -x-u>=-(r+d)):
ans += 1
print(ans)
```
| 0
|
|
79
|
A
|
Bus Game
|
PROGRAMMING
| 1,200
|
[
"greedy"
] |
A. Bus Game
|
2
|
256
|
After Fox Ciel won an onsite round of a programming contest, she took a bus to return to her castle. The fee of the bus was 220 yen. She met Rabbit Hanako in the bus. They decided to play the following game because they got bored in the bus.
- Initially, there is a pile that contains *x* 100-yen coins and *y* 10-yen coins. - They take turns alternatively. Ciel takes the first turn. - In each turn, they must take exactly 220 yen from the pile. In Ciel's turn, if there are multiple ways to take 220 yen, she will choose the way that contains the maximal number of 100-yen coins. In Hanako's turn, if there are multiple ways to take 220 yen, she will choose the way that contains the maximal number of 10-yen coins. - If Ciel or Hanako can't take exactly 220 yen from the pile, she loses.
Determine the winner of the game.
|
The first line contains two integers *x* (0<=≤<=*x*<=≤<=106) and *y* (0<=≤<=*y*<=≤<=106), separated by a single space.
|
If Ciel wins, print "Ciel". Otherwise, print "Hanako".
|
[
"2 2\n",
"3 22\n"
] |
[
"Ciel\n",
"Hanako\n"
] |
In the first turn (Ciel's turn), she will choose 2 100-yen coins and 2 10-yen coins. In the second turn (Hanako's turn), she will choose 1 100-yen coin and 12 10-yen coins. In the third turn (Ciel's turn), she can't pay exactly 220 yen, so Ciel will lose.
| 500
|
[
{
"input": "2 2",
"output": "Ciel"
},
{
"input": "3 22",
"output": "Hanako"
},
{
"input": "0 22",
"output": "Ciel"
},
{
"input": "1000 1000",
"output": "Ciel"
},
{
"input": "0 0",
"output": "Hanako"
},
{
"input": "0 21",
"output": "Hanako"
},
{
"input": "1 11",
"output": "Hanako"
},
{
"input": "1 12",
"output": "Ciel"
},
{
"input": "2 1",
"output": "Hanako"
},
{
"input": "2 23",
"output": "Ciel"
},
{
"input": "2 24",
"output": "Hanako"
},
{
"input": "3 1",
"output": "Hanako"
},
{
"input": "3 2",
"output": "Ciel"
},
{
"input": "3 13",
"output": "Ciel"
},
{
"input": "3 14",
"output": "Hanako"
},
{
"input": "4 1",
"output": "Hanako"
},
{
"input": "4 2",
"output": "Ciel"
},
{
"input": "4 25",
"output": "Hanako"
},
{
"input": "4 26",
"output": "Ciel"
},
{
"input": "5 1",
"output": "Hanako"
},
{
"input": "5 2",
"output": "Ciel"
},
{
"input": "5 15",
"output": "Hanako"
},
{
"input": "5 16",
"output": "Ciel"
},
{
"input": "5 23",
"output": "Ciel"
},
{
"input": "5 24",
"output": "Hanako"
},
{
"input": "6 1",
"output": "Hanako"
},
{
"input": "6 2",
"output": "Ciel"
},
{
"input": "6 13",
"output": "Ciel"
},
{
"input": "6 14",
"output": "Hanako"
},
{
"input": "6 23",
"output": "Ciel"
},
{
"input": "6 24",
"output": "Hanako"
},
{
"input": "7 1",
"output": "Hanako"
},
{
"input": "7 2",
"output": "Ciel"
},
{
"input": "7 13",
"output": "Ciel"
},
{
"input": "7 14",
"output": "Hanako"
},
{
"input": "7 25",
"output": "Hanako"
},
{
"input": "7 26",
"output": "Ciel"
},
{
"input": "8 1",
"output": "Hanako"
},
{
"input": "8 2",
"output": "Ciel"
},
{
"input": "8 15",
"output": "Hanako"
},
{
"input": "8 16",
"output": "Ciel"
},
{
"input": "8 25",
"output": "Hanako"
},
{
"input": "8 26",
"output": "Ciel"
},
{
"input": "9 1",
"output": "Hanako"
},
{
"input": "9 2",
"output": "Ciel"
},
{
"input": "9 15",
"output": "Hanako"
},
{
"input": "9 16",
"output": "Ciel"
},
{
"input": "9 23",
"output": "Ciel"
},
{
"input": "9 24",
"output": "Hanako"
},
{
"input": "10 12",
"output": "Ciel"
},
{
"input": "10 13",
"output": "Ciel"
},
{
"input": "10 22",
"output": "Ciel"
},
{
"input": "10 23",
"output": "Ciel"
},
{
"input": "11 12",
"output": "Ciel"
},
{
"input": "11 13",
"output": "Ciel"
},
{
"input": "11 24",
"output": "Hanako"
},
{
"input": "11 25",
"output": "Hanako"
},
{
"input": "12 14",
"output": "Hanako"
},
{
"input": "12 15",
"output": "Hanako"
},
{
"input": "12 24",
"output": "Hanako"
},
{
"input": "12 25",
"output": "Hanako"
},
{
"input": "0 1000000",
"output": "Hanako"
},
{
"input": "1000000 0",
"output": "Hanako"
},
{
"input": "1000000 1000000",
"output": "Ciel"
},
{
"input": "178087 42116",
"output": "Ciel"
},
{
"input": "378897 104123",
"output": "Ciel"
},
{
"input": "61207 166129",
"output": "Hanako"
},
{
"input": "743519 228136",
"output": "Ciel"
},
{
"input": "425829 771644",
"output": "Ciel"
},
{
"input": "626640 833651",
"output": "Ciel"
},
{
"input": "308950 895657",
"output": "Hanako"
},
{
"input": "991262 957664",
"output": "Ciel"
},
{
"input": "192071 19670",
"output": "Hanako"
},
{
"input": "874382 81677",
"output": "Hanako"
},
{
"input": "202081 745873",
"output": "Hanako"
},
{
"input": "233663 723781",
"output": "Hanako"
},
{
"input": "783744 701689",
"output": "Hanako"
},
{
"input": "333825 679597",
"output": "Hanako"
},
{
"input": "365407 657504",
"output": "Hanako"
},
{
"input": "915488 635412",
"output": "Hanako"
},
{
"input": "947070 613320",
"output": "Hanako"
},
{
"input": "497151 591228",
"output": "Hanako"
},
{
"input": "528732 87635",
"output": "Ciel"
},
{
"input": "78813 65543",
"output": "Hanako"
},
{
"input": "6 4",
"output": "Hanako"
},
{
"input": "3 5",
"output": "Ciel"
},
{
"input": "4 24",
"output": "Hanako"
},
{
"input": "2 14",
"output": "Ciel"
},
{
"input": "2 26",
"output": "Hanako"
},
{
"input": "5 26",
"output": "Ciel"
},
{
"input": "5 5",
"output": "Hanako"
},
{
"input": "359 479",
"output": "Hanako"
},
{
"input": "1 20",
"output": "Ciel"
},
{
"input": "11 0",
"output": "Hanako"
},
{
"input": "3 0",
"output": "Hanako"
},
{
"input": "0 30",
"output": "Ciel"
},
{
"input": "0 35",
"output": "Ciel"
},
{
"input": "100 99",
"output": "Ciel"
},
{
"input": "3 4",
"output": "Ciel"
}
] | 1,692,266,203
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 99
| 184
| 1,228,800
|
x,y = map(int, input().split())
n = ['Ciel','Hanako']
while True:
if x>=2 and y>=2:
x,y = x-2,y-2
elif x>=1 and y>=12:
x,y = x-1,y-12
elif x>=0 and y>=22:
x,y = x,y-22
else:
s = 1
break
if x>=0 and y>=22:
x,y = x,y-22
elif x>=1 and y>=12:
x,y = x-1,y-12
elif x>=2 and y>=2:
x,y = x-2,y-2
else:
s = 0
break
print(n[s])
|
Title: Bus Game
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
After Fox Ciel won an onsite round of a programming contest, she took a bus to return to her castle. The fee of the bus was 220 yen. She met Rabbit Hanako in the bus. They decided to play the following game because they got bored in the bus.
- Initially, there is a pile that contains *x* 100-yen coins and *y* 10-yen coins. - They take turns alternatively. Ciel takes the first turn. - In each turn, they must take exactly 220 yen from the pile. In Ciel's turn, if there are multiple ways to take 220 yen, she will choose the way that contains the maximal number of 100-yen coins. In Hanako's turn, if there are multiple ways to take 220 yen, she will choose the way that contains the maximal number of 10-yen coins. - If Ciel or Hanako can't take exactly 220 yen from the pile, she loses.
Determine the winner of the game.
Input Specification:
The first line contains two integers *x* (0<=≤<=*x*<=≤<=106) and *y* (0<=≤<=*y*<=≤<=106), separated by a single space.
Output Specification:
If Ciel wins, print "Ciel". Otherwise, print "Hanako".
Demo Input:
['2 2\n', '3 22\n']
Demo Output:
['Ciel\n', 'Hanako\n']
Note:
In the first turn (Ciel's turn), she will choose 2 100-yen coins and 2 10-yen coins. In the second turn (Hanako's turn), she will choose 1 100-yen coin and 12 10-yen coins. In the third turn (Ciel's turn), she can't pay exactly 220 yen, so Ciel will lose.
|
```python
x,y = map(int, input().split())
n = ['Ciel','Hanako']
while True:
if x>=2 and y>=2:
x,y = x-2,y-2
elif x>=1 and y>=12:
x,y = x-1,y-12
elif x>=0 and y>=22:
x,y = x,y-22
else:
s = 1
break
if x>=0 and y>=22:
x,y = x,y-22
elif x>=1 and y>=12:
x,y = x-1,y-12
elif x>=2 and y>=2:
x,y = x-2,y-2
else:
s = 0
break
print(n[s])
```
| 3.951711
|
386
|
A
|
Second-Price Auction
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different.
|
The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder.
|
The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based.
|
[
"2\n5 7\n",
"3\n10 2 8\n",
"6\n3 8 2 9 4 14\n"
] |
[
"2 5\n",
"1 8\n",
"6 9\n"
] |
none
| 500
|
[
{
"input": "2\n5 7",
"output": "2 5"
},
{
"input": "3\n10 2 8",
"output": "1 8"
},
{
"input": "6\n3 8 2 9 4 14",
"output": "6 9"
},
{
"input": "4\n4707 7586 4221 5842",
"output": "2 5842"
},
{
"input": "5\n3304 4227 4869 6937 6002",
"output": "4 6002"
},
{
"input": "6\n5083 3289 7708 5362 9031 7458",
"output": "5 7708"
},
{
"input": "7\n9038 6222 3392 1706 3778 1807 2657",
"output": "1 6222"
},
{
"input": "8\n7062 2194 4481 3864 7470 1814 8091 733",
"output": "7 7470"
},
{
"input": "9\n2678 5659 9199 2628 7906 7496 4524 2663 3408",
"output": "3 7906"
},
{
"input": "2\n3458 1504",
"output": "1 1504"
},
{
"input": "50\n9237 3904 407 9052 6657 9229 9752 3888 7732 2512 4614 1055 2355 7108 6506 6849 2529 8862 159 8630 7906 7941 960 8470 333 8659 54 9475 3163 5625 6393 6814 2656 3388 169 7918 4881 8468 9983 6281 6340 280 5108 2996 101 7617 3313 8172 326 1991",
"output": "39 9752"
},
{
"input": "100\n2515 3324 7975 6171 4240 1217 4829 5203 8603 6900 3031 4699 4732 6070 4221 3228 6497 7359 9130 4346 4619 1109 3945 5442 3271 16 9711 2045 6410 2301 3406 8125 6003 1892 1260 9661 3940 6692 4708 7027 4930 6925 1979 5361 4263 3144 867 8639 6230 5562 9714 3676 4231 3347 4733 4920 4881 3431 1059 7313 8912 3038 9308 72 9583 7009 3034 7425 2398 6141 3245 2495 2933 6710 8289 9806 1226 8393 7349 6462 1618 9613 3546 6012 2964 9995 1578 210 2123 4874 1252 8625 348 8020 803 7244 9080 5088 706 2602",
"output": "86 9806"
},
{
"input": "2\n2 1",
"output": "1 1"
},
{
"input": "2\n1 2",
"output": "2 1"
},
{
"input": "3\n10 20 30",
"output": "3 20"
},
{
"input": "3\n10 30 20",
"output": "2 20"
},
{
"input": "3\n20 10 30",
"output": "3 20"
},
{
"input": "3\n20 30 10",
"output": "2 20"
},
{
"input": "3\n30 10 20",
"output": "1 20"
},
{
"input": "3\n30 20 10",
"output": "1 20"
},
{
"input": "2\n1 10000",
"output": "2 1"
},
{
"input": "2\n10000 999",
"output": "1 999"
},
{
"input": "3\n3 4 1",
"output": "2 3"
},
{
"input": "6\n1 2 24 6 5 7",
"output": "3 7"
}
] | 1,627,571,372
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 62
| 6,963,200
|
n = input()
p = list(map(int, input().split()))
mp = max (p)
mid = p.index(mp) + 1
p.remove(mp)
print(mid, max(p))
|
Title: Second-Price Auction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem we consider a special type of an auction, which is called the second-price auction. As in regular auction *n* bidders place a bid which is price a bidder ready to pay. The auction is closed, that is, each bidder secretly informs the organizer of the auction price he is willing to pay. After that, the auction winner is the participant who offered the highest price. However, he pay not the price he offers, but the highest price among the offers of other participants (hence the name: the second-price auction).
Write a program that reads prices offered by bidders and finds the winner and the price he will pay. Consider that all of the offered prices are different.
Input Specification:
The first line of the input contains *n* (2<=≤<=*n*<=≤<=1000) — number of bidders. The second line contains *n* distinct integer numbers *p*1,<=*p*2,<=... *p**n*, separated by single spaces (1<=≤<=*p**i*<=≤<=10000), where *p**i* stands for the price offered by the *i*-th bidder.
Output Specification:
The single output line should contain two integers: index of the winner and the price he will pay. Indices are 1-based.
Demo Input:
['2\n5 7\n', '3\n10 2 8\n', '6\n3 8 2 9 4 14\n']
Demo Output:
['2 5\n', '1 8\n', '6 9\n']
Note:
none
|
```python
n = input()
p = list(map(int, input().split()))
mp = max (p)
mid = p.index(mp) + 1
p.remove(mp)
print(mid, max(p))
```
| 3
|
|
804
|
B
|
Minimum number of steps
|
PROGRAMMING
| 1,400
|
[
"combinatorics",
"greedy",
"implementation",
"math"
] | null | null |
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109<=+<=7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
|
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
|
Print the minimum number of steps modulo 109<=+<=7.
|
[
"ab\n",
"aab\n"
] |
[
"1\n",
"3\n"
] |
The first example: "ab" → "bba".
The second example: "aab" → "abba" → "bbaba" → "bbbbaa".
| 1,000
|
[
{
"input": "ab",
"output": "1"
},
{
"input": "aab",
"output": "3"
},
{
"input": "aaaaabaabababaaaaaba",
"output": "17307"
},
{
"input": "abaabaaabbabaabab",
"output": "1795"
},
{
"input": "abbaa",
"output": "2"
},
{
"input": "abbaaabaabaaaaabbbbaababaaaaabaabbaaaaabbaabbaaaabbbabbbabb",
"output": "690283580"
},
{
"input": "aababbaaaabbaabbbbbbbbabbababbbaaabbaaabbabbba",
"output": "2183418"
},
{
"input": "aabbaababbabbbaabbaababaaaabbaaaabaaaaaababbaaaabaababbabbbb",
"output": "436420225"
},
{
"input": "aaabaaaabbababbaabbababbbbaaaaaaabbabbba",
"output": "8431094"
},
{
"input": "abbbbababbabbbbbabaabbbaabbbbbbbaaab",
"output": "8180"
},
{
"input": "bbababbbaabaaaaaaaabbabbbb",
"output": "40979"
},
{
"input": "abbbaaabbbbbabaabbaaabbbababbbaabaabababababa",
"output": "2065758"
},
{
"input": "abaaaaaabaaaabbabbaaabbbbabababaaaaabbaabbaaaaabbbaababaaaaaaabbbbbaaaaabaababbabababbabbbbaabbaabbabbbabaabbaabbaaaaaab",
"output": "235606597"
},
{
"input": "abbbbbbbbbbbbbbbbbbbbbbbbbbaababaaaaaaabaabaaababaabaababaaabababaababab",
"output": "7"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbaaaaaaaaabaabaaababaabaababaaabababaabbbbbbb",
"output": "557763786"
}
] | 1,656,900,809
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 5
| 124
| 512,000
|
M = 10 ** 9 + 7
s = input()
n = len(s)
l = []
c = 0
for i in range(n):
if(s[i] == 'a'):
c += 1
else:
l.append(c)
ans = 0
for i in l:
ans += (pow(2, i, M) - 1) % M
print(ans)
|
Title: Minimum number of steps
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109<=+<=7.
The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.
Input Specification:
The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.
Output Specification:
Print the minimum number of steps modulo 109<=+<=7.
Demo Input:
['ab\n', 'aab\n']
Demo Output:
['1\n', '3\n']
Note:
The first example: "ab" → "bba".
The second example: "aab" → "abba" → "bbaba" → "bbbbaa".
|
```python
M = 10 ** 9 + 7
s = input()
n = len(s)
l = []
c = 0
for i in range(n):
if(s[i] == 'a'):
c += 1
else:
l.append(c)
ans = 0
for i in l:
ans += (pow(2, i, M) - 1) % M
print(ans)
```
| 0
|
|
1,006
|
B
|
Polycarp's Practice
|
PROGRAMMING
| 1,200
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
Polycarp is practicing his problem solving skill. He has a list of $n$ problems with difficulties $a_1, a_2, \dots, a_n$, respectively. His plan is to practice for exactly $k$ days. Each day he has to solve at least one problem from his list. Polycarp solves the problems in the order they are given in his list, he cannot skip any problem from his list. He has to solve all $n$ problems in exactly $k$ days.
Thus, each day Polycarp solves a contiguous sequence of (consecutive) problems from the start of the list. He can't skip problems or solve them multiple times. As a result, in $k$ days he will solve all the $n$ problems.
The profit of the $j$-th day of Polycarp's practice is the maximum among all the difficulties of problems Polycarp solves during the $j$-th day (i.e. if he solves problems with indices from $l$ to $r$ during a day, then the profit of the day is $\max\limits_{l \le i \le r}a_i$). The total profit of his practice is the sum of the profits over all $k$ days of his practice.
You want to help Polycarp to get the maximum possible total profit over all valid ways to solve problems. Your task is to distribute all $n$ problems between $k$ days satisfying the conditions above in such a way, that the total profit is maximum.
For example, if $n = 8, k = 3$ and $a = [5, 4, 2, 6, 5, 1, 9, 2]$, one of the possible distributions with maximum total profit is: $[5, 4, 2], [6, 5], [1, 9, 2]$. Here the total profit equals $5 + 6 + 9 = 20$.
|
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2000$) — the number of problems and the number of days, respectively.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2000$) — difficulties of problems in Polycarp's list, in the order they are placed in the list (i.e. in the order Polycarp will solve them).
|
In the first line of the output print the maximum possible total profit.
In the second line print exactly $k$ positive integers $t_1, t_2, \dots, t_k$ ($t_1 + t_2 + \dots + t_k$ must equal $n$), where $t_j$ means the number of problems Polycarp will solve during the $j$-th day in order to achieve the maximum possible total profit of his practice.
If there are many possible answers, you may print any of them.
|
[
"8 3\n5 4 2 6 5 1 9 2\n",
"5 1\n1 1 1 1 1\n",
"4 2\n1 2000 2000 2\n"
] |
[
"20\n3 2 3",
"1\n5\n",
"4000\n2 2\n"
] |
The first example is described in the problem statement.
In the second example there is only one possible distribution.
In the third example the best answer is to distribute problems in the following way: $[1, 2000], [2000, 2]$. The total profit of this distribution is $2000 + 2000 = 4000$.
| 0
|
[
{
"input": "8 3\n5 4 2 6 5 1 9 2",
"output": "20\n4 1 3"
},
{
"input": "5 1\n1 1 1 1 1",
"output": "1\n5"
},
{
"input": "4 2\n1 2000 2000 2",
"output": "4000\n2 2"
},
{
"input": "1 1\n2000",
"output": "2000\n1"
},
{
"input": "1 1\n1234",
"output": "1234\n1"
},
{
"input": "3 2\n1 1 1",
"output": "2\n2 1"
},
{
"input": "4 2\n3 5 1 1",
"output": "8\n1 3"
},
{
"input": "5 3\n5 5 6 7 1",
"output": "18\n2 1 2"
},
{
"input": "6 4\n1 1 1 1 2 2",
"output": "6\n3 1 1 1"
},
{
"input": "5 3\n5 5 6 6 4",
"output": "17\n2 1 2"
},
{
"input": "16 15\n14 4 9 12 17 1 1 8 12 13 6 9 17 2 18 12",
"output": "154\n1 1 1 1 1 2 1 1 1 1 1 1 1 1 1"
},
{
"input": "1 1\n1996",
"output": "1996\n1"
},
{
"input": "5 3\n5 5 5 9 10",
"output": "24\n3 1 1"
},
{
"input": "18 15\n18 2 13 1 18 3 2 18 18 20 9 2 20 20 4 20 9 12",
"output": "204\n1 2 2 1 2 1 1 1 1 1 1 1 1 1 1"
},
{
"input": "5 3\n1 20 20 50 50",
"output": "120\n3 1 1"
},
{
"input": "8 3\n15 14 11 19 17 14 14 8",
"output": "51\n1 3 4"
},
{
"input": "5 2\n15 20 6 19 6",
"output": "39\n2 3"
},
{
"input": "6 3\n5 5 5 5 6 9",
"output": "20\n4 1 1"
},
{
"input": "5 3\n2 2 2 3 3",
"output": "8\n3 1 1"
},
{
"input": "7 3\n2 2 2 2 2 3 3",
"output": "8\n5 1 1"
},
{
"input": "6 5\n1 1 6 6 6 6",
"output": "25\n2 1 1 1 1"
},
{
"input": "8 4\n1 2 2 2 2 3 4 5",
"output": "14\n5 1 1 1"
},
{
"input": "6 4\n1 1 1 5 5 5",
"output": "16\n3 1 1 1"
},
{
"input": "6 3\n1 2 2 2 4 5",
"output": "11\n4 1 1"
},
{
"input": "18 6\n17 17 19 14 10 20 18 16 6 7 2 15 14 16 13 6 12 11",
"output": "107\n1 1 1 3 1 11"
},
{
"input": "6 3\n1 1 2 2 3 4",
"output": "9\n4 1 1"
},
{
"input": "8 3\n5 4 2 5 6 1 9 2",
"output": "20\n4 1 3"
}
] | 1,602,726,431
| 731
|
PyPy 3
|
OK
|
TESTS
| 37
| 171
| 2,560,000
|
n,k = map(int, input().split())
A = list(map(int, input().split()))
B = []
for i, a in enumerate(A):
B.append((a, i))
B.sort(reverse=True)
B = B[0:k]
idx = []
ans = 0
for a, i in B:
idx.append(i)
ans += a
idx.sort()
#print(idx)
s = []
pre = -1
for j in range(len(idx)):
i = idx[j]
if j != len(idx)-1:
s.append(i-pre)
pre = i
else:
s.append(n-1-pre)
print(ans)
print(*s)
|
Title: Polycarp's Practice
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is practicing his problem solving skill. He has a list of $n$ problems with difficulties $a_1, a_2, \dots, a_n$, respectively. His plan is to practice for exactly $k$ days. Each day he has to solve at least one problem from his list. Polycarp solves the problems in the order they are given in his list, he cannot skip any problem from his list. He has to solve all $n$ problems in exactly $k$ days.
Thus, each day Polycarp solves a contiguous sequence of (consecutive) problems from the start of the list. He can't skip problems or solve them multiple times. As a result, in $k$ days he will solve all the $n$ problems.
The profit of the $j$-th day of Polycarp's practice is the maximum among all the difficulties of problems Polycarp solves during the $j$-th day (i.e. if he solves problems with indices from $l$ to $r$ during a day, then the profit of the day is $\max\limits_{l \le i \le r}a_i$). The total profit of his practice is the sum of the profits over all $k$ days of his practice.
You want to help Polycarp to get the maximum possible total profit over all valid ways to solve problems. Your task is to distribute all $n$ problems between $k$ days satisfying the conditions above in such a way, that the total profit is maximum.
For example, if $n = 8, k = 3$ and $a = [5, 4, 2, 6, 5, 1, 9, 2]$, one of the possible distributions with maximum total profit is: $[5, 4, 2], [6, 5], [1, 9, 2]$. Here the total profit equals $5 + 6 + 9 = 20$.
Input Specification:
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2000$) — the number of problems and the number of days, respectively.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2000$) — difficulties of problems in Polycarp's list, in the order they are placed in the list (i.e. in the order Polycarp will solve them).
Output Specification:
In the first line of the output print the maximum possible total profit.
In the second line print exactly $k$ positive integers $t_1, t_2, \dots, t_k$ ($t_1 + t_2 + \dots + t_k$ must equal $n$), where $t_j$ means the number of problems Polycarp will solve during the $j$-th day in order to achieve the maximum possible total profit of his practice.
If there are many possible answers, you may print any of them.
Demo Input:
['8 3\n5 4 2 6 5 1 9 2\n', '5 1\n1 1 1 1 1\n', '4 2\n1 2000 2000 2\n']
Demo Output:
['20\n3 2 3', '1\n5\n', '4000\n2 2\n']
Note:
The first example is described in the problem statement.
In the second example there is only one possible distribution.
In the third example the best answer is to distribute problems in the following way: $[1, 2000], [2000, 2]$. The total profit of this distribution is $2000 + 2000 = 4000$.
|
```python
n,k = map(int, input().split())
A = list(map(int, input().split()))
B = []
for i, a in enumerate(A):
B.append((a, i))
B.sort(reverse=True)
B = B[0:k]
idx = []
ans = 0
for a, i in B:
idx.append(i)
ans += a
idx.sort()
#print(idx)
s = []
pre = -1
for j in range(len(idx)):
i = idx[j]
if j != len(idx)-1:
s.append(i-pre)
pre = i
else:
s.append(n-1-pre)
print(ans)
print(*s)
```
| 3
|
|
705
|
A
|
Hulk
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Dr. Bruce Banner hates his enemies (like others don't). As we all know, he can barely talk when he turns into the incredible Hulk. That's why he asked you to help him to express his feelings.
Hulk likes the Inception so much, and like that his feelings are complicated. They have *n* layers. The first layer is hate, second one is love, third one is hate and so on...
For example if *n*<==<=1, then his feeling is "I hate it" or if *n*<==<=2 it's "I hate that I love it", and if *n*<==<=3 it's "I hate that I love that I hate it" and so on.
Please help Dr. Banner.
|
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of layers of love and hate.
|
Print Dr.Banner's feeling in one line.
|
[
"1\n",
"2\n",
"3\n"
] |
[
"I hate it\n",
"I hate that I love it\n",
"I hate that I love that I hate it\n"
] |
none
| 500
|
[
{
"input": "1",
"output": "I hate it"
},
{
"input": "2",
"output": "I hate that I love it"
},
{
"input": "3",
"output": "I hate that I love that I hate it"
},
{
"input": "4",
"output": "I hate that I love that I hate that I love it"
},
{
"input": "5",
"output": "I hate that I love that I hate that I love that I hate it"
},
{
"input": "10",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love it"
},
{
"input": "18",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love it"
},
{
"input": "19",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate it"
},
{
"input": "76",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "77",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "81",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "82",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "85",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "86",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "100",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "99",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "98",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "33",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate it"
},
{
"input": "34",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love it"
},
{
"input": "57",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
},
{
"input": "100",
"output": "I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate that I love that I hate ..."
}
] | 1,695,769,907
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 31
| 5,632,000
|
x=int(input())
if x==1:print('I hate it')
if x==2:print('I hate that I love it')
if x==3:print('I hate that I love that I hate it')
|
Title: Hulk
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dr. Bruce Banner hates his enemies (like others don't). As we all know, he can barely talk when he turns into the incredible Hulk. That's why he asked you to help him to express his feelings.
Hulk likes the Inception so much, and like that his feelings are complicated. They have *n* layers. The first layer is hate, second one is love, third one is hate and so on...
For example if *n*<==<=1, then his feeling is "I hate it" or if *n*<==<=2 it's "I hate that I love it", and if *n*<==<=3 it's "I hate that I love that I hate it" and so on.
Please help Dr. Banner.
Input Specification:
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of layers of love and hate.
Output Specification:
Print Dr.Banner's feeling in one line.
Demo Input:
['1\n', '2\n', '3\n']
Demo Output:
['I hate it\n', 'I hate that I love it\n', 'I hate that I love that I hate it\n']
Note:
none
|
```python
x=int(input())
if x==1:print('I hate it')
if x==2:print('I hate that I love it')
if x==3:print('I hate that I love that I hate it')
```
| 0
|
|
5
|
A
|
Chat Servers Outgoing Traffic
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
A. Chat Server's Outgoing Traffic
|
1
|
64
|
Polycarp is working on a new project called "Polychat". Following modern tendencies in IT, he decided, that this project should contain chat as well. To achieve this goal, Polycarp has spent several hours in front of his laptop and implemented a chat server that can process three types of commands:
- Include a person to the chat ('Add' command). - Remove a person from the chat ('Remove' command). - Send a message from a person to all people, who are currently in the chat, including the one, who sends the message ('Send' command).
Now Polycarp wants to find out the amount of outgoing traffic that the server will produce while processing a particular set of commands.
Polycarp knows that chat server sends no traffic for 'Add' and 'Remove' commands. When 'Send' command is processed, server sends *l* bytes to each participant of the chat, where *l* is the length of the message.
As Polycarp has no time, he is asking for your help in solving this problem.
|
Input file will contain not more than 100 commands, each in its own line. No line will exceed 100 characters. Formats of the commands will be the following:
- +<name> for 'Add' command. - -<name> for 'Remove' command. - <sender_name>:<message_text> for 'Send' command.
<name> and <sender_name> is a non-empty sequence of Latin letters and digits. <message_text> can contain letters, digits and spaces, but can't start or end with a space. <message_text> can be an empty line.
It is guaranteed, that input data are correct, i.e. there will be no 'Add' command if person with such a name is already in the chat, there will be no 'Remove' command if there is no person with such a name in the chat etc.
All names are case-sensitive.
|
Print a single number — answer to the problem.
|
[
"+Mike\nMike:hello\n+Kate\n+Dmitry\n-Dmitry\nKate:hi\n-Kate\n",
"+Mike\n-Mike\n+Mike\nMike:Hi I am here\n-Mike\n+Kate\n-Kate\n"
] |
[
"9\n",
"14\n"
] |
none
| 0
|
[
{
"input": "+Mike\nMike:hello\n+Kate\n+Dmitry\n-Dmitry\nKate:hi\n-Kate",
"output": "9"
},
{
"input": "+Mike\n-Mike\n+Mike\nMike:Hi I am here\n-Mike\n+Kate\n-Kate",
"output": "14"
},
{
"input": "+Dmitry\n+Mike\nDmitry:All letters will be used\nDmitry:qwertyuiopasdfghjklzxcvbnm QWERTYUIOPASDFGHJKLZXCVBNM\nDmitry:And digits too\nDmitry:1234567890 0987654321\n-Dmitry",
"output": "224"
},
{
"input": "+Dmitry\n+Mike\n+Kate\nDmitry:",
"output": "0"
},
{
"input": "+Dmitry\nDmitry:No phrases with spaces at the beginning and at the end\n+Mike\nDmitry:spaces spaces\n-Dmitry",
"output": "86"
},
{
"input": "+XqD\n+aT537\nXqD:x6ZPjMR1DDKG2\nXqD:lLCriywPnB\n-XqD",
"output": "46"
},
{
"input": "+8UjgAJ\n8UjgAJ:02hR7UBc1tqqfL\n-8UjgAJ\n+zdi\n-zdi",
"output": "14"
},
{
"input": "+6JPKkgXDrA\n+j6JHjv70An\n+QGtsceK0zJ\n6JPKkgXDrA:o4\n+CSmwi9zDra\nQGtsceK0zJ:Zl\nQGtsceK0zJ:0\nj6JHjv70An:7\nj6JHjv70An:B\nQGtsceK0zJ:OO",
"output": "34"
},
{
"input": "+1aLNq9S7uLV\n-1aLNq9S7uLV\n+O9ykq3xDJv\n-O9ykq3xDJv\n+54Yq1xJq14F\n+0zJ5Vo0RDZ\n-54Yq1xJq14F\n-0zJ5Vo0RDZ\n+lxlH7sdolyL\n-lxlH7sdolyL",
"output": "0"
},
{
"input": "+qlHEc2AuYy\nqlHEc2AuYy:YYRwD0 edNZgpE nGfOguRWnMYpTpGUVM aXDKGXo1Gv1tHL9\nqlHEc2AuYy:yvh3GsPcImqrvoUcBNQcP6ezwpU0 xAVltaKZp94VKiNao\nqlHEc2AuYy:zuCO6Opey L eu7lTwysaSk00zjpv zrDfbt8l hpHfu\n+pErDMxgVgh\nqlHEc2AuYy:I1FLis mmQbZtd8Ui7y 1vcax6yZBMhVRdD6Ahlq7MNCw\nqlHEc2AuYy:lz MFUNJZhlqBYckHUDlNhLiEkmecRh1o0t7alXBvCRVEFVx\npErDMxgVgh:jCyMbu1dkuEj5TzbBOjyUhpfC50cL8R900Je3R KxRgAI dT\nqlHEc2AuYy:62b47eabo2hf vSUD7KioN ZHki6WB6gh3u GKv5rgwyfF\npErDMxgVgh:zD5 9 ympl4wR gy7a7eAGAn5xVdGP9FbL6hRCZAR6O4pT6zb",
"output": "615"
},
{
"input": "+adabacaba0",
"output": "0"
},
{
"input": "+acabadab\n+caba0aba",
"output": "0"
},
{
"input": "+dabaca\n-dabaca\n+aba0ab",
"output": "0"
},
{
"input": "+cab\n+abac\n-abac\n+baca",
"output": "0"
},
{
"input": "+cabadabac\n-cabadabac\n+abacaba1ab\n-abacaba1ab\n+ba0abaca",
"output": "0"
},
{
"input": "+adabacaba\n-adabacaba\n+aca\naca:caba\n-aca\n+bacaba\n-bacaba\n+aba\n-aba\n+bad",
"output": "4"
},
{
"input": "+acabadab\n-acabadab\n+aba0abacab\n+baca\n+abacaba0ab\n-baca\n-abacaba0ab\n-aba0abacab\n+cab\n-cab\n+abacabada\n-abacabada\n+badabaca\n-badabaca\n+badaba",
"output": "0"
},
{
"input": "+badabac\nbadabac:abacabad\n-badabac\n+0ab\n-0ab\n+dabacab\n-dabacab\n+a0ab\n-a0ab\n+0abaca\n-0abaca\n+dabac\n-dabac\n+abaca\n-abaca\n+bacabada\n-bacabada\n+aca\n-aca\n+abadabaca\n-abadabaca\n+acaba\n-acaba\n+abacabadab\n-abacabadab",
"output": "8"
}
] | 1,519,865,793
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 92
| 5,632,000
|
todo = []
chatbits = int(0)
while True:
keyboard = input()
try:
if keyboard.startswith('+'):
todo.append(keyboard[1:len(keyboard)])
elif keyboard.startswith('-'):
todo.remove(keyboard[1:len(keyboard)])
elif(':' in keyboard == 1):
td = int(keyboard.index(':'))
lk = int(len(keyboard))
slicer = keyboard[td:lk]
chatbits += (len(todo)*len(slicer))
except EOFError:
break
print(chatbits)
|
Title: Chat Servers Outgoing Traffic
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Polycarp is working on a new project called "Polychat". Following modern tendencies in IT, he decided, that this project should contain chat as well. To achieve this goal, Polycarp has spent several hours in front of his laptop and implemented a chat server that can process three types of commands:
- Include a person to the chat ('Add' command). - Remove a person from the chat ('Remove' command). - Send a message from a person to all people, who are currently in the chat, including the one, who sends the message ('Send' command).
Now Polycarp wants to find out the amount of outgoing traffic that the server will produce while processing a particular set of commands.
Polycarp knows that chat server sends no traffic for 'Add' and 'Remove' commands. When 'Send' command is processed, server sends *l* bytes to each participant of the chat, where *l* is the length of the message.
As Polycarp has no time, he is asking for your help in solving this problem.
Input Specification:
Input file will contain not more than 100 commands, each in its own line. No line will exceed 100 characters. Formats of the commands will be the following:
- +<name> for 'Add' command. - -<name> for 'Remove' command. - <sender_name>:<message_text> for 'Send' command.
<name> and <sender_name> is a non-empty sequence of Latin letters and digits. <message_text> can contain letters, digits and spaces, but can't start or end with a space. <message_text> can be an empty line.
It is guaranteed, that input data are correct, i.e. there will be no 'Add' command if person with such a name is already in the chat, there will be no 'Remove' command if there is no person with such a name in the chat etc.
All names are case-sensitive.
Output Specification:
Print a single number — answer to the problem.
Demo Input:
['+Mike\nMike:hello\n+Kate\n+Dmitry\n-Dmitry\nKate:hi\n-Kate\n', '+Mike\n-Mike\n+Mike\nMike:Hi I am here\n-Mike\n+Kate\n-Kate\n']
Demo Output:
['9\n', '14\n']
Note:
none
|
```python
todo = []
chatbits = int(0)
while True:
keyboard = input()
try:
if keyboard.startswith('+'):
todo.append(keyboard[1:len(keyboard)])
elif keyboard.startswith('-'):
todo.remove(keyboard[1:len(keyboard)])
elif(':' in keyboard == 1):
td = int(keyboard.index(':'))
lk = int(len(keyboard))
slicer = keyboard[td:lk]
chatbits += (len(todo)*len(slicer))
except EOFError:
break
print(chatbits)
```
| -1
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,569,489,915
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 80
| 248
| 0
|
n=int(input())
'''x_count=0
y_count=0
z_count=0
while n:
li=list(map(int,input().split()))
x_count+=li[0]
y_count+=li[1]
z_count+=li[2]
n-=1
if(x_count+y_count+z_count==0):
print("YES")
else:
print("NO")
'''
count=0
while n:
count+=sum(list(map(int,input().split())))
n-=1
if(count==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())
'''x_count=0
y_count=0
z_count=0
while n:
li=list(map(int,input().split()))
x_count+=li[0]
y_count+=li[1]
z_count+=li[2]
n-=1
if(x_count+y_count+z_count==0):
print("YES")
else:
print("NO")
'''
count=0
while n:
count+=sum(list(map(int,input().split())))
n-=1
if(count==0):
print("YES")
else:
print("NO")
```
| 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": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,549,923,761
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 77
| 0
|
phrase = list(input())
hello = 'hello'
for char in 'hello':
if char in phrase:
while hello.index(char) != hello.index(char):
del phrase[hello.index(char)]
if phrase == hello:
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
phrase = list(input())
hello = 'hello'
for char in 'hello':
if char in phrase:
while hello.index(char) != hello.index(char):
del phrase[hello.index(char)]
if phrase == hello:
print("YES")
else:
print("NO")
```
| 0
|
112
|
A
|
Petya and Strings
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Petya and Strings
|
2
|
256
|
Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.
|
Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.
|
If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.
|
[
"aaaa\naaaA\n",
"abs\nAbz\n",
"abcdefg\nAbCdEfF\n"
] |
[
"0\n",
"-1\n",
"1\n"
] |
If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
- http://en.wikipedia.org/wiki/Lexicographical_order
| 500
|
[
{
"input": "aaaa\naaaA",
"output": "0"
},
{
"input": "abs\nAbz",
"output": "-1"
},
{
"input": "abcdefg\nAbCdEfF",
"output": "1"
},
{
"input": "asadasdasd\nasdwasdawd",
"output": "-1"
},
{
"input": "aslkjlkasdd\nasdlkjdajwi",
"output": "1"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "0"
},
{
"input": "aAaaaAAaAaaAzZsssSsdDfeEaeqZlpP\nAaaaAaaAaaAaZzSSSSsDdFeeAeQZLpp",
"output": "0"
},
{
"input": "bwuEhEveouaTECagLZiqmUdxEmhRSOzMauJRWLQMppZOumxhAmwuGeDIkvkBLvMXwUoFmpAfDprBcFtEwOULcZWRQhcTbTbX\nHhoDWbcxwiMnCNexOsKsujLiSGcLllXOkRSbnOzThAjnnliLYFFmsYkOfpTxRNEfBsoUHfoLTiqAINRPxWRqrTJhgfkKcDOH",
"output": "-1"
},
{
"input": "kGWUuguKzcvxqKTNpxeDWXpXkrXDvGMFGoXKDfPBZvWSDUyIYBynbKOUonHvmZaKeirUhfmVRKtGhAdBfKMWXDUoqvbfpfHYcg\ncvOULleuIIiYVVxcLZmHVpNGXuEpzcWZZWyMOwIwbpkKPwCfkVbKkUuosvxYCKjqfVmHfJKbdrsAcatPYgrCABaFcoBuOmMfFt",
"output": "1"
},
{
"input": "nCeNVIzHqPceNhjHeHvJvgBsNFiXBATRrjSTXJzhLMDMxiJztphxBRlDlqwDFImWeEPkggZCXSRwelOdpNrYnTepiOqpvkr\nHJbjJFtlvNxIbkKlxQUwmZHJFVNMwPAPDRslIoXISBYHHfymyIaQHLgECPxAmqnOCizwXnIUBRmpYUBVPenoUKhCobKdOjL",
"output": "1"
},
{
"input": "ttXjenUAlfixytHEOrPkgXmkKTSGYuyVXGIHYmWWYGlBYpHkujueqBSgjLguSgiMGJWATIGEUjjAjKXdMiVbHozZUmqQtFrT\nJziDBFBDmDJCcGqFsQwDFBYdOidLxxhBCtScznnDgnsiStlWFnEXQrJxqTXKPxZyIGfLIToETKWZBPUIBmLeImrlSBWCkTNo",
"output": "1"
},
{
"input": "AjQhPqSVhwQQjcgCycjKorWBgFCRuQBwgdVuAPSMJAvTyxGVuFHjfJzkKfsmfhFbKqFrFIohSZBbpjgEHebezmVlGLTPSCTMf\nXhxWuSnMmKFrCUOwkTUmvKAfbTbHWzzOTzxJatLLCdlGnHVaBUnxDlsqpvjLHMThOPAFBggVKDyKBrZAmjnjrhHlrnSkyzBja",
"output": "-1"
},
{
"input": "HCIgYtnqcMyjVngziNflxKHtdTmcRJhzMAjFAsNdWXFJYEhiTzsQUtFNkAbdrFBRmvLirkuirqTDvIpEfyiIqkrwsjvpPWTEdI\nErqiiWKsmIjyZuzgTlTqxYZwlrpvRyaVhRTOYUqtPMVGGtWOkDCOOQRKrkkRzPftyQCkYkzKkzTPqqXmeZhvvEEiEhkdOmoMvy",
"output": "1"
},
{
"input": "mtBeJYILXcECGyEVSyzLFdQJbiVnnfkbsYYsdUJSIRmyzLfTTtFwIBmRLVnwcewIqcuydkcLpflHAFyDaToLiFMgeHvQorTVbI\nClLvyejznjbRfCDcrCzkLvqQaGzTjwmWONBdCctJAPJBcQrcYvHaSLQgPIJbmkFBhFzuQLBiRzAdNHulCjIAkBvZxxlkdzUWLR",
"output": "1"
},
{
"input": "tjucSbGESVmVridTBjTmpVBCwwdWKBPeBvmgdxgIVLwQxveETnSdxkTVJpXoperWSgdpPMKNmwDiGeHfxnuqaDissgXPlMuNZIr\nHfjOOJhomqNIKHvqSgfySjlsWJQBuWYwhLQhlZYlpZwboMpoLoluGsBmhhlYgeIouwdkPfiaAIrkYRlxtiFazOPOllPsNZHcIZd",
"output": "1"
},
{
"input": "AanbDfbZNlUodtBQlvPMyomStKNhgvSGhSbTdabxGFGGXCdpsJDimsAykKjfBDPMulkhBMsqLmVKLDoesHZsRAEEdEzqigueXInY\ncwfyjoppiJNrjrOLNZkqcGimrpTsiyFBVgMWEPXsMrxLJDDbtYzerXiFGuLBcQYitLdqhGHBpdjRnkUegmnwhGHAKXGyFtscWDSI",
"output": "-1"
},
{
"input": "HRfxniwuJCaHOcaOVgjOGHXKrwxrDQxJpppeGDXnTAowyKbCsCQPbchCKeTWOcKbySSYnoaTJDnmRcyGPbfXJyZoPcARHBu\nxkLXvwkvGIWSQaFTznLOctUXNuzzBBOlqvzmVfTSejekTAlwidRrsxkbZTsGGeEWxCXHzqWVuLGoCyrGjKkQoHqduXwYQKC",
"output": "-1"
},
{
"input": "OjYwwNuPESIazoyLFREpObIaMKhCaKAMWMfRGgucEuyNYRantwdwQkmflzfqbcFRaXBnZoIUGsFqXZHGKwlaBUXABBcQEWWPvkjW\nRxLqGcTTpBwHrHltCOllnTpRKLDofBUqqHxnOtVWPgvGaeHIevgUSOeeDOJubfqonFpVNGVbHFcAhjnyFvrrqnRgKhkYqQZmRfUl",
"output": "-1"
},
{
"input": "tatuhQPIzjptlzzJpCAPXSRTKZRlwgfoCIsFjJquRoIDyZZYRSPdFUTjjUPhLBBfeEIfLQpygKXRcyQFiQsEtRtLnZErBqW\ntkHUjllbafLUWhVCnvblKjgYIEoHhsjVmrDBmAWbvtkHxDbRFvsXAjHIrujaDbYwOZmacknhZPeCcorbRgHjjgAgoJdjvLo",
"output": "-1"
},
{
"input": "cymCPGqdXKUdADEWDdUaLEEMHiXHsdAZuDnJDMUvxvrLRBrPSDpXPAgMRoGplLtniFRTomDTAHXWAdgUveTxaqKVSvnOyhOwiRN\nuhmyEWzapiRNPFDisvHTbenXMfeZaHqOFlKjrfQjUBwdFktNpeiRoDWuBftZLcCZZAVfioOihZVNqiNCNDIsUdIhvbcaxpTRWoV",
"output": "-1"
},
{
"input": "sSvpcITJAwghVfJaLKBmyjOkhltTGjYJVLWCYMFUomiJaKQYhXTajvZVHIMHbyckYROGQZzjWyWCcnmDmrkvTKfHSSzCIhsXgEZa\nvhCXkCwAmErGVBPBAnkSYEYvseFKbWSktoqaHYXUmYkHfOkRwuEyBRoGoBrOXBKVxXycjZGStuvDarnXMbZLWrbjrisDoJBdSvWJ",
"output": "-1"
},
{
"input": "hJDANKUNBisOOINDsTixJmYgHNogtpwswwcvVMptfGwIjvqgwTYFcqTdyAqaqlnhOCMtsnWXQqtjFwQlEcBtMFAtSqnqthVb\nrNquIcjNWESjpPVWmzUJFrelpUZeGDmSvCurCqVmKHKVAAPkaHksniOlzjiKYIJtvbuQWZRufMebpTFPqyxIWWjfPaWYiNlK",
"output": "-1"
},
{
"input": "ycLoapxsfsDTHMSfAAPIUpiEhQKUIXUcXEiopMBuuZLHtfPpLmCHwNMNQUwsEXxCEmKHTBSnKhtQhGWUvppUFZUgSpbeChX\ndCZhgVXofkGousCzObxZSJwXcHIaqUDSCPKzXntcVmPxtNcXmVcjsetZYxedmgQzXTZHMvzjoaXCMKsncGciSDqQWIIRlys",
"output": "1"
},
{
"input": "nvUbnrywIePXcoukIhwTfUVcHUEgXcsMyNQhmMlTltZiCooyZiIKRIGVHMCnTKgzXXIuvoNDEZswKoACOBGSyVNqTNQqMhAG\nplxuGSsyyJjdvpddrSebOARSAYcZKEaKjqbCwvjhNykuaECoQVHTVFMKXwvrQXRaqXsHsBaGVhCxGRxNyGUbMlxOarMZNXxy",
"output": "-1"
},
{
"input": "EncmXtAblQzcVRzMQqdDqXfAhXbtJKQwZVWyHoWUckohnZqfoCmNJDzexFgFJYrwNHGgzCJTzQQFnxGlhmvQTpicTkEeVICKac\nNIUNZoMLFMyAjVgQLITELJSodIXcGSDWfhFypRoGYuogJpnqGTotWxVqpvBHjFOWcDRDtARsaHarHaOkeNWEHGTaGOFCOFEwvK",
"output": "-1"
},
{
"input": "UG\nak",
"output": "1"
},
{
"input": "JZR\nVae",
"output": "-1"
},
{
"input": "a\nZ",
"output": "-1"
},
{
"input": "rk\nkv",
"output": "1"
},
{
"input": "RvuT\nbJzE",
"output": "1"
},
{
"input": "PPS\nydq",
"output": "-1"
},
{
"input": "q\nq",
"output": "0"
},
{
"input": "peOw\nIgSJ",
"output": "1"
},
{
"input": "PyK\noKN",
"output": "1"
},
{
"input": "O\ni",
"output": "1"
},
{
"input": "NmGY\npDlP",
"output": "-1"
},
{
"input": "nG\nZf",
"output": "-1"
},
{
"input": "m\na",
"output": "1"
},
{
"input": "MWyB\nWZEV",
"output": "-1"
},
{
"input": "Gre\nfxc",
"output": "1"
},
{
"input": "Ooq\nwap",
"output": "-1"
},
{
"input": "XId\nlbB",
"output": "1"
},
{
"input": "lfFpECEqUMEOJhipvkZjDPcpDNJedOVXiSMgBvBZbtfzIKekcvpWPCazKAhJyHircRtgcBIJwwstpHaLAgxFOngAWUZRgCef\nLfFPEcequmeojHIpVkzjDPcpdNJEDOVXiSmGBVBZBtfZikEKcvPwpCAzKAHJyHIrCRTgCbIJWwSTphALagXfOnGAwUzRGcEF",
"output": "0"
},
{
"input": "DQBdtSEDtFGiNRUeJNbOIfDZnsryUlzJHGTXGFXnwsVyxNtLgmklmFvRCzYETBVdmkpJJIvIOkMDgCFHZOTODiYrkwXd\nDQbDtsEdTFginRUEJNBOIfdZnsryulZJHGtxGFxnwSvYxnTLgmKlmFVRCzyEtBVdmKpJjiVioKMDgCFhzoTODiYrKwXD",
"output": "0"
},
{
"input": "tYWRijFQSzHBpCjUzqBtNvBKyzZRnIdWEuyqnORBQTLyOQglIGfYJIRjuxnbLvkqZakNqPiGDvgpWYkfxYNXsdoKXZtRkSasfa\nTYwRiJfqsZHBPcJuZQBTnVbkyZZRnidwEuYQnorbQTLYOqGligFyjirJUxnblVKqZaknQpigDVGPwyKfxyNXSDoKxztRKSaSFA",
"output": "0"
},
{
"input": "KhScXYiErQIUtmVhNTCXSLAviefIeHIIdiGhsYnPkSBaDTvMkyanfMLBOvDWgRybLtDqvXVdVjccNunDyijhhZEAKBrdz\nkHsCXyiErqIuTMVHNTCxSLaViEFIEhIIDiGHsYNpKsBAdTvMKyANFMLBovdwGRYbLtdQVxvDVJCcNUndYiJHhzeakBrdZ",
"output": "0"
},
{
"input": "cpPQMpjRQJKQVXjWDYECXbagSmNcVfOuBWNZxihdERraVuiOpSVDCPgTGuSQALNoVjySceHcKXwOEpSzXrEqWwwrYeppNiWhDVg\nCPPqmPjRqJkQvxJwdyECXBAGsMNcVfOuBWNzxIhderRavUiOpSvDCpGTgusqAlNovjyScEhCKXwoePSZxrEQwWwryEPPniWHDvG",
"output": "0"
},
{
"input": "SajcCGMepaLjZIWLRBGFcrZRCRvvoCsIyKsQerbrwsIamxxpRmQSZSalasJLVFbCHCuXJlubciQAvLxXYBazLsMKLHLdDQ\nsaJcCgmEpaLJziWlrBgFcRzrCrVVOcSIykSQerBrwSIamxxPrMqSzSalASjLVFbChCUxjLUbCIQAVlxxybAZLsmkLhLDdQ",
"output": "0"
},
{
"input": "kigPrWNTOUNDBskAfefjhHYZNYdnfZWuXWzHiBxFQryBbAkPtenFwWvCSTYGpzOntUNzNUhxRWjKmicTwLwJAnbAxj\nkigpRWntOUNdBsKaFEFjhhYZnYDNfzWuXwZhibxFQRybbakPteNfwwvcStyGPzoNTunznuHXrWjKMIctWLWJANBAxJ",
"output": "0"
},
{
"input": "nTomZZuTTRTAAPoUsySVFGElrpQRNLjqvFmcYytiheQnjUhPLnqNBiYtQkljbcvmjuNAVKbvQOWpqqFlQhAhULIhquoCnjUI\nntOmzZuttrtAAPOUSySVFgeLRPQrNLjQvfmCyYTiHEQnjuHPlNQNbIYtqKLJBCVmjunavkbvQOWPQQFlqHaHULIHQuOcnJUi",
"output": "0"
},
{
"input": "abac\nadaa",
"output": "-1"
},
{
"input": "Bbc\nabc",
"output": "1"
},
{
"input": "aaaba\naaaab",
"output": "1"
}
] | 1,695,189,544
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 53
| 92
| 0
|
s1=input()
s2=input()
s1=s1.lower()
s2=s2.lower()
#print(ord(s2[0]))
s1=list(s1)
s2=list(s2)
#print(ord(s2[0]))
f=0
for i in range(len(s1)):
if s1[i]!=s2[i]:
if ord(s1[i])<ord(s2[i]):
f=-1
break
elif ord(s1[i])>ord(s2[i]):
f=1
break
else:
f=0
break
print(f)
|
Title: Petya and Strings
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.
Input Specification:
Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.
Output Specification:
If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.
Demo Input:
['aaaa\naaaA\n', 'abs\nAbz\n', 'abcdefg\nAbCdEfF\n']
Demo Output:
['0\n', '-1\n', '1\n']
Note:
If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
- http://en.wikipedia.org/wiki/Lexicographical_order
|
```python
s1=input()
s2=input()
s1=s1.lower()
s2=s2.lower()
#print(ord(s2[0]))
s1=list(s1)
s2=list(s2)
#print(ord(s2[0]))
f=0
for i in range(len(s1)):
if s1[i]!=s2[i]:
if ord(s1[i])<ord(s2[i]):
f=-1
break
elif ord(s1[i])>ord(s2[i]):
f=1
break
else:
f=0
break
print(f)
```
| 3.977
|
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",
"output": "NO"
},
{
"input": "1010010100000000010",
"output": "YES"
},
{
"input": "101010101",
"output": "NO"
},
{
"input": "000000000100000000000110101100000",
"output": "YES"
},
{
"input": "100001000000110101100000",
"output": "NO"
},
{
"input": "100001000011010110000",
"output": "NO"
},
{
"input": "010",
"output": "NO"
},
{
"input": "10101011111111111111111111111100",
"output": "YES"
},
{
"input": "1001101100",
"output": "NO"
},
{
"input": "1001101010",
"output": "NO"
},
{
"input": "1111100111",
"output": "NO"
},
{
"input": "00110110001110001111",
"output": "NO"
},
{
"input": "11110001001111110001",
"output": "NO"
},
{
"input": "10001111001011111101",
"output": "NO"
},
{
"input": "10000010100000001000110001010100001001001010011",
"output": "YES"
},
{
"input": "01111011111010111100101100001011001010111110000010",
"output": "NO"
},
{
"input": "00100000100100101110011001011011101110110110010100",
"output": "NO"
},
{
"input": "10110100110001001011110101110010100010000000000100101010111110111110100011",
"output": "YES"
},
{
"input": "00011101010101111001011011001101101011111101000010100000111000011100101011",
"output": "NO"
},
{
"input": "01110000110100110101110100111000101101011101011110110100100111100001110111",
"output": "NO"
},
{
"input": "11110110011000100111100111101101011111110100010101011011111101110110110111",
"output": "YES"
},
{
"input": "100100010101110010001011001110100011100010011110100101100011010001001010001001101111001100",
"output": "NO"
},
{
"input": "111110010001011010010011111100110110001111000010100011011100111101111101110010101111011110000001010",
"output": "NO"
},
{
"input": "111110111100010100000100001010111011101011000111011011011010110010100010000101011111000011010011110",
"output": "NO"
},
{
"input": "1011110110111010110111111010010010100011111011110000011000110010011110111010110100011010100010111000",
"output": "NO"
},
{
"input": "0010100111100010110110000011100111110100111110001010000100111111111010111100101101010101001011010110",
"output": "YES"
},
{
"input": "1111010100010100101011101100101101110011000010100010000001111100010011100101010001101111000001011000",
"output": "NO"
},
{
"input": "0101100011001110001110100111100011010101011000000000110110010010111100101111010111100011101100100101",
"output": "YES"
},
{
"input": "0001101110011101110000000010011111101001101111100001001010110000110001100000010001111011011110001101",
"output": "YES"
},
{
"input": "1000010000100000100010000100001000010000100001000010000100001000010000100001000010000100001000010000",
"output": "NO"
},
{
"input": "1000000000100000000010000000001000000000100000000010000000001000000000100000000010000000001000000000",
"output": "YES"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "YES"
},
{
"input": "1111111111111111111111111111111111111111011111111111111111111111111111111111111111111111111111111111",
"output": "YES"
},
{
"input": "10100101000",
"output": "NO"
},
{
"input": "11110111011101",
"output": "NO"
},
{
"input": "10000000",
"output": "YES"
},
{
"input": "00000001",
"output": "YES"
},
{
"input": "01111111",
"output": "YES"
},
{
"input": "11111110",
"output": "YES"
}
] | 1,670,742,609
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 60
| 0
|
x = str(input())
zer = '0000000'
one = '1111111'
if zer in x or one in x:
print("Yes")
else:
print('No')
|
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 = str(input())
zer = '0000000'
one = '1111111'
if zer in x or one in x:
print("Yes")
else:
print('No')
```
| 0
|
854
|
A
|
Fraction
|
PROGRAMMING
| 800
|
[
"brute force",
"constructive algorithms",
"math"
] | null | null |
Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem.
|
In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction.
|
Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.
|
[
"3\n",
"4\n",
"12\n"
] |
[
"1 2\n",
"1 3\n",
"5 7\n"
] |
none
| 500
|
[
{
"input": "3",
"output": "1 2"
},
{
"input": "4",
"output": "1 3"
},
{
"input": "12",
"output": "5 7"
},
{
"input": "34",
"output": "15 19"
},
{
"input": "13",
"output": "6 7"
},
{
"input": "11",
"output": "5 6"
},
{
"input": "24",
"output": "11 13"
},
{
"input": "17",
"output": "8 9"
},
{
"input": "10",
"output": "3 7"
},
{
"input": "69",
"output": "34 35"
},
{
"input": "100",
"output": "49 51"
},
{
"input": "57",
"output": "28 29"
},
{
"input": "1000",
"output": "499 501"
},
{
"input": "999",
"output": "499 500"
},
{
"input": "998",
"output": "497 501"
},
{
"input": "997",
"output": "498 499"
},
{
"input": "996",
"output": "497 499"
},
{
"input": "995",
"output": "497 498"
},
{
"input": "994",
"output": "495 499"
},
{
"input": "5",
"output": "2 3"
},
{
"input": "6",
"output": "1 5"
},
{
"input": "8",
"output": "3 5"
},
{
"input": "9",
"output": "4 5"
},
{
"input": "423",
"output": "211 212"
},
{
"input": "876",
"output": "437 439"
},
{
"input": "29",
"output": "14 15"
}
] | 1,594,136,722
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 3
| 140
| 20,172,800
|
from math import *
n = int(input())
if (n%2==0):
print(n//2-1,n//2+1)
else:
print(n//2,n//2+1)
|
Title: Fraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem.
Input Specification:
In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction.
Output Specification:
Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.
Demo Input:
['3\n', '4\n', '12\n']
Demo Output:
['1 2\n', '1 3\n', '5 7\n']
Note:
none
|
```python
from math import *
n = int(input())
if (n%2==0):
print(n//2-1,n//2+1)
else:
print(n//2,n//2+1)
```
| 0
|
|
370
|
A
|
Rook, Bishop and King
|
PROGRAMMING
| 1,100
|
[
"graphs",
"math",
"shortest paths"
] | null | null |
Little Petya is learning to play chess. He has already learned how to move a king, a rook and a bishop. Let us remind you the rules of moving chess pieces. A chessboard is 64 square fields organized into an 8<=×<=8 table. A field is represented by a pair of integers (*r*,<=*c*) — the number of the row and the number of the column (in a classical game the columns are traditionally indexed by letters). Each chess piece takes up exactly one field. To make a move is to move a chess piece, the pieces move by the following rules:
- A rook moves any number of fields horizontally or vertically. - A bishop moves any number of fields diagonally. - A king moves one field in any direction — horizontally, vertically or diagonally.
Petya is thinking about the following problem: what minimum number of moves is needed for each of these pieces to move from field (*r*1,<=*c*1) to field (*r*2,<=*c*2)? At that, we assume that there are no more pieces besides this one on the board. Help him solve this problem.
|
The input contains four integers *r*1,<=*c*1,<=*r*2,<=*c*2 (1<=≤<=*r*1,<=*c*1,<=*r*2,<=*c*2<=≤<=8) — the coordinates of the starting and the final field. The starting field doesn't coincide with the final one.
You can assume that the chessboard rows are numbered from top to bottom 1 through 8, and the columns are numbered from left to right 1 through 8.
|
Print three space-separated integers: the minimum number of moves the rook, the bishop and the king (in this order) is needed to move from field (*r*1,<=*c*1) to field (*r*2,<=*c*2). If a piece cannot make such a move, print a 0 instead of the corresponding number.
|
[
"4 3 1 6\n",
"5 5 5 6\n"
] |
[
"2 1 3\n",
"1 0 1\n"
] |
none
| 500
|
[
{
"input": "4 3 1 6",
"output": "2 1 3"
},
{
"input": "5 5 5 6",
"output": "1 0 1"
},
{
"input": "1 1 8 8",
"output": "2 1 7"
},
{
"input": "1 1 8 1",
"output": "1 0 7"
},
{
"input": "1 1 1 8",
"output": "1 0 7"
},
{
"input": "8 1 1 1",
"output": "1 0 7"
},
{
"input": "8 1 1 8",
"output": "2 1 7"
},
{
"input": "7 7 6 6",
"output": "2 1 1"
},
{
"input": "8 1 8 8",
"output": "1 0 7"
},
{
"input": "1 8 1 1",
"output": "1 0 7"
},
{
"input": "1 8 8 1",
"output": "2 1 7"
},
{
"input": "1 8 8 8",
"output": "1 0 7"
},
{
"input": "8 8 1 1",
"output": "2 1 7"
},
{
"input": "8 8 1 8",
"output": "1 0 7"
},
{
"input": "8 8 8 1",
"output": "1 0 7"
},
{
"input": "1 3 1 6",
"output": "1 0 3"
},
{
"input": "1 3 1 4",
"output": "1 0 1"
},
{
"input": "1 3 1 5",
"output": "1 2 2"
},
{
"input": "3 3 2 4",
"output": "2 1 1"
},
{
"input": "3 3 1 5",
"output": "2 1 2"
},
{
"input": "1 6 2 1",
"output": "2 2 5"
},
{
"input": "1 5 6 4",
"output": "2 2 5"
},
{
"input": "1 3 3 7",
"output": "2 2 4"
},
{
"input": "1 1 8 1",
"output": "1 0 7"
},
{
"input": "1 7 5 4",
"output": "2 0 4"
},
{
"input": "1 5 2 7",
"output": "2 0 2"
},
{
"input": "1 4 6 2",
"output": "2 0 5"
},
{
"input": "1 2 3 5",
"output": "2 0 3"
},
{
"input": "1 8 8 7",
"output": "2 2 7"
},
{
"input": "6 5 6 2",
"output": "1 0 3"
},
{
"input": "6 3 3 5",
"output": "2 0 3"
},
{
"input": "6 1 7 8",
"output": "2 2 7"
},
{
"input": "1 2 3 2",
"output": "1 2 2"
},
{
"input": "3 8 7 2",
"output": "2 2 6"
},
{
"input": "4 2 6 4",
"output": "2 1 2"
},
{
"input": "1 1 1 3",
"output": "1 2 2"
},
{
"input": "6 8 8 6",
"output": "2 1 2"
},
{
"input": "6 7 4 1",
"output": "2 2 6"
},
{
"input": "6 5 1 4",
"output": "2 2 5"
},
{
"input": "3 2 7 6",
"output": "2 1 4"
},
{
"input": "3 8 4 1",
"output": "2 2 7"
},
{
"input": "3 6 1 4",
"output": "2 1 2"
}
] | 1,670,493,171
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 0
|
x1,y1,x2,y2 = map(int, input().split())
a1, a2,a3 = 0,0,0
if x1-x2 != 0:
a1 += 1
if y1 - y2 != 0:
a1 += 1
if y2%2 + x2%2 != y1%2 + x1%2:
a2 = 0
elif abs(y1-y2) == abs(x1-x2):
a2 = 1
else:
a2 = 2
q = min(abs(x2-x1), abs(y2-y1))
a3 = abs(abs(y2-y1)-q) + abs(abs(x2-x1)-q)+q
print(a1,a2,a3)
|
Title: Rook, Bishop and King
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Petya is learning to play chess. He has already learned how to move a king, a rook and a bishop. Let us remind you the rules of moving chess pieces. A chessboard is 64 square fields organized into an 8<=×<=8 table. A field is represented by a pair of integers (*r*,<=*c*) — the number of the row and the number of the column (in a classical game the columns are traditionally indexed by letters). Each chess piece takes up exactly one field. To make a move is to move a chess piece, the pieces move by the following rules:
- A rook moves any number of fields horizontally or vertically. - A bishop moves any number of fields diagonally. - A king moves one field in any direction — horizontally, vertically or diagonally.
Petya is thinking about the following problem: what minimum number of moves is needed for each of these pieces to move from field (*r*1,<=*c*1) to field (*r*2,<=*c*2)? At that, we assume that there are no more pieces besides this one on the board. Help him solve this problem.
Input Specification:
The input contains four integers *r*1,<=*c*1,<=*r*2,<=*c*2 (1<=≤<=*r*1,<=*c*1,<=*r*2,<=*c*2<=≤<=8) — the coordinates of the starting and the final field. The starting field doesn't coincide with the final one.
You can assume that the chessboard rows are numbered from top to bottom 1 through 8, and the columns are numbered from left to right 1 through 8.
Output Specification:
Print three space-separated integers: the minimum number of moves the rook, the bishop and the king (in this order) is needed to move from field (*r*1,<=*c*1) to field (*r*2,<=*c*2). If a piece cannot make such a move, print a 0 instead of the corresponding number.
Demo Input:
['4 3 1 6\n', '5 5 5 6\n']
Demo Output:
['2 1 3\n', '1 0 1\n']
Note:
none
|
```python
x1,y1,x2,y2 = map(int, input().split())
a1, a2,a3 = 0,0,0
if x1-x2 != 0:
a1 += 1
if y1 - y2 != 0:
a1 += 1
if y2%2 + x2%2 != y1%2 + x1%2:
a2 = 0
elif abs(y1-y2) == abs(x1-x2):
a2 = 1
else:
a2 = 2
q = min(abs(x2-x1), abs(y2-y1))
a3 = abs(abs(y2-y1)-q) + abs(abs(x2-x1)-q)+q
print(a1,a2,a3)
```
| 0
|
|
53
|
A
|
Autocomplete
|
PROGRAMMING
| 1,100
|
[
"implementation"
] |
A. Autocomplete
|
2
|
256
|
Autocomplete is a program function that enables inputting the text (in editors, command line shells, browsers etc.) completing the text by its inputted part. Vasya is busy working on a new browser called 'BERowser'. He happens to be working on the autocomplete function in the address line at this very moment. A list consisting of *n* last visited by the user pages and the inputted part *s* are known. Your task is to complete *s* to make it an address of one of the pages from the list. You have to find the lexicographically smallest address having a prefix *s*.
|
The first line contains the *s* line which is the inputted part. The second line contains an integer *n* (1<=≤<=*n*<=≤<=100) which is the number of visited pages. Then follow *n* lines which are the visited pages, one on each line. All the lines have lengths of from 1 to 100 symbols inclusively and consist of lowercase Latin letters only.
|
If *s* is not the beginning of any of *n* addresses of the visited pages, print *s*. Otherwise, print the lexicographically minimal address of one of the visited pages starting from *s*.
The lexicographical order is the order of words in a dictionary. The lexicographical comparison of lines is realized by the '<' operator in the modern programming languages.
|
[
"next\n2\nnextpermutation\nnextelement\n",
"find\n4\nfind\nfindfirstof\nfindit\nfand\n",
"find\n4\nfondfind\nfondfirstof\nfondit\nfand\n"
] |
[
"nextelement\n",
"find\n",
"find\n"
] |
none
| 500
|
[
{
"input": "next\n2\nnextpermutation\nnextelement",
"output": "nextelement"
},
{
"input": "find\n4\nfind\nfindfirstof\nfindit\nfand",
"output": "find"
},
{
"input": "find\n4\nfondfind\nfondfirstof\nfondit\nfand",
"output": "find"
},
{
"input": "kudljmxcse\n4\nkudljmxcse\nszjebdoad\nchz\na",
"output": "kudljmxcse"
},
{
"input": "ntqwpa\n5\nvvepyowvn\nntqwpakay\nhh\nygiafasda\nntqwpadm",
"output": "ntqwpadm"
},
{
"input": "aflb\n6\nsaej\nujxsiijg\npp\nhgoprw\ncp\nnt",
"output": "aflb"
},
{
"input": "dzwzyj\n7\nwvixktp\ndzwzyjuhn\ndzwzyjqrbd\ndzwzyji\ndzwzyjyfys\ndzwzyjrcb\nxptb",
"output": "dzwzyji"
},
{
"input": "wmblbphwdjjskzmlsyiznluiudelhlvcpyrooajvbwudnnstdhesauyxjugdwhrrwg\n1\nwjhsbxrrhadgtnybsugdtprncwerwezxuaxnqfpnosbispmnymnaqssdkjeynrnn",
"output": "wmblbphwdjjskzmlsyiznluiudelhlvcpyrooajvbwudnnstdhesauyxjugdwhrrwg"
},
{
"input": "hzkqvwliymwjbejfpnydrbwskhyrtrlsdinfrgwmnbdpwytcnjeoowxrfgfuaffzayjylvzu\n1\nhzkqvwliymwjbejfpnydrbwskhyrtrlsdinfrgwmnbdpwytcnjeoowxrfgfuaffzayjylvzubwjlvhhsfurqb",
"output": "hzkqvwliymwjbejfpnydrbwskhyrtrlsdinfrgwmnbdpwytcnjeoowxrfgfuaffzayjylvzubwjlvhhsfurqb"
},
{
"input": "msjnqudojxtzvpc\n2\nvlxclsvqbucmbrkwwtoxek\nmsjnqudojxtzvpcldwjyystsxrtexfhllzhnkidmhmyxpld",
"output": "msjnqudojxtzvpcldwjyystsxrtexfhllzhnkidmhmyxpld"
}
] | 1,569,452,223
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 216
| 0
|
arr=[]
s=input()
n=int(input())
for i in range(n):
a=input()
arr.append(a)
def isPrefixe(word,str):
for i in range(0,len(word)):
if (word[i]!=str[i]):
return False
return True
min=s
c=0
for x in arr:
if isPrefixe(s,x):
if c==0:
min=x
else:
if len(x)<len(min):
min=x
if len(min)==0:
print(s)
else:
print(min)
|
Title: Autocomplete
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Autocomplete is a program function that enables inputting the text (in editors, command line shells, browsers etc.) completing the text by its inputted part. Vasya is busy working on a new browser called 'BERowser'. He happens to be working on the autocomplete function in the address line at this very moment. A list consisting of *n* last visited by the user pages and the inputted part *s* are known. Your task is to complete *s* to make it an address of one of the pages from the list. You have to find the lexicographically smallest address having a prefix *s*.
Input Specification:
The first line contains the *s* line which is the inputted part. The second line contains an integer *n* (1<=≤<=*n*<=≤<=100) which is the number of visited pages. Then follow *n* lines which are the visited pages, one on each line. All the lines have lengths of from 1 to 100 symbols inclusively and consist of lowercase Latin letters only.
Output Specification:
If *s* is not the beginning of any of *n* addresses of the visited pages, print *s*. Otherwise, print the lexicographically minimal address of one of the visited pages starting from *s*.
The lexicographical order is the order of words in a dictionary. The lexicographical comparison of lines is realized by the '<' operator in the modern programming languages.
Demo Input:
['next\n2\nnextpermutation\nnextelement\n', 'find\n4\nfind\nfindfirstof\nfindit\nfand\n', 'find\n4\nfondfind\nfondfirstof\nfondit\nfand\n']
Demo Output:
['nextelement\n', 'find\n', 'find\n']
Note:
none
|
```python
arr=[]
s=input()
n=int(input())
for i in range(n):
a=input()
arr.append(a)
def isPrefixe(word,str):
for i in range(0,len(word)):
if (word[i]!=str[i]):
return False
return True
min=s
c=0
for x in arr:
if isPrefixe(s,x):
if c==0:
min=x
else:
if len(x)<len(min):
min=x
if len(min)==0:
print(s)
else:
print(min)
```
| 0
|
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"
},
{
"input": "AMVFNFJIAVNQJWIVONQOAOOQSNQSONOASONAONQINAONAOIQONANOIQOANOQINAONOQINAONOXJCOIAQOAOQAQAQAQAQWWWAQQAQ",
"output": "1077"
},
{
"input": "AAQQAXBQQBQQXBNQRJAQKQNAQNQVDQASAGGANQQQQTJFFQQQTQQA",
"output": "568"
},
{
"input": "KAZXAVLPJQBQVQQQQQAPAQQGQTQVZQAAAOYA",
"output": "70"
},
{
"input": "W",
"output": "0"
},
{
"input": "DBA",
"output": "0"
},
{
"input": "RQAWNACASAAKAGAAAAQ",
"output": "10"
},
{
"input": "QJAWZAAOAAGIAAAAAOQATASQAEAAAAQFQQHPA",
"output": "111"
},
{
"input": "QQKWQAQAAAAAAAAGAAVAQUEQQUMQMAQQQNQLAMAAAUAEAAEMAAA",
"output": "411"
},
{
"input": "QQUMQAYAUAAGWAAAQSDAVAAQAAAASKQJJQQQQMAWAYYAAAAAAEAJAXWQQ",
"output": "625"
},
{
"input": "QORZOYAQ",
"output": "1"
},
{
"input": "QCQAQAGAWAQQQAQAVQAQQQQAQAQQQAQAAATQAAVAAAQQQQAAAUUQAQQNQQWQQWAQAAQQKQYAQAAQQQAAQRAQQQWBQQQQAPBAQGQA",
"output": "13174"
},
{
"input": "QQAQQAKQFAQLQAAWAMQAZQAJQAAQQOACQQAAAYANAQAQQAQAAQQAOBQQJQAQAQAQQQAAAAABQQQAVNZAQQQQAMQQAFAAEAQAQHQT",
"output": "10420"
},
{
"input": "AQEGQHQQKQAQQPQKAQQQAAAAQQQAQEQAAQAAQAQFSLAAQQAQOQQAVQAAAPQQAWAQAQAFQAXAQQQQTRLOQAQQJQNQXQQQQSQVDQQQ",
"output": "12488"
},
{
"input": "QNQKQQQLASQBAVQQQQAAQQOQRJQQAQQQEQZUOANAADAAQQJAQAQARAAAQQQEQBHTQAAQAAAAQQMKQQQIAOJJQQAQAAADADQUQQQA",
"output": "9114"
},
{
"input": "QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ",
"output": "35937"
},
{
"input": "AMQQAAQAAQAAAAAAQQQBOAAANAAKQJCYQAE",
"output": "254"
},
{
"input": "AYQBAEQGAQEOAKGIXLQJAIAKQAAAQPUAJAKAATFWQQAOQQQUFQYAQQMQHOKAAJXGFCARAQSATHAUQQAATQJJQDQRAANQQAE",
"output": "2174"
},
{
"input": "AAQXAAQAYQAAAAGAQHVQYAGIVACADFAAQAAAAQZAAQMAKZAADQAQDAAQDAAAMQQOXYAQQQAKQBAAQQKAXQBJZDDLAAHQQ",
"output": "2962"
},
{
"input": "AYQQYAVAMNIAUAAKBBQVACWKTQSAQZAAQAAASZJAWBCAALAARHACQAKQQAQAARPAQAAQAQAAZQUSHQAMFVFZQQQQSAQQXAA",
"output": "2482"
},
{
"input": "LQMAQQARQAQBJQQQAGAAZQQXALQQAARQAQQQQAAQQAQQQAQQCAQQAQQAYQQQRAAZATQALYQQAAHHAAQHAAAAAAAAQQMAAQNAKQ",
"output": "7768"
},
{
"input": "MAQQWAQOYQMAAAQAQPQZAOAAQAUAQNAAQAAAITQSAQAKAQKAQQWSQAAQQAGUCDQMQWKQUXKWQQAAQQAAQQZQDQQQAABXQUUXQOA",
"output": "5422"
},
{
"input": "QTAAQDAQXAQQJQQQGAAAQQQQSBQZKAQQAQQQQEAQNUQBZCQLYQZQEQQAAQHQVAORKQVAQYQNASZQAARZAAGAAAAOQDCQ",
"output": "3024"
},
{
"input": "QQWAQQGQQUZQQQLZAAQYQXQVAQFQUAQZUQZZQUKBHSHTQYLQAOQXAQQGAQQTQOAQARQADAJRAAQPQAQQUQAUAMAUVQAAAQQAWQ",
"output": "4527"
},
{
"input": "QQAAQQAQVAQZQQQQAOEAQZPQIBQZACQQAFQQLAAQDATZQANHKYQQAQTAAFQRQAIQAJPWQAQTEIRXAEQQAYWAAAUKQQAQAQQQSQQH",
"output": "6416"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQAQAAAA",
"output": "14270"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQ",
"output": "13136"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQAQAA",
"output": "14270"
},
{
"input": "AQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQQAA",
"output": "14231"
},
{
"input": "QQQQQAQAAQQAQAQAAAAAAAAAQAQAAAAAQAQAQQQAQQQAAAQQQAAAAAAAQAAAAQQQQQQQAQQQQAQAAAQAAAAAQAQAAAAAQAQAA",
"output": "15296"
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ",
"output": "0"
},
{
"input": "QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "0"
},
{
"input": "QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ",
"output": "0"
},
{
"input": "QAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQA",
"output": "20825"
},
{
"input": "AQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQAQ",
"output": "20825"
},
{
"input": "Q",
"output": "0"
},
{
"input": "A",
"output": "0"
},
{
"input": "FFF",
"output": "0"
},
{
"input": "AAAAAA",
"output": "0"
}
] | 1,593,595,416
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 93
| 0
|
a=input()
b=a.count('Q')
c=a.count('A')
if b%2==0:
print((b*c)-1)
else:
print((b*c)-2)
|
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
a=input()
b=a.count('Q')
c=a.count('A')
if b%2==0:
print((b*c)-1)
else:
print((b*c)-2)
```
| 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",
"output": "NO"
},
{
"input": "1010010100000000010",
"output": "YES"
},
{
"input": "101010101",
"output": "NO"
},
{
"input": "000000000100000000000110101100000",
"output": "YES"
},
{
"input": "100001000000110101100000",
"output": "NO"
},
{
"input": "100001000011010110000",
"output": "NO"
},
{
"input": "010",
"output": "NO"
},
{
"input": "10101011111111111111111111111100",
"output": "YES"
},
{
"input": "1001101100",
"output": "NO"
},
{
"input": "1001101010",
"output": "NO"
},
{
"input": "1111100111",
"output": "NO"
},
{
"input": "00110110001110001111",
"output": "NO"
},
{
"input": "11110001001111110001",
"output": "NO"
},
{
"input": "10001111001011111101",
"output": "NO"
},
{
"input": "10000010100000001000110001010100001001001010011",
"output": "YES"
},
{
"input": "01111011111010111100101100001011001010111110000010",
"output": "NO"
},
{
"input": "00100000100100101110011001011011101110110110010100",
"output": "NO"
},
{
"input": "10110100110001001011110101110010100010000000000100101010111110111110100011",
"output": "YES"
},
{
"input": "00011101010101111001011011001101101011111101000010100000111000011100101011",
"output": "NO"
},
{
"input": "01110000110100110101110100111000101101011101011110110100100111100001110111",
"output": "NO"
},
{
"input": "11110110011000100111100111101101011111110100010101011011111101110110110111",
"output": "YES"
},
{
"input": "100100010101110010001011001110100011100010011110100101100011010001001010001001101111001100",
"output": "NO"
},
{
"input": "111110010001011010010011111100110110001111000010100011011100111101111101110010101111011110000001010",
"output": "NO"
},
{
"input": "111110111100010100000100001010111011101011000111011011011010110010100010000101011111000011010011110",
"output": "NO"
},
{
"input": "1011110110111010110111111010010010100011111011110000011000110010011110111010110100011010100010111000",
"output": "NO"
},
{
"input": "0010100111100010110110000011100111110100111110001010000100111111111010111100101101010101001011010110",
"output": "YES"
},
{
"input": "1111010100010100101011101100101101110011000010100010000001111100010011100101010001101111000001011000",
"output": "NO"
},
{
"input": "0101100011001110001110100111100011010101011000000000110110010010111100101111010111100011101100100101",
"output": "YES"
},
{
"input": "0001101110011101110000000010011111101001101111100001001010110000110001100000010001111011011110001101",
"output": "YES"
},
{
"input": "1000010000100000100010000100001000010000100001000010000100001000010000100001000010000100001000010000",
"output": "NO"
},
{
"input": "1000000000100000000010000000001000000000100000000010000000001000000000100000000010000000001000000000",
"output": "YES"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "YES"
},
{
"input": "1111111111111111111111111111111111111111011111111111111111111111111111111111111111111111111111111111",
"output": "YES"
},
{
"input": "10100101000",
"output": "NO"
},
{
"input": "11110111011101",
"output": "NO"
},
{
"input": "10000000",
"output": "YES"
},
{
"input": "00000001",
"output": "YES"
},
{
"input": "01111111",
"output": "YES"
},
{
"input": "11111110",
"output": "YES"
}
] | 1,676,969,051
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 9
| 92
| 0
|
def Team(s):
c = 0
for i in range(len(s)-1):
if s[i] == s[i+1]:
c += 1
if c >= 7 :
return "YES"
return "NO"
s = input()
ans = Team(s)
print(ans)
|
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
def Team(s):
c = 0
for i in range(len(s)-1):
if s[i] == s[i+1]:
c += 1
if c >= 7 :
return "YES"
return "NO"
s = input()
ans = Team(s)
print(ans)
```
| 0
|
864
|
A
|
Fair Game
|
PROGRAMMING
| 1,000
|
[
"implementation",
"sortings"
] | null | null |
Petya and Vasya decided to play a game. They have *n* cards (*n* is an even number). A single integer is written on each card.
Before the game Petya will choose an integer and after that Vasya will choose another integer (different from the number that Petya chose). During the game each player takes all the cards with number he chose. For example, if Petya chose number 5 before the game he will take all cards on which 5 is written and if Vasya chose number 10 before the game he will take all cards on which 10 is written.
The game is considered fair if Petya and Vasya can take all *n* cards, and the number of cards each player gets is the same.
Determine whether Petya and Vasya can choose integer numbers before the game so that the game is fair.
|
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=100) — number of cards. It is guaranteed that *n* is an even number.
The following *n* lines contain a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (one integer per line, 1<=≤<=*a**i*<=≤<=100) — numbers written on the *n* cards.
|
If it is impossible for Petya and Vasya to choose numbers in such a way that the game will be fair, print "NO" (without quotes) in the first line. In this case you should not print anything more.
In the other case print "YES" (without quotes) in the first line. In the second line print two distinct integers — number that Petya should choose and the number that Vasya should choose to make the game fair. If there are several solutions, print any of them.
|
[
"4\n11\n27\n27\n11\n",
"2\n6\n6\n",
"6\n10\n20\n30\n20\n10\n20\n",
"6\n1\n1\n2\n2\n3\n3\n"
] |
[
"YES\n11 27\n",
"NO\n",
"NO\n",
"NO\n"
] |
In the first example the game will be fair if, for example, Petya chooses number 11, and Vasya chooses number 27. Then the will take all cards — Petya will take cards 1 and 4, and Vasya will take cards 2 and 3. Thus, each of them will take exactly two cards.
In the second example fair game is impossible because the numbers written on the cards are equal, but the numbers that Petya and Vasya should choose should be distinct.
In the third example it is impossible to take all cards. Petya and Vasya can take at most five cards — for example, Petya can choose number 10 and Vasya can choose number 20. But for the game to be fair it is necessary to take 6 cards.
| 500
|
[
{
"input": "4\n11\n27\n27\n11",
"output": "YES\n11 27"
},
{
"input": "2\n6\n6",
"output": "NO"
},
{
"input": "6\n10\n20\n30\n20\n10\n20",
"output": "NO"
},
{
"input": "6\n1\n1\n2\n2\n3\n3",
"output": "NO"
},
{
"input": "2\n1\n100",
"output": "YES\n1 100"
},
{
"input": "2\n1\n1",
"output": "NO"
},
{
"input": "2\n100\n100",
"output": "NO"
},
{
"input": "14\n43\n43\n43\n43\n43\n43\n43\n43\n43\n43\n43\n43\n43\n43",
"output": "NO"
},
{
"input": "100\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n14\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32\n32",
"output": "YES\n14 32"
},
{
"input": "2\n50\n100",
"output": "YES\n50 100"
},
{
"input": "2\n99\n100",
"output": "YES\n99 100"
},
{
"input": "4\n4\n4\n5\n5",
"output": "YES\n4 5"
},
{
"input": "10\n10\n10\n10\n10\n10\n23\n23\n23\n23\n23",
"output": "YES\n10 23"
},
{
"input": "20\n34\n34\n34\n34\n34\n34\n34\n34\n34\n34\n11\n11\n11\n11\n11\n11\n11\n11\n11\n11",
"output": "YES\n11 34"
},
{
"input": "40\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n20\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30\n30",
"output": "YES\n20 30"
},
{
"input": "58\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1",
"output": "YES\n1 100"
},
{
"input": "98\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99\n99",
"output": "YES\n2 99"
},
{
"input": "100\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100",
"output": "YES\n1 100"
},
{
"input": "100\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2\n2",
"output": "YES\n1 2"
},
{
"input": "100\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n49\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12",
"output": "YES\n12 49"
},
{
"input": "100\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n15\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94\n94",
"output": "YES\n15 94"
},
{
"input": "100\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42\n42",
"output": "YES\n33 42"
},
{
"input": "100\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n16\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35\n35",
"output": "YES\n16 35"
},
{
"input": "100\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n33\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44\n44",
"output": "YES\n33 44"
},
{
"input": "100\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n54\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98\n98",
"output": "YES\n54 98"
},
{
"input": "100\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n81\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12\n12",
"output": "YES\n12 81"
},
{
"input": "100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100",
"output": "NO"
},
{
"input": "100\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1",
"output": "NO"
},
{
"input": "40\n20\n20\n30\n30\n20\n20\n20\n30\n30\n20\n20\n30\n30\n30\n30\n20\n30\n30\n30\n30\n20\n20\n30\n30\n30\n20\n30\n20\n30\n20\n30\n20\n20\n20\n30\n20\n20\n20\n30\n30",
"output": "NO"
},
{
"input": "58\n100\n100\n100\n100\n100\n1\n1\n1\n1\n1\n1\n100\n100\n1\n100\n1\n100\n100\n1\n1\n100\n100\n1\n100\n1\n100\n100\n1\n1\n100\n1\n1\n1\n100\n1\n1\n1\n1\n100\n1\n100\n100\n100\n100\n100\n1\n1\n100\n100\n100\n100\n1\n100\n1\n1\n1\n1\n1",
"output": "NO"
},
{
"input": "98\n2\n99\n99\n99\n99\n2\n99\n99\n99\n2\n2\n99\n2\n2\n2\n2\n99\n99\n2\n99\n2\n2\n99\n99\n99\n99\n2\n2\n99\n2\n99\n99\n2\n2\n99\n2\n99\n2\n99\n2\n2\n2\n99\n2\n2\n2\n2\n99\n99\n99\n99\n2\n2\n2\n2\n2\n2\n2\n2\n99\n2\n99\n99\n2\n2\n99\n99\n99\n99\n99\n99\n99\n99\n2\n99\n2\n99\n2\n2\n2\n99\n99\n99\n99\n99\n99\n2\n99\n99\n2\n2\n2\n2\n2\n99\n99\n99\n2",
"output": "NO"
},
{
"input": "100\n100\n1\n100\n1\n1\n100\n1\n1\n1\n100\n100\n1\n100\n1\n100\n100\n1\n1\n1\n100\n1\n100\n1\n100\n100\n1\n100\n1\n100\n1\n1\n1\n1\n1\n100\n1\n100\n100\n100\n1\n100\n100\n1\n100\n1\n1\n100\n100\n100\n1\n100\n100\n1\n1\n100\n100\n1\n100\n1\n100\n1\n1\n100\n100\n100\n100\n100\n100\n1\n100\n100\n1\n100\n100\n1\n100\n1\n1\n1\n100\n100\n1\n100\n1\n100\n1\n1\n1\n1\n100\n1\n1\n100\n1\n100\n100\n1\n100\n1\n100",
"output": "NO"
},
{
"input": "100\n100\n100\n100\n1\n100\n1\n1\n1\n100\n1\n1\n1\n1\n100\n1\n100\n1\n100\n1\n100\n100\n100\n1\n100\n1\n1\n1\n100\n1\n1\n1\n1\n1\n100\n100\n1\n100\n1\n1\n100\n1\n1\n100\n1\n100\n100\n100\n1\n100\n100\n100\n1\n100\n1\n100\n100\n100\n1\n1\n100\n100\n100\n100\n1\n100\n36\n100\n1\n100\n1\n100\n100\n100\n1\n1\n1\n1\n1\n1\n1\n1\n1\n100\n1\n1\n100\n100\n100\n100\n100\n1\n100\n1\n100\n1\n1\n100\n100\n1\n100",
"output": "NO"
},
{
"input": "100\n2\n1\n1\n2\n2\n1\n1\n1\n1\n2\n1\n1\n1\n2\n2\n2\n1\n1\n1\n2\n1\n2\n2\n2\n2\n1\n1\n2\n1\n1\n2\n1\n27\n1\n1\n1\n2\n2\n2\n1\n2\n1\n2\n1\n1\n2\n2\n2\n2\n2\n2\n2\n2\n1\n2\n2\n2\n2\n1\n2\n1\n1\n1\n1\n1\n2\n1\n1\n1\n2\n2\n2\n2\n2\n2\n1\n1\n1\n1\n2\n2\n1\n2\n2\n1\n1\n1\n2\n1\n2\n2\n1\n1\n2\n1\n1\n1\n2\n2\n1",
"output": "NO"
},
{
"input": "100\n99\n99\n100\n99\n99\n100\n100\n100\n99\n100\n99\n99\n100\n99\n99\n99\n99\n99\n99\n100\n100\n100\n99\n100\n100\n99\n100\n99\n100\n100\n99\n100\n99\n99\n99\n100\n99\n10\n99\n100\n100\n100\n99\n100\n100\n100\n100\n100\n100\n100\n99\n100\n100\n100\n99\n99\n100\n99\n100\n99\n100\n100\n99\n99\n99\n99\n100\n99\n100\n100\n100\n100\n100\n100\n99\n99\n100\n100\n99\n99\n99\n99\n99\n99\n100\n99\n99\n100\n100\n99\n100\n99\n99\n100\n99\n99\n99\n99\n100\n100",
"output": "NO"
},
{
"input": "100\n29\n43\n43\n29\n43\n29\n29\n29\n43\n29\n29\n29\n29\n43\n29\n29\n29\n29\n43\n29\n29\n29\n43\n29\n29\n29\n43\n43\n43\n43\n43\n43\n29\n29\n43\n43\n43\n29\n43\n43\n43\n29\n29\n29\n43\n29\n29\n29\n43\n43\n43\n43\n29\n29\n29\n29\n43\n29\n43\n43\n29\n29\n43\n43\n29\n29\n95\n29\n29\n29\n43\n43\n29\n29\n29\n29\n29\n43\n43\n43\n43\n29\n29\n43\n43\n43\n43\n43\n43\n29\n43\n43\n43\n43\n43\n43\n29\n43\n29\n43",
"output": "NO"
},
{
"input": "100\n98\n98\n98\n88\n88\n88\n88\n98\n98\n88\n98\n88\n98\n88\n88\n88\n88\n88\n98\n98\n88\n98\n98\n98\n88\n88\n88\n98\n98\n88\n88\n88\n98\n88\n98\n88\n98\n88\n88\n98\n98\n98\n88\n88\n98\n98\n88\n88\n88\n88\n88\n98\n98\n98\n88\n98\n88\n88\n98\n98\n88\n98\n88\n88\n98\n88\n88\n98\n27\n88\n88\n88\n98\n98\n88\n88\n98\n98\n98\n98\n98\n88\n98\n88\n98\n98\n98\n98\n88\n88\n98\n88\n98\n88\n98\n98\n88\n98\n98\n88",
"output": "NO"
},
{
"input": "100\n50\n1\n1\n50\n50\n50\n50\n1\n50\n100\n50\n50\n50\n100\n1\n100\n1\n100\n50\n50\n50\n50\n50\n1\n50\n1\n100\n1\n1\n50\n100\n50\n50\n100\n50\n50\n100\n1\n50\n50\n100\n1\n1\n50\n1\n100\n50\n50\n100\n100\n1\n100\n1\n50\n100\n50\n50\n1\n1\n50\n100\n50\n100\n100\n100\n50\n50\n1\n1\n50\n100\n1\n50\n100\n100\n1\n50\n50\n50\n100\n50\n50\n100\n1\n50\n50\n50\n50\n1\n50\n50\n50\n50\n1\n50\n50\n100\n1\n50\n100",
"output": "NO"
},
{
"input": "100\n45\n45\n45\n45\n45\n45\n44\n44\n44\n43\n45\n44\n44\n45\n44\n44\n45\n44\n43\n44\n43\n43\n43\n45\n43\n45\n44\n45\n43\n44\n45\n45\n45\n45\n45\n45\n45\n45\n43\n45\n43\n43\n45\n44\n45\n45\n45\n44\n45\n45\n45\n45\n45\n45\n44\n43\n45\n45\n43\n44\n45\n45\n45\n45\n44\n45\n45\n45\n43\n43\n44\n44\n43\n45\n43\n45\n45\n45\n44\n44\n43\n43\n44\n44\n44\n43\n45\n43\n44\n43\n45\n43\n43\n45\n45\n44\n45\n43\n43\n45",
"output": "NO"
},
{
"input": "100\n12\n12\n97\n15\n97\n12\n15\n97\n12\n97\n12\n12\n97\n12\n15\n12\n12\n15\n12\n12\n97\n12\n12\n15\n15\n12\n97\n15\n12\n97\n15\n12\n12\n15\n15\n15\n97\n15\n97\n12\n12\n12\n12\n12\n97\n12\n97\n12\n15\n15\n12\n15\n12\n15\n12\n12\n12\n12\n12\n12\n12\n12\n97\n97\n12\n12\n97\n12\n97\n97\n15\n97\n12\n97\n97\n12\n12\n12\n97\n97\n15\n12\n12\n15\n12\n15\n97\n97\n12\n15\n12\n12\n97\n12\n15\n15\n15\n15\n12\n12",
"output": "NO"
},
{
"input": "12\n2\n3\n1\n3\n3\n1\n2\n1\n2\n1\n3\n2",
"output": "NO"
},
{
"input": "48\n99\n98\n100\n100\n99\n100\n99\n100\n100\n98\n99\n98\n98\n99\n98\n99\n98\n100\n100\n98\n100\n98\n99\n100\n98\n99\n98\n99\n99\n100\n98\n99\n99\n98\n100\n99\n98\n99\n98\n100\n100\n100\n99\n98\n99\n98\n100\n100",
"output": "NO"
},
{
"input": "4\n1\n3\n3\n3",
"output": "NO"
},
{
"input": "6\n1\n1\n1\n1\n2\n2",
"output": "NO"
},
{
"input": "4\n1\n1\n1\n2",
"output": "NO"
},
{
"input": "4\n1\n2\n2\n2",
"output": "NO"
},
{
"input": "4\n1\n2\n3\n4",
"output": "NO"
},
{
"input": "8\n1\n1\n2\n2\n3\n3\n4\n4",
"output": "NO"
},
{
"input": "4\n1\n3\n2\n4",
"output": "NO"
},
{
"input": "4\n10\n10\n10\n20",
"output": "NO"
},
{
"input": "4\n11\n12\n13\n13",
"output": "NO"
},
{
"input": "4\n1\n1\n1\n3",
"output": "NO"
},
{
"input": "6\n1\n1\n2\n2\n2\n2",
"output": "NO"
},
{
"input": "10\n1\n1\n2\n2\n2\n3\n3\n4\n4\n4",
"output": "NO"
}
] | 1,698,484,930
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 46
| 77
| 0
|
from collections import defaultdict
n=int(input())
list1=[int(input()) for i in range (n)]
d=defaultdict(set)
for i in list1:
d[list1.count(i)].add(i)
flag=True
for i in d:
if len(d[i])%2==1:
flag=False
if flag:
print('YES')
for i in d:
for j in d[i]:
print(j,end=' ')
else:
print('NO')
|
Title: Fair Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya and Vasya decided to play a game. They have *n* cards (*n* is an even number). A single integer is written on each card.
Before the game Petya will choose an integer and after that Vasya will choose another integer (different from the number that Petya chose). During the game each player takes all the cards with number he chose. For example, if Petya chose number 5 before the game he will take all cards on which 5 is written and if Vasya chose number 10 before the game he will take all cards on which 10 is written.
The game is considered fair if Petya and Vasya can take all *n* cards, and the number of cards each player gets is the same.
Determine whether Petya and Vasya can choose integer numbers before the game so that the game is fair.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=100) — number of cards. It is guaranteed that *n* is an even number.
The following *n* lines contain a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (one integer per line, 1<=≤<=*a**i*<=≤<=100) — numbers written on the *n* cards.
Output Specification:
If it is impossible for Petya and Vasya to choose numbers in such a way that the game will be fair, print "NO" (without quotes) in the first line. In this case you should not print anything more.
In the other case print "YES" (without quotes) in the first line. In the second line print two distinct integers — number that Petya should choose and the number that Vasya should choose to make the game fair. If there are several solutions, print any of them.
Demo Input:
['4\n11\n27\n27\n11\n', '2\n6\n6\n', '6\n10\n20\n30\n20\n10\n20\n', '6\n1\n1\n2\n2\n3\n3\n']
Demo Output:
['YES\n11 27\n', 'NO\n', 'NO\n', 'NO\n']
Note:
In the first example the game will be fair if, for example, Petya chooses number 11, and Vasya chooses number 27. Then the will take all cards — Petya will take cards 1 and 4, and Vasya will take cards 2 and 3. Thus, each of them will take exactly two cards.
In the second example fair game is impossible because the numbers written on the cards are equal, but the numbers that Petya and Vasya should choose should be distinct.
In the third example it is impossible to take all cards. Petya and Vasya can take at most five cards — for example, Petya can choose number 10 and Vasya can choose number 20. But for the game to be fair it is necessary to take 6 cards.
|
```python
from collections import defaultdict
n=int(input())
list1=[int(input()) for i in range (n)]
d=defaultdict(set)
for i in list1:
d[list1.count(i)].add(i)
flag=True
for i in d:
if len(d[i])%2==1:
flag=False
if flag:
print('YES')
for i in d:
for j in d[i]:
print(j,end=' ')
else:
print('NO')
```
| 0
|
|
499
|
B
|
Lecture
|
PROGRAMMING
| 1,000
|
[
"implementation",
"strings"
] | null | null |
You have a new professor of graph theory and he speaks very quickly. You come up with the following plan to keep up with his lecture and make notes.
You know two languages, and the professor is giving the lecture in the first one. The words in both languages consist of lowercase English characters, each language consists of several words. For each language, all words are distinct, i.e. they are spelled differently. Moreover, the words of these languages have a one-to-one correspondence, that is, for each word in each language, there exists exactly one word in the other language having has the same meaning.
You can write down every word the professor says in either the first language or the second language. Of course, during the lecture you write down each word in the language in which the word is shorter. In case of equal lengths of the corresponding words you prefer the word of the first language.
You are given the text of the lecture the professor is going to read. Find out how the lecture will be recorded in your notes.
|
The first line contains two integers, *n* and *m* (1<=≤<=*n*<=≤<=3000, 1<=≤<=*m*<=≤<=3000) — the number of words in the professor's lecture and the number of words in each of these languages.
The following *m* lines contain the words. The *i*-th line contains two strings *a**i*, *b**i* meaning that the word *a**i* belongs to the first language, the word *b**i* belongs to the second language, and these two words have the same meaning. It is guaranteed that no word occurs in both languages, and each word occurs in its language exactly once.
The next line contains *n* space-separated strings *c*1,<=*c*2,<=...,<=*c**n* — the text of the lecture. It is guaranteed that each of the strings *c**i* belongs to the set of strings {*a*1,<=*a*2,<=... *a**m*}.
All the strings in the input are non-empty, each consisting of no more than 10 lowercase English letters.
|
Output exactly *n* words: how you will record the lecture in your notebook. Output the words of the lecture in the same order as in the input.
|
[
"4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest\n",
"5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll\n"
] |
[
"codeforces round letter round\n",
"hbnyiyc joll joll un joll\n"
] |
none
| 500
|
[
{
"input": "4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest",
"output": "codeforces round letter round"
},
{
"input": "5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll",
"output": "hbnyiyc joll joll un joll"
},
{
"input": "5 5\nqueyqj f\nb vn\ntabzvk qpfzoqx\nytnyonoc hnxsd\njpggvr lchinjmt\nqueyqj jpggvr b ytnyonoc b",
"output": "f jpggvr b hnxsd b"
},
{
"input": "10 22\nazbrll oen\ngh vdyayei\njphveblohx vfglv\nmfyxib jepnvhcuwo\nrpikazqj uam\nl rx\nokjenof qpnyi\nj tixqrno\nod itozmfct\nikkdxmirx ev\nqexftojc p\nkdazb zjs\nmbk ykvqjrxaxu\nhbcwhouzq pwt\nmirpsz zfaegpl\nuhkkvcj rlvwj\nef iqnnwtolrc\npjzfcpmeag ecdayth\nsa qcthz\ncbfhfxi qrnbvdryz\nwqel tj\natx smkbid\nef hbcwhouzq cbfhfxi hbcwhouzq mirpsz cbfhfxi cbfhfxi okjenof pjzfcpmeag kdazb",
"output": "ef pwt cbfhfxi pwt mirpsz cbfhfxi cbfhfxi qpnyi ecdayth zjs"
},
{
"input": "1 1\namit am\namit",
"output": "am"
},
{
"input": "1 1\na c\na",
"output": "a"
}
] | 1,697,353,894
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 31
| 0
|
# Read input
n, m = map(int, input().split())
words = {}
for _ in range(m):
a, b = input().split()
words[a] = b
lecture_words = input().split()
# Process the lecture
output = []
for word in lecture_words:
if len(words[word]) < len(word) or (len(words[word]) == len(word) and words[word] < word):
output.append(words[word])
else:
output.append(word)
# Print the result
print(" ".join(output))
|
Title: Lecture
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a new professor of graph theory and he speaks very quickly. You come up with the following plan to keep up with his lecture and make notes.
You know two languages, and the professor is giving the lecture in the first one. The words in both languages consist of lowercase English characters, each language consists of several words. For each language, all words are distinct, i.e. they are spelled differently. Moreover, the words of these languages have a one-to-one correspondence, that is, for each word in each language, there exists exactly one word in the other language having has the same meaning.
You can write down every word the professor says in either the first language or the second language. Of course, during the lecture you write down each word in the language in which the word is shorter. In case of equal lengths of the corresponding words you prefer the word of the first language.
You are given the text of the lecture the professor is going to read. Find out how the lecture will be recorded in your notes.
Input Specification:
The first line contains two integers, *n* and *m* (1<=≤<=*n*<=≤<=3000, 1<=≤<=*m*<=≤<=3000) — the number of words in the professor's lecture and the number of words in each of these languages.
The following *m* lines contain the words. The *i*-th line contains two strings *a**i*, *b**i* meaning that the word *a**i* belongs to the first language, the word *b**i* belongs to the second language, and these two words have the same meaning. It is guaranteed that no word occurs in both languages, and each word occurs in its language exactly once.
The next line contains *n* space-separated strings *c*1,<=*c*2,<=...,<=*c**n* — the text of the lecture. It is guaranteed that each of the strings *c**i* belongs to the set of strings {*a*1,<=*a*2,<=... *a**m*}.
All the strings in the input are non-empty, each consisting of no more than 10 lowercase English letters.
Output Specification:
Output exactly *n* words: how you will record the lecture in your notebook. Output the words of the lecture in the same order as in the input.
Demo Input:
['4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest\n', '5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll\n']
Demo Output:
['codeforces round letter round\n', 'hbnyiyc joll joll un joll\n']
Note:
none
|
```python
# Read input
n, m = map(int, input().split())
words = {}
for _ in range(m):
a, b = input().split()
words[a] = b
lecture_words = input().split()
# Process the lecture
output = []
for word in lecture_words:
if len(words[word]) < len(word) or (len(words[word]) == len(word) and words[word] < word):
output.append(words[word])
else:
output.append(word)
# Print the result
print(" ".join(output))
```
| 0
|
|
651
|
A
|
Joysticks
|
PROGRAMMING
| 1,100
|
[
"dp",
"greedy",
"implementation",
"math"
] | null | null |
Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at *a*1 percent and second one is charged at *a*2 percent. You can connect charger to a joystick only at the beginning of each minute. In one minute joystick either discharges by 2 percent (if not connected to a charger) or charges by 1 percent (if connected to a charger).
Game continues while both joysticks have a positive charge. Hence, if at the beginning of minute some joystick is charged by 1 percent, it has to be connected to a charger, otherwise the game stops. If some joystick completely discharges (its charge turns to 0), the game also stops.
Determine the maximum number of minutes that game can last. It is prohibited to pause the game, i. e. at each moment both joysticks should be enabled. It is allowed for joystick to be charged by more than 100 percent.
|
The first line of the input contains two positive integers *a*1 and *a*2 (1<=≤<=*a*1,<=*a*2<=≤<=100), the initial charge level of first and second joystick respectively.
|
Output the only integer, the maximum number of minutes that the game can last. Game continues until some joystick is discharged.
|
[
"3 5\n",
"4 4\n"
] |
[
"6\n",
"5\n"
] |
In the first sample game lasts for 6 minute by using the following algorithm:
- at the beginning of the first minute connect first joystick to the charger, by the end of this minute first joystick is at 4%, second is at 3%; - continue the game without changing charger, by the end of the second minute the first joystick is at 5%, second is at 1%; - at the beginning of the third minute connect second joystick to the charger, after this minute the first joystick is at 3%, the second one is at 2%; - continue the game without changing charger, by the end of the fourth minute first joystick is at 1%, second one is at 3%; - at the beginning of the fifth minute connect first joystick to the charger, after this minute the first joystick is at 2%, the second one is at 1%; - at the beginning of the sixth minute connect second joystick to the charger, after this minute the first joystick is at 0%, the second one is at 2%.
After that the first joystick is completely discharged and the game is stopped.
| 500
|
[
{
"input": "3 5",
"output": "6"
},
{
"input": "4 4",
"output": "5"
},
{
"input": "100 100",
"output": "197"
},
{
"input": "1 100",
"output": "98"
},
{
"input": "100 1",
"output": "98"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "8 8",
"output": "13"
},
{
"input": "7 2",
"output": "7"
},
{
"input": "24 15",
"output": "36"
},
{
"input": "19 30",
"output": "47"
},
{
"input": "15 31",
"output": "44"
},
{
"input": "14 15",
"output": "27"
},
{
"input": "58 33",
"output": "89"
},
{
"input": "15 25",
"output": "38"
},
{
"input": "59 45",
"output": "102"
},
{
"input": "3 73",
"output": "74"
},
{
"input": "48 1",
"output": "47"
},
{
"input": "100 25",
"output": "122"
},
{
"input": "40 49",
"output": "86"
},
{
"input": "85 73",
"output": "155"
},
{
"input": "29 1",
"output": "28"
},
{
"input": "74 25",
"output": "97"
},
{
"input": "24 57",
"output": "78"
},
{
"input": "23 12",
"output": "33"
},
{
"input": "2 99",
"output": "99"
},
{
"input": "98 2",
"output": "97"
},
{
"input": "2 97",
"output": "97"
},
{
"input": "30 54",
"output": "81"
},
{
"input": "32 53",
"output": "82"
},
{
"input": "32 54",
"output": "84"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "2 1",
"output": "1"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "1 3",
"output": "2"
},
{
"input": "3 1",
"output": "2"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "3 2",
"output": "3"
}
] | 1,584,111,006
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 39
| 156
| 0
|
a1,a2=map(int,input().split())
ans=0
while (a1 and a2):
if(a1<=a2):
a1+=1
if(a2>=2):
a2-=2
ans+=1
else:
a2=0
else:
a2+=1
if(a1>=2):
a1-=2
ans+=1
else:
a1=0
print(ans)
|
Title: Joysticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at *a*1 percent and second one is charged at *a*2 percent. You can connect charger to a joystick only at the beginning of each minute. In one minute joystick either discharges by 2 percent (if not connected to a charger) or charges by 1 percent (if connected to a charger).
Game continues while both joysticks have a positive charge. Hence, if at the beginning of minute some joystick is charged by 1 percent, it has to be connected to a charger, otherwise the game stops. If some joystick completely discharges (its charge turns to 0), the game also stops.
Determine the maximum number of minutes that game can last. It is prohibited to pause the game, i. e. at each moment both joysticks should be enabled. It is allowed for joystick to be charged by more than 100 percent.
Input Specification:
The first line of the input contains two positive integers *a*1 and *a*2 (1<=≤<=*a*1,<=*a*2<=≤<=100), the initial charge level of first and second joystick respectively.
Output Specification:
Output the only integer, the maximum number of minutes that the game can last. Game continues until some joystick is discharged.
Demo Input:
['3 5\n', '4 4\n']
Demo Output:
['6\n', '5\n']
Note:
In the first sample game lasts for 6 minute by using the following algorithm:
- at the beginning of the first minute connect first joystick to the charger, by the end of this minute first joystick is at 4%, second is at 3%; - continue the game without changing charger, by the end of the second minute the first joystick is at 5%, second is at 1%; - at the beginning of the third minute connect second joystick to the charger, after this minute the first joystick is at 3%, the second one is at 2%; - continue the game without changing charger, by the end of the fourth minute first joystick is at 1%, second one is at 3%; - at the beginning of the fifth minute connect first joystick to the charger, after this minute the first joystick is at 2%, the second one is at 1%; - at the beginning of the sixth minute connect second joystick to the charger, after this minute the first joystick is at 0%, the second one is at 2%.
After that the first joystick is completely discharged and the game is stopped.
|
```python
a1,a2=map(int,input().split())
ans=0
while (a1 and a2):
if(a1<=a2):
a1+=1
if(a2>=2):
a2-=2
ans+=1
else:
a2=0
else:
a2+=1
if(a1>=2):
a1-=2
ans+=1
else:
a1=0
print(ans)
```
| 3
|
|
439
|
B
|
Devu, the Dumb Guy
|
PROGRAMMING
| 1,200
|
[
"implementation",
"sortings"
] | null | null |
Devu is a dumb guy, his learning curve is very slow. You are supposed to teach him *n* subjects, the *i**th* subject has *c**i* chapters. When you teach him, you are supposed to teach all the chapters of a subject continuously.
Let us say that his initial per chapter learning power of a subject is *x* hours. In other words he can learn a chapter of a particular subject in *x* hours.
Well Devu is not complete dumb, there is a good thing about him too. If you teach him a subject, then time required to teach any chapter of the next subject will require exactly 1 hour less than previously required (see the examples to understand it more clearly). Note that his per chapter learning power can not be less than 1 hour.
You can teach him the *n* subjects in any possible order. Find out minimum amount of time (in hours) Devu will take to understand all the subjects and you will be free to do some enjoying task rather than teaching a dumb guy.
Please be careful that answer might not fit in 32 bit data type.
|
The first line will contain two space separated integers *n*, *x* (1<=≤<=*n*,<=*x*<=≤<=105). The next line will contain *n* space separated integers: *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=105).
|
Output a single integer representing the answer to the problem.
|
[
"2 3\n4 1\n",
"4 2\n5 1 2 1\n",
"3 3\n1 1 1\n"
] |
[
"11\n",
"10\n",
"6\n"
] |
Look at the first example. Consider the order of subjects: 1, 2. When you teach Devu the first subject, it will take him 3 hours per chapter, so it will take 12 hours to teach first subject. After teaching first subject, his per chapter learning time will be 2 hours. Now teaching him second subject will take 2 × 1 = 2 hours. Hence you will need to spend 12 + 2 = 14 hours.
Consider the order of subjects: 2, 1. When you teach Devu the second subject, then it will take him 3 hours per chapter, so it will take 3 × 1 = 3 hours to teach the second subject. After teaching the second subject, his per chapter learning time will be 2 hours. Now teaching him the first subject will take 2 × 4 = 8 hours. Hence you will need to spend 11 hours.
So overall, minimum of both the cases is 11 hours.
Look at the third example. The order in this example doesn't matter. When you teach Devu the first subject, it will take him 3 hours per chapter. When you teach Devu the second subject, it will take him 2 hours per chapter. When you teach Devu the third subject, it will take him 1 hours per chapter. In total it takes 6 hours.
| 1,000
|
[
{
"input": "2 3\n4 1",
"output": "11"
},
{
"input": "4 2\n5 1 2 1",
"output": "10"
},
{
"input": "3 3\n1 1 1",
"output": "6"
},
{
"input": "20 4\n1 1 3 5 5 1 3 4 2 5 2 4 3 1 3 3 3 3 4 3",
"output": "65"
},
{
"input": "20 10\n6 6 1 2 6 4 5 3 6 5 4 5 6 5 4 6 6 2 3 3",
"output": "196"
},
{
"input": "1 1\n9273",
"output": "9273"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "1 2\n1",
"output": "2"
},
{
"input": "1 2\n2",
"output": "4"
},
{
"input": "2 1\n1 2",
"output": "3"
}
] | 1,604,156,537
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 124
| 0
|
n, x = map(int, input().split())
list = list(map(int, input().split()))
list.sort()
ans = 0
for i in range(n):
ans += i*max(x, 1)
x -= 1
print(ans)
|
Title: Devu, the Dumb Guy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Devu is a dumb guy, his learning curve is very slow. You are supposed to teach him *n* subjects, the *i**th* subject has *c**i* chapters. When you teach him, you are supposed to teach all the chapters of a subject continuously.
Let us say that his initial per chapter learning power of a subject is *x* hours. In other words he can learn a chapter of a particular subject in *x* hours.
Well Devu is not complete dumb, there is a good thing about him too. If you teach him a subject, then time required to teach any chapter of the next subject will require exactly 1 hour less than previously required (see the examples to understand it more clearly). Note that his per chapter learning power can not be less than 1 hour.
You can teach him the *n* subjects in any possible order. Find out minimum amount of time (in hours) Devu will take to understand all the subjects and you will be free to do some enjoying task rather than teaching a dumb guy.
Please be careful that answer might not fit in 32 bit data type.
Input Specification:
The first line will contain two space separated integers *n*, *x* (1<=≤<=*n*,<=*x*<=≤<=105). The next line will contain *n* space separated integers: *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=105).
Output Specification:
Output a single integer representing the answer to the problem.
Demo Input:
['2 3\n4 1\n', '4 2\n5 1 2 1\n', '3 3\n1 1 1\n']
Demo Output:
['11\n', '10\n', '6\n']
Note:
Look at the first example. Consider the order of subjects: 1, 2. When you teach Devu the first subject, it will take him 3 hours per chapter, so it will take 12 hours to teach first subject. After teaching first subject, his per chapter learning time will be 2 hours. Now teaching him second subject will take 2 × 1 = 2 hours. Hence you will need to spend 12 + 2 = 14 hours.
Consider the order of subjects: 2, 1. When you teach Devu the second subject, then it will take him 3 hours per chapter, so it will take 3 × 1 = 3 hours to teach the second subject. After teaching the second subject, his per chapter learning time will be 2 hours. Now teaching him the first subject will take 2 × 4 = 8 hours. Hence you will need to spend 11 hours.
So overall, minimum of both the cases is 11 hours.
Look at the third example. The order in this example doesn't matter. When you teach Devu the first subject, it will take him 3 hours per chapter. When you teach Devu the second subject, it will take him 2 hours per chapter. When you teach Devu the third subject, it will take him 1 hours per chapter. In total it takes 6 hours.
|
```python
n, x = map(int, input().split())
list = list(map(int, input().split()))
list.sort()
ans = 0
for i in range(n):
ans += i*max(x, 1)
x -= 1
print(ans)
```
| 0
|
|
362
|
B
|
Petya and Staircases
|
PROGRAMMING
| 1,100
|
[
"implementation",
"sortings"
] | null | null |
Little boy Petya loves stairs very much. But he is bored from simple going up and down them — he loves jumping over several stairs at a time. As he stands on some stair, he can either jump to the next one or jump over one or two stairs at a time. But some stairs are too dirty and Petya doesn't want to step on them.
Now Petya is on the first stair of the staircase, consisting of *n* stairs. He also knows the numbers of the dirty stairs of this staircase. Help Petya find out if he can jump through the entire staircase and reach the last stair number *n* without touching a dirty stair once.
One has to note that anyway Petya should step on the first and last stairs, so if the first or the last stair is dirty, then Petya cannot choose a path with clean steps only.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=109, 0<=≤<=*m*<=≤<=3000) — the number of stairs in the staircase and the number of dirty stairs, correspondingly. The second line contains *m* different space-separated integers *d*1,<=*d*2,<=...,<=*d**m* (1<=≤<=*d**i*<=≤<=*n*) — the numbers of the dirty stairs (in an arbitrary order).
|
Print "YES" if Petya can reach stair number *n*, stepping only on the clean stairs. Otherwise print "NO".
|
[
"10 5\n2 4 8 3 6\n",
"10 5\n2 4 5 7 9\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "10 5\n2 4 8 3 6",
"output": "NO"
},
{
"input": "10 5\n2 4 5 7 9",
"output": "YES"
},
{
"input": "10 9\n2 3 4 5 6 7 8 9 10",
"output": "NO"
},
{
"input": "5 2\n4 5",
"output": "NO"
},
{
"input": "123 13\n36 73 111 2 92 5 47 55 48 113 7 78 37",
"output": "YES"
},
{
"input": "10 10\n7 6 4 2 5 10 8 3 9 1",
"output": "NO"
},
{
"input": "12312 0",
"output": "YES"
},
{
"input": "9817239 1\n6323187",
"output": "YES"
},
{
"input": "1 1\n1",
"output": "NO"
},
{
"input": "5 4\n4 2 5 1",
"output": "NO"
},
{
"input": "5 3\n4 3 5",
"output": "NO"
},
{
"input": "500 3\n18 62 445",
"output": "YES"
},
{
"input": "500 50\n72 474 467 241 442 437 336 234 410 120 438 164 405 177 142 114 27 20 445 235 46 176 88 488 242 391 28 414 145 92 206 334 152 343 367 254 100 243 155 348 148 450 461 483 97 34 471 69 416 362",
"output": "NO"
},
{
"input": "500 8\n365 313 338 410 482 417 325 384",
"output": "YES"
},
{
"input": "1000000000 10\n2 3 5 6 8 9 123 874 1230 1000000000",
"output": "NO"
},
{
"input": "1000000000 10\n1 2 3 5 6 8 9 123 874 1230",
"output": "NO"
},
{
"input": "10 1\n1",
"output": "NO"
},
{
"input": "10 4\n1 2 4 5",
"output": "NO"
},
{
"input": "50 20\n22 33 17 23 27 5 26 31 41 20 8 24 6 3 4 29 40 25 13 16",
"output": "NO"
},
{
"input": "50 40\n14 27 19 30 31 20 28 11 37 29 23 33 7 26 22 16 1 6 18 3 47 36 38 2 48 9 41 8 5 50 4 45 44 25 39 12 43 42 40 46",
"output": "NO"
},
{
"input": "123 12\n35 95 47 99 79 122 58 94 31 57 18 10",
"output": "YES"
},
{
"input": "10 5\n1 3 5 7 9",
"output": "NO"
},
{
"input": "100 7\n2 3 5 6 8 9 100",
"output": "NO"
},
{
"input": "100 3\n98 99 100",
"output": "NO"
},
{
"input": "100 3\n97 98 99",
"output": "NO"
},
{
"input": "100 3\n96 98 99",
"output": "YES"
},
{
"input": "10 6\n2 3 5 6 8 9",
"output": "YES"
},
{
"input": "1000000000 10\n2 4 10 18 40 42 49 58 59 60",
"output": "NO"
},
{
"input": "10 3\n1 4 6",
"output": "NO"
},
{
"input": "8 3\n2 3 4",
"output": "NO"
},
{
"input": "100 3\n4 5 6",
"output": "NO"
},
{
"input": "10 2\n10 1",
"output": "NO"
},
{
"input": "10 1\n10",
"output": "NO"
},
{
"input": "4 2\n2 3",
"output": "YES"
},
{
"input": "2 1\n1",
"output": "NO"
},
{
"input": "2 0",
"output": "YES"
},
{
"input": "4 3\n2 3 4",
"output": "NO"
},
{
"input": "5 3\n4 2 3",
"output": "NO"
}
] | 1,629,722,962
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 61
| 6,656,000
|
def main():
stairs, dirty_stairs = input().split()
dirty_stairs_idx = set(int(x) for x in input().split())
current_stair = 1
for i in range(int(stairs)):
if current_stair + 1 not in dirty_stairs_idx:
current_stair += 1
continue
if current_stair + 2 not in dirty_stairs_idx:
current_stair += 2
continue
if current_stair + 3 not in dirty_stairs_idx:
current_stair += 3
continue
return "NO"
return "YES"
print(main())
|
Title: Petya and Staircases
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little boy Petya loves stairs very much. But he is bored from simple going up and down them — he loves jumping over several stairs at a time. As he stands on some stair, he can either jump to the next one or jump over one or two stairs at a time. But some stairs are too dirty and Petya doesn't want to step on them.
Now Petya is on the first stair of the staircase, consisting of *n* stairs. He also knows the numbers of the dirty stairs of this staircase. Help Petya find out if he can jump through the entire staircase and reach the last stair number *n* without touching a dirty stair once.
One has to note that anyway Petya should step on the first and last stairs, so if the first or the last stair is dirty, then Petya cannot choose a path with clean steps only.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=109, 0<=≤<=*m*<=≤<=3000) — the number of stairs in the staircase and the number of dirty stairs, correspondingly. The second line contains *m* different space-separated integers *d*1,<=*d*2,<=...,<=*d**m* (1<=≤<=*d**i*<=≤<=*n*) — the numbers of the dirty stairs (in an arbitrary order).
Output Specification:
Print "YES" if Petya can reach stair number *n*, stepping only on the clean stairs. Otherwise print "NO".
Demo Input:
['10 5\n2 4 8 3 6\n', '10 5\n2 4 5 7 9\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
def main():
stairs, dirty_stairs = input().split()
dirty_stairs_idx = set(int(x) for x in input().split())
current_stair = 1
for i in range(int(stairs)):
if current_stair + 1 not in dirty_stairs_idx:
current_stair += 1
continue
if current_stair + 2 not in dirty_stairs_idx:
current_stair += 2
continue
if current_stair + 3 not in dirty_stairs_idx:
current_stair += 3
continue
return "NO"
return "YES"
print(main())
```
| 0
|
|
155
|
A
|
I_love_\%username\%
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him.
|
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000.
|
Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests.
|
[
"5\n100 50 200 150 200\n",
"10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n"
] |
[
"2\n",
"4\n"
] |
In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing.
| 500
|
[
{
"input": "5\n100 50 200 150 200",
"output": "2"
},
{
"input": "10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242",
"output": "4"
},
{
"input": "1\n6",
"output": "0"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "5\n100 36 53 7 81",
"output": "2"
},
{
"input": "5\n7 36 53 81 100",
"output": "4"
},
{
"input": "5\n100 81 53 36 7",
"output": "4"
},
{
"input": "10\n8 6 3 4 9 10 7 7 1 3",
"output": "5"
},
{
"input": "10\n1627 1675 1488 1390 1812 1137 1746 1324 1952 1862",
"output": "6"
},
{
"input": "10\n1 3 3 4 6 7 7 8 9 10",
"output": "7"
},
{
"input": "10\n1952 1862 1812 1746 1675 1627 1488 1390 1324 1137",
"output": "9"
},
{
"input": "25\n1448 4549 2310 2725 2091 3509 1565 2475 2232 3989 4231 779 2967 2702 608 3739 721 1552 2767 530 3114 665 1940 48 4198",
"output": "5"
},
{
"input": "33\n1097 1132 1091 1104 1049 1038 1023 1080 1104 1029 1035 1061 1049 1060 1088 1106 1105 1087 1063 1076 1054 1103 1047 1041 1028 1120 1126 1063 1117 1110 1044 1093 1101",
"output": "5"
},
{
"input": "34\n821 5536 2491 6074 7216 9885 764 1603 778 8736 8987 771 617 1587 8943 7922 439 7367 4115 8886 7878 6899 8811 5752 3184 3401 9760 9400 8995 4681 1323 6637 6554 6498",
"output": "7"
},
{
"input": "68\n6764 6877 6950 6768 6839 6755 6726 6778 6699 6805 6777 6985 6821 6801 6791 6805 6940 6761 6677 6999 6911 6699 6959 6933 6903 6843 6972 6717 6997 6756 6789 6668 6735 6852 6735 6880 6723 6834 6810 6694 6780 6679 6698 6857 6826 6896 6979 6968 6957 6988 6960 6700 6919 6892 6984 6685 6813 6678 6715 6857 6976 6902 6780 6686 6777 6686 6842 6679",
"output": "9"
},
{
"input": "60\n9000 9014 9034 9081 9131 9162 9174 9199 9202 9220 9221 9223 9229 9235 9251 9260 9268 9269 9270 9298 9307 9309 9313 9323 9386 9399 9407 9495 9497 9529 9531 9544 9614 9615 9627 9627 9643 9654 9656 9657 9685 9699 9701 9736 9745 9758 9799 9827 9843 9845 9854 9854 9885 9891 9896 9913 9942 9963 9986 9992",
"output": "57"
},
{
"input": "100\n7 61 12 52 41 16 34 99 30 44 48 89 31 54 21 1 48 52 61 15 35 87 21 76 64 92 44 81 16 93 84 92 32 15 68 76 53 39 26 4 11 26 7 4 99 99 61 65 55 85 65 67 47 39 2 74 63 49 98 87 5 94 22 30 25 42 31 84 49 23 89 60 16 26 92 27 9 57 75 61 94 35 83 47 99 100 63 24 91 88 79 10 15 45 22 64 3 11 89 83",
"output": "4"
},
{
"input": "100\n9999 9999 9999 9998 9998 9998 9997 9996 9996 9995 9993 9993 9991 9990 9989 9986 9984 9984 9983 9981 9981 9980 9980 9980 9979 9977 9977 9977 9977 9977 9976 9976 9975 9975 9973 9972 9972 9972 9972 9971 9969 9969 9969 9968 9967 9965 9962 9962 9962 9962 9960 9958 9958 9955 9954 9954 9954 9954 9950 9949 9949 9947 9943 9943 9940 9940 9939 9937 9937 9937 9936 9935 9932 9932 9931 9929 9927 9927 9926 9923 9919 9919 9918 9918 9918 9917 9917 9916 9916 9915 9914 9911 9910 9907 9907 9906 9905 9905 9904 9901",
"output": "57"
},
{
"input": "100\n158 159 157 156 155 154 160 153 161 152 162 163 151 164 150 165 149 148 147 166 146 167 145 144 143 142 168 141 169 170 140 139 138 137 171 136 135 134 172 173 174 133 175 132 177 178 131 179 180 130 129 181 128 127 182 126 183 125 124 184 123 122 121 120 119 118 117 185 186 187 188 189 116 190 115 191 192 193 114 113 112 111 110 109 108 194 195 107 106 105 196 197 104 198 199 103 102 200 101 100",
"output": "99"
},
{
"input": "2\n0 10000",
"output": "1"
},
{
"input": "2\n5 5",
"output": "0"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n10 10",
"output": "0"
},
{
"input": "1\n0",
"output": "0"
}
] | 1,676,126,042
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 92
| 0
|
n = int(input())
l = []
c = 0
a = input().split()
b = int(a[0])
for i in a:
ii = int(i)
if ii > b:
c += 1
b = ii
else:
b = ii
continue
print(c)
|
Title: I_love_\%username\%
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him.
Input Specification:
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000.
Output Specification:
Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests.
Demo Input:
['5\n100 50 200 150 200\n', '10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n']
Demo Output:
['2\n', '4\n']
Note:
In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing.
|
```python
n = int(input())
l = []
c = 0
a = input().split()
b = int(a[0])
for i in a:
ii = int(i)
if ii > b:
c += 1
b = ii
else:
b = ii
continue
print(c)
```
| 0
|
|
181
|
A
|
Series of Crimes
|
PROGRAMMING
| 800
|
[
"brute force",
"geometry",
"implementation"
] | null | null |
The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed.
|
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements.
|
Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right.
|
[
"3 2\n.*\n..\n**\n",
"3 3\n*.*\n*..\n...\n"
] |
[
"1 1\n",
"2 3\n"
] |
none
| 500
|
[
{
"input": "3 2\n.*\n..\n**",
"output": "1 1"
},
{
"input": "2 5\n*....\n*...*",
"output": "1 5"
},
{
"input": "7 2\n..\n**\n..\n..\n..\n..\n.*",
"output": "7 1"
},
{
"input": "7 2\n*.\n..\n..\n..\n..\n..\n**",
"output": "1 2"
},
{
"input": "2 10\n*......*..\n.......*..",
"output": "2 1"
},
{
"input": "10 3\n*..\n...\n...\n...\n...\n...\n...\n...\n...\n**.",
"output": "1 2"
},
{
"input": "100 2\n*.\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n**",
"output": "1 2"
},
{
"input": "99 3\n**.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n*..",
"output": "99 2"
},
{
"input": "98 3\n...\n.*.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n.**",
"output": "2 3"
},
{
"input": "99 3\n**.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n*..",
"output": "99 2"
},
{
"input": "98 3\n...\n.*.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n.**",
"output": "2 3"
},
{
"input": "2 100\n...................................................................................................*\n*..................................................................................................*",
"output": "1 1"
},
{
"input": "2 2\n*.\n**",
"output": "1 2"
},
{
"input": "2 2\n**\n.*",
"output": "2 1"
},
{
"input": "2 3\n*.*\n..*",
"output": "2 1"
}
] | 1,664,888,675
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 92
| 0
|
x = y = 0
n, _ = map(int, input().split())
for i in range(n):
for j, ch in enumerate(input()):
if ch == '*':
x, y = x^i, y^j
print(x+1, y+1)
|
Title: Series of Crimes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed.
Input Specification:
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements.
Output Specification:
Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right.
Demo Input:
['3 2\n.*\n..\n**\n', '3 3\n*.*\n*..\n...\n']
Demo Output:
['1 1\n', '2 3\n']
Note:
none
|
```python
x = y = 0
n, _ = map(int, input().split())
for i in range(n):
for j, ch in enumerate(input()):
if ch == '*':
x, y = x^i, y^j
print(x+1, y+1)
```
| 3
|
|
863
|
B
|
Kayaking
|
PROGRAMMING
| 1,500
|
[
"brute force",
"greedy",
"sortings"
] | null | null |
Vadim is really keen on travelling. Recently he heard about kayaking activity near his town and became very excited about it, so he joined a party of kayakers.
Now the party is ready to start its journey, but firstly they have to choose kayaks. There are 2·*n* people in the group (including Vadim), and they have exactly *n*<=-<=1 tandem kayaks (each of which, obviously, can carry two people) and 2 single kayaks. *i*-th person's weight is *w**i*, and weight is an important matter in kayaking — if the difference between the weights of two people that sit in the same tandem kayak is too large, then it can crash. And, of course, people want to distribute their seats in kayaks in order to minimize the chances that kayaks will crash.
Formally, the instability of a single kayak is always 0, and the instability of a tandem kayak is the absolute difference between weights of the people that are in this kayak. Instability of the whole journey is the total instability of all kayaks.
Help the party to determine minimum possible total instability!
|
The first line contains one number *n* (2<=≤<=*n*<=≤<=50).
The second line contains 2·*n* integer numbers *w*1, *w*2, ..., *w*2*n*, where *w**i* is weight of person *i* (1<=≤<=*w**i*<=≤<=1000).
|
Print minimum possible total instability.
|
[
"2\n1 2 3 4\n",
"4\n1 3 4 6 3 4 100 200\n"
] |
[
"1\n",
"5\n"
] |
none
| 0
|
[
{
"input": "2\n1 2 3 4",
"output": "1"
},
{
"input": "4\n1 3 4 6 3 4 100 200",
"output": "5"
},
{
"input": "3\n305 139 205 406 530 206",
"output": "102"
},
{
"input": "3\n610 750 778 6 361 407",
"output": "74"
},
{
"input": "5\n97 166 126 164 154 98 221 7 51 47",
"output": "35"
},
{
"input": "50\n1 1 2 2 1 3 2 2 1 1 1 1 2 3 3 1 2 1 3 3 2 1 2 3 1 1 2 1 3 1 3 1 3 3 3 1 1 1 3 3 2 2 2 2 3 2 2 2 2 3 1 3 3 3 3 1 3 3 1 3 3 3 3 2 3 1 3 3 1 1 1 3 1 2 2 2 1 1 1 3 1 2 3 2 1 3 3 2 2 1 3 1 3 1 2 2 1 2 3 2",
"output": "0"
},
{
"input": "50\n5 5 5 5 4 2 2 3 2 2 4 1 5 5 1 2 4 2 4 2 5 2 2 2 2 3 2 4 2 5 5 4 3 1 2 3 3 5 4 2 2 5 2 4 5 5 4 4 1 5 5 3 2 2 5 1 3 3 2 4 4 5 1 2 3 4 4 1 3 3 3 5 1 2 4 4 4 4 2 5 2 5 3 2 4 5 5 2 1 1 2 4 5 3 2 1 2 4 4 4",
"output": "1"
},
{
"input": "50\n499 780 837 984 481 526 944 482 862 136 265 605 5 631 974 967 574 293 969 467 573 845 102 224 17 873 648 120 694 996 244 313 404 129 899 583 541 314 525 496 443 857 297 78 575 2 430 137 387 319 382 651 594 411 845 746 18 232 6 289 889 81 174 175 805 1000 799 950 475 713 951 685 729 925 262 447 139 217 788 514 658 572 784 185 112 636 10 251 621 218 210 89 597 553 430 532 264 11 160 476",
"output": "368"
},
{
"input": "50\n873 838 288 87 889 364 720 410 565 651 577 356 740 99 549 592 994 385 777 435 486 118 887 440 749 533 356 790 413 681 267 496 475 317 88 660 374 186 61 437 729 860 880 538 277 301 667 180 60 393 955 540 896 241 362 146 74 680 734 767 851 337 751 860 542 735 444 793 340 259 495 903 743 961 964 966 87 275 22 776 368 701 835 732 810 735 267 988 352 647 924 183 1 924 217 944 322 252 758 597",
"output": "393"
},
{
"input": "50\n297 787 34 268 439 629 600 398 425 833 721 908 830 636 64 509 420 647 499 675 427 599 396 119 798 742 577 355 22 847 389 574 766 453 196 772 808 261 106 844 726 975 173 992 874 89 775 616 678 52 69 591 181 573 258 381 665 301 589 379 362 146 790 842 765 100 229 916 938 97 340 793 758 177 736 396 247 562 571 92 923 861 165 748 345 703 431 930 101 761 862 595 505 393 126 846 431 103 596 21",
"output": "387"
},
{
"input": "50\n721 631 587 746 692 406 583 90 388 16 161 948 921 70 387 426 39 398 517 724 879 377 906 502 359 950 798 408 846 718 911 845 57 886 9 668 537 632 344 762 19 193 658 447 870 173 98 156 592 519 183 539 274 393 962 615 551 626 148 183 769 763 829 120 796 761 14 744 537 231 696 284 581 688 611 826 703 145 224 600 965 613 791 275 984 375 402 281 851 580 992 8 816 454 35 532 347 250 242 637",
"output": "376"
},
{
"input": "50\n849 475 37 120 754 183 758 374 543 198 896 691 11 607 198 343 761 660 239 669 628 259 223 182 216 158 20 565 454 884 137 923 156 22 310 77 267 707 582 169 120 308 439 309 59 152 206 696 210 177 296 887 559 22 154 553 142 247 491 692 473 572 461 206 532 319 503 164 328 365 541 366 300 392 486 257 863 432 877 404 520 69 418 99 519 239 374 927 601 103 226 316 423 219 240 26 455 101 184 61",
"output": "351"
},
{
"input": "3\n1 2 10 11 100 100",
"output": "1"
},
{
"input": "17\n814 744 145 886 751 1000 272 914 270 529 467 164 410 369 123 424 991 12 702 582 561 858 746 950 598 393 606 498 648 686 455 873 728 858",
"output": "318"
},
{
"input": "45\n476 103 187 696 463 457 588 632 763 77 391 721 95 124 378 812 980 193 694 898 859 572 721 274 605 264 929 615 257 918 42 493 1 3 697 349 990 800 82 535 382 816 943 735 11 272 562 323 653 370 766 332 666 130 704 604 645 717 267 255 37 470 925 941 376 611 332 758 504 40 477 263 708 434 38 596 650 990 714 662 572 467 949 799 648 581 545 828 508 636",
"output": "355"
},
{
"input": "2\n55 5 25 51",
"output": "4"
},
{
"input": "25\n89 50 640 463 858 301 522 241 923 378 892 822 550 17 42 66 706 779 657 840 273 222 444 459 94 925 437 159 182 727 92 851 742 215 653 891 782 533 29 128 133 883 317 475 165 994 802 434 744 973",
"output": "348"
},
{
"input": "4\n35 48 71 44 78 79 57 48",
"output": "10"
},
{
"input": "3\n58 89 73 15 5 47",
"output": "21"
},
{
"input": "2\n1 20 99 100",
"output": "1"
}
] | 1,672,467,388
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 0
|
n=input()
n=int(n)*2
weights=[]
weight=input()
x=''
answer=0
answer2=float("inf")
for letter in weight:
if letter==" ":
weights.append(int(x))
x=''
else:
x+=letter
weights.append(int(x))
weights.sort()
weights2=weights
for i in range(n):
for e in range(n):
if e<=i:
continue
weights2.append(weights[e])
weights2.append(weights[i])
del weights2[e]
del weights2[i]
weights2.sort()
for a in range(0,n-2,2):
answer+=weights[a+1]-weights[a]
answer2=min(answer2,answer)
weights2=weights
print(answer2)
|
Title: Kayaking
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vadim is really keen on travelling. Recently he heard about kayaking activity near his town and became very excited about it, so he joined a party of kayakers.
Now the party is ready to start its journey, but firstly they have to choose kayaks. There are 2·*n* people in the group (including Vadim), and they have exactly *n*<=-<=1 tandem kayaks (each of which, obviously, can carry two people) and 2 single kayaks. *i*-th person's weight is *w**i*, and weight is an important matter in kayaking — if the difference between the weights of two people that sit in the same tandem kayak is too large, then it can crash. And, of course, people want to distribute their seats in kayaks in order to minimize the chances that kayaks will crash.
Formally, the instability of a single kayak is always 0, and the instability of a tandem kayak is the absolute difference between weights of the people that are in this kayak. Instability of the whole journey is the total instability of all kayaks.
Help the party to determine minimum possible total instability!
Input Specification:
The first line contains one number *n* (2<=≤<=*n*<=≤<=50).
The second line contains 2·*n* integer numbers *w*1, *w*2, ..., *w*2*n*, where *w**i* is weight of person *i* (1<=≤<=*w**i*<=≤<=1000).
Output Specification:
Print minimum possible total instability.
Demo Input:
['2\n1 2 3 4\n', '4\n1 3 4 6 3 4 100 200\n']
Demo Output:
['1\n', '5\n']
Note:
none
|
```python
n=input()
n=int(n)*2
weights=[]
weight=input()
x=''
answer=0
answer2=float("inf")
for letter in weight:
if letter==" ":
weights.append(int(x))
x=''
else:
x+=letter
weights.append(int(x))
weights.sort()
weights2=weights
for i in range(n):
for e in range(n):
if e<=i:
continue
weights2.append(weights[e])
weights2.append(weights[i])
del weights2[e]
del weights2[i]
weights2.sort()
for a in range(0,n-2,2):
answer+=weights[a+1]-weights[a]
answer2=min(answer2,answer)
weights2=weights
print(answer2)
```
| 0
|
|
49
|
A
|
Sleuth
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Sleuth
|
2
|
256
|
Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z
|
The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter.
|
Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters.
|
[
"Is it a melon?\n",
"Is it an apple?\n",
"Is it a banana ?\n",
"Is it an apple and a banana simultaneouSLY?\n"
] |
[
"NO\n",
"YES\n",
"YES\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "Is it a melon?",
"output": "NO"
},
{
"input": "Is it an apple?",
"output": "YES"
},
{
"input": " Is it a banana ?",
"output": "YES"
},
{
"input": "Is it an apple and a banana simultaneouSLY?",
"output": "YES"
},
{
"input": "oHtSbDwzHb?",
"output": "NO"
},
{
"input": "sZecYdUvZHrXx?",
"output": "NO"
},
{
"input": "uMtXK?",
"output": "NO"
},
{
"input": "U?",
"output": "YES"
},
{
"input": "aqFDkCUKeHMyvZFcAyWlMUSQTFomtaWjoKLVyxLCw vcufPBFbaljOuHWiDCROYTcmbgzbaqHXKPOYEbuEtRqqoxBbOETCsQzhw?",
"output": "NO"
},
{
"input": "dJcNqQiFXzcbsj fItCpBLyXOnrSBPebwyFHlxUJHqCUzzCmcAvMiKL NunwOXnKeIxUZmBVwiCUfPkjRAkTPbkYCmwRRnDSLaz?",
"output": "NO"
},
{
"input": "gxzXbdcAQMuFKuuiPohtMgeypr wpDIoDSyOYTdvylcg SoEBZjnMHHYZGEqKgCgBeTbyTwyGuPZxkxsnSczotBdYyfcQsOVDVC?",
"output": "NO"
},
{
"input": "FQXBisXaJFMiHFQlXjixBDMaQuIbyqSBKGsBfTmBKCjszlGVZxEOqYYqRTUkGpSDDAoOXyXcQbHcPaegeOUBNeSD JiKOdECPOF?",
"output": "NO"
},
{
"input": "YhCuZnrWUBEed?",
"output": "NO"
},
{
"input": "hh?",
"output": "NO"
},
{
"input": "whU?",
"output": "YES"
},
{
"input": "fgwg?",
"output": "NO"
},
{
"input": "GlEmEPKrYcOnBNJUIFjszWUyVdvWw DGDjoCMtRJUburkPToCyDrOtMr?",
"output": "NO"
},
{
"input": "n?",
"output": "NO"
},
{
"input": "BueDOlxgzeNlxrzRrMbKiQdmGujEKmGxclvaPpTuHmTqBp?",
"output": "NO"
},
{
"input": "iehvZNQXDGCuVmJPOEysLyUryTdfaIxIuTzTadDbqRQGoCLXkxnyfWSGoLXebNnQQNTqAQJebbyYvHOfpUnXeWdjx?",
"output": "NO"
},
{
"input": " J ?",
"output": "NO"
},
{
"input": " j ?",
"output": "NO"
},
{
"input": " o ?",
"output": "YES"
},
{
"input": " T ?",
"output": "NO"
},
{
"input": " q ?",
"output": "NO"
},
{
"input": " j ?",
"output": "NO"
},
{
"input": " c ?",
"output": "NO"
},
{
"input": " B ?",
"output": "NO"
},
{
"input": "LuhxDHVwMPTtUIUMIQTuQETgXCOQPsfdFlyHvpfOVedjUTpGLAZGOHloIjJJtOLAlHPivzA?",
"output": "YES"
},
{
"input": "wmztmzFfwbGyOmNHENUFMTsFEMWYA?",
"output": "YES"
},
{
"input": "wGsfZCSwN PEUhNUrLfABrxA?",
"output": "YES"
},
{
"input": "mCDHENXjYbgMdBimAdPnewaHfpGWowjWrVAdvWczjw iDcUbyzMsmsnwbviiKiAyGVA?",
"output": "YES"
},
{
"input": "ARIWnwqFqxsQXsXXzHqvFjxOCttAGPUzDtWzsenPYdNXuFOIUGYZsLLK IaoxiyjBBRThoelwdPTkuCQfcBLUEJpCPIrVZlvUWA?",
"output": "YES"
},
{
"input": " PslvVpgpN BXkMFBEVXsyZFIQbBEFxGkYTeXKrOdcmhbiTUatYRUoYAayrchqbksswIlfIjerZPqptvCGnMUhyrQSvwltRhFzA?",
"output": "YES"
},
{
"input": "HpBkttwSjBXDmyleGiRWNUMPaAIE uzTrp KJDzaUiCdsMYOoWKHoUhWUoecCPmACymMUUbGav UMRpCytPETwNFAObZJA?",
"output": "YES"
}
] | 1,621,240,865
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 92
| 0
|
n = input()
#print(n)
a = ['A', 'E', 'I', 'O', 'U', 'Y', 'a', 'e', 'i', 'o', 'u', 'y']
if n[-2] in a:
print("YES")
else:
print('NO')
|
Title: Sleuth
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z
Input Specification:
The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter.
Output Specification:
Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters.
Demo Input:
['Is it a melon?\n', 'Is it an apple?\n', 'Is it a banana ?\n', 'Is it an apple and a banana simultaneouSLY?\n']
Demo Output:
['NO\n', 'YES\n', 'YES\n', 'YES\n']
Note:
none
|
```python
n = input()
#print(n)
a = ['A', 'E', 'I', 'O', 'U', 'Y', 'a', 'e', 'i', 'o', 'u', 'y']
if n[-2] in a:
print("YES")
else:
print('NO')
```
| 0
|
371
|
D
|
Vessels
|
PROGRAMMING
| 1,800
|
[
"data structures",
"dsu",
"implementation",
"trees"
] | null | null |
There is a system of *n* vessels arranged one above the other as shown in the figure below. Assume that the vessels are numbered from 1 to *n*, in the order from the highest to the lowest, the volume of the *i*-th vessel is *a**i* liters.
Initially, all the vessels are empty. In some vessels water is poured. All the water that overflows from the *i*-th vessel goes to the (*i*<=+<=1)-th one. The liquid that overflows from the *n*-th vessel spills on the floor.
Your task is to simulate pouring water into the vessels. To do this, you will need to handle two types of queries:
1. Add *x**i* liters of water to the *p**i*-th vessel; 1. Print the number of liters of water in the *k**i*-th vessel.
When you reply to the second request you can assume that all the water poured up to this point, has already overflown between the vessels.
|
The first line contains integer *n* — the number of vessels (1<=≤<=*n*<=≤<=2·105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* — the vessels' capacities (1<=≤<=*a**i*<=≤<=109). The vessels' capacities do not necessarily increase from the top vessels to the bottom ones (see the second sample). The third line contains integer *m* — the number of queries (1<=≤<=*m*<=≤<=2·105). Each of the next *m* lines contains the description of one query. The query of the first type is represented as "1 *p**i* *x**i*", the query of the second type is represented as "2 *k**i*" (1<=≤<=*p**i*<=≤<=*n*, 1<=≤<=*x**i*<=≤<=109, 1<=≤<=*k**i*<=≤<=*n*).
|
For each query, print on a single line the number of liters of water in the corresponding vessel.
|
[
"2\n5 10\n6\n1 1 4\n2 1\n1 2 5\n1 1 4\n2 1\n2 2\n",
"3\n5 10 8\n6\n1 1 12\n2 2\n1 1 6\n1 3 2\n2 2\n2 3\n"
] |
[
"4\n5\n8\n",
"7\n10\n5\n"
] |
none
| 2,000
|
[
{
"input": "2\n5 10\n6\n1 1 4\n2 1\n1 2 5\n1 1 4\n2 1\n2 2",
"output": "4\n5\n8"
},
{
"input": "3\n5 10 8\n6\n1 1 12\n2 2\n1 1 6\n1 3 2\n2 2\n2 3",
"output": "7\n10\n5"
},
{
"input": "10\n71 59 88 55 18 98 38 73 53 58\n20\n1 5 93\n1 7 69\n2 3\n1 1 20\n2 10\n1 6 74\n1 7 100\n1 9 14\n2 3\n2 4\n2 7\n1 3 31\n2 4\n1 6 64\n2 2\n2 2\n1 3 54\n2 9\n2 1\n1 6 86",
"output": "0\n0\n0\n0\n38\n0\n0\n0\n53\n20"
},
{
"input": "10\n3 7 10 1 5 4 4 3 3 1\n20\n2 4\n2 4\n1 1 10\n1 1 10\n2 4\n2 3\n1 4 2\n1 4 6\n2 2\n1 8 9\n2 2\n2 5\n2 9\n1 2 1\n1 6 9\n1 1 6\n2 5\n2 2\n2 3\n1 4 10",
"output": "0\n0\n0\n10\n7\n7\n5\n3\n5\n7\n10"
},
{
"input": "50\n57 63 98 44 22 63 5 65 36 69 49 54 61 15 29 79 50 30 43 93 18 94 46 92 72 67 67 51 34 40 50 77 58 53 79 72 72 34 91 75 83 67 71 80 11 51 85 20 6 3\n20\n2 40\n1 14 102\n2 22\n2 15\n2 43\n1 29 532\n2 27\n2 47\n1 24 107\n1 20 720\n1 21 315\n2 20\n1 2 787\n1 27 532\n2 38\n1 32 445\n1 38 17\n1 26 450\n2 40\n1 45 192",
"output": "0\n0\n29\n0\n0\n0\n93\n34\n75"
},
{
"input": "1\n1\n1\n2 1",
"output": "0"
}
] | 1,593,377,412
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 44
| 1,357
| 24,166,400
|
#kylin1993
n=int(input());
water=(n+2)*[0];
vol=[int(i) for i in input().split()];
trace=(n+2)*[0];
next = [i+1 for i in range(n+2)];
m=int(input());
out=[]
for i in range(m):
c=[int(i) for i in input().split()];
if c[0]==1 :
w=c[2];
k=c[1]-1;
r=0;
while((w>0)and(k<n)):
if(w<=vol[k]-water[k]):
water[k]=water[k]+w;
break;
else:
w=w-(vol[k]-water[k]);
water[k]=vol[k];
trace[r]=k;
r=r+1;
k=next[k];
for j in range(r):
next[trace[j]]=k;
if c[0]==2:
out.append(water[c[1]-1]);
for i in out:
print(i)
|
Title: Vessels
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a system of *n* vessels arranged one above the other as shown in the figure below. Assume that the vessels are numbered from 1 to *n*, in the order from the highest to the lowest, the volume of the *i*-th vessel is *a**i* liters.
Initially, all the vessels are empty. In some vessels water is poured. All the water that overflows from the *i*-th vessel goes to the (*i*<=+<=1)-th one. The liquid that overflows from the *n*-th vessel spills on the floor.
Your task is to simulate pouring water into the vessels. To do this, you will need to handle two types of queries:
1. Add *x**i* liters of water to the *p**i*-th vessel; 1. Print the number of liters of water in the *k**i*-th vessel.
When you reply to the second request you can assume that all the water poured up to this point, has already overflown between the vessels.
Input Specification:
The first line contains integer *n* — the number of vessels (1<=≤<=*n*<=≤<=2·105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* — the vessels' capacities (1<=≤<=*a**i*<=≤<=109). The vessels' capacities do not necessarily increase from the top vessels to the bottom ones (see the second sample). The third line contains integer *m* — the number of queries (1<=≤<=*m*<=≤<=2·105). Each of the next *m* lines contains the description of one query. The query of the first type is represented as "1 *p**i* *x**i*", the query of the second type is represented as "2 *k**i*" (1<=≤<=*p**i*<=≤<=*n*, 1<=≤<=*x**i*<=≤<=109, 1<=≤<=*k**i*<=≤<=*n*).
Output Specification:
For each query, print on a single line the number of liters of water in the corresponding vessel.
Demo Input:
['2\n5 10\n6\n1 1 4\n2 1\n1 2 5\n1 1 4\n2 1\n2 2\n', '3\n5 10 8\n6\n1 1 12\n2 2\n1 1 6\n1 3 2\n2 2\n2 3\n']
Demo Output:
['4\n5\n8\n', '7\n10\n5\n']
Note:
none
|
```python
#kylin1993
n=int(input());
water=(n+2)*[0];
vol=[int(i) for i in input().split()];
trace=(n+2)*[0];
next = [i+1 for i in range(n+2)];
m=int(input());
out=[]
for i in range(m):
c=[int(i) for i in input().split()];
if c[0]==1 :
w=c[2];
k=c[1]-1;
r=0;
while((w>0)and(k<n)):
if(w<=vol[k]-water[k]):
water[k]=water[k]+w;
break;
else:
w=w-(vol[k]-water[k]);
water[k]=vol[k];
trace[r]=k;
r=r+1;
k=next[k];
for j in range(r):
next[trace[j]]=k;
if c[0]==2:
out.append(water[c[1]-1]);
for i in out:
print(i)
```
| 3
|
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
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{
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{
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},
{
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"output": "100011101001001000011011011001111000100000010100100100"
},
{
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"output": "1100110010000101101010111111101001001001110101110010110"
},
{
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},
{
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"output": "110001010001000011000101110101000100001011111001011001001"
},
{
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"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
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"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
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"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
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},
{
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{
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{
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"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
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"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
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"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
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"output": "00011111011111001000011100010011100011010100101011011000001001111110"
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{
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"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
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"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
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"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
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"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
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"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
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"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
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"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
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"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
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"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
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"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
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"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
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"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
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},
{
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},
{
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},
{
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},
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},
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},
{
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},
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{
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},
{
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"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
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"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
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"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,602,748,385
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 93
| 0
|
n = str(input())
m = str(input())
length = len(n)
r = ""
for i in range(length - 1):
if n[i] == m[i]:
r = r.replace("", "0")
else:
r = r.replace("", "1")
print(r)
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
n = str(input())
m = str(input())
length = len(n)
r = ""
for i in range(length - 1):
if n[i] == m[i]:
r = r.replace("", "0")
else:
r = r.replace("", "1")
print(r)
```
| 0
|
686
|
A
|
Free Ice Cream
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
|
The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
|
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
|
[
"5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n",
"5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n"
] |
[
"22 1\n",
"3 2\n"
] |
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.
| 500
|
[
{
"input": "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20",
"output": "22 1"
},
{
"input": "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98",
"output": "3 2"
},
{
"input": "6 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000",
"output": "7000000000 0"
},
{
"input": "5 12\n- 12\n+ 7\n- 6\n- 1\n+ 46",
"output": "46 0"
},
{
"input": "11 1000\n- 100\n+ 100\n+ 100\n+ 100\n+ 100\n- 100\n- 100\n- 100\n- 100\n- 100\n- 100",
"output": "700 0"
},
{
"input": "1 0\n- 526403222",
"output": "0 1"
},
{
"input": "1 897986543\n- 371188251",
"output": "526798292 0"
},
{
"input": "1 0\n+ 1",
"output": "1 0"
},
{
"input": "1 0\n- 1",
"output": "0 1"
},
{
"input": "1 10\n+ 10",
"output": "20 0"
},
{
"input": "1 3\n- 5",
"output": "3 1"
},
{
"input": "1 0\n- 5",
"output": "0 1"
},
{
"input": "1 0\n+ 5",
"output": "5 0"
}
] | 1,593,181,493
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 109
| 6,656,000
|
n,x = [int(j) for j in input().split()]
ans = 0
for i in range(n):
c,d = input().split()
if (c == '+'):
k = int(d)
x += int(d)
elif (c == '-'):
k = int(d)
if (x - k < 0):
ans += 1
else:
x -= k
print(x,ans)
|
Title: Free Ice Cream
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input Specification:
The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
Output Specification:
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Demo Input:
['5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n', '5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n']
Demo Output:
['22 1\n', '3 2\n']
Note:
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.
|
```python
n,x = [int(j) for j in input().split()]
ans = 0
for i in range(n):
c,d = input().split()
if (c == '+'):
k = int(d)
x += int(d)
elif (c == '-'):
k = int(d)
if (x - k < 0):
ans += 1
else:
x -= k
print(x,ans)
```
| 3
|
|
540
|
A
|
Combination Lock
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock.
The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that?
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of disks on the combination lock.
The second line contains a string of *n* digits — the original state of the disks.
The third line contains a string of *n* digits — Scrooge McDuck's combination that opens the lock.
|
Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock.
|
[
"5\n82195\n64723\n"
] |
[
"13\n"
] |
In the sample he needs 13 moves:
- 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 500
|
[
{
"input": "5\n82195\n64723",
"output": "13"
},
{
"input": "12\n102021090898\n010212908089",
"output": "16"
},
{
"input": "1\n8\n1",
"output": "3"
},
{
"input": "2\n83\n57",
"output": "7"
},
{
"input": "10\n0728592530\n1362615763",
"output": "27"
},
{
"input": "100\n4176196363694273682807653052945037727131821799902563705176501742060696655282954944720643131654235909\n3459912084922154505910287499879975659298239371519889866585472674423008837878123067103005344986554746",
"output": "245"
},
{
"input": "1\n8\n1",
"output": "3"
},
{
"input": "2\n83\n57",
"output": "7"
},
{
"input": "3\n607\n684",
"output": "5"
},
{
"input": "4\n0809\n0636",
"output": "8"
},
{
"input": "5\n84284\n08941",
"output": "16"
},
{
"input": "25\n8037856825987124762280548\n9519431339078678836940020",
"output": "72"
},
{
"input": "125\n23269567683904664184142384849516523616863461607751021071772615078579713054027902974007001544768640273491193035874486891541257\n47635110303703399505805044019026243695451609639556649012447370081552870340011971572363458960190590266459684717415349529509024",
"output": "305"
},
{
"input": "5\n84284\n08941",
"output": "16"
},
{
"input": "25\n8037856825987124762285484\n9519431339078678836940202",
"output": "74"
},
{
"input": "125\n23269567689466418414238845152168634610771021717726157879713054270294007001544768647391193035874486891412573389247025830678706\n47635110307339950580504010224954516093956649124473708152870340117152363458960190596659684717415349529090241694059599629136831",
"output": "357"
},
{
"input": "5\n84284\n08941",
"output": "16"
},
{
"input": "25\n8378525987476228048406972\n9194339078883694020217816",
"output": "55"
},
{
"input": "125\n23269576839046618414238484916523616863461607750210717761078579713054027902974007015447686027349193035874486891541257338624472\n47635103037033950580504401926243695451609639556490124437081552870340011971572363489601905026645984717415349529509024169604599",
"output": "274"
},
{
"input": "1\n0\n0",
"output": "0"
},
{
"input": "1\n7\n7",
"output": "0"
},
{
"input": "1\n0\n5",
"output": "5"
},
{
"input": "1\n2\n7",
"output": "5"
},
{
"input": "1\n7\n9",
"output": "2"
},
{
"input": "1\n9\n7",
"output": "2"
},
{
"input": "1\n2\n9",
"output": "3"
},
{
"input": "1\n9\n2",
"output": "3"
},
{
"input": "25\n3164978461316464614169874\n9413979197249127496597357",
"output": "66"
},
{
"input": "4\n9999\n9999",
"output": "0"
},
{
"input": "2\n11\n11",
"output": "0"
}
] | 1,655,625,836
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 46
| 0
|
n = int(input())
a = input()
s = input()
count = 0
for i in range(n):
if a[i] != s[i]:
if abs(int(a[i])-int(s[i])) < 5:
count += abs(int(a[i])-int(s[i]))
else:
count +=10 - abs(int(a[i])-int(s[i]))
print(count)
|
Title: Combination Lock
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock.
The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of disks on the combination lock.
The second line contains a string of *n* digits — the original state of the disks.
The third line contains a string of *n* digits — Scrooge McDuck's combination that opens the lock.
Output Specification:
Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock.
Demo Input:
['5\n82195\n64723\n']
Demo Output:
['13\n']
Note:
In the sample he needs 13 moves:
- 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
n = int(input())
a = input()
s = input()
count = 0
for i in range(n):
if a[i] != s[i]:
if abs(int(a[i])-int(s[i])) < 5:
count += abs(int(a[i])-int(s[i]))
else:
count +=10 - abs(int(a[i])-int(s[i]))
print(count)
```
| 3
|
|
923
|
A
|
Primal Sport
|
PROGRAMMING
| 1,700
|
[
"math",
"number theory"
] | null | null |
Alice and Bob begin their day with a quick game. They first choose a starting number *X*0<=≥<=3 and try to reach one million by the process described below.
Alice goes first and then they take alternating turns. In the *i*-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.
Formally, he or she selects a prime *p*<=<<=*X**i*<=-<=1 and then finds the minimum *X**i*<=≥<=*X**i*<=-<=1 such that *p* divides *X**i*. Note that if the selected prime *p* already divides *X**i*<=-<=1, then the number does not change.
Eve has witnessed the state of the game after two turns. Given *X*2, help her determine what is the smallest possible starting number *X*0. Note that the players don't necessarily play optimally. You should consider all possible game evolutions.
|
The input contains a single integer *X*2 (4<=≤<=*X*2<=≤<=106). It is guaranteed that the integer *X*2 is composite, that is, is not prime.
|
Output a single integer — the minimum possible *X*0.
|
[
"14\n",
"20\n",
"8192\n"
] |
[
"6\n",
"15\n",
"8191\n"
] |
In the first test, the smallest possible starting number is *X*<sub class="lower-index">0</sub> = 6. One possible course of the game is as follows:
- Alice picks prime 5 and announces *X*<sub class="lower-index">1</sub> = 10 - Bob picks prime 7 and announces *X*<sub class="lower-index">2</sub> = 14.
In the second case, let *X*<sub class="lower-index">0</sub> = 15.
- Alice picks prime 2 and announces *X*<sub class="lower-index">1</sub> = 16 - Bob picks prime 5 and announces *X*<sub class="lower-index">2</sub> = 20.
| 500
|
[
{
"input": "14",
"output": "6"
},
{
"input": "20",
"output": "15"
},
{
"input": "8192",
"output": "8191"
},
{
"input": "1000000",
"output": "998677"
},
{
"input": "959806",
"output": "239958"
},
{
"input": "1452",
"output": "1206"
},
{
"input": "4",
"output": "3"
},
{
"input": "6",
"output": "3"
},
{
"input": "8",
"output": "7"
},
{
"input": "9",
"output": "7"
},
{
"input": "10",
"output": "4"
},
{
"input": "12",
"output": "6"
},
{
"input": "15",
"output": "8"
},
{
"input": "16",
"output": "11"
},
{
"input": "110880",
"output": "55440"
},
{
"input": "166320",
"output": "110879"
},
{
"input": "221760",
"output": "110880"
},
{
"input": "277200",
"output": "138600"
},
{
"input": "332640",
"output": "166320"
},
{
"input": "498960",
"output": "332639"
},
{
"input": "554400",
"output": "415798"
},
{
"input": "665280",
"output": "498958"
},
{
"input": "720720",
"output": "540538"
},
{
"input": "510510",
"output": "255248"
},
{
"input": "570570",
"output": "285282"
},
{
"input": "690690",
"output": "460455"
},
{
"input": "959818",
"output": "239958"
},
{
"input": "959878",
"output": "239978"
},
{
"input": "959902",
"output": "239978"
},
{
"input": "974847",
"output": "324954"
},
{
"input": "974859",
"output": "324978"
},
{
"input": "974931",
"output": "324980"
},
{
"input": "885481",
"output": "442272"
},
{
"input": "896809",
"output": "447944"
},
{
"input": "908209",
"output": "453632"
},
{
"input": "935089",
"output": "467064"
},
{
"input": "720721",
"output": "355298"
},
{
"input": "690691",
"output": "342864"
},
{
"input": "959903",
"output": "479702"
},
{
"input": "974932",
"output": "470060"
},
{
"input": "935090",
"output": "463950"
},
{
"input": "524288",
"output": "524287"
},
{
"input": "524289",
"output": "174768"
},
{
"input": "524286",
"output": "262110"
},
{
"input": "531441",
"output": "526737"
},
{
"input": "531442",
"output": "262490"
},
{
"input": "531440",
"output": "265704"
},
{
"input": "81",
"output": "76"
},
{
"input": "999958",
"output": "250008"
},
{
"input": "2048",
"output": "1959"
}
] | 1,520,702,104
| 6,004
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
PRETESTS
| 2
| 1,500
| 5,632,000
|
from math import sqrt
def is_prime(x):
prime = False if x != 2 else True
if x > 2 and x % 2 != 0:
prime = True
k = 3
n = sqrt(x)
while k <= n and prime == True:
if x % k == 0:
prime = False
k += 2
return prime
def play_game(x):
max_possible = 0
for q in range(3,x):
num = q
for k in range(2,q):
if not is_prime(k):
continue
num_p = q
while num ==q:
num_p += 1
if (num_p%k)==0:
break
for second_k in range(2,x):
if k==second_k:
continue
if not is_prime(second_k):
continue
if (q%k==0) and (num_p%k==0):
continue
if (num_p%second_k==0) and (x%second_k==0):
continue
if (x%second_k==0) and (num_p%second_k!=0) and (find_next_num(num_p,second_k)==x):
print(q)
return
def find_next_num(x,y):
cont = True
num = x
while cont:
if (num%y==0):
cont = False
continue
num += 1
return num
def find_lower_prime(x):
highest = 1
for q in range(1,x):
if is_prime(q):
highest = q
return highest
a = int(input())
play_game(a)
|
Title: Primal Sport
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Alice and Bob begin their day with a quick game. They first choose a starting number *X*0<=≥<=3 and try to reach one million by the process described below.
Alice goes first and then they take alternating turns. In the *i*-th turn, the player whose turn it is selects a prime number smaller than the current number, and announces the smallest multiple of this prime number that is not smaller than the current number.
Formally, he or she selects a prime *p*<=<<=*X**i*<=-<=1 and then finds the minimum *X**i*<=≥<=*X**i*<=-<=1 such that *p* divides *X**i*. Note that if the selected prime *p* already divides *X**i*<=-<=1, then the number does not change.
Eve has witnessed the state of the game after two turns. Given *X*2, help her determine what is the smallest possible starting number *X*0. Note that the players don't necessarily play optimally. You should consider all possible game evolutions.
Input Specification:
The input contains a single integer *X*2 (4<=≤<=*X*2<=≤<=106). It is guaranteed that the integer *X*2 is composite, that is, is not prime.
Output Specification:
Output a single integer — the minimum possible *X*0.
Demo Input:
['14\n', '20\n', '8192\n']
Demo Output:
['6\n', '15\n', '8191\n']
Note:
In the first test, the smallest possible starting number is *X*<sub class="lower-index">0</sub> = 6. One possible course of the game is as follows:
- Alice picks prime 5 and announces *X*<sub class="lower-index">1</sub> = 10 - Bob picks prime 7 and announces *X*<sub class="lower-index">2</sub> = 14.
In the second case, let *X*<sub class="lower-index">0</sub> = 15.
- Alice picks prime 2 and announces *X*<sub class="lower-index">1</sub> = 16 - Bob picks prime 5 and announces *X*<sub class="lower-index">2</sub> = 20.
|
```python
from math import sqrt
def is_prime(x):
prime = False if x != 2 else True
if x > 2 and x % 2 != 0:
prime = True
k = 3
n = sqrt(x)
while k <= n and prime == True:
if x % k == 0:
prime = False
k += 2
return prime
def play_game(x):
max_possible = 0
for q in range(3,x):
num = q
for k in range(2,q):
if not is_prime(k):
continue
num_p = q
while num ==q:
num_p += 1
if (num_p%k)==0:
break
for second_k in range(2,x):
if k==second_k:
continue
if not is_prime(second_k):
continue
if (q%k==0) and (num_p%k==0):
continue
if (num_p%second_k==0) and (x%second_k==0):
continue
if (x%second_k==0) and (num_p%second_k!=0) and (find_next_num(num_p,second_k)==x):
print(q)
return
def find_next_num(x,y):
cont = True
num = x
while cont:
if (num%y==0):
cont = False
continue
num += 1
return num
def find_lower_prime(x):
highest = 1
for q in range(1,x):
if is_prime(q):
highest = q
return highest
a = int(input())
play_game(a)
```
| 0
|
|
862
|
B
|
Mahmoud and Ehab and the bipartiteness
|
PROGRAMMING
| 1,300
|
[
"dfs and similar",
"graphs",
"trees"
] | null | null |
Mahmoud and Ehab continue their adventures! As everybody in the evil land knows, Dr. Evil likes bipartite graphs, especially trees.
A tree is a connected acyclic graph. A bipartite graph is a graph, whose vertices can be partitioned into 2 sets in such a way, that for each edge (*u*,<=*v*) that belongs to the graph, *u* and *v* belong to different sets. You can find more formal definitions of a tree and a bipartite graph in the notes section below.
Dr. Evil gave Mahmoud and Ehab a tree consisting of *n* nodes and asked them to add edges to it in such a way, that the graph is still bipartite. Besides, after adding these edges the graph should be simple (doesn't contain loops or multiple edges). What is the maximum number of edges they can add?
A loop is an edge, which connects a node with itself. Graph doesn't contain multiple edges when for each pair of nodes there is no more than one edge between them. A cycle and a loop aren't the same .
|
The first line of input contains an integer *n* — the number of nodes in the tree (1<=≤<=*n*<=≤<=105).
The next *n*<=-<=1 lines contain integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*) — the description of the edges of the tree.
It's guaranteed that the given graph is a tree.
|
Output one integer — the maximum number of edges that Mahmoud and Ehab can add to the tree while fulfilling the conditions.
|
[
"3\n1 2\n1 3\n",
"5\n1 2\n2 3\n3 4\n4 5\n"
] |
[
"0\n",
"2\n"
] |
Tree definition: [https://en.wikipedia.org/wiki/Tree_(graph_theory)](https://en.wikipedia.org/wiki/Tree_(graph_theory))
Bipartite graph definition: [https://en.wikipedia.org/wiki/Bipartite_graph](https://en.wikipedia.org/wiki/Bipartite_graph)
In the first test case the only edge that can be added in such a way, that graph won't contain loops or multiple edges is (2, 3), but adding this edge will make the graph non-bipartite so the answer is 0.
In the second test case Mahmoud and Ehab can add edges (1, 4) and (2, 5).
| 1,000
|
[
{
"input": "3\n1 2\n1 3",
"output": "0"
},
{
"input": "5\n1 2\n2 3\n3 4\n4 5",
"output": "2"
},
{
"input": "10\n3 8\n6 2\n9 7\n10 1\n3 5\n1 3\n6 7\n5 4\n3 6",
"output": "16"
},
{
"input": "10\n7 6\n2 7\n4 1\n8 5\n9 4\n5 3\n8 7\n10 8\n10 4",
"output": "16"
},
{
"input": "10\n2 6\n3 7\n8 4\n4 10\n6 9\n9 7\n3 10\n1 2\n5 8",
"output": "16"
},
{
"input": "10\n6 9\n9 7\n9 4\n10 9\n9 1\n9 8\n9 2\n9 5\n3 9",
"output": "0"
},
{
"input": "2\n1 2",
"output": "0"
}
] | 1,587,170,673
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 529
| 23,756,800
|
import math
import sys
from collections import defaultdict
from collections import Counter
from collections import deque
import bisect
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
ilele = lambda: map(int,input().split())
alele = lambda: list(map(int, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
INF = 10 ** 18
MOD = 10 ** 9 + 7
first = 1 #if first node is 1
G = defaultdict(list)
def addEdge(a,b):
G[a].append(b)
G[b].append(a)
def BFSUtil(queue,vis,level):
while queue:
s = queue.popleft()
for child in G.get(s,[]):
if vis[child] == False:
queue.append(child)
level[child] = level[s] + 1
vis[child] = True
def nodesatlevel(node,N):
vis = [False for i in range(N+1)]
level ={}
queue = deque()
level[node] = 0
queue.append(node)
vis[node] = True
BFSUtil(queue,vis,level)
return level
mark ={}
N = int(input())
for i in range(N-1):
u,v = ilele()
addEdge(u,v)
root = 1
mark = nodesatlevel(root,N)
#print(mark,G)
len1 = 0;len2 =0
for j in mark.values():
if j%2==0:
len1+=1
else:
len2+=1
Ans = 0
#print(len1,len2)
for i,j in G.items():
if mark.get(i,-1) % 2== 0:
Ans += max(0,len2 -len(j))
else:
Ans += max(0,len1-len(j))
print(Ans//2)
|
Title: Mahmoud and Ehab and the bipartiteness
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mahmoud and Ehab continue their adventures! As everybody in the evil land knows, Dr. Evil likes bipartite graphs, especially trees.
A tree is a connected acyclic graph. A bipartite graph is a graph, whose vertices can be partitioned into 2 sets in such a way, that for each edge (*u*,<=*v*) that belongs to the graph, *u* and *v* belong to different sets. You can find more formal definitions of a tree and a bipartite graph in the notes section below.
Dr. Evil gave Mahmoud and Ehab a tree consisting of *n* nodes and asked them to add edges to it in such a way, that the graph is still bipartite. Besides, after adding these edges the graph should be simple (doesn't contain loops or multiple edges). What is the maximum number of edges they can add?
A loop is an edge, which connects a node with itself. Graph doesn't contain multiple edges when for each pair of nodes there is no more than one edge between them. A cycle and a loop aren't the same .
Input Specification:
The first line of input contains an integer *n* — the number of nodes in the tree (1<=≤<=*n*<=≤<=105).
The next *n*<=-<=1 lines contain integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*) — the description of the edges of the tree.
It's guaranteed that the given graph is a tree.
Output Specification:
Output one integer — the maximum number of edges that Mahmoud and Ehab can add to the tree while fulfilling the conditions.
Demo Input:
['3\n1 2\n1 3\n', '5\n1 2\n2 3\n3 4\n4 5\n']
Demo Output:
['0\n', '2\n']
Note:
Tree definition: [https://en.wikipedia.org/wiki/Tree_(graph_theory)](https://en.wikipedia.org/wiki/Tree_(graph_theory))
Bipartite graph definition: [https://en.wikipedia.org/wiki/Bipartite_graph](https://en.wikipedia.org/wiki/Bipartite_graph)
In the first test case the only edge that can be added in such a way, that graph won't contain loops or multiple edges is (2, 3), but adding this edge will make the graph non-bipartite so the answer is 0.
In the second test case Mahmoud and Ehab can add edges (1, 4) and (2, 5).
|
```python
import math
import sys
from collections import defaultdict
from collections import Counter
from collections import deque
import bisect
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
ilele = lambda: map(int,input().split())
alele = lambda: list(map(int, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
INF = 10 ** 18
MOD = 10 ** 9 + 7
first = 1 #if first node is 1
G = defaultdict(list)
def addEdge(a,b):
G[a].append(b)
G[b].append(a)
def BFSUtil(queue,vis,level):
while queue:
s = queue.popleft()
for child in G.get(s,[]):
if vis[child] == False:
queue.append(child)
level[child] = level[s] + 1
vis[child] = True
def nodesatlevel(node,N):
vis = [False for i in range(N+1)]
level ={}
queue = deque()
level[node] = 0
queue.append(node)
vis[node] = True
BFSUtil(queue,vis,level)
return level
mark ={}
N = int(input())
for i in range(N-1):
u,v = ilele()
addEdge(u,v)
root = 1
mark = nodesatlevel(root,N)
#print(mark,G)
len1 = 0;len2 =0
for j in mark.values():
if j%2==0:
len1+=1
else:
len2+=1
Ans = 0
#print(len1,len2)
for i,j in G.items():
if mark.get(i,-1) % 2== 0:
Ans += max(0,len2 -len(j))
else:
Ans += max(0,len1-len(j))
print(Ans//2)
```
| 3
|
|
263
|
A
|
Beautiful Matrix
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
|
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
|
Print a single integer — the minimum number of moves needed to make the matrix beautiful.
|
[
"0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n",
"0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n"
] |
[
"3\n",
"1\n"
] |
none
| 500
|
[
{
"input": "0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1",
"output": "4"
},
{
"input": "0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0",
"output": "3"
}
] | 1,698,700,591
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 25
| 154
| 0
|
l1=[]
for i in range(5):
n=input().split()
n1=list(map(int,n))
l1.append(n1)
for i in range(5):
for j in range(5):
if(l1[i][j]==1):
row=i
column=j
break
for k in range(5):
if(row==2 and column==2):
print(k)
break
elif(row>2):
row=row-1
elif(column>2):
column=column-1
elif(row<2):
row=row+1
elif(column<2):
column=column+1
|
Title: Beautiful Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
Input Specification:
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
Output Specification:
Print a single integer — the minimum number of moves needed to make the matrix beautiful.
Demo Input:
['0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n', '0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n']
Demo Output:
['3\n', '1\n']
Note:
none
|
```python
l1=[]
for i in range(5):
n=input().split()
n1=list(map(int,n))
l1.append(n1)
for i in range(5):
for j in range(5):
if(l1[i][j]==1):
row=i
column=j
break
for k in range(5):
if(row==2 and column==2):
print(k)
break
elif(row>2):
row=row-1
elif(column>2):
column=column-1
elif(row<2):
row=row+1
elif(column<2):
column=column+1
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question.
|
The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants
|
Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
|
[
"1 3 2 1 2 1\n",
"1 1 1 1 1 99\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater.
| 0
|
[
{
"input": "1 3 2 1 2 1",
"output": "YES"
},
{
"input": "1 1 1 1 1 99",
"output": "NO"
},
{
"input": "1000 1000 1000 1000 1000 1000",
"output": "YES"
},
{
"input": "0 0 0 0 0 0",
"output": "YES"
},
{
"input": "633 609 369 704 573 416",
"output": "NO"
},
{
"input": "353 313 327 470 597 31",
"output": "NO"
},
{
"input": "835 638 673 624 232 266",
"output": "NO"
},
{
"input": "936 342 19 398 247 874",
"output": "NO"
},
{
"input": "417 666 978 553 271 488",
"output": "NO"
},
{
"input": "71 66 124 199 67 147",
"output": "YES"
},
{
"input": "54 26 0 171 239 12",
"output": "YES"
},
{
"input": "72 8 186 92 267 69",
"output": "YES"
},
{
"input": "180 179 188 50 75 214",
"output": "YES"
},
{
"input": "16 169 110 136 404 277",
"output": "YES"
},
{
"input": "101 400 9 200 300 10",
"output": "YES"
},
{
"input": "101 400 200 9 300 10",
"output": "YES"
},
{
"input": "101 200 400 9 300 10",
"output": "YES"
},
{
"input": "101 400 200 300 9 10",
"output": "YES"
},
{
"input": "101 200 400 300 9 10",
"output": "YES"
},
{
"input": "4 4 4 4 5 4",
"output": "NO"
},
{
"input": "2 2 2 2 2 1",
"output": "NO"
},
{
"input": "1000 1000 999 1000 1000 1000",
"output": "NO"
},
{
"input": "129 1 10 29 8 111",
"output": "NO"
},
{
"input": "1000 1000 1000 999 999 1000",
"output": "YES"
},
{
"input": "101 200 300 400 9 10",
"output": "YES"
},
{
"input": "101 400 200 300 10 9",
"output": "YES"
},
{
"input": "101 200 400 300 10 9",
"output": "YES"
},
{
"input": "101 200 300 400 10 9",
"output": "YES"
},
{
"input": "101 200 300 10 400 9",
"output": "YES"
},
{
"input": "1 1 1 1 1 5",
"output": "NO"
},
{
"input": "8 1 1 3 3 0",
"output": "NO"
},
{
"input": "1 1 2 2 3 3",
"output": "YES"
},
{
"input": "1 2 2 5 2 5",
"output": "NO"
},
{
"input": "1 2 3 6 6 6",
"output": "NO"
},
{
"input": "36 91 7 86 51 89",
"output": "NO"
},
{
"input": "10 1 1 1 23 24",
"output": "NO"
},
{
"input": "1 1 1 10 23 24",
"output": "NO"
},
{
"input": "20 10 1 2 3 44",
"output": "NO"
},
{
"input": "7 0 14 11 8 6",
"output": "NO"
},
{
"input": "100 496 1 1 1 1",
"output": "NO"
},
{
"input": "5 4 2 5 11 3",
"output": "NO"
},
{
"input": "1 3 7 8 8 9",
"output": "YES"
},
{
"input": "1 3 4 5 18 19",
"output": "YES"
},
{
"input": "5 5 1 2 2 15",
"output": "NO"
},
{
"input": "2 1 0 0 0 5",
"output": "NO"
},
{
"input": "1 6 6 1 20 2",
"output": "NO"
},
{
"input": "2 10 0 0 0 0",
"output": "NO"
},
{
"input": "1 1 3 1 1 11",
"output": "NO"
},
{
"input": "10 10 1 1 1 37",
"output": "NO"
},
{
"input": "1 1 0 0 0 4",
"output": "NO"
},
{
"input": "1 1 10 1 1 28",
"output": "NO"
},
{
"input": "1 5 5 5 6 8",
"output": "YES"
},
{
"input": "0 2 3 4 4 5",
"output": "YES"
}
] | 1,510,503,023
| 323
|
Python 3
|
OK
|
TESTS
| 53
| 77
| 0
|
ai = list(map(int,input().split()))
summ = sum(ai)
ans = 0
for i in range(6):
for j in range(i+1,6):
for z in range(j+1,6):
if summ == (ai[i]+ai[j]+ai[z])*2:
ans = 1
print("YES"*ans+"NO"*(1-ans))
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question.
Input Specification:
The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants
Output Specification:
Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
Demo Input:
['1 3 2 1 2 1\n', '1 1 1 1 1 99\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater.
|
```python
ai = list(map(int,input().split()))
summ = sum(ai)
ans = 0
for i in range(6):
for j in range(i+1,6):
for z in range(j+1,6):
if summ == (ai[i]+ai[j]+ai[z])*2:
ans = 1
print("YES"*ans+"NO"*(1-ans))
```
| 3
|
|
348
|
A
|
Mafia
|
PROGRAMMING
| 1,600
|
[
"binary search",
"math",
"sortings"
] | null | null |
One day *n* friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and other *n*<=-<=1 people take part in the game. For each person we know in how many rounds he wants to be a player, not the supervisor: the *i*-th person wants to play *a**i* rounds. What is the minimum number of rounds of the "Mafia" game they need to play to let each person play at least as many rounds as they want?
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=105). The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the *i*-th number in the list is the number of rounds the *i*-th person wants to play.
|
In a single line print a single integer — the minimum number of game rounds the friends need to let the *i*-th person play at least *a**i* rounds.
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.
|
[
"3\n3 2 2\n",
"4\n2 2 2 2\n"
] |
[
"4\n",
"3\n"
] |
You don't need to know the rules of "Mafia" to solve this problem. If you're curious, it's a game Russia got from the Soviet times: http://en.wikipedia.org/wiki/Mafia_(party_game).
| 500
|
[
{
"input": "3\n3 2 2",
"output": "4"
},
{
"input": "4\n2 2 2 2",
"output": "3"
},
{
"input": "7\n9 7 7 8 8 7 8",
"output": "9"
},
{
"input": "10\n13 12 10 13 13 14 10 10 12 12",
"output": "14"
},
{
"input": "10\n94 96 91 95 99 94 96 92 95 99",
"output": "106"
},
{
"input": "100\n1 555 876 444 262 234 231 598 416 261 206 165 181 988 469 123 602 592 533 97 864 716 831 156 962 341 207 377 892 51 866 96 757 317 832 476 549 472 770 1000 887 145 956 515 992 653 972 677 973 527 984 559 280 346 580 30 372 547 209 929 492 520 446 726 47 170 699 560 814 206 688 955 308 287 26 102 77 430 262 71 415 586 532 562 419 615 732 658 108 315 268 574 86 12 23 429 640 995 342 305",
"output": "1000"
},
{
"input": "3\n1 1 1",
"output": "2"
},
{
"input": "30\n94 93 90 94 90 91 93 91 93 94 93 90 100 94 97 94 94 95 94 96 94 98 97 95 97 91 91 95 98 96",
"output": "100"
},
{
"input": "5\n1000000000 5 5 4 4",
"output": "1000000000"
},
{
"input": "3\n1 2 1",
"output": "2"
},
{
"input": "3\n2 1 1",
"output": "2"
},
{
"input": "4\n1 2 3 4",
"output": "4"
},
{
"input": "3\n1000000000 1000000000 10000000",
"output": "1005000000"
},
{
"input": "3\n677876423 834056477 553175531",
"output": "1032554216"
},
{
"input": "5\n1000000000 1 1 1 1",
"output": "1000000000"
},
{
"input": "4\n1000000000 1000000000 1000000000 1000000000",
"output": "1333333334"
},
{
"input": "3\n4 10 11",
"output": "13"
},
{
"input": "5\n1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "1250000000"
}
] | 1,683,661,350
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 4
| 92
| 0
|
import sys
input = lambda: sys.stdin.readline().rstrip()
N = int(input())
A = list(map(int, input().split()))
A.sort()
t = A[-1]//(N-1)
m = A[-1]%(N-1)
done = A[-1]-t
if m:
done-=1
ans = A[-1]+ max(0, A[0]-done)
print(ans)
|
Title: Mafia
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day *n* friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and other *n*<=-<=1 people take part in the game. For each person we know in how many rounds he wants to be a player, not the supervisor: the *i*-th person wants to play *a**i* rounds. What is the minimum number of rounds of the "Mafia" game they need to play to let each person play at least as many rounds as they want?
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=105). The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the *i*-th number in the list is the number of rounds the *i*-th person wants to play.
Output Specification:
In a single line print a single integer — the minimum number of game rounds the friends need to let the *i*-th person play at least *a**i* rounds.
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.
Demo Input:
['3\n3 2 2\n', '4\n2 2 2 2\n']
Demo Output:
['4\n', '3\n']
Note:
You don't need to know the rules of "Mafia" to solve this problem. If you're curious, it's a game Russia got from the Soviet times: http://en.wikipedia.org/wiki/Mafia_(party_game).
|
```python
import sys
input = lambda: sys.stdin.readline().rstrip()
N = int(input())
A = list(map(int, input().split()))
A.sort()
t = A[-1]//(N-1)
m = A[-1]%(N-1)
done = A[-1]-t
if m:
done-=1
ans = A[-1]+ max(0, A[0]-done)
print(ans)
```
| 0
|
|
43
|
A
|
Football
|
PROGRAMMING
| 1,000
|
[
"strings"
] |
A. Football
|
2
|
256
|
One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
|
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
|
[
"1\nABC\n",
"5\nA\nABA\nABA\nA\nA\n"
] |
[
"ABC\n",
"A\n"
] |
none
| 500
|
[
{
"input": "1\nABC",
"output": "ABC"
},
{
"input": "5\nA\nABA\nABA\nA\nA",
"output": "A"
},
{
"input": "2\nXTSJEP\nXTSJEP",
"output": "XTSJEP"
},
{
"input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ",
"output": "XZYDJAEDZ"
},
{
"input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD",
"output": "QCCYXL"
},
{
"input": "3\nAZID\nEERWBC\nEERWBC",
"output": "EERWBC"
},
{
"input": "3\nHNCGYL\nHNCGYL\nHNCGYL",
"output": "HNCGYL"
},
{
"input": "4\nZZWZTG\nZZWZTG\nZZWZTG\nZZWZTG",
"output": "ZZWZTG"
},
{
"input": "4\nA\nA\nKUDLJMXCSE\nA",
"output": "A"
},
{
"input": "5\nPHBTW\nPHBTW\nPHBTW\nPHBTW\nPHBTW",
"output": "PHBTW"
},
{
"input": "5\nPKUZYTFYWN\nPKUZYTFYWN\nSTC\nPKUZYTFYWN\nPKUZYTFYWN",
"output": "PKUZYTFYWN"
},
{
"input": "5\nHH\nHH\nNTQWPA\nNTQWPA\nHH",
"output": "HH"
},
{
"input": "10\nW\nW\nW\nW\nW\nD\nW\nD\nD\nW",
"output": "W"
},
{
"input": "19\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nTGACNIH\nTGACNIH\nXBCP",
"output": "XBCP"
},
{
"input": "33\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS",
"output": "PYPAS"
},
{
"input": "51\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC",
"output": "NC"
},
{
"input": "89\nH\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nVOCI\nVOCI\nH\nH\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nH\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI",
"output": "VOCI"
},
{
"input": "100\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nM\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA",
"output": "HA"
},
{
"input": "100\nG\nG\nS\nS\nG\nG\nS\nS\nG\nS\nS\nS\nG\nS\nG\nG\nS\nG\nS\nS\nG\nS\nS\nS\nS\nS\nG\nS\nG\nS\nS\nG\nG\nG\nS\nS\nS\nS\nG\nS\nS\nG\nG\nG\nG\nG\nS\nG\nG\nS\nS\nS\nS\nS\nG\nG\nS\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nS\nS\nG\nS\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nG\nG\nS\nG\nG\nG\nS",
"output": "G"
},
{
"input": "100\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nOBH\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL",
"output": "WL"
}
] | 1,656,583,425
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 372
| 3,788,800
|
import collections,math
x=int(input())
arr=[]
for i in range(x):
z=input()
arr.append(z)
elem=collections.Counter(arr)
elem=list(elem.items())
i=0
if elem[1][1]>elem[0][1]:
i=1
print(elem[i][0])
|
Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
Output Specification:
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
Demo Input:
['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n']
Demo Output:
['ABC\n', 'A\n']
Note:
none
|
```python
import collections,math
x=int(input())
arr=[]
for i in range(x):
z=input()
arr.append(z)
elem=collections.Counter(arr)
elem=list(elem.items())
i=0
if elem[1][1]>elem[0][1]:
i=1
print(elem[i][0])
```
| -1
|
59
|
A
|
Word
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Word
|
2
|
256
|
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
|
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
|
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
|
[
"HoUse\n",
"ViP\n",
"maTRIx\n"
] |
[
"house\n",
"VIP\n",
"matrix\n"
] |
none
| 500
|
[
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,675,285,307
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 92
| 0
|
M = 0
m = 0
s = input()
l = list(s)
for i in l:
if i.isupper():
M = M + 1
else:
m = m + 1
if m < M:
print(s.upper())
else:
print(s.lower())
|
Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none
|
```python
M = 0
m = 0
s = input()
l = list(s)
for i in l:
if i.isupper():
M = M + 1
else:
m = m + 1
if m < M:
print(s.upper())
else:
print(s.lower())
```
| 3.977
|
723
|
A
|
The New Year: Meeting Friends
|
PROGRAMMING
| 800
|
[
"implementation",
"math",
"sortings"
] | null | null |
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
|
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
|
Print one integer — the minimum total distance the friends need to travel in order to meet together.
|
[
"7 1 4\n",
"30 20 10\n"
] |
[
"6\n",
"20\n"
] |
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
| 500
|
[
{
"input": "7 1 4",
"output": "6"
},
{
"input": "30 20 10",
"output": "20"
},
{
"input": "1 4 100",
"output": "99"
},
{
"input": "100 1 91",
"output": "99"
},
{
"input": "1 45 100",
"output": "99"
},
{
"input": "1 2 3",
"output": "2"
},
{
"input": "71 85 88",
"output": "17"
},
{
"input": "30 38 99",
"output": "69"
},
{
"input": "23 82 95",
"output": "72"
},
{
"input": "22 41 47",
"output": "25"
},
{
"input": "9 94 77",
"output": "85"
},
{
"input": "1 53 51",
"output": "52"
},
{
"input": "25 97 93",
"output": "72"
},
{
"input": "42 53 51",
"output": "11"
},
{
"input": "81 96 94",
"output": "15"
},
{
"input": "21 5 93",
"output": "88"
},
{
"input": "50 13 75",
"output": "62"
},
{
"input": "41 28 98",
"output": "70"
},
{
"input": "69 46 82",
"output": "36"
},
{
"input": "87 28 89",
"output": "61"
},
{
"input": "44 45 40",
"output": "5"
},
{
"input": "86 97 68",
"output": "29"
},
{
"input": "43 92 30",
"output": "62"
},
{
"input": "16 70 1",
"output": "69"
},
{
"input": "40 46 19",
"output": "27"
},
{
"input": "71 38 56",
"output": "33"
},
{
"input": "82 21 80",
"output": "61"
},
{
"input": "75 8 35",
"output": "67"
},
{
"input": "75 24 28",
"output": "51"
},
{
"input": "78 23 56",
"output": "55"
},
{
"input": "85 31 10",
"output": "75"
},
{
"input": "76 50 9",
"output": "67"
},
{
"input": "95 37 34",
"output": "61"
},
{
"input": "84 61 35",
"output": "49"
},
{
"input": "87 85 37",
"output": "50"
},
{
"input": "1 3 2",
"output": "2"
},
{
"input": "4 2 6",
"output": "4"
},
{
"input": "6 9 3",
"output": "6"
},
{
"input": "12 4 8",
"output": "8"
},
{
"input": "15 10 5",
"output": "10"
},
{
"input": "1 50 17",
"output": "49"
},
{
"input": "10 5 15",
"output": "10"
},
{
"input": "8 1 9",
"output": "8"
},
{
"input": "3 5 4",
"output": "2"
},
{
"input": "2 1 3",
"output": "2"
},
{
"input": "1 8 2",
"output": "7"
},
{
"input": "1 100 2",
"output": "99"
},
{
"input": "1 4 6",
"output": "5"
}
] | 1,688,381,340
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 30
| 0
|
a,b,c=map(int,input().split())
min(a,min(b,c))
max(a,max(b,c))
print(max-min)
|
Title: The New Year: Meeting Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
Input Specification:
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
Output Specification:
Print one integer — the minimum total distance the friends need to travel in order to meet together.
Demo Input:
['7 1 4\n', '30 20 10\n']
Demo Output:
['6\n', '20\n']
Note:
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
|
```python
a,b,c=map(int,input().split())
min(a,min(b,c))
max(a,max(b,c))
print(max-min)
```
| -1
|
|
763
|
A
|
Timofey and a tree
|
PROGRAMMING
| 1,600
|
[
"dfs and similar",
"dp",
"dsu",
"graphs",
"implementation",
"trees"
] | null | null |
Each New Year Timofey and his friends cut down a tree of *n* vertices and bring it home. After that they paint all the *n* its vertices, so that the *i*-th vertex gets color *c**i*.
Now it's time for Timofey birthday, and his mother asked him to remove the tree. Timofey removes the tree in the following way: he takes some vertex in hands, while all the other vertices move down so that the tree becomes rooted at the chosen vertex. After that Timofey brings the tree to a trash can.
Timofey doesn't like it when many colors are mixing together. A subtree annoys him if there are vertices of different color in it. Timofey wants to find a vertex which he should take in hands so that there are no subtrees that annoy him. He doesn't consider the whole tree as a subtree since he can't see the color of the root vertex.
A subtree of some vertex is a subgraph containing that vertex and all its descendants.
Your task is to determine if there is a vertex, taking which in hands Timofey wouldn't be annoyed.
|
The first line contains single integer *n* (2<=≤<=*n*<=≤<=105) — the number of vertices in the tree.
Each of the next *n*<=-<=1 lines contains two integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*), denoting there is an edge between vertices *u* and *v*. It is guaranteed that the given graph is a tree.
The next line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=105), denoting the colors of the vertices.
|
Print "NO" in a single line, if Timofey can't take the tree in such a way that it doesn't annoy him.
Otherwise print "YES" in the first line. In the second line print the index of the vertex which Timofey should take in hands. If there are multiple answers, print any of them.
|
[
"4\n1 2\n2 3\n3 4\n1 2 1 1\n",
"3\n1 2\n2 3\n1 2 3\n",
"4\n1 2\n2 3\n3 4\n1 2 1 2\n"
] |
[
"YES\n2",
"YES\n2",
"NO"
] |
none
| 500
|
[
{
"input": "4\n1 2\n2 3\n3 4\n1 2 1 1",
"output": "YES\n2"
},
{
"input": "3\n1 2\n2 3\n1 2 3",
"output": "YES\n2"
},
{
"input": "4\n1 2\n2 3\n3 4\n1 2 1 2",
"output": "NO"
},
{
"input": "3\n2 1\n2 3\n1 2 3",
"output": "YES\n2"
},
{
"input": "4\n1 2\n2 4\n4 3\n1 1 3 2",
"output": "YES\n4"
},
{
"input": "2\n1 2\n1 1",
"output": "YES\n1"
},
{
"input": "10\n5 7\n4 5\n10 2\n3 6\n1 2\n3 4\n8 5\n4 9\n2 3\n15 15 15 15 5 15 26 18 15 15",
"output": "YES\n5"
},
{
"input": "8\n1 2\n1 3\n3 5\n3 6\n1 4\n4 7\n4 8\n1 3 1 1 1 1 1 2",
"output": "NO"
},
{
"input": "3\n2 1\n2 3\n4 4 4",
"output": "YES\n1"
},
{
"input": "3\n1 2\n1 3\n1 2 2",
"output": "YES\n1"
},
{
"input": "4\n1 4\n2 4\n3 4\n1 2 3 1",
"output": "YES\n4"
},
{
"input": "4\n1 2\n1 3\n1 4\n1 2 3 4",
"output": "YES\n1"
},
{
"input": "9\n1 2\n2 3\n3 4\n4 5\n2 7\n7 6\n2 8\n8 9\n1 1 2 2 2 3 3 4 4",
"output": "YES\n2"
},
{
"input": "3\n2 1\n2 3\n4 4 5",
"output": "YES\n2"
},
{
"input": "4\n1 2\n2 3\n3 4\n1 2 2 1",
"output": "NO"
}
] | 1,649,960,763
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 451
| 16,384,000
|
n=int(input())
graph=[[] for _ in range(n)]
for _ in range(n-1):
a,b=map(int,input().split())
a-=1
b-=1
graph[a].append(b)
graph[b].append(a)
c=list(map(int,input().split()))
if len(set(c))==1:
print("YES")
print(1)
else :
q=[]
seen=set()
res=set()
for el in range(n):
if len(graph[el])==1:
q.append(el)
seen.add(el)
while q:
node=q.pop()
for nie in graph[node]:
if nie not in seen:
if c[nie]==c[node]:
q.append(nie)
seen.add(nie)
else :
res.add(nie)
if len(res)==1:
print("YES")
print(list(res)[0]+1)
else :
print("NO")
|
Title: Timofey and a tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Each New Year Timofey and his friends cut down a tree of *n* vertices and bring it home. After that they paint all the *n* its vertices, so that the *i*-th vertex gets color *c**i*.
Now it's time for Timofey birthday, and his mother asked him to remove the tree. Timofey removes the tree in the following way: he takes some vertex in hands, while all the other vertices move down so that the tree becomes rooted at the chosen vertex. After that Timofey brings the tree to a trash can.
Timofey doesn't like it when many colors are mixing together. A subtree annoys him if there are vertices of different color in it. Timofey wants to find a vertex which he should take in hands so that there are no subtrees that annoy him. He doesn't consider the whole tree as a subtree since he can't see the color of the root vertex.
A subtree of some vertex is a subgraph containing that vertex and all its descendants.
Your task is to determine if there is a vertex, taking which in hands Timofey wouldn't be annoyed.
Input Specification:
The first line contains single integer *n* (2<=≤<=*n*<=≤<=105) — the number of vertices in the tree.
Each of the next *n*<=-<=1 lines contains two integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*), denoting there is an edge between vertices *u* and *v*. It is guaranteed that the given graph is a tree.
The next line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=105), denoting the colors of the vertices.
Output Specification:
Print "NO" in a single line, if Timofey can't take the tree in such a way that it doesn't annoy him.
Otherwise print "YES" in the first line. In the second line print the index of the vertex which Timofey should take in hands. If there are multiple answers, print any of them.
Demo Input:
['4\n1 2\n2 3\n3 4\n1 2 1 1\n', '3\n1 2\n2 3\n1 2 3\n', '4\n1 2\n2 3\n3 4\n1 2 1 2\n']
Demo Output:
['YES\n2', 'YES\n2', 'NO']
Note:
none
|
```python
n=int(input())
graph=[[] for _ in range(n)]
for _ in range(n-1):
a,b=map(int,input().split())
a-=1
b-=1
graph[a].append(b)
graph[b].append(a)
c=list(map(int,input().split()))
if len(set(c))==1:
print("YES")
print(1)
else :
q=[]
seen=set()
res=set()
for el in range(n):
if len(graph[el])==1:
q.append(el)
seen.add(el)
while q:
node=q.pop()
for nie in graph[node]:
if nie not in seen:
if c[nie]==c[node]:
q.append(nie)
seen.add(nie)
else :
res.add(nie)
if len(res)==1:
print("YES")
print(list(res)[0]+1)
else :
print("NO")
```
| 0
|
|
365
|
A
|
Good Number
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*).
|
The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109).
|
Print a single integer — the number of *k*-good numbers in *a*.
|
[
"10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n",
"2 1\n1\n10\n"
] |
[
"10\n",
"1\n"
] |
none
| 500
|
[
{
"input": "10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560",
"output": "10"
},
{
"input": "2 1\n1\n10",
"output": "1"
},
{
"input": "1 0\n1000000000",
"output": "1"
},
{
"input": "1 1\n1000000000",
"output": "1"
},
{
"input": "6 0\n10\n102\n120\n1032\n1212103\n1999999",
"output": "5"
},
{
"input": "1 3\n1000000000",
"output": "0"
},
{
"input": "1 9\n1000000000",
"output": "0"
},
{
"input": "1 0\n1",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 3\n1",
"output": "0"
},
{
"input": "1 9\n1",
"output": "0"
},
{
"input": "2 8\n123456780\n123",
"output": "1"
},
{
"input": "2 8\n12345678\n1230",
"output": "0"
},
{
"input": "6 1\n10\n102\n120\n1032\n1212103\n1999999",
"output": "5"
},
{
"input": "6 2\n10\n102\n120\n1032\n1212103\n1999999",
"output": "4"
},
{
"input": "6 3\n10\n102\n120\n1032\n1212103\n1999999",
"output": "2"
},
{
"input": "6 9\n10\n102\n120\n1032\n1212103\n1999999",
"output": "0"
},
{
"input": "1 2\n2220",
"output": "0"
}
] | 1,673,419,160
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 61
| 2,867,200
|
a = int(input())
count = 0
for x in range(a) :
k = int(input())
list_k = [str(i) for i in range(k)]
list_k.sort()
list_k.append(str(k))
list_a = [(x) for x in input().split()]
list_a.sort()
str_1 = "".join(list_a)
str_2 = "".join(list_k)
if str_1 == str_2:
count += 1
print(count)
|
Title: Good Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*).
Input Specification:
The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print a single integer — the number of *k*-good numbers in *a*.
Demo Input:
['10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n', '2 1\n1\n10\n']
Demo Output:
['10\n', '1\n']
Note:
none
|
```python
a = int(input())
count = 0
for x in range(a) :
k = int(input())
list_k = [str(i) for i in range(k)]
list_k.sort()
list_k.append(str(k))
list_a = [(x) for x in input().split()]
list_a.sort()
str_1 = "".join(list_a)
str_2 = "".join(list_k)
if str_1 == str_2:
count += 1
print(count)
```
| -1
|
|
723
|
A
|
The New Year: Meeting Friends
|
PROGRAMMING
| 800
|
[
"implementation",
"math",
"sortings"
] | null | null |
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
|
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
|
Print one integer — the minimum total distance the friends need to travel in order to meet together.
|
[
"7 1 4\n",
"30 20 10\n"
] |
[
"6\n",
"20\n"
] |
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
| 500
|
[
{
"input": "7 1 4",
"output": "6"
},
{
"input": "30 20 10",
"output": "20"
},
{
"input": "1 4 100",
"output": "99"
},
{
"input": "100 1 91",
"output": "99"
},
{
"input": "1 45 100",
"output": "99"
},
{
"input": "1 2 3",
"output": "2"
},
{
"input": "71 85 88",
"output": "17"
},
{
"input": "30 38 99",
"output": "69"
},
{
"input": "23 82 95",
"output": "72"
},
{
"input": "22 41 47",
"output": "25"
},
{
"input": "9 94 77",
"output": "85"
},
{
"input": "1 53 51",
"output": "52"
},
{
"input": "25 97 93",
"output": "72"
},
{
"input": "42 53 51",
"output": "11"
},
{
"input": "81 96 94",
"output": "15"
},
{
"input": "21 5 93",
"output": "88"
},
{
"input": "50 13 75",
"output": "62"
},
{
"input": "41 28 98",
"output": "70"
},
{
"input": "69 46 82",
"output": "36"
},
{
"input": "87 28 89",
"output": "61"
},
{
"input": "44 45 40",
"output": "5"
},
{
"input": "86 97 68",
"output": "29"
},
{
"input": "43 92 30",
"output": "62"
},
{
"input": "16 70 1",
"output": "69"
},
{
"input": "40 46 19",
"output": "27"
},
{
"input": "71 38 56",
"output": "33"
},
{
"input": "82 21 80",
"output": "61"
},
{
"input": "75 8 35",
"output": "67"
},
{
"input": "75 24 28",
"output": "51"
},
{
"input": "78 23 56",
"output": "55"
},
{
"input": "85 31 10",
"output": "75"
},
{
"input": "76 50 9",
"output": "67"
},
{
"input": "95 37 34",
"output": "61"
},
{
"input": "84 61 35",
"output": "49"
},
{
"input": "87 85 37",
"output": "50"
},
{
"input": "1 3 2",
"output": "2"
},
{
"input": "4 2 6",
"output": "4"
},
{
"input": "6 9 3",
"output": "6"
},
{
"input": "12 4 8",
"output": "8"
},
{
"input": "15 10 5",
"output": "10"
},
{
"input": "1 50 17",
"output": "49"
},
{
"input": "10 5 15",
"output": "10"
},
{
"input": "8 1 9",
"output": "8"
},
{
"input": "3 5 4",
"output": "2"
},
{
"input": "2 1 3",
"output": "2"
},
{
"input": "1 8 2",
"output": "7"
},
{
"input": "1 100 2",
"output": "99"
},
{
"input": "1 4 6",
"output": "5"
}
] | 1,683,231,753
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 61
| 0
|
def calc(n, m, o):
centroid = (n + m + o) / 3
dist = abs(n + m + o - 3 * centroid)
return dist
n, m, o = input().split()
n, m, o = int(n), int(m), int(o)
|
Title: The New Year: Meeting Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
Input Specification:
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
Output Specification:
Print one integer — the minimum total distance the friends need to travel in order to meet together.
Demo Input:
['7 1 4\n', '30 20 10\n']
Demo Output:
['6\n', '20\n']
Note:
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
|
```python
def calc(n, m, o):
centroid = (n + m + o) / 3
dist = abs(n + m + o - 3 * centroid)
return dist
n, m, o = input().split()
n, m, o = int(n), int(m), int(o)
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Pavel loves grid mazes. A grid maze is an *n*<=×<=*m* rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.
Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly *k* empty cells into walls so that all the remaining cells still formed a connected area. Help him.
|
The first line contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*<=≤<=500, 0<=≤<=*k*<=<<=*s*), where *n* and *m* are the maze's height and width, correspondingly, *k* is the number of walls Pavel wants to add and letter *s* represents the number of empty cells in the original maze.
Each of the next *n* lines contains *m* characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.
|
Print *n* lines containing *m* characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").
It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.
|
[
"3 4 2\n#..#\n..#.\n#...\n",
"5 4 5\n#...\n#.#.\n.#..\n...#\n.#.#\n"
] |
[
"#.X#\nX.#.\n#...\n",
"#XXX\n#X#.\nX#..\n...#\n.#.#\n"
] |
none
| 0
|
[
{
"input": "5 4 5\n#...\n#.#.\n.#..\n...#\n.#.#",
"output": "#XXX\n#X#.\nX#..\n...#\n.#.#"
},
{
"input": "3 3 2\n#.#\n...\n#.#",
"output": "#X#\nX..\n#.#"
},
{
"input": "7 7 18\n#.....#\n..#.#..\n.#...#.\n...#...\n.#...#.\n..#.#..\n#.....#",
"output": "#XXXXX#\nXX#X#X.\nX#XXX#.\nXXX#...\nX#...#.\nX.#.#..\n#.....#"
},
{
"input": "1 1 0\n.",
"output": "."
},
{
"input": "2 3 1\n..#\n#..",
"output": "X.#\n#.."
},
{
"input": "2 3 1\n#..\n..#",
"output": "#.X\n..#"
},
{
"input": "3 3 1\n...\n.#.\n..#",
"output": "...\n.#X\n..#"
},
{
"input": "3 3 1\n...\n.#.\n#..",
"output": "...\nX#.\n#.."
},
{
"input": "5 4 4\n#..#\n....\n.##.\n....\n#..#",
"output": "#XX#\nXX..\n.##.\n....\n#..#"
},
{
"input": "5 5 2\n.#..#\n..#.#\n#....\n##.#.\n###..",
"output": "X#..#\nX.#.#\n#....\n##.#.\n###.."
},
{
"input": "4 6 3\n#.....\n#.#.#.\n.#...#\n...#.#",
"output": "#.....\n#X#.#X\nX#...#\n...#.#"
},
{
"input": "7 5 4\n.....\n.#.#.\n#...#\n.#.#.\n.#...\n..#..\n....#",
"output": "X...X\nX#.#X\n#...#\n.#.#.\n.#...\n..#..\n....#"
},
{
"input": "16 14 19\n##############\n..############\n#.############\n#..###########\n....##########\n..############\n.#############\n.#.###########\n....##########\n###..#########\n##...#########\n###....#######\n###.##.......#\n###..###.#..#.\n###....#......\n#...#...##.###",
"output": "##############\nXX############\n#X############\n#XX###########\nXXXX##########\nXX############\nX#############\nX#.###########\nX...##########\n###..#########\n##...#########\n###....#######\n###.##.......#\n###..###.#..#.\n###...X#......\n#X..#XXX##.###"
},
{
"input": "10 17 32\n######.##########\n####.#.##########\n...#....#########\n.........########\n##.......########\n........#########\n#.....###########\n#################\n#################\n#################",
"output": "######X##########\n####X#X##########\nXXX#XXXX#########\nXXXXXXXXX########\n##XXX.XXX########\nXXXX...X#########\n#XX...###########\n#################\n#################\n#################"
},
{
"input": "16 10 38\n##########\n##########\n##########\n..########\n...#######\n...#######\n...#######\n....######\n.....####.\n......###.\n......##..\n.......#..\n.........#\n.........#\n.........#\n.........#",
"output": "##########\n##########\n##########\nXX########\nXXX#######\nXXX#######\nXXX#######\nXXXX######\nXXXXX####.\nXXXXX.###.\nXXXX..##..\nXXX....#..\nXXX......#\nXX.......#\nX........#\n.........#"
},
{
"input": "15 16 19\n########.....###\n########.....###\n############.###\n############.###\n############.###\n############.###\n############.###\n############.###\n############.###\n############.###\n.....#####.#..##\n................\n.#...........###\n###.########.###\n###.########.###",
"output": "########XXXXX###\n########XXXXX###\n############.###\n############.###\n############.###\n############.###\n############.###\n############.###\n############.###\n############.###\nXXXX.#####.#..##\nXXX.............\nX#...........###\n###.########.###\n###X########.###"
},
{
"input": "12 19 42\n.........##########\n...................\n.##.##############.\n..################.\n..#################\n..#################\n..#################\n..#################\n..#################\n..#################\n..##########.######\n.............######",
"output": "XXXXXXXXX##########\nXXXXXXXXXXXXXXXXXXX\nX##X##############X\nXX################X\nXX#################\nXX#################\nXX#################\nX.#################\nX.#################\n..#################\n..##########.######\n.............######"
},
{
"input": "3 5 1\n#...#\n..#..\n..#..",
"output": "#...#\n..#..\nX.#.."
},
{
"input": "4 5 10\n.....\n.....\n..#..\n..#..",
"output": "XXX..\nXXX..\nXX#..\nXX#.."
},
{
"input": "3 5 3\n.....\n..#..\n..#..",
"output": ".....\nX.#..\nXX#.."
},
{
"input": "3 5 1\n#....\n..#..\n..###",
"output": "#....\n..#.X\n..###"
},
{
"input": "4 5 1\n.....\n.##..\n..#..\n..###",
"output": ".....\n.##..\n..#.X\n..###"
},
{
"input": "3 5 2\n..#..\n..#..\n....#",
"output": "X.#..\nX.#..\n....#"
},
{
"input": "10 10 1\n##########\n##......##\n#..#..#..#\n#..####..#\n#######.##\n#######.##\n#..####..#\n#..#..#..#\n##......##\n##########",
"output": "##########\n##......##\n#..#..#..#\n#X.####..#\n#######.##\n#######.##\n#..####..#\n#..#..#..#\n##......##\n##########"
},
{
"input": "10 10 3\n..........\n.########.\n.########.\n.########.\n.########.\n.########.\n.#######..\n.#######..\n.####..###\n.......###",
"output": "..........\n.########.\n.########.\n.########.\n.########.\n.########.\n.#######X.\n.#######XX\n.####..###\n.......###"
},
{
"input": "5 7 10\n..#....\n..#.#..\n.##.#..\n..#.#..\n....#..",
"output": "XX#....\nXX#.#..\nX##.#..\nXX#.#..\nXXX.#.."
},
{
"input": "5 7 10\n..#....\n..#.##.\n.##.##.\n..#.#..\n....#..",
"output": "XX#....\nXX#.##.\nX##.##.\nXX#.#..\nXXX.#.."
},
{
"input": "10 10 1\n##########\n##..##..##\n#...##...#\n#.######.#\n#..####..#\n#..####..#\n#.######.#\n#........#\n##..##..##\n##########",
"output": "##########\n##.X##..##\n#...##...#\n#.######.#\n#..####..#\n#..####..#\n#.######.#\n#........#\n##..##..##\n##########"
},
{
"input": "4 5 1\n.....\n.###.\n..#..\n..#..",
"output": ".....\n.###.\n..#..\n.X#.."
},
{
"input": "2 5 2\n###..\n###..",
"output": "###X.\n###X."
},
{
"input": "2 5 3\n.....\n..#..",
"output": "X....\nXX#.."
},
{
"input": "12 12 3\n############\n#..........#\n#.########.#\n#.########.#\n#.########.#\n#.########.#\n#.########.#\n#.#######..#\n#.#######..#\n#.####..####\n#.......####\n############",
"output": "############\n#..........#\n#.########.#\n#.########.#\n#.########.#\n#.########.#\n#.########.#\n#.#######X.#\n#.#######XX#\n#.####..####\n#.......####\n############"
},
{
"input": "5 5 1\n.....\n.##..\n..###\n..###\n#####",
"output": ".....\n.##.X\n..###\n..###\n#####"
},
{
"input": "4 4 1\n....\n.#..\n..##\n..##",
"output": "....\n.#.X\n..##\n..##"
},
{
"input": "5 5 1\n....#\n.##..\n.##..\n...##\n...##",
"output": "....#\n.##..\n.##.X\n...##\n...##"
},
{
"input": "5 5 1\n.....\n.##..\n..###\n..###\n..###",
"output": ".....\n.##.X\n..###\n..###\n..###"
},
{
"input": "4 5 1\n#....\n#.#..\n..###\n..###",
"output": "#....\n#.#.X\n..###\n..###"
},
{
"input": "4 4 3\n....\n.#..\n..##\n..##",
"output": "...X\n.#XX\n..##\n..##"
},
{
"input": "4 7 6\n.......\n....#..\n.##.#..\n....#..",
"output": "X......\nX...#..\nX##.#..\nXXX.#.."
},
{
"input": "8 8 7\n........\n.##.....\n.#######\n..######\n..######\n..######\n..######\n..######",
"output": ".....XXX\n.##.XXXX\n.#######\n..######\n..######\n..######\n..######\n..######"
}
] | 1,668,874,815
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 4
| 46
| 0
|
matr = []
n,m,k = map(int,input().split())
for i in range(n):
matr.append(input())
used = [[0]*(m+2) for i in range(n+2)]
count = int(0)
def dfs(i, j):
global count
if count==k:
return
flag = int(0)
used[i+1][j+1]=1
if (i >= 1):
if (used[i][j+1] == 0 and matr[i-1][j]=="."):
flag = 1
dfs(i-1,j)
if (i < n-1):
if (used[i+2][j+1] == 0 and matr[i+1][j]=="."):
flag = 1
dfs(i+1,j)
if (j >= 1):
if (used[i+1][j] == 0 and matr[i][j-1]=="."):
flag = 1
dfs(i,j-1)
if (j <m-1):
if (used[i+1][j+2] == 0 and matr[i][j+1]=="."):
flag = 1
dfs(i,j+1)
if flag == 0:
matr[i] = matr[i][0:j] + "X" + matr[i][j+1:]
used[i + 1][j + 1] = 0
count+=1
j_2 = int(0)
flag_2 = int(0)
i = int(0)
j = int(0)
while (i != n):
j=0
while (j != m):
if matr[i][j]=="." and used[i+1][j]+used[i+1][j+2]+used[i][j+1]+used[i+2][j+1]<=1:
dfs(i, j)
if count<k:
j-=1
if count == k:
flag_2 = 1
break
j+=1
i+=1
if flag_2 == 1:
break
for i in range(n):
for j in range(m):
print(matr[i][j], end="")
print()
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pavel loves grid mazes. A grid maze is an *n*<=×<=*m* rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.
Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly *k* empty cells into walls so that all the remaining cells still formed a connected area. Help him.
Input Specification:
The first line contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*<=≤<=500, 0<=≤<=*k*<=<<=*s*), where *n* and *m* are the maze's height and width, correspondingly, *k* is the number of walls Pavel wants to add and letter *s* represents the number of empty cells in the original maze.
Each of the next *n* lines contains *m* characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.
Output Specification:
Print *n* lines containing *m* characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").
It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.
Demo Input:
['3 4 2\n#..#\n..#.\n#...\n', '5 4 5\n#...\n#.#.\n.#..\n...#\n.#.#\n']
Demo Output:
['#.X#\nX.#.\n#...\n', '#XXX\n#X#.\nX#..\n...#\n.#.#\n']
Note:
none
|
```python
matr = []
n,m,k = map(int,input().split())
for i in range(n):
matr.append(input())
used = [[0]*(m+2) for i in range(n+2)]
count = int(0)
def dfs(i, j):
global count
if count==k:
return
flag = int(0)
used[i+1][j+1]=1
if (i >= 1):
if (used[i][j+1] == 0 and matr[i-1][j]=="."):
flag = 1
dfs(i-1,j)
if (i < n-1):
if (used[i+2][j+1] == 0 and matr[i+1][j]=="."):
flag = 1
dfs(i+1,j)
if (j >= 1):
if (used[i+1][j] == 0 and matr[i][j-1]=="."):
flag = 1
dfs(i,j-1)
if (j <m-1):
if (used[i+1][j+2] == 0 and matr[i][j+1]=="."):
flag = 1
dfs(i,j+1)
if flag == 0:
matr[i] = matr[i][0:j] + "X" + matr[i][j+1:]
used[i + 1][j + 1] = 0
count+=1
j_2 = int(0)
flag_2 = int(0)
i = int(0)
j = int(0)
while (i != n):
j=0
while (j != m):
if matr[i][j]=="." and used[i+1][j]+used[i+1][j+2]+used[i][j+1]+used[i+2][j+1]<=1:
dfs(i, j)
if count<k:
j-=1
if count == k:
flag_2 = 1
break
j+=1
i+=1
if flag_2 == 1:
break
for i in range(n):
for j in range(m):
print(matr[i][j], end="")
print()
```
| 0
|
|
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"
},
{
"input": "999",
"output": "2889"
},
{
"input": "55",
"output": "101"
},
{
"input": "222222222",
"output": "1888888896"
},
{
"input": "8",
"output": "8"
},
{
"input": "13",
"output": "17"
},
{
"input": "313",
"output": "831"
},
{
"input": "1342",
"output": "4261"
},
{
"input": "30140",
"output": "139594"
},
{
"input": "290092",
"output": "1629447"
},
{
"input": "2156660",
"output": "13985516"
},
{
"input": "96482216",
"output": "760746625"
},
{
"input": "943006819",
"output": "8375950269"
},
{
"input": "1",
"output": "1"
},
{
"input": "7",
"output": "7"
},
{
"input": "35",
"output": "61"
},
{
"input": "996",
"output": "2880"
},
{
"input": "6120",
"output": "23373"
},
{
"input": "30660",
"output": "142194"
},
{
"input": "349463",
"output": "1985673"
},
{
"input": "8171970",
"output": "56092686"
},
{
"input": "36123011",
"output": "277872985"
},
{
"input": "986747865",
"output": "8769619683"
},
{
"input": "9",
"output": "9"
},
{
"input": "10",
"output": "11"
},
{
"input": "11",
"output": "13"
},
{
"input": "101",
"output": "195"
},
{
"input": "1000",
"output": "2893"
},
{
"input": "1001",
"output": "2897"
},
{
"input": "9999",
"output": "38889"
},
{
"input": "10000",
"output": "38894"
},
{
"input": "10001",
"output": "38899"
},
{
"input": "99999",
"output": "488889"
},
{
"input": "100000",
"output": "488895"
},
{
"input": "100001",
"output": "488901"
},
{
"input": "999999",
"output": "5888889"
},
{
"input": "1000001",
"output": "5888903"
},
{
"input": "9999999",
"output": "68888889"
},
{
"input": "10000000",
"output": "68888897"
},
{
"input": "10000001",
"output": "68888905"
},
{
"input": "99999999",
"output": "788888889"
},
{
"input": "100000000",
"output": "788888898"
},
{
"input": "100000001",
"output": "788888907"
},
{
"input": "999999999",
"output": "8888888889"
},
{
"input": "95",
"output": "181"
},
{
"input": "995",
"output": "2877"
},
{
"input": "9995",
"output": "38873"
},
{
"input": "99995",
"output": "488869"
},
{
"input": "999999995",
"output": "8888888853"
},
{
"input": "999999998",
"output": "8888888880"
}
] | 1,658,921,637
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 1
| 31
| 0
|
a = input()
b = len(a)
d = int(a)
c = 0;s = '';t = 0
s = (b-1)*'9'
t = int(s)
c += b*(d-t)
d = d-t
c = c + int((b-1)*'9')
print(c)
|
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
a = input()
b = len(a)
d = int(a)
c = 0;s = '';t = 0
s = (b-1)*'9'
t = int(s)
c += b*(d-t)
d = d-t
c = c + int((b-1)*'9')
print(c)
```
| -1
|
|
622
|
B
|
The Time
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
You are given the current time in 24-hour format hh:mm. Find and print the time after *a* minutes.
Note that you should find only the time after *a* minutes, see the examples to clarify the problem statement.
You can read more about 24-hour format here [https://en.wikipedia.org/wiki/24-hour_clock](https://en.wikipedia.org/wiki/24-hour_clock).
|
The first line contains the current time in the format hh:mm (0<=≤<=*hh*<=<<=24,<=0<=≤<=*mm*<=<<=60). The hours and the minutes are given with two digits (the hours or the minutes less than 10 are given with the leading zeroes).
The second line contains integer *a* (0<=≤<=*a*<=≤<=104) — the number of the minutes passed.
|
The only line should contain the time after *a* minutes in the format described in the input. Note that you should print exactly two digits for the hours and the minutes (add leading zeroes to the numbers if needed).
See the examples to check the input/output format.
|
[
"23:59\n10\n",
"20:20\n121\n",
"10:10\n0\n"
] |
[
"00:09\n",
"22:21\n",
"10:10\n"
] |
none
| 0
|
[
{
"input": "23:59\n10",
"output": "00:09"
},
{
"input": "20:20\n121",
"output": "22:21"
},
{
"input": "10:10\n0",
"output": "10:10"
},
{
"input": "12:34\n10000",
"output": "11:14"
},
{
"input": "00:00\n10000",
"output": "22:40"
},
{
"input": "00:00\n1440",
"output": "00:00"
},
{
"input": "23:59\n8640",
"output": "23:59"
},
{
"input": "10:01\n0",
"output": "10:01"
},
{
"input": "04:05\n0",
"output": "04:05"
},
{
"input": "02:59\n1",
"output": "03:00"
},
{
"input": "05:15\n10",
"output": "05:25"
},
{
"input": "03:10\n20",
"output": "03:30"
},
{
"input": "09:11\n0",
"output": "09:11"
},
{
"input": "19:00\n0",
"output": "19:00"
},
{
"input": "23:59\n1",
"output": "00:00"
},
{
"input": "11:59\n1",
"output": "12:00"
},
{
"input": "19:34\n566",
"output": "05:00"
},
{
"input": "00:01\n59",
"output": "01:00"
},
{
"input": "03:30\n0",
"output": "03:30"
},
{
"input": "22:30\n30",
"output": "23:00"
},
{
"input": "22:50\n70",
"output": "00:00"
},
{
"input": "05:12\n0",
"output": "05:12"
},
{
"input": "09:20\n40",
"output": "10:00"
},
{
"input": "15:04\n36",
"output": "15:40"
},
{
"input": "05:37\n23",
"output": "06:00"
},
{
"input": "23:59\n59",
"output": "00:58"
},
{
"input": "21:09\n9997",
"output": "19:46"
},
{
"input": "11:00\n1",
"output": "11:01"
},
{
"input": "20:01\n2699",
"output": "17:00"
},
{
"input": "01:00\n59",
"output": "01:59"
},
{
"input": "07:09\n6538",
"output": "20:07"
},
{
"input": "00:00\n10",
"output": "00:10"
},
{
"input": "02:09\n2074",
"output": "12:43"
},
{
"input": "01:10\n1",
"output": "01:11"
},
{
"input": "23:01\n59",
"output": "00:00"
},
{
"input": "08:50\n20",
"output": "09:10"
},
{
"input": "13:18\n5121",
"output": "02:39"
},
{
"input": "18:31\n2677",
"output": "15:08"
},
{
"input": "14:17\n108",
"output": "16:05"
},
{
"input": "02:45\n5617",
"output": "00:22"
},
{
"input": "00:00\n3600",
"output": "12:00"
},
{
"input": "19:01\n59",
"output": "20:00"
},
{
"input": "19:02\n59",
"output": "20:01"
},
{
"input": "14:00\n2880",
"output": "14:00"
},
{
"input": "01:10\n44",
"output": "01:54"
},
{
"input": "02:01\n59",
"output": "03:00"
},
{
"input": "07:02\n121",
"output": "09:03"
},
{
"input": "10:00\n61",
"output": "11:01"
},
{
"input": "23:59\n61",
"output": "01:00"
},
{
"input": "00:00\n0",
"output": "00:00"
},
{
"input": "23:59\n121",
"output": "02:00"
},
{
"input": "00:00\n60",
"output": "01:00"
},
{
"input": "15:52\n60",
"output": "16:52"
},
{
"input": "00:39\n6525",
"output": "13:24"
},
{
"input": "00:30\n30",
"output": "01:00"
},
{
"input": "00:59\n1",
"output": "01:00"
},
{
"input": "00:55\n4321",
"output": "00:56"
},
{
"input": "10:05\n1",
"output": "10:06"
},
{
"input": "23:00\n60",
"output": "00:00"
}
] | 1,646,968,704
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 59
| 46
| 0
|
h, m = map(int, input().split(':'))
a = int(input())
b = (h * 60 + m + a) % (24 * 60)
print(f'{b // 60:02d}:{b % 60:02d}')
|
Title: The Time
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given the current time in 24-hour format hh:mm. Find and print the time after *a* minutes.
Note that you should find only the time after *a* minutes, see the examples to clarify the problem statement.
You can read more about 24-hour format here [https://en.wikipedia.org/wiki/24-hour_clock](https://en.wikipedia.org/wiki/24-hour_clock).
Input Specification:
The first line contains the current time in the format hh:mm (0<=≤<=*hh*<=<<=24,<=0<=≤<=*mm*<=<<=60). The hours and the minutes are given with two digits (the hours or the minutes less than 10 are given with the leading zeroes).
The second line contains integer *a* (0<=≤<=*a*<=≤<=104) — the number of the minutes passed.
Output Specification:
The only line should contain the time after *a* minutes in the format described in the input. Note that you should print exactly two digits for the hours and the minutes (add leading zeroes to the numbers if needed).
See the examples to check the input/output format.
Demo Input:
['23:59\n10\n', '20:20\n121\n', '10:10\n0\n']
Demo Output:
['00:09\n', '22:21\n', '10:10\n']
Note:
none
|
```python
h, m = map(int, input().split(':'))
a = int(input())
b = (h * 60 + m + a) % (24 * 60)
print(f'{b // 60:02d}:{b % 60:02d}')
```
| 3
|
|
527
|
A
|
Playing with Paper
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math"
] | null | null |
One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular *a* mm <=×<= *b* mm sheet of paper (*a*<=><=*b*). Usually the first step in making an origami is making a square piece of paper from the rectangular sheet by folding the sheet along the bisector of the right angle, and cutting the excess part.
After making a paper ship from the square piece, Vasya looked on the remaining (*a*<=-<=*b*) mm <=×<= *b* mm strip of paper. He got the idea to use this strip of paper in the same way to make an origami, and then use the remainder (if it exists) and so on. At the moment when he is left with a square piece of paper, he will make the last ship from it and stop.
Can you determine how many ships Vasya will make during the lesson?
|
The first line of the input contains two integers *a*, *b* (1<=≤<=*b*<=<<=*a*<=≤<=1012) — the sizes of the original sheet of paper.
|
Print a single integer — the number of ships that Vasya will make.
|
[
"2 1\n",
"10 7\n",
"1000000000000 1\n"
] |
[
"2\n",
"6\n",
"1000000000000\n"
] |
Pictures to the first and second sample test.
| 500
|
[
{
"input": "2 1",
"output": "2"
},
{
"input": "10 7",
"output": "6"
},
{
"input": "1000000000000 1",
"output": "1000000000000"
},
{
"input": "3 1",
"output": "3"
},
{
"input": "4 1",
"output": "4"
},
{
"input": "3 2",
"output": "3"
},
{
"input": "4 2",
"output": "2"
},
{
"input": "1000 700",
"output": "6"
},
{
"input": "959986566087 524054155168",
"output": "90"
},
{
"input": "4 3",
"output": "4"
},
{
"input": "7 6",
"output": "7"
},
{
"input": "1000 999",
"output": "1000"
},
{
"input": "1000 998",
"output": "500"
},
{
"input": "1000 997",
"output": "336"
},
{
"input": "42 1",
"output": "42"
},
{
"input": "1000 1",
"output": "1000"
},
{
"input": "8 5",
"output": "5"
},
{
"input": "13 8",
"output": "6"
},
{
"input": "987 610",
"output": "15"
},
{
"input": "442 42",
"output": "22"
},
{
"input": "754 466",
"output": "13"
},
{
"input": "1000000000000 999999999999",
"output": "1000000000000"
},
{
"input": "1000000000000 999999999998",
"output": "500000000000"
},
{
"input": "941 14",
"output": "74"
},
{
"input": "998 2",
"output": "499"
},
{
"input": "1000 42",
"output": "32"
},
{
"input": "1000 17",
"output": "66"
},
{
"input": "5 1",
"output": "5"
},
{
"input": "5 2",
"output": "4"
},
{
"input": "5 3",
"output": "4"
},
{
"input": "5 4",
"output": "5"
},
{
"input": "293 210",
"output": "17"
},
{
"input": "787878787878 424242424242",
"output": "8"
},
{
"input": "956722026041 591286729879",
"output": "58"
},
{
"input": "956722026041 365435296162",
"output": "58"
},
{
"input": "628625247282 464807889701",
"output": "102"
},
{
"input": "695928431619 424778620208",
"output": "167"
},
{
"input": "1000000000000 42",
"output": "23809523821"
},
{
"input": "987654345678 23",
"output": "42941493300"
},
{
"input": "10000000001 2",
"output": "5000000002"
},
{
"input": "1000000000000 2",
"output": "500000000000"
},
{
"input": "1000000000000 3",
"output": "333333333336"
},
{
"input": "100000000000 3",
"output": "33333333336"
},
{
"input": "100000000000 23",
"output": "4347826109"
},
{
"input": "999999999997 7",
"output": "142857142861"
},
{
"input": "8589934592 4294967296",
"output": "2"
}
] | 1,602,658,831
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 46
| 109
| 0
|
a,b = [int(x) for x in input().split()]
ans = 0
while a > 0 and b > 0:
if a >= b:
ans += int(a/b)
a = a%b
else:
ans += int(b/a)
b = b%a
print(ans)
|
Title: Playing with Paper
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular *a* mm <=×<= *b* mm sheet of paper (*a*<=><=*b*). Usually the first step in making an origami is making a square piece of paper from the rectangular sheet by folding the sheet along the bisector of the right angle, and cutting the excess part.
After making a paper ship from the square piece, Vasya looked on the remaining (*a*<=-<=*b*) mm <=×<= *b* mm strip of paper. He got the idea to use this strip of paper in the same way to make an origami, and then use the remainder (if it exists) and so on. At the moment when he is left with a square piece of paper, he will make the last ship from it and stop.
Can you determine how many ships Vasya will make during the lesson?
Input Specification:
The first line of the input contains two integers *a*, *b* (1<=≤<=*b*<=<<=*a*<=≤<=1012) — the sizes of the original sheet of paper.
Output Specification:
Print a single integer — the number of ships that Vasya will make.
Demo Input:
['2 1\n', '10 7\n', '1000000000000 1\n']
Demo Output:
['2\n', '6\n', '1000000000000\n']
Note:
Pictures to the first and second sample test.
|
```python
a,b = [int(x) for x in input().split()]
ans = 0
while a > 0 and b > 0:
if a >= b:
ans += int(a/b)
a = a%b
else:
ans += int(b/a)
b = b%a
print(ans)
```
| 3
|
|
500
|
A
|
New Year Transportation
|
PROGRAMMING
| 1,000
|
[
"dfs and similar",
"graphs",
"implementation"
] | null | null |
New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system.
|
The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
|
If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO".
|
[
"8 4\n1 2 1 2 1 2 1\n",
"8 5\n1 2 1 2 1 1 1\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit.
| 500
|
[
{
"input": "8 4\n1 2 1 2 1 2 1",
"output": "YES"
},
{
"input": "8 5\n1 2 1 2 1 1 1",
"output": "NO"
},
{
"input": "20 19\n13 16 7 6 12 1 5 7 8 6 5 7 5 5 3 3 2 2 1",
"output": "YES"
},
{
"input": "50 49\n11 7 1 41 26 36 19 16 38 14 36 35 37 27 20 27 3 6 21 2 27 11 18 17 19 16 22 8 8 9 1 7 5 12 5 6 13 6 11 2 6 3 1 5 1 1 2 2 1",
"output": "YES"
},
{
"input": "120 104\n41 15 95 85 34 11 25 42 65 39 77 80 74 17 66 73 21 14 36 63 63 79 45 24 65 7 63 80 51 21 2 19 78 28 71 2 15 23 17 68 62 18 54 39 43 70 3 46 34 23 41 65 32 10 13 18 10 3 16 48 54 18 57 28 3 24 44 50 15 2 20 22 45 44 3 29 2 27 11 2 12 25 25 31 1 2 32 4 11 30 13 16 26 21 1 13 21 8 15 5 18 13 5 15 3 8 13 6 5 1 9 7 1 2 4 1 1 2 1",
"output": "NO"
},
{
"input": "10 3\n8 3 5 4 2 3 2 2 1",
"output": "NO"
},
{
"input": "10 9\n8 3 5 4 2 3 2 2 1",
"output": "YES"
},
{
"input": "3 2\n1 1",
"output": "YES"
},
{
"input": "3 2\n2 1",
"output": "NO"
},
{
"input": "4 2\n2 1 1",
"output": "NO"
},
{
"input": "4 4\n2 2 1",
"output": "YES"
},
{
"input": "8 8\n1 2 1 2 1 2 1",
"output": "YES"
},
{
"input": "3 3\n1 1",
"output": "YES"
},
{
"input": "8 8\n1 2 1 2 1 1 1",
"output": "YES"
},
{
"input": "3 3\n2 1",
"output": "YES"
},
{
"input": "4 4\n1 1 1",
"output": "YES"
},
{
"input": "8 8\n1 1 1 1 1 1 1",
"output": "YES"
},
{
"input": "5 5\n1 1 1 1",
"output": "YES"
}
] | 1,686,337,179
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 34
| 61
| 4,812,800
|
#!/usr/bin/env python3
import math
import sys
input = lambda: sys.stdin.readline().rstrip("\r\n")
def test_case():
n, t = map(int, input().split())
t -= 1
a = list(map(int, input().split()))
a.append(0)
ans = [False for _ in range(n)]
ans[0] = True
for i in range(n):
if ans[i]:
ans[i+a[i]] = True
if ans[t]:
print("YES")
else:
print("NO")
def main():
t = 1
# t = int(input())
for _ in range(t):
test_case()
if __name__ == '__main__':
main()
|
Title: New Year Transportation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system.
Input Specification:
The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output Specification:
If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO".
Demo Input:
['8 4\n1 2 1 2 1 2 1\n', '8 5\n1 2 1 2 1 1 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit.
|
```python
#!/usr/bin/env python3
import math
import sys
input = lambda: sys.stdin.readline().rstrip("\r\n")
def test_case():
n, t = map(int, input().split())
t -= 1
a = list(map(int, input().split()))
a.append(0)
ans = [False for _ in range(n)]
ans[0] = True
for i in range(n):
if ans[i]:
ans[i+a[i]] = True
if ans[t]:
print("YES")
else:
print("NO")
def main():
t = 1
# t = int(input())
for _ in range(t):
test_case()
if __name__ == '__main__':
main()
```
| 3
|
|
978
|
C
|
Letters
|
PROGRAMMING
| 1,000
|
[
"binary search",
"implementation",
"two pointers"
] | null | null |
There are $n$ dormitories in Berland State University, they are numbered with integers from $1$ to $n$. Each dormitory consists of rooms, there are $a_i$ rooms in $i$-th dormitory. The rooms in $i$-th dormitory are numbered from $1$ to $a_i$.
A postman delivers letters. Sometimes there is no specific dormitory and room number in it on an envelope. Instead of it only a room number among all rooms of all $n$ dormitories is written on an envelope. In this case, assume that all the rooms are numbered from $1$ to $a_1 + a_2 + \dots + a_n$ and the rooms of the first dormitory go first, the rooms of the second dormitory go after them and so on.
For example, in case $n=2$, $a_1=3$ and $a_2=5$ an envelope can have any integer from $1$ to $8$ written on it. If the number $7$ is written on an envelope, it means that the letter should be delivered to the room number $4$ of the second dormitory.
For each of $m$ letters by the room number among all $n$ dormitories, determine the particular dormitory and the room number in a dormitory where this letter should be delivered.
|
The first line contains two integers $n$ and $m$ $(1 \le n, m \le 2 \cdot 10^{5})$ — the number of dormitories and the number of letters.
The second line contains a sequence $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^{10})$, where $a_i$ equals to the number of rooms in the $i$-th dormitory. The third line contains a sequence $b_1, b_2, \dots, b_m$ $(1 \le b_j \le a_1 + a_2 + \dots + a_n)$, where $b_j$ equals to the room number (among all rooms of all dormitories) for the $j$-th letter. All $b_j$ are given in increasing order.
|
Print $m$ lines. For each letter print two integers $f$ and $k$ — the dormitory number $f$ $(1 \le f \le n)$ and the room number $k$ in this dormitory $(1 \le k \le a_f)$ to deliver the letter.
|
[
"3 6\n10 15 12\n1 9 12 23 26 37\n",
"2 3\n5 10000000000\n5 6 9999999999\n"
] |
[
"1 1\n1 9\n2 2\n2 13\n3 1\n3 12\n",
"1 5\n2 1\n2 9999999994\n"
] |
In the first example letters should be delivered in the following order:
- the first letter in room $1$ of the first dormitory - the second letter in room $9$ of the first dormitory - the third letter in room $2$ of the second dormitory - the fourth letter in room $13$ of the second dormitory - the fifth letter in room $1$ of the third dormitory - the sixth letter in room $12$ of the third dormitory
| 0
|
[
{
"input": "3 6\n10 15 12\n1 9 12 23 26 37",
"output": "1 1\n1 9\n2 2\n2 13\n3 1\n3 12"
},
{
"input": "2 3\n5 10000000000\n5 6 9999999999",
"output": "1 5\n2 1\n2 9999999994"
},
{
"input": "1 1\n1\n1",
"output": "1 1"
},
{
"input": "5 15\n10 20 30 20 10\n1 6 10 11 15 30 31 54 60 61 76 80 81 84 90",
"output": "1 1\n1 6\n1 10\n2 1\n2 5\n2 20\n3 1\n3 24\n3 30\n4 1\n4 16\n4 20\n5 1\n5 4\n5 10"
},
{
"input": "1 10\n10\n1 2 3 4 5 6 7 8 9 10",
"output": "1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10"
},
{
"input": "5 8\n10 1 1 1 10\n9 10 11 12 13 14 15 23",
"output": "1 9\n1 10\n2 1\n3 1\n4 1\n5 1\n5 2\n5 10"
},
{
"input": "1 3\n10000\n1 4325 10000",
"output": "1 1\n1 4325\n1 10000"
},
{
"input": "4 18\n5 6 3 4\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18",
"output": "1 1\n1 2\n1 3\n1 4\n1 5\n2 1\n2 2\n2 3\n2 4\n2 5\n2 6\n3 1\n3 2\n3 3\n4 1\n4 2\n4 3\n4 4"
},
{
"input": "3 10\n1000000000 1000000000 1000000000\n543678543 567869543 1000000000 1000000001 1500000000 1999999999 2000000000 2000000001 2754432345 3000000000",
"output": "1 543678543\n1 567869543\n1 1000000000\n2 1\n2 500000000\n2 999999999\n2 1000000000\n3 1\n3 754432345\n3 1000000000"
}
] | 1,668,016,618
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 2
| 62
| 2,867,200
|
n,m=list(map(int,input().split()))
a=list(map(int,input().split()))
b=list(map(int,input().split()))
x=[]
for j in range(n):
for i in b:
if i>=1 and i<=a[j]:
print(j+1,i)
if i>a[j] and i<=a[j]+a[j+1]:
print(j+2,i-a[j])
if i>a[j]+a[j+1] and i<=a[j]+a[j+1]+a[j+2]:
print(j+3,i-a[j]-a[j+1])
break
|
Title: Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are $n$ dormitories in Berland State University, they are numbered with integers from $1$ to $n$. Each dormitory consists of rooms, there are $a_i$ rooms in $i$-th dormitory. The rooms in $i$-th dormitory are numbered from $1$ to $a_i$.
A postman delivers letters. Sometimes there is no specific dormitory and room number in it on an envelope. Instead of it only a room number among all rooms of all $n$ dormitories is written on an envelope. In this case, assume that all the rooms are numbered from $1$ to $a_1 + a_2 + \dots + a_n$ and the rooms of the first dormitory go first, the rooms of the second dormitory go after them and so on.
For example, in case $n=2$, $a_1=3$ and $a_2=5$ an envelope can have any integer from $1$ to $8$ written on it. If the number $7$ is written on an envelope, it means that the letter should be delivered to the room number $4$ of the second dormitory.
For each of $m$ letters by the room number among all $n$ dormitories, determine the particular dormitory and the room number in a dormitory where this letter should be delivered.
Input Specification:
The first line contains two integers $n$ and $m$ $(1 \le n, m \le 2 \cdot 10^{5})$ — the number of dormitories and the number of letters.
The second line contains a sequence $a_1, a_2, \dots, a_n$ $(1 \le a_i \le 10^{10})$, where $a_i$ equals to the number of rooms in the $i$-th dormitory. The third line contains a sequence $b_1, b_2, \dots, b_m$ $(1 \le b_j \le a_1 + a_2 + \dots + a_n)$, where $b_j$ equals to the room number (among all rooms of all dormitories) for the $j$-th letter. All $b_j$ are given in increasing order.
Output Specification:
Print $m$ lines. For each letter print two integers $f$ and $k$ — the dormitory number $f$ $(1 \le f \le n)$ and the room number $k$ in this dormitory $(1 \le k \le a_f)$ to deliver the letter.
Demo Input:
['3 6\n10 15 12\n1 9 12 23 26 37\n', '2 3\n5 10000000000\n5 6 9999999999\n']
Demo Output:
['1 1\n1 9\n2 2\n2 13\n3 1\n3 12\n', '1 5\n2 1\n2 9999999994\n']
Note:
In the first example letters should be delivered in the following order:
- the first letter in room $1$ of the first dormitory - the second letter in room $9$ of the first dormitory - the third letter in room $2$ of the second dormitory - the fourth letter in room $13$ of the second dormitory - the fifth letter in room $1$ of the third dormitory - the sixth letter in room $12$ of the third dormitory
|
```python
n,m=list(map(int,input().split()))
a=list(map(int,input().split()))
b=list(map(int,input().split()))
x=[]
for j in range(n):
for i in b:
if i>=1 and i<=a[j]:
print(j+1,i)
if i>a[j] and i<=a[j]+a[j+1]:
print(j+2,i-a[j])
if i>a[j]+a[j+1] and i<=a[j]+a[j+1]+a[j+2]:
print(j+3,i-a[j]-a[j+1])
break
```
| -1
|
|
137
|
B
|
Permutation
|
PROGRAMMING
| 1,000
|
[
"greedy"
] | null | null |
"Hey, it's homework time" — thought Polycarpus and of course he started with his favourite subject, IT. Polycarpus managed to solve all tasks but for the last one in 20 minutes. However, as he failed to solve the last task after some considerable time, the boy asked you to help him.
The sequence of *n* integers is called a permutation if it contains all integers from 1 to *n* exactly once.
You are given an arbitrary sequence *a*1,<=*a*2,<=...,<=*a**n* containing *n* integers. Each integer is not less than 1 and not greater than 5000. Determine what minimum number of elements Polycarpus needs to change to get a permutation (he should not delete or add numbers). In a single change he can modify any single sequence element (i. e. replace it with another integer).
|
The first line of the input data contains an integer *n* (1<=≤<=*n*<=≤<=5000) which represents how many numbers are in the sequence. The second line contains a sequence of integers *a**i* (1<=≤<=*a**i*<=≤<=5000,<=1<=≤<=*i*<=≤<=*n*).
|
Print the only number — the minimum number of changes needed to get the permutation.
|
[
"3\n3 1 2\n",
"2\n2 2\n",
"5\n5 3 3 3 1\n"
] |
[
"0\n",
"1\n",
"2\n"
] |
The first sample contains the permutation, which is why no replacements are required.
In the second sample it is enough to replace the first element with the number 1 and that will make the sequence the needed permutation.
In the third sample we can replace the second element with number 4 and the fourth element with number 2.
| 1,000
|
[
{
"input": "3\n3 1 2",
"output": "0"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "5\n5 3 3 3 1",
"output": "2"
},
{
"input": "5\n6 6 6 6 6",
"output": "5"
},
{
"input": "10\n1 1 2 2 8 8 7 7 9 9",
"output": "5"
},
{
"input": "8\n9 8 7 6 5 4 3 2",
"output": "1"
},
{
"input": "15\n1 2 3 4 5 5 4 3 2 1 1 2 3 4 5",
"output": "10"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n5000",
"output": "1"
},
{
"input": "4\n5000 5000 5000 5000",
"output": "4"
},
{
"input": "5\n3366 3461 4 5 4370",
"output": "3"
},
{
"input": "10\n8 2 10 3 4 6 1 7 9 5",
"output": "0"
},
{
"input": "10\n551 3192 3213 2846 3068 1224 3447 1 10 9",
"output": "7"
},
{
"input": "15\n4 1459 12 4281 3241 2748 10 3590 14 845 3518 1721 2 2880 1974",
"output": "10"
},
{
"input": "15\n15 1 8 2 13 11 12 7 3 14 6 10 9 4 5",
"output": "0"
},
{
"input": "15\n2436 2354 4259 1210 2037 2665 700 3578 2880 973 1317 1024 24 3621 4142",
"output": "15"
},
{
"input": "30\n28 1 3449 9 3242 4735 26 3472 15 21 2698 7 4073 3190 10 3 29 1301 4526 22 345 3876 19 12 4562 2535 2 630 18 27",
"output": "14"
},
{
"input": "100\n50 39 95 30 66 78 2169 4326 81 31 74 34 80 40 19 48 97 63 82 6 88 16 21 57 92 77 10 1213 17 93 32 91 38 4375 29 75 44 22 4 45 14 2395 3254 59 3379 2 85 96 8 83 27 94 1512 2960 100 9 73 79 7 25 55 69 90 99 51 87 98 62 18 35 43 4376 4668 28 72 56 4070 61 65 36 54 4106 11 24 15 86 70 71 4087 23 13 76 20 4694 26 4962 4726 37 14 64",
"output": "18"
},
{
"input": "100\n340 14 3275 2283 2673 1107 817 2243 1226 32 2382 3638 4652 418 68 4962 387 764 4647 159 1846 225 2760 4904 3150 403 3 2439 91 4428 92 4705 75 348 1566 1465 69 6 49 4 62 4643 564 1090 3447 1871 2255 139 24 99 2669 969 86 61 4550 158 4537 3993 1589 872 2907 1888 401 80 1825 1483 63 1 2264 4068 4113 2548 41 885 4806 36 67 167 4447 34 1248 2593 82 202 81 1783 1284 4973 16 43 95 7 865 2091 3008 1793 20 947 4912 3604",
"output": "70"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "2\n5000 5000",
"output": "2"
},
{
"input": "2\n1 2",
"output": "0"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "2\n2 3",
"output": "1"
},
{
"input": "2\n3 4",
"output": "2"
},
{
"input": "10\n1 2 3 4 5 6 7 1000 10 10",
"output": "2"
}
] | 1,481,271,474
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 48
| 124
| 409,600
|
n = int(input())
ans = n
A = set(map(int, input().split()))
for i in range(1, n + 1):
if i in A:
ans -= 1
print(ans)
|
Title: Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Hey, it's homework time" — thought Polycarpus and of course he started with his favourite subject, IT. Polycarpus managed to solve all tasks but for the last one in 20 minutes. However, as he failed to solve the last task after some considerable time, the boy asked you to help him.
The sequence of *n* integers is called a permutation if it contains all integers from 1 to *n* exactly once.
You are given an arbitrary sequence *a*1,<=*a*2,<=...,<=*a**n* containing *n* integers. Each integer is not less than 1 and not greater than 5000. Determine what minimum number of elements Polycarpus needs to change to get a permutation (he should not delete or add numbers). In a single change he can modify any single sequence element (i. e. replace it with another integer).
Input Specification:
The first line of the input data contains an integer *n* (1<=≤<=*n*<=≤<=5000) which represents how many numbers are in the sequence. The second line contains a sequence of integers *a**i* (1<=≤<=*a**i*<=≤<=5000,<=1<=≤<=*i*<=≤<=*n*).
Output Specification:
Print the only number — the minimum number of changes needed to get the permutation.
Demo Input:
['3\n3 1 2\n', '2\n2 2\n', '5\n5 3 3 3 1\n']
Demo Output:
['0\n', '1\n', '2\n']
Note:
The first sample contains the permutation, which is why no replacements are required.
In the second sample it is enough to replace the first element with the number 1 and that will make the sequence the needed permutation.
In the third sample we can replace the second element with number 4 and the fourth element with number 2.
|
```python
n = int(input())
ans = n
A = set(map(int, input().split()))
for i in range(1, n + 1):
if i in A:
ans -= 1
print(ans)
```
| 3
|
|
665
|
F
|
Four Divisors
|
PROGRAMMING
| 2,400
|
[
"data structures",
"dp",
"math",
"number theory",
"sortings",
"two pointers"
] | null | null |
If an integer *a* is divisible by another integer *b*, then *b* is called the divisor of *a*.
For example: 12 has positive 6 divisors. They are 1, 2, 3, 4, 6 and 12.
Let’s define a function *D*(*n*) — number of integers between 1 and *n* (inclusive) which has exactly four positive divisors.
Between 1 and 10 only the integers 6, 8 and 10 has exactly four positive divisors. So, *D*(10)<==<=3.
You are given an integer *n*. You have to calculate *D*(*n*).
|
The only line contains integer *n* (1<=≤<=*n*<=≤<=1011) — the parameter from the problem statement.
|
Print the only integer *c* — the number of integers between 1 and *n* with exactly four divisors.
|
[
"10\n",
"20\n"
] |
[
"3\n",
"5\n"
] |
none
| 0
|
[
{
"input": "10",
"output": "3"
},
{
"input": "20",
"output": "5"
},
{
"input": "1",
"output": "0"
},
{
"input": "27",
"output": "9"
},
{
"input": "100",
"output": "32"
},
{
"input": "1000",
"output": "292"
},
{
"input": "10000",
"output": "2608"
},
{
"input": "100000",
"output": "23327"
},
{
"input": "1000000",
"output": "209892"
},
{
"input": "100000000",
"output": "17426119"
},
{
"input": "1000000000",
"output": "160785303"
},
{
"input": "10000000000",
"output": "1493767176"
},
{
"input": "100000000000",
"output": "13959963675"
},
{
"input": "353964",
"output": "77787"
},
{
"input": "437388",
"output": "95228"
},
{
"input": "553516",
"output": "119256"
},
{
"input": "636940",
"output": "136364"
},
{
"input": "753068",
"output": "160058"
},
{
"input": "803788",
"output": "170304"
},
{
"input": "24403439",
"output": "4484483"
},
{
"input": "907519567",
"output": "146391993"
},
{
"input": "790635695",
"output": "128128652"
},
{
"input": "968719119",
"output": "155921638"
},
{
"input": "146802543",
"output": "25227297"
},
{
"input": "324885967",
"output": "54280184"
},
{
"input": "448485",
"output": "97564"
},
{
"input": "324885968",
"output": "54280184"
}
] | 1,464,125,371
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 46
| 4,915,200
|
import sys
def divi(n,k):
if k==1:
return 0
elif n%k==0:
return 1+divi(n,k-1)
else:
return 0+divi(n,k-1)
n=int(sys.stdin.readline().strip())
print(divi(n,n))
|
Title: Four Divisors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
If an integer *a* is divisible by another integer *b*, then *b* is called the divisor of *a*.
For example: 12 has positive 6 divisors. They are 1, 2, 3, 4, 6 and 12.
Let’s define a function *D*(*n*) — number of integers between 1 and *n* (inclusive) which has exactly four positive divisors.
Between 1 and 10 only the integers 6, 8 and 10 has exactly four positive divisors. So, *D*(10)<==<=3.
You are given an integer *n*. You have to calculate *D*(*n*).
Input Specification:
The only line contains integer *n* (1<=≤<=*n*<=≤<=1011) — the parameter from the problem statement.
Output Specification:
Print the only integer *c* — the number of integers between 1 and *n* with exactly four divisors.
Demo Input:
['10\n', '20\n']
Demo Output:
['3\n', '5\n']
Note:
none
|
```python
import sys
def divi(n,k):
if k==1:
return 0
elif n%k==0:
return 1+divi(n,k-1)
else:
return 0+divi(n,k-1)
n=int(sys.stdin.readline().strip())
print(divi(n,n))
```
| 0
|
|
615
|
D
|
Multipliers
|
PROGRAMMING
| 2,000
|
[
"math",
"number theory"
] | null | null |
Ayrat has number *n*, represented as it's prime factorization *p**i* of size *m*, i.e. *n*<==<=*p*1·*p*2·...·*p**m*. Ayrat got secret information that that the product of all divisors of *n* taken modulo 109<=+<=7 is the password to the secret data base. Now he wants to calculate this value.
|
The first line of the input contains a single integer *m* (1<=≤<=*m*<=≤<=200<=000) — the number of primes in factorization of *n*.
The second line contains *m* primes numbers *p**i* (2<=≤<=*p**i*<=≤<=200<=000).
|
Print one integer — the product of all divisors of *n* modulo 109<=+<=7.
|
[
"2\n2 3\n",
"3\n2 3 2\n"
] |
[
"36\n",
"1728\n"
] |
In the first sample *n* = 2·3 = 6. The divisors of 6 are 1, 2, 3 and 6, their product is equal to 1·2·3·6 = 36.
In the second sample 2·3·2 = 12. The divisors of 12 are 1, 2, 3, 4, 6 and 12. 1·2·3·4·6·12 = 1728.
| 2,000
|
[
{
"input": "2\n2 3",
"output": "36"
},
{
"input": "3\n2 3 2",
"output": "1728"
},
{
"input": "1\n2017",
"output": "2017"
},
{
"input": "2\n63997 63997",
"output": "135893224"
},
{
"input": "5\n11 7 11 7 11",
"output": "750455957"
},
{
"input": "5\n2 2 2 2 2",
"output": "32768"
},
{
"input": "4\n3 3 3 5",
"output": "332150625"
},
{
"input": "6\n101 103 107 109 101 103",
"output": "760029909"
},
{
"input": "10\n3 3 3 3 3 3 3 3 3 3",
"output": "555340537"
},
{
"input": "5\n7 5 2 3 13",
"output": "133580280"
},
{
"input": "23\n190979 191627 93263 72367 52561 188317 198397 24979 70313 105239 86263 78697 6163 7673 84137 199967 14657 84391 101009 16231 175103 24239 123289",
"output": "727083628"
},
{
"input": "7\n34429 104287 171293 101333 104287 34429 104287",
"output": "249330396"
},
{
"input": "27\n151153 29429 91411 91411 194507 194819 91411 91411 194507 181211 194507 131363 9371 194819 181211 194507 151153 91411 91411 192391 192391 151153 151153 194507 192391 192391 194819",
"output": "132073405"
},
{
"input": "47\n9041 60013 53609 82939 160861 123377 74383 74383 184039 19867 123377 101879 74383 193603 123377 115331 101879 53609 74383 115331 51869 51869 184039 193603 91297 160861 160861 115331 184039 51869 123377 74383 160861 74383 115331 115331 51869 74383 19867 193603 193603 115331 184039 9041 53609 53609 193603",
"output": "648634399"
},
{
"input": "67\n98929 19079 160079 181891 17599 91807 19079 98929 182233 92647 77477 98929 98639 182233 181891 182233 160079 98929 19079 98639 114941 98929 161341 91807 160079 22777 132361 92647 98929 77477 182233 103913 160079 77477 55711 77477 77477 182233 114941 91807 98929 19079 104393 182233 182233 131009 132361 16883 161341 103913 16883 98929 182233 114941 92647 92647 104393 132361 181891 114941 19079 91807 114941 132361 98639 161341 182233",
"output": "5987226"
},
{
"input": "44\n73 59 17 41 37 7 71 47 29 83 67 17 53 61 43 43 3 23 29 11 7 83 61 79 31 37 37 83 41 71 11 19 83 2 83 73 7 67 83 13 2 53 31 47",
"output": "464170294"
},
{
"input": "100\n2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541",
"output": "72902143"
},
{
"input": "130\n2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 5 5 5 5 5 5 5 5 5 5 7 7 7 7 7 7 7 7 7 7 11 11 11 11 11 11 11 11 11 11 13 13 13 13 13 13 13 13 13 13 17 17 17 17 17 17 17 17 17 17 19 19 19 19 19 19 19 19 19 19 23 23 23 23 23 23 23 23 23 23 29 29 29 29 29 29 29 29 29 29 31 31 31 31 31 31 31 31 31 31 37 37 37 37 37 37 37 37 37 37 41 41 41 41 41 41 41 41 41 41",
"output": "869075922"
},
{
"input": "101\n2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 2",
"output": "918713851"
},
{
"input": "42\n1657 1871 2423 3037 5023 5099 5449 5701 6361 6619 7393 7489 8179 9743 9791 9907 12289 12457 13063 13933 14947 16141 16829 16943 17191 17863 20161 20947 21661 22727 23197 23201 23813 24023 24181 24223 24391 26479 28619 30529 32441 32611",
"output": "468186759"
}
] | 1,687,873,518
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 20
| 124
| 18,841,600
|
M=10**9+7
def modpow(x, p, mod):
if p==0:
return 1
if p<0:
return modpow(x, (2-mod)*p, mod)
y=modpow(x, p//2, mod)
if p%2:
return (y*y*x)%mod
return (y*y)%mod
m=int(input())
R=2*10**5
c=[0]*(R+1)
n=1
for e in input().split():
c[int(e)]+=1
ppref=[1]*(R+1)
for i in range(1, R+1):
ppref[i]=ppref[i-1]*(c[i]+1)
ppref[i]%=M
psuf=[1]*(R+2)
for i in range(R, 0, -1):
psuf[i]=psuf[i+1]*(c[i]+1)
psuf[i]%=M
for i in range(1, R+1):
n*=modpow(i, (c[i]*(c[i]+1)//2)*ppref[i-1]*psuf[i+1], M)
n%=M
print(n)
|
Title: Multipliers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ayrat has number *n*, represented as it's prime factorization *p**i* of size *m*, i.e. *n*<==<=*p*1·*p*2·...·*p**m*. Ayrat got secret information that that the product of all divisors of *n* taken modulo 109<=+<=7 is the password to the secret data base. Now he wants to calculate this value.
Input Specification:
The first line of the input contains a single integer *m* (1<=≤<=*m*<=≤<=200<=000) — the number of primes in factorization of *n*.
The second line contains *m* primes numbers *p**i* (2<=≤<=*p**i*<=≤<=200<=000).
Output Specification:
Print one integer — the product of all divisors of *n* modulo 109<=+<=7.
Demo Input:
['2\n2 3\n', '3\n2 3 2\n']
Demo Output:
['36\n', '1728\n']
Note:
In the first sample *n* = 2·3 = 6. The divisors of 6 are 1, 2, 3 and 6, their product is equal to 1·2·3·6 = 36.
In the second sample 2·3·2 = 12. The divisors of 12 are 1, 2, 3, 4, 6 and 12. 1·2·3·4·6·12 = 1728.
|
```python
M=10**9+7
def modpow(x, p, mod):
if p==0:
return 1
if p<0:
return modpow(x, (2-mod)*p, mod)
y=modpow(x, p//2, mod)
if p%2:
return (y*y*x)%mod
return (y*y)%mod
m=int(input())
R=2*10**5
c=[0]*(R+1)
n=1
for e in input().split():
c[int(e)]+=1
ppref=[1]*(R+1)
for i in range(1, R+1):
ppref[i]=ppref[i-1]*(c[i]+1)
ppref[i]%=M
psuf=[1]*(R+2)
for i in range(R, 0, -1):
psuf[i]=psuf[i+1]*(c[i]+1)
psuf[i]%=M
for i in range(1, R+1):
n*=modpow(i, (c[i]*(c[i]+1)//2)*ppref[i-1]*psuf[i+1], M)
n%=M
print(n)
```
| 0
|
|
884
|
B
|
Japanese Crosswords Strike Back
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
A one-dimensional Japanese crossword can be represented as a binary string of length *x*. An encoding of this crossword is an array *a* of size *n*, where *n* is the number of segments formed completely of 1's, and *a**i* is the length of *i*-th segment. No two segments touch or intersect.
For example:
- If *x*<==<=6 and the crossword is 111011, then its encoding is an array {3,<=2}; - If *x*<==<=8 and the crossword is 01101010, then its encoding is an array {2,<=1,<=1}; - If *x*<==<=5 and the crossword is 11111, then its encoding is an array {5}; - If *x*<==<=5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
|
The first line contains two integer numbers *n* and *x* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*x*<=≤<=109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10000) — the encoding.
|
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
|
[
"2 4\n1 3\n",
"3 10\n3 3 2\n",
"2 10\n1 3\n"
] |
[
"NO\n",
"YES\n",
"NO\n"
] |
none
| 0
|
[
{
"input": "2 4\n1 3",
"output": "NO"
},
{
"input": "3 10\n3 3 2",
"output": "YES"
},
{
"input": "2 10\n1 3",
"output": "NO"
},
{
"input": "1 1\n1",
"output": "YES"
},
{
"input": "1 10\n10",
"output": "YES"
},
{
"input": "1 10000\n10000",
"output": "YES"
},
{
"input": "10 1\n5 78 3 87 4 9 5 8 9 1235",
"output": "NO"
},
{
"input": "3 12\n3 3 3",
"output": "NO"
},
{
"input": "3 9\n2 2 2",
"output": "NO"
},
{
"input": "2 5\n1 1",
"output": "NO"
},
{
"input": "1 2\n1",
"output": "NO"
},
{
"input": "3 13\n3 3 3",
"output": "NO"
},
{
"input": "3 6\n1 1 1",
"output": "NO"
},
{
"input": "1 6\n5",
"output": "NO"
},
{
"input": "3 11\n3 3 2",
"output": "NO"
},
{
"input": "2 6\n1 3",
"output": "NO"
},
{
"input": "3 10\n2 2 2",
"output": "NO"
},
{
"input": "3 8\n2 1 1",
"output": "NO"
},
{
"input": "1 5\n2",
"output": "NO"
},
{
"input": "1 3\n1",
"output": "NO"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "NO"
},
{
"input": "2 10\n4 4",
"output": "NO"
},
{
"input": "2 8\n2 3",
"output": "NO"
},
{
"input": "2 4\n1 1",
"output": "NO"
},
{
"input": "3 10\n1 2 4",
"output": "NO"
},
{
"input": "3 10\n2 1 3",
"output": "NO"
},
{
"input": "2 6\n1 2",
"output": "NO"
},
{
"input": "3 4\n1 1 1",
"output": "NO"
},
{
"input": "3 11\n1 2 4",
"output": "NO"
},
{
"input": "3 12\n3 3 2",
"output": "NO"
},
{
"input": "4 9\n1 1 1 1",
"output": "NO"
},
{
"input": "1 10\n9",
"output": "NO"
},
{
"input": "1 7\n5",
"output": "NO"
},
{
"input": "2 5\n1 2",
"output": "NO"
},
{
"input": "3 8\n1 1 2",
"output": "NO"
},
{
"input": "3 7\n1 1 1",
"output": "NO"
},
{
"input": "1 10\n1",
"output": "NO"
},
{
"input": "3 10\n2 3 4",
"output": "NO"
},
{
"input": "3 9\n1 2 3",
"output": "NO"
},
{
"input": "3 9\n3 3 2",
"output": "NO"
},
{
"input": "3 6\n3 3 2",
"output": "NO"
},
{
"input": "1 1\n3",
"output": "NO"
},
{
"input": "1 3\n2",
"output": "NO"
},
{
"input": "3 10\n3 3 3",
"output": "NO"
},
{
"input": "3 5\n1 1 1",
"output": "YES"
},
{
"input": "2 1\n100 100",
"output": "NO"
},
{
"input": "3 3\n3 3 3",
"output": "NO"
},
{
"input": "4 17\n3 3 9 1",
"output": "NO"
},
{
"input": "4 1660\n1505 13 37 100",
"output": "NO"
},
{
"input": "3 5\n3 3 2",
"output": "NO"
},
{
"input": "4 10\n3 3 2 5",
"output": "NO"
},
{
"input": "5 5\n5 5 5 5 5",
"output": "NO"
},
{
"input": "1 1\n2",
"output": "NO"
},
{
"input": "5 10\n1 2 2 4 5",
"output": "NO"
},
{
"input": "2 1\n1 1",
"output": "NO"
}
] | 1,592,486,641
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 66
| 171
| 9,830,400
|
n,m=[int(i) for i in input().split()]
a=[int(i) for i in input().split()]
d=0
for i in range(0,len(a)):
d=d+a[i]
d=d+(len(a)-1)
if(d==m):
print('YES')
else:
print("NO")
|
Title: Japanese Crosswords Strike Back
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A one-dimensional Japanese crossword can be represented as a binary string of length *x*. An encoding of this crossword is an array *a* of size *n*, where *n* is the number of segments formed completely of 1's, and *a**i* is the length of *i*-th segment. No two segments touch or intersect.
For example:
- If *x*<==<=6 and the crossword is 111011, then its encoding is an array {3,<=2}; - If *x*<==<=8 and the crossword is 01101010, then its encoding is an array {2,<=1,<=1}; - If *x*<==<=5 and the crossword is 11111, then its encoding is an array {5}; - If *x*<==<=5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
Input Specification:
The first line contains two integer numbers *n* and *x* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*x*<=≤<=109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10000) — the encoding.
Output Specification:
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
Demo Input:
['2 4\n1 3\n', '3 10\n3 3 2\n', '2 10\n1 3\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
none
|
```python
n,m=[int(i) for i in input().split()]
a=[int(i) for i in input().split()]
d=0
for i in range(0,len(a)):
d=d+a[i]
d=d+(len(a)-1)
if(d==m):
print('YES')
else:
print("NO")
```
| 3
|
|
29
|
A
|
Spit Problem
|
PROGRAMMING
| 1,000
|
[
"brute force"
] |
A. Spit Problem
|
2
|
256
|
In a Berland's zoo there is an enclosure with camels. It is known that camels like to spit. Bob watched these interesting animals for the whole day and registered in his notepad where each animal spitted. Now he wants to know if in the zoo there are two camels, which spitted at each other. Help him to solve this task.
The trajectory of a camel's spit is an arc, i.e. if the camel in position *x* spits *d* meters right, he can hit only the camel in position *x*<=+<=*d*, if such a camel exists.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the amount of camels in the zoo. Each of the following *n* lines contains two integers *x**i* and *d**i* (<=-<=104<=≤<=*x**i*<=≤<=104,<=1<=≤<=|*d**i*|<=≤<=2·104) — records in Bob's notepad. *x**i* is a position of the *i*-th camel, and *d**i* is a distance at which the *i*-th camel spitted. Positive values of *d**i* correspond to the spits right, negative values correspond to the spits left. No two camels may stand in the same position.
|
If there are two camels, which spitted at each other, output YES. Otherwise, output NO.
|
[
"2\n0 1\n1 -1\n",
"3\n0 1\n1 1\n2 -2\n",
"5\n2 -10\n3 10\n0 5\n5 -5\n10 1\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "2\n0 1\n1 -1",
"output": "YES"
},
{
"input": "3\n0 1\n1 1\n2 -2",
"output": "NO"
},
{
"input": "5\n2 -10\n3 10\n0 5\n5 -5\n10 1",
"output": "YES"
},
{
"input": "10\n-9897 -1144\n-4230 -6350\n2116 -3551\n-3635 4993\n3907 -9071\n-2362 4120\n-6542 984\n5807 3745\n7594 7675\n-5412 -6872",
"output": "NO"
},
{
"input": "11\n-1536 3809\n-2406 -8438\n-1866 395\n5636 -490\n-6867 -7030\n7525 3575\n-6796 2908\n3884 4629\n-2862 -6122\n-8984 6122\n7137 -326",
"output": "YES"
},
{
"input": "12\n-9765 1132\n-1382 -215\n-9405 7284\n-2040 3947\n-9360 3150\n6425 9386\n806 -2278\n-2121 -7284\n5663 -1608\n-8377 9297\n6245 708\n8470 6024",
"output": "YES"
},
{
"input": "15\n8122 -9991\n-4068 -3386\n8971 3731\n3458 5161\n-8700 7562\n2691 8735\n-1510 -3892\n5183 -3753\n-7018 6637\n-7454 3386\n-818 -6377\n6771 -8647\n-7357 -1246\n-6186 1922\n9889 -3627",
"output": "YES"
},
{
"input": "20\n-5264 6424\n-3664 -7459\n-2780 -9859\n-3317 6842\n5681 -8092\n1555 1904\n-6684 1414\n6593 -1253\n-5708 -1202\n335 1733\n-926 7579\n3459 -1904\n-4486 4006\n6201 3616\n2847 -5255\n8438 7057\n8171 6042\n-9102 3545\n7731 -233\n6264 6563",
"output": "YES"
},
{
"input": "30\n-398 -1774\n313 -6974\n2346 -4657\n8552 -9647\n-5265 1538\n8195 4864\n-5641 -5219\n-1394 8563\n-1190 1992\n-4669 -1156\n7574 256\n9206 -2414\n4140 -549\n-294 2169\n7029 -2871\n3808 -9799\n3141 5690\n4648 -2680\n-5990 9800\n-2299 1697\n6077 -7177\n-400 -9724\n-4644 -2392\n-2198 -9531\n-2105 9386\n-8165 -4201\n-1589 -7916\n2518 -7840\n4173 -6949\n-3368 -9943",
"output": "NO"
}
] | 1,651,255,596
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 122
| 102,400
|
from collections import Counter
t = int(input())
p = Counter()
ans = "NO"
while t > 0:
x, d = map(int,input().split())
if p[x + d,x] == 1 :
ans = "YES"
else:
p[x,x + d] = 1
t -= 1
print(ans)
|
Title: Spit Problem
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
In a Berland's zoo there is an enclosure with camels. It is known that camels like to spit. Bob watched these interesting animals for the whole day and registered in his notepad where each animal spitted. Now he wants to know if in the zoo there are two camels, which spitted at each other. Help him to solve this task.
The trajectory of a camel's spit is an arc, i.e. if the camel in position *x* spits *d* meters right, he can hit only the camel in position *x*<=+<=*d*, if such a camel exists.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the amount of camels in the zoo. Each of the following *n* lines contains two integers *x**i* and *d**i* (<=-<=104<=≤<=*x**i*<=≤<=104,<=1<=≤<=|*d**i*|<=≤<=2·104) — records in Bob's notepad. *x**i* is a position of the *i*-th camel, and *d**i* is a distance at which the *i*-th camel spitted. Positive values of *d**i* correspond to the spits right, negative values correspond to the spits left. No two camels may stand in the same position.
Output Specification:
If there are two camels, which spitted at each other, output YES. Otherwise, output NO.
Demo Input:
['2\n0 1\n1 -1\n', '3\n0 1\n1 1\n2 -2\n', '5\n2 -10\n3 10\n0 5\n5 -5\n10 1\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none
|
```python
from collections import Counter
t = int(input())
p = Counter()
ans = "NO"
while t > 0:
x, d = map(int,input().split())
if p[x + d,x] == 1 :
ans = "YES"
else:
p[x,x + d] = 1
t -= 1
print(ans)
```
| 3.969309
|
995
|
D
|
Game
|
PROGRAMMING
| 2,500
|
[
"math"
] | null | null |
Allen and Bessie are playing a simple number game. They both know a function $f: \{0, 1\}^n \to \mathbb{R}$, i. e. the function takes $n$ binary arguments and returns a real value. At the start of the game, the variables $x_1, x_2, \dots, x_n$ are all set to $-1$. Each round, with equal probability, one of Allen or Bessie gets to make a move. A move consists of picking an $i$ such that $x_i = -1$ and either setting $x_i \to 0$ or $x_i \to 1$.
After $n$ rounds all variables are set, and the game value resolves to $f(x_1, x_2, \dots, x_n)$. Allen wants to maximize the game value, and Bessie wants to minimize it.
Your goal is to help Allen and Bessie find the expected game value! They will play $r+1$ times though, so between each game, exactly one value of $f$ changes. In other words, between rounds $i$ and $i+1$ for $1 \le i \le r$, $f(z_1, \dots, z_n) \to g_i$ for some $(z_1, \dots, z_n) \in \{0, 1\}^n$. You are to find the expected game value in the beginning and after each change.
|
The first line contains two integers $n$ and $r$ ($1 \le n \le 18$, $0 \le r \le 2^{18}$).
The next line contains $2^n$ integers $c_0, c_1, \dots, c_{2^n-1}$ ($0 \le c_i \le 10^9$), denoting the initial values of $f$. More specifically, $f(x_0, x_1, \dots, x_{n-1}) = c_x$, if $x = \overline{x_{n-1} \ldots x_0}$ in binary.
Each of the next $r$ lines contains two integers $z$ and $g$ ($0 \le z \le 2^n - 1$, $0 \le g \le 10^9$). If $z = \overline{z_{n-1} \dots z_0}$ in binary, then this means to set $f(z_0, \dots, z_{n-1}) \to g$.
|
Print $r+1$ lines, the $i$-th of which denotes the value of the game $f$ during the $i$-th round. Your answer must have absolute or relative error within $10^{-6}$.
Formally, let your answer be $a$, and the jury's answer be $b$. Your answer is considered correct if $\frac{|a - b|}{\max{(1, |b|)}} \le 10^{-6}$.
|
[
"2 2\n0 1 2 3\n2 5\n0 4\n",
"1 0\n2 3\n",
"2 0\n1 1 1 1\n"
] |
[
"1.500000\n2.250000\n3.250000\n",
"2.500000\n",
"1.000000\n"
] |
Consider the second test case. If Allen goes first, he will set $x_1 \to 1$, so the final value will be $3$. If Bessie goes first, then she will set $x_1 \to 0$ so the final value will be $2$. Thus the answer is $2.5$.
In the third test case, the game value will always be $1$ regardless of Allen and Bessie's play.
| 1,750
|
[
{
"input": "2 2\n0 1 2 3\n2 5\n0 4",
"output": "1.500000\n2.250000\n3.250000"
},
{
"input": "1 0\n2 3",
"output": "2.500000"
},
{
"input": "2 0\n1 1 1 1",
"output": "1.000000"
}
] | 1,689,344,541
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
print("_RANDOM_GUESS_1689344541.728262")# 1689344541.728322
|
Title: Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Allen and Bessie are playing a simple number game. They both know a function $f: \{0, 1\}^n \to \mathbb{R}$, i. e. the function takes $n$ binary arguments and returns a real value. At the start of the game, the variables $x_1, x_2, \dots, x_n$ are all set to $-1$. Each round, with equal probability, one of Allen or Bessie gets to make a move. A move consists of picking an $i$ such that $x_i = -1$ and either setting $x_i \to 0$ or $x_i \to 1$.
After $n$ rounds all variables are set, and the game value resolves to $f(x_1, x_2, \dots, x_n)$. Allen wants to maximize the game value, and Bessie wants to minimize it.
Your goal is to help Allen and Bessie find the expected game value! They will play $r+1$ times though, so between each game, exactly one value of $f$ changes. In other words, between rounds $i$ and $i+1$ for $1 \le i \le r$, $f(z_1, \dots, z_n) \to g_i$ for some $(z_1, \dots, z_n) \in \{0, 1\}^n$. You are to find the expected game value in the beginning and after each change.
Input Specification:
The first line contains two integers $n$ and $r$ ($1 \le n \le 18$, $0 \le r \le 2^{18}$).
The next line contains $2^n$ integers $c_0, c_1, \dots, c_{2^n-1}$ ($0 \le c_i \le 10^9$), denoting the initial values of $f$. More specifically, $f(x_0, x_1, \dots, x_{n-1}) = c_x$, if $x = \overline{x_{n-1} \ldots x_0}$ in binary.
Each of the next $r$ lines contains two integers $z$ and $g$ ($0 \le z \le 2^n - 1$, $0 \le g \le 10^9$). If $z = \overline{z_{n-1} \dots z_0}$ in binary, then this means to set $f(z_0, \dots, z_{n-1}) \to g$.
Output Specification:
Print $r+1$ lines, the $i$-th of which denotes the value of the game $f$ during the $i$-th round. Your answer must have absolute or relative error within $10^{-6}$.
Formally, let your answer be $a$, and the jury's answer be $b$. Your answer is considered correct if $\frac{|a - b|}{\max{(1, |b|)}} \le 10^{-6}$.
Demo Input:
['2 2\n0 1 2 3\n2 5\n0 4\n', '1 0\n2 3\n', '2 0\n1 1 1 1\n']
Demo Output:
['1.500000\n2.250000\n3.250000\n', '2.500000\n', '1.000000\n']
Note:
Consider the second test case. If Allen goes first, he will set $x_1 \to 1$, so the final value will be $3$. If Bessie goes first, then she will set $x_1 \to 0$ so the final value will be $2$. Thus the answer is $2.5$.
In the third test case, the game value will always be $1$ regardless of Allen and Bessie's play.
|
```python
print("_RANDOM_GUESS_1689344541.728262")# 1689344541.728322
```
| 0
|
|
573
|
A
|
Bear and Poker
|
PROGRAMMING
| 1,300
|
[
"implementation",
"math",
"number theory"
] | null | null |
Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are *n* players (including Limak himself) and right now all of them have bids on the table. *i*-th of them has bid with size *a**i* dollars.
Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot?
|
First line of input contains an integer *n* (2<=≤<=*n*<=≤<=105), the number of players.
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the bids of players.
|
Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise.
|
[
"4\n75 150 75 50\n",
"3\n100 150 250\n"
] |
[
"Yes\n",
"No\n"
] |
In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid.
It can be shown that in the second sample test there is no way to make all bids equal.
| 500
|
[
{
"input": "4\n75 150 75 50",
"output": "Yes"
},
{
"input": "3\n100 150 250",
"output": "No"
},
{
"input": "7\n34 34 68 34 34 68 34",
"output": "Yes"
},
{
"input": "10\n72 96 12 18 81 20 6 2 54 1",
"output": "No"
},
{
"input": "20\n958692492 954966768 77387000 724664764 101294996 614007760 202904092 555293973 707655552 108023967 73123445 612562357 552908390 914853758 915004122 466129205 122853497 814592742 373389439 818473058",
"output": "No"
},
{
"input": "2\n1 1",
"output": "Yes"
},
{
"input": "2\n72 72",
"output": "Yes"
},
{
"input": "2\n49 42",
"output": "No"
},
{
"input": "3\n1000000000 1000000000 1000000000",
"output": "Yes"
},
{
"input": "6\n162000 96000 648000 1000 864000 432000",
"output": "Yes"
},
{
"input": "8\n600000 100000 100000 100000 900000 600000 900000 600000",
"output": "Yes"
},
{
"input": "12\n2048 1024 6144 1024 3072 3072 6144 1024 4096 2048 6144 3072",
"output": "Yes"
},
{
"input": "20\n246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246 246",
"output": "Yes"
},
{
"input": "50\n840868705 387420489 387420489 795385082 634350497 206851546 536870912 536870912 414927754 387420489 387420489 536870912 387420489 149011306 373106005 536870912 700746206 387420489 777952883 847215247 176645254 576664386 387420489 230876513 536870912 536870912 536870912 387420489 387420489 536870912 460495524 528643722 387420489 536870912 470369206 899619085 387420489 631148352 387420489 387420489 536870912 414666674 521349938 776784669 387420489 102428009 536870912 387420489 536870912 718311009",
"output": "No"
},
{
"input": "2\n5 6",
"output": "No"
},
{
"input": "3\n536870912 387420489 257407169",
"output": "No"
},
{
"input": "4\n2 2 5 2",
"output": "No"
},
{
"input": "2\n33554432 59049",
"output": "Yes"
},
{
"input": "3\n536870912 387420489 387420489",
"output": "Yes"
},
{
"input": "2\n1 5",
"output": "No"
},
{
"input": "18\n2 3 5 7 11 13 17 19 23 29 31 37 43 47 53 59 67 71",
"output": "No"
},
{
"input": "2\n1 30",
"output": "No"
},
{
"input": "3\n335544320 71744535 71744535",
"output": "Yes"
},
{
"input": "5\n1000000000 999999999 999999998 999999997 999999996",
"output": "No"
},
{
"input": "2\n25 5",
"output": "No"
},
{
"input": "4\n75 150 75 5",
"output": "No"
},
{
"input": "3\n536870912 387420489 362797056",
"output": "Yes"
},
{
"input": "3\n536870912 387420489 89",
"output": "No"
},
{
"input": "4\n547 2606459 222763549 143466789",
"output": "No"
},
{
"input": "3\n129140163 33554432 1",
"output": "Yes"
},
{
"input": "10\n244140625 244140625 244140625 244140625 244140625 244140625 244140625 244140625 536870912 387420489",
"output": "No"
},
{
"input": "3\n5 5 1",
"output": "No"
},
{
"input": "5\n3 7 29 36760123 823996703",
"output": "No"
}
] | 1,441,397,912
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 70
| 1,606
| 9,113,600
|
n = int(input())
s = list(map(int, input().split()))
for i in range(n):
flag = True
while flag:
if s[i] % 2 == 0:
s[i] = s[i] // 2
if s[i] % 3 == 0:
s[i] = s[i] // 3
if (s[i] % 2 != 0) and (s[i] % 3 != 0):
flag = False
flag = True
if n > 1:
for i in range(1, n):
if s[i] != s[i - 1]:
flag = False
if flag:
print('YES')
else:
print('NO')
else:
print('YES')
|
Title: Bear and Poker
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are *n* players (including Limak himself) and right now all of them have bids on the table. *i*-th of them has bid with size *a**i* dollars.
Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot?
Input Specification:
First line of input contains an integer *n* (2<=≤<=*n*<=≤<=105), the number of players.
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the bids of players.
Output Specification:
Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise.
Demo Input:
['4\n75 150 75 50\n', '3\n100 150 250\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid.
It can be shown that in the second sample test there is no way to make all bids equal.
|
```python
n = int(input())
s = list(map(int, input().split()))
for i in range(n):
flag = True
while flag:
if s[i] % 2 == 0:
s[i] = s[i] // 2
if s[i] % 3 == 0:
s[i] = s[i] // 3
if (s[i] % 2 != 0) and (s[i] % 3 != 0):
flag = False
flag = True
if n > 1:
for i in range(1, n):
if s[i] != s[i - 1]:
flag = False
if flag:
print('YES')
else:
print('NO')
else:
print('YES')
```
| 3
|
|
466
|
A
|
Cheap Travel
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
Ann has recently started commuting by subway. We know that a one ride subway ticket costs *a* rubles. Besides, Ann found out that she can buy a special ticket for *m* rides (she can buy it several times). It costs *b* rubles. Ann did the math; she will need to use subway *n* times. Help Ann, tell her what is the minimum sum of money she will have to spend to make *n* rides?
|
The single line contains four space-separated integers *n*, *m*, *a*, *b* (1<=≤<=*n*,<=*m*,<=*a*,<=*b*<=≤<=1000) — the number of rides Ann has planned, the number of rides covered by the *m* ride ticket, the price of a one ride ticket and the price of an *m* ride ticket.
|
Print a single integer — the minimum sum in rubles that Ann will need to spend.
|
[
"6 2 1 2\n",
"5 2 2 3\n"
] |
[
"6\n",
"8\n"
] |
In the first sample one of the optimal solutions is: each time buy a one ride ticket. There are other optimal solutions. For example, buy three *m* ride tickets.
| 500
|
[
{
"input": "6 2 1 2",
"output": "6"
},
{
"input": "5 2 2 3",
"output": "8"
},
{
"input": "10 3 5 1",
"output": "4"
},
{
"input": "1000 1 1000 1000",
"output": "1000000"
},
{
"input": "1000 3 1000 1000",
"output": "334000"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "10 2 1 1",
"output": "5"
},
{
"input": "1 1000 1 2",
"output": "1"
},
{
"input": "1 1000 3 2",
"output": "2"
},
{
"input": "10 3 1 2",
"output": "7"
},
{
"input": "995 1 2 1",
"output": "995"
},
{
"input": "556 2 16 15",
"output": "4170"
},
{
"input": "477 2 16 14",
"output": "3346"
},
{
"input": "101 110 1 100",
"output": "100"
},
{
"input": "9 3 3 10",
"output": "27"
},
{
"input": "100 8 10 1",
"output": "13"
},
{
"input": "6 4 1 3",
"output": "5"
},
{
"input": "8 5 2 8",
"output": "14"
},
{
"input": "1000 2 1 1000",
"output": "1000"
}
] | 1,690,141,468
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 14
| 77
| 0
|
inpu = input().split()
nrides = int(inpu[0])
mrides = int(inpu[1])
oneRidCost = int(inpu[2])
mRideCost = int(inpu[3])
cost = 0
while nrides > 0:
if (mRideCost < nrides * oneRidCost):
cost += mRideCost
nrides -= mrides
else:
i = 0
while i < mrides and nrides > 0:
cost += oneRidCost
nrides -= 1
i += 1
print(cost)
|
Title: Cheap Travel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ann has recently started commuting by subway. We know that a one ride subway ticket costs *a* rubles. Besides, Ann found out that she can buy a special ticket for *m* rides (she can buy it several times). It costs *b* rubles. Ann did the math; she will need to use subway *n* times. Help Ann, tell her what is the minimum sum of money she will have to spend to make *n* rides?
Input Specification:
The single line contains four space-separated integers *n*, *m*, *a*, *b* (1<=≤<=*n*,<=*m*,<=*a*,<=*b*<=≤<=1000) — the number of rides Ann has planned, the number of rides covered by the *m* ride ticket, the price of a one ride ticket and the price of an *m* ride ticket.
Output Specification:
Print a single integer — the minimum sum in rubles that Ann will need to spend.
Demo Input:
['6 2 1 2\n', '5 2 2 3\n']
Demo Output:
['6\n', '8\n']
Note:
In the first sample one of the optimal solutions is: each time buy a one ride ticket. There are other optimal solutions. For example, buy three *m* ride tickets.
|
```python
inpu = input().split()
nrides = int(inpu[0])
mrides = int(inpu[1])
oneRidCost = int(inpu[2])
mRideCost = int(inpu[3])
cost = 0
while nrides > 0:
if (mRideCost < nrides * oneRidCost):
cost += mRideCost
nrides -= mrides
else:
i = 0
while i < mrides and nrides > 0:
cost += oneRidCost
nrides -= 1
i += 1
print(cost)
```
| 0
|
|
139
|
A
|
Petr and Book
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book.
|
The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero.
|
Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
|
[
"100\n15 20 20 15 10 30 45\n",
"2\n1 0 0 0 0 0 0\n"
] |
[
"6\n",
"1\n"
] |
Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book.
| 500
|
[
{
"input": "100\n15 20 20 15 10 30 45",
"output": "6"
},
{
"input": "2\n1 0 0 0 0 0 0",
"output": "1"
},
{
"input": "100\n100 200 100 200 300 400 500",
"output": "1"
},
{
"input": "3\n1 1 1 1 1 1 1",
"output": "3"
},
{
"input": "1\n1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "20\n5 3 7 2 1 6 4",
"output": "6"
},
{
"input": "10\n5 1 1 1 1 1 5",
"output": "6"
},
{
"input": "50\n10 1 10 1 10 1 10",
"output": "1"
},
{
"input": "77\n11 11 11 11 11 11 10",
"output": "1"
},
{
"input": "1\n1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "1000\n100 100 100 100 100 100 100",
"output": "3"
},
{
"input": "999\n10 20 10 20 30 20 10",
"output": "3"
},
{
"input": "433\n109 58 77 10 39 125 15",
"output": "7"
},
{
"input": "1\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "5\n1 0 1 0 1 0 1",
"output": "1"
},
{
"input": "997\n1 1 0 0 1 0 1",
"output": "1"
},
{
"input": "1000\n1 1 1 1 1 1 1",
"output": "6"
},
{
"input": "1000\n1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "1000\n1 0 0 0 0 0 0",
"output": "1"
},
{
"input": "1000\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "1000\n1 0 0 1 0 0 1",
"output": "1"
},
{
"input": "509\n105 23 98 0 7 0 155",
"output": "2"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "7"
},
{
"input": "2\n1 1 0 0 0 0 0",
"output": "2"
},
{
"input": "1\n0 0 0 0 0 1 0",
"output": "6"
},
{
"input": "10\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "5\n0 0 0 0 0 6 0",
"output": "6"
},
{
"input": "3\n0 1 0 0 0 0 0",
"output": "2"
},
{
"input": "10\n0 0 0 0 0 0 10",
"output": "7"
},
{
"input": "28\n1 2 3 4 5 6 7",
"output": "7"
},
{
"input": "100\n5 5 5 5 5 5 5",
"output": "6"
},
{
"input": "4\n1 0 0 0 0 0 1",
"output": "7"
},
{
"input": "2\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "7\n0 0 0 0 0 0 7",
"output": "7"
},
{
"input": "7\n2 1 1 1 1 1 0",
"output": "6"
},
{
"input": "2\n0 0 1 1 0 0 0",
"output": "4"
},
{
"input": "6\n1 1 1 1 1 1 0",
"output": "6"
},
{
"input": "5\n1 1 1 0 0 1 1",
"output": "7"
},
{
"input": "100\n10 20 30 10 10 10 10",
"output": "7"
},
{
"input": "1\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "70\n10 10 10 10 10 10 10",
"output": "7"
},
{
"input": "22\n1 2 3 4 5 6 10",
"output": "7"
},
{
"input": "5\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "2\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "6\n1 0 0 0 0 0 2",
"output": "7"
},
{
"input": "10\n1 2 2 1 2 1 1",
"output": "7"
},
{
"input": "5\n0 0 0 0 0 0 10",
"output": "7"
},
{
"input": "4\n0 1 1 0 0 0 0",
"output": "3"
},
{
"input": "100\n0 0 0 0 0 1 0",
"output": "6"
}
] | 1,693,800,273
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 62
| 0
|
n = int(input())
l = list(map(int,input().split()))
s = sum(l)
if(n<s):
p = 0
e = 0
for i,j in enumerate(l):
p+=j
e = i
if(p>=n):
break
print(e+1)
if(n>s):
r = n%s
week = 0
while(r>0):
r -= l[week]
week = (week+1)%7
print(week+1)
if(n==s):
print(7)
|
Title: Petr and Book
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book.
Input Specification:
The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero.
Output Specification:
Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
Demo Input:
['100\n15 20 20 15 10 30 45\n', '2\n1 0 0 0 0 0 0\n']
Demo Output:
['6\n', '1\n']
Note:
Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book.
|
```python
n = int(input())
l = list(map(int,input().split()))
s = sum(l)
if(n<s):
p = 0
e = 0
for i,j in enumerate(l):
p+=j
e = i
if(p>=n):
break
print(e+1)
if(n>s):
r = n%s
week = 0
while(r>0):
r -= l[week]
week = (week+1)%7
print(week+1)
if(n==s):
print(7)
```
| 0
|
|
959
|
B
|
Mahmoud and Ehab and the message
|
PROGRAMMING
| 1,200
|
[
"dsu",
"greedy",
"implementation"
] | null | null |
Mahmoud wants to send a message to his friend Ehab. Their language consists of *n* words numbered from 1 to *n*. Some words have the same meaning so there are *k* groups of words such that all the words in some group have the same meaning.
Mahmoud knows that the *i*-th word can be sent with cost *a**i*. For each word in his message, Mahmoud can either replace it with another word of the same meaning or leave it as it is. Can you help Mahmoud determine the minimum cost of sending the message?
The cost of sending the message is the sum of the costs of sending every word in it.
|
The first line of input contains integers *n*, *k* and *m* (1<=≤<=*k*<=≤<=*n*<=≤<=105,<=1<=≤<=*m*<=≤<=105) — the number of words in their language, the number of groups of words, and the number of words in Mahmoud's message respectively.
The second line contains *n* strings consisting of lowercase English letters of length not exceeding 20 which represent the words. It's guaranteed that the words are distinct.
The third line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109) where *a**i* is the cost of sending the *i*-th word.
The next *k* lines describe the groups of words of same meaning. The next *k* lines each start with an integer *x* (1<=≤<=*x*<=≤<=*n*) which means that there are *x* words in this group, followed by *x* integers which represent the indices of words in this group. It's guaranteed that each word appears in exactly one group.
The next line contains *m* space-separated words which represent Mahmoud's message. Each of these words appears in the list of language's words.
|
The only line should contain the minimum cost to send the message after replacing some words (maybe none) with some words of the same meaning.
|
[
"5 4 4\ni loser am the second\n100 1 1 5 10\n1 1\n1 3\n2 2 5\n1 4\ni am the second\n",
"5 4 4\ni loser am the second\n100 20 1 5 10\n1 1\n1 3\n2 2 5\n1 4\ni am the second\n"
] |
[
"107",
"116"
] |
In the first sample, Mahmoud should replace the word "second" with the word "loser" because it has less cost so the cost will be 100+1+5+1=107.
In the second sample, Mahmoud shouldn't do any replacement so the cost will be 100+1+5+10=116.
| 1,000
|
[
{
"input": "5 4 4\ni loser am the second\n100 1 1 5 10\n1 1\n1 3\n2 2 5\n1 4\ni am the second",
"output": "107"
},
{
"input": "5 4 4\ni loser am the second\n100 20 1 5 10\n1 1\n1 3\n2 2 5\n1 4\ni am the second",
"output": "116"
},
{
"input": "1 1 1\na\n1000000000\n1 1\na",
"output": "1000000000"
},
{
"input": "1 1 10\na\n1000000000\n1 1\na a a a a a a a a a",
"output": "10000000000"
}
] | 1,661,330,123
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 22
| 967
| 67,788,800
|
# Problem Link: https://codeforces.com/problemset/problem/959/B
# Problem Status:
# -------------------------- SEPARATOR --------------------------
# IDEA:
# -------------------------- SEPARATOR --------------------------
N, K, M = list(map(int, input().split()))
Lang = input().split()
Values = list(map(int, input().split()))
Table = dict()
Groups = []
GroupsPointers = dict()
Answer = 0
# -------------------------- SEPARATOR --------------------------
for i in range(N):
Table.update({Lang[i]: Values[i]})
for i in range(K):
Text = list(map(int, input().split()))
Groups.append([])
for j in range(1, Text[0]+1):
Groups[-1].append(Table.get(Lang[Text[j]-1]))
GroupsPointers.update({Lang[Text[j]-1]: len(Groups)-1})
Groups[-1].sort()
Message = input().split()
# -------------------------- SEPARATOR --------------------------
for i in range(M):
Answer += Groups[GroupsPointers.get(str(Message[i]))][0]
# -------------------------- SEPARATOR --------------------------
print(Answer)
# -------------------------- SEPARATOR --------------------------
|
Title: Mahmoud and Ehab and the message
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mahmoud wants to send a message to his friend Ehab. Their language consists of *n* words numbered from 1 to *n*. Some words have the same meaning so there are *k* groups of words such that all the words in some group have the same meaning.
Mahmoud knows that the *i*-th word can be sent with cost *a**i*. For each word in his message, Mahmoud can either replace it with another word of the same meaning or leave it as it is. Can you help Mahmoud determine the minimum cost of sending the message?
The cost of sending the message is the sum of the costs of sending every word in it.
Input Specification:
The first line of input contains integers *n*, *k* and *m* (1<=≤<=*k*<=≤<=*n*<=≤<=105,<=1<=≤<=*m*<=≤<=105) — the number of words in their language, the number of groups of words, and the number of words in Mahmoud's message respectively.
The second line contains *n* strings consisting of lowercase English letters of length not exceeding 20 which represent the words. It's guaranteed that the words are distinct.
The third line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109) where *a**i* is the cost of sending the *i*-th word.
The next *k* lines describe the groups of words of same meaning. The next *k* lines each start with an integer *x* (1<=≤<=*x*<=≤<=*n*) which means that there are *x* words in this group, followed by *x* integers which represent the indices of words in this group. It's guaranteed that each word appears in exactly one group.
The next line contains *m* space-separated words which represent Mahmoud's message. Each of these words appears in the list of language's words.
Output Specification:
The only line should contain the minimum cost to send the message after replacing some words (maybe none) with some words of the same meaning.
Demo Input:
['5 4 4\ni loser am the second\n100 1 1 5 10\n1 1\n1 3\n2 2 5\n1 4\ni am the second\n', '5 4 4\ni loser am the second\n100 20 1 5 10\n1 1\n1 3\n2 2 5\n1 4\ni am the second\n']
Demo Output:
['107', '116']
Note:
In the first sample, Mahmoud should replace the word "second" with the word "loser" because it has less cost so the cost will be 100+1+5+1=107.
In the second sample, Mahmoud shouldn't do any replacement so the cost will be 100+1+5+10=116.
|
```python
# Problem Link: https://codeforces.com/problemset/problem/959/B
# Problem Status:
# -------------------------- SEPARATOR --------------------------
# IDEA:
# -------------------------- SEPARATOR --------------------------
N, K, M = list(map(int, input().split()))
Lang = input().split()
Values = list(map(int, input().split()))
Table = dict()
Groups = []
GroupsPointers = dict()
Answer = 0
# -------------------------- SEPARATOR --------------------------
for i in range(N):
Table.update({Lang[i]: Values[i]})
for i in range(K):
Text = list(map(int, input().split()))
Groups.append([])
for j in range(1, Text[0]+1):
Groups[-1].append(Table.get(Lang[Text[j]-1]))
GroupsPointers.update({Lang[Text[j]-1]: len(Groups)-1})
Groups[-1].sort()
Message = input().split()
# -------------------------- SEPARATOR --------------------------
for i in range(M):
Answer += Groups[GroupsPointers.get(str(Message[i]))][0]
# -------------------------- SEPARATOR --------------------------
print(Answer)
# -------------------------- SEPARATOR --------------------------
```
| 3
|
|
950
|
A
|
Left-handers, Right-handers and Ambidexters
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand.
The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands.
Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand.
Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively.
|
The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training.
|
Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players.
|
[
"1 4 2\n",
"5 5 5\n",
"0 2 0\n"
] |
[
"6\n",
"14\n",
"0\n"
] |
In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team.
In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand.
| 500
|
[
{
"input": "1 4 2",
"output": "6"
},
{
"input": "5 5 5",
"output": "14"
},
{
"input": "0 2 0",
"output": "0"
},
{
"input": "30 70 34",
"output": "128"
},
{
"input": "89 32 24",
"output": "112"
},
{
"input": "89 44 77",
"output": "210"
},
{
"input": "0 0 0",
"output": "0"
},
{
"input": "100 100 100",
"output": "300"
},
{
"input": "1 1 1",
"output": "2"
},
{
"input": "30 70 35",
"output": "130"
},
{
"input": "89 44 76",
"output": "208"
},
{
"input": "0 100 100",
"output": "200"
},
{
"input": "100 0 100",
"output": "200"
},
{
"input": "100 1 100",
"output": "200"
},
{
"input": "1 100 100",
"output": "200"
},
{
"input": "100 100 0",
"output": "200"
},
{
"input": "100 100 1",
"output": "200"
},
{
"input": "1 2 1",
"output": "4"
},
{
"input": "0 0 100",
"output": "100"
},
{
"input": "0 100 0",
"output": "0"
},
{
"input": "100 0 0",
"output": "0"
},
{
"input": "10 8 7",
"output": "24"
},
{
"input": "45 47 16",
"output": "108"
},
{
"input": "59 43 100",
"output": "202"
},
{
"input": "34 1 30",
"output": "62"
},
{
"input": "14 81 1",
"output": "30"
},
{
"input": "53 96 94",
"output": "242"
},
{
"input": "62 81 75",
"output": "218"
},
{
"input": "21 71 97",
"output": "188"
},
{
"input": "49 82 73",
"output": "204"
},
{
"input": "88 19 29",
"output": "96"
},
{
"input": "89 4 62",
"output": "132"
},
{
"input": "58 3 65",
"output": "126"
},
{
"input": "27 86 11",
"output": "76"
},
{
"input": "35 19 80",
"output": "134"
},
{
"input": "4 86 74",
"output": "156"
},
{
"input": "32 61 89",
"output": "182"
},
{
"input": "68 60 98",
"output": "226"
},
{
"input": "37 89 34",
"output": "142"
},
{
"input": "92 9 28",
"output": "74"
},
{
"input": "79 58 98",
"output": "234"
},
{
"input": "35 44 88",
"output": "166"
},
{
"input": "16 24 19",
"output": "58"
},
{
"input": "74 71 75",
"output": "220"
},
{
"input": "83 86 99",
"output": "268"
},
{
"input": "97 73 15",
"output": "176"
},
{
"input": "77 76 73",
"output": "226"
},
{
"input": "48 85 55",
"output": "188"
},
{
"input": "1 2 2",
"output": "4"
},
{
"input": "2 2 2",
"output": "6"
},
{
"input": "2 1 2",
"output": "4"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "3 2 1",
"output": "6"
},
{
"input": "1 2 3",
"output": "6"
},
{
"input": "1 3 2",
"output": "6"
},
{
"input": "2 1 3",
"output": "6"
},
{
"input": "2 3 1",
"output": "6"
},
{
"input": "3 1 2",
"output": "6"
},
{
"input": "99 99 99",
"output": "296"
},
{
"input": "99 99 100",
"output": "298"
},
{
"input": "99 100 99",
"output": "298"
},
{
"input": "99 100 100",
"output": "298"
},
{
"input": "100 99 99",
"output": "298"
},
{
"input": "100 99 100",
"output": "298"
},
{
"input": "100 100 99",
"output": "298"
},
{
"input": "89 32 23",
"output": "110"
},
{
"input": "4 5 0",
"output": "8"
},
{
"input": "3 0 3",
"output": "6"
},
{
"input": "0 0 2",
"output": "2"
},
{
"input": "97 97 0",
"output": "194"
},
{
"input": "1 4 0",
"output": "2"
},
{
"input": "5 2 0",
"output": "4"
},
{
"input": "0 5 10",
"output": "14"
},
{
"input": "0 1 2",
"output": "2"
},
{
"input": "5 2 3",
"output": "10"
},
{
"input": "5 5 0",
"output": "10"
},
{
"input": "0 0 10",
"output": "10"
},
{
"input": "0 1 1",
"output": "2"
},
{
"input": "0 0 1",
"output": "0"
}
] | 1,646,302,907
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 5
| 46
| 0
|
l,r,a=map(int,input().split())
if r>l:
l=l+a
print(min(l,r)*2)
if r<l:
r=r+a
print(min(l,r)*2)
if r==l:
if a%2==0:
r=r+(a//2)+1
print(r*2)
else:
r=r+(a//2)
print(r*2)
|
Title: Left-handers, Right-handers and Ambidexters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are at a water bowling training. There are *l* people who play with their left hand, *r* people, who play with their right hand, and *a* ambidexters, who can play with left or right hand.
The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands.
Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand.
Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively.
Input Specification:
The only line contains three integers *l*, *r* and *a* (0<=≤<=*l*,<=*r*,<=*a*<=≤<=100) — the number of left-handers, the number of right-handers and the number of ambidexters at the training.
Output Specification:
Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zero number of players.
Demo Input:
['1 4 2\n', '5 5 5\n', '0 2 0\n']
Demo Output:
['6\n', '14\n', '0\n']
Note:
In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and three right-handers to play with right hand. The only person left can't be taken into the team.
In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play with left hand and two ambidexters to play with right hand.
|
```python
l,r,a=map(int,input().split())
if r>l:
l=l+a
print(min(l,r)*2)
if r<l:
r=r+a
print(min(l,r)*2)
if r==l:
if a%2==0:
r=r+(a//2)+1
print(r*2)
else:
r=r+(a//2)
print(r*2)
```
| 0
|
|
794
|
B
|
Cutting Carrot
|
PROGRAMMING
| 1,200
|
[
"geometry",
"math"
] | null | null |
Igor the analyst has adopted *n* little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into *n* pieces of equal area.
Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to *h*. Igor wants to make *n*<=-<=1 cuts parallel to the base to cut the carrot into *n* pieces. He wants to make sure that all *n* pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area?
|
The first and only line of input contains two space-separated integers, *n* and *h* (2<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=105).
|
The output should contain *n*<=-<=1 real numbers *x*1,<=*x*2,<=...,<=*x**n*<=-<=1. The number *x**i* denotes that the *i*-th cut must be made *x**i* units away from the apex of the carrot. In addition, 0<=<<=*x*1<=<<=*x*2<=<<=...<=<<=*x**n*<=-<=1<=<<=*h* must hold.
Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10<=-<=6.
Formally, let your answer be *a*, and the jury's answer be *b*. Your answer is considered correct if .
|
[
"3 2\n",
"2 100000\n"
] |
[
"1.154700538379 1.632993161855\n",
"70710.678118654752\n"
] |
Definition of isosceles triangle: [https://en.wikipedia.org/wiki/Isosceles_triangle](https://en.wikipedia.org/wiki/Isosceles_triangle).
| 1,000
|
[
{
"input": "3 2",
"output": "1.154700538379 1.632993161855"
},
{
"input": "2 100000",
"output": "70710.678118654752"
},
{
"input": "1000 100000",
"output": "3162.277660168379 4472.135954999579 5477.225575051661 6324.555320336759 7071.067811865475 7745.966692414834 8366.600265340755 8944.271909999159 9486.832980505138 10000.000000000000 10488.088481701515 10954.451150103322 11401.754250991380 11832.159566199232 12247.448713915890 12649.110640673517 13038.404810405297 13416.407864998738 13784.048752090222 14142.135623730950 14491.376746189439 14832.396974191326 15165.750888103101 15491.933384829668 15811.388300841897 16124.515496597099 16431.676725154983 16733.2..."
},
{
"input": "2 1",
"output": "0.707106781187"
},
{
"input": "1000 1",
"output": "0.031622776602 0.044721359550 0.054772255751 0.063245553203 0.070710678119 0.077459666924 0.083666002653 0.089442719100 0.094868329805 0.100000000000 0.104880884817 0.109544511501 0.114017542510 0.118321595662 0.122474487139 0.126491106407 0.130384048104 0.134164078650 0.137840487521 0.141421356237 0.144913767462 0.148323969742 0.151657508881 0.154919333848 0.158113883008 0.161245154966 0.164316767252 0.167332005307 0.170293863659 0.173205080757 0.176068168617 0.178885438200 0.181659021246 0.184390889146 0..."
},
{
"input": "20 17",
"output": "3.801315561750 5.375872022286 6.584071688553 7.602631123499 8.500000000000 9.311283477588 10.057335631269 10.751744044572 11.403946685249 12.020815280171 12.607537428063 13.168143377105 13.705838172108 14.223220451079 14.722431864335 15.205262246999 15.673225577398 16.127616066859 16.569550386175"
},
{
"input": "999 1",
"output": "0.031638599858 0.044743737014 0.054799662435 0.063277199717 0.070746059996 0.077498425829 0.083707867056 0.089487474029 0.094915799575 0.100050037531 0.104933364623 0.109599324870 0.114074594073 0.118380800867 0.122535770349 0.126554399434 0.130449289063 0.134231211043 0.137909459498 0.141492119993 0.144986278734 0.148398187395 0.151733394554 0.154996851658 0.158192999292 0.161325838061 0.164398987305 0.167415734111 0.170379074505 0.173291748303 0.176156268782 0.178974948057 0.181749918935 0.184483153795 0..."
},
{
"input": "998 99999",
"output": "3165.413034717700 4476.570044210349 5482.656203071844 6330.826069435401 7078.078722492680 7753.646760213179 8374.895686665300 8953.140088420697 9496.239104153101 10009.914924893578 10498.487342658843 10965.312406143687 11413.059004696742 11843.891063542002 12259.591967329534 12661.652138870802 13051.332290848021 13429.710132631046 13797.715532900862 14156.157444985360 14505.744837393740 14847.103184390411 15180.787616204127 15507.293520426358 15827.065173588502 16140.502832606510 16447.968609215531 16749.7..."
},
{
"input": "574 29184",
"output": "1218.116624752432 1722.677051277028 2109.839883615525 2436.233249504864 2723.791577469041 2983.764177844748 3222.833656968322 3445.354102554056 3654.349874257297 3852.022989934325 4040.035795197963 4219.679767231051 4391.981950040022 4557.775066957079 4717.745401404559 4872.466499009729 5022.423508175150 5168.031153831084 5309.647268742708 5447.583154938083 5582.111638212139 5713.473414041731 5841.882108059006 5967.528355689497 6090.583123762161 6211.200439444432 6329.519650846576 6445.667313936643 6559.75..."
},
{
"input": "2 5713",
"output": "4039.701040918746"
},
{
"input": "937 23565",
"output": "769.834993893392 1088.711089153444 1333.393322867831 1539.669987786784 1721.403377803760 1885.702921177414 2036.791944396843 2177.422178306887 2309.504981680176 2434.432003204934 2553.253825229922 2666.786645735663 2775.679544129132 2880.458791498282 2981.558110676796 3079.339975573568 3174.110994119182 3266.133267460331 3355.632941582547 3442.806755607520 3527.827132142336 3610.846187821139 3691.998931463184 3771.405842354828 3849.174969466960 3925.403656108988 4000.179968603494 4073.583888793686 4145.688..."
},
{
"input": "693 39706",
"output": "1508.306216302128 2133.067107306117 2612.463000007259 3016.612432604256 3372.675230537060 3694.580605808168 3990.603149268227 4266.134214612233 4524.918648906384 4769.683052505315 5002.485788434792 5224.926000014517 5438.275401978402 5643.565095743912 5841.644856719264 6033.224865208513 6218.905845589392 6399.201321918350 6574.554372775177 6745.350461074120 6911.927407376938 7074.583247583148 7233.582498950279 7389.161211616337 7541.531081510641 7690.882829397851 7837.389000021776 7981.206298536455 8122.47..."
},
{
"input": "449 88550",
"output": "4178.932872810542 5909.903544975429 7238.124057127628 8357.865745621084 9344.377977012855 10236.253207728862 11056.417127089408 11819.807089950858 12536.798618431626 13214.946067032045 13859.952363194553 14476.248114255256 15067.356749640443 15636.135052384012 16184.937421313947 16715.731491242168 17230.181636963718 17729.710634926286 18215.546084421264 18688.755954025709 19150.276213793575 19600.932605874766 20041.458005232581 20472.506415457724 20894.664364052710 21308.460264455309 21714.372171382883 221..."
},
{
"input": "642 37394",
"output": "1475.823459881026 2087.129552632132 2556.201215516026 2951.646919762052 3300.041579082908 3615.014427137354 3904.661853880105 4174.259105264265 4427.470379643078 4666.963557534173 4894.752673229489 5112.402431032051 5321.157158133711 5522.025750238117 5715.839682061424 5903.293839524104 6084.976009853978 6261.388657896397 6432.965320127946 6600.083158165816 6763.072717296425 6922.225614943105 7077.800671741869 7230.028854274709 7379.117299405130 7525.252620551370 7668.603646548077 7809.323707760210 7947.55..."
},
{
"input": "961 53535",
"output": "1726.935483870968 2442.255582633666 2991.139999458060 3453.870967741935 3861.545134691976 4230.110754190240 4569.041820575576 4884.511165267332 5180.806451612903 5461.049501197232 5727.597037150849 5982.279998916119 6226.554436514989 6461.600909707837 6688.392369006905 6907.741935483871 7120.337408627144 7326.766747900998 7527.537256208063 7723.090269383951 7913.812575143900 8100.045409746687 8282.091632275692 8460.221508380480 8634.677419354839 8805.677730973862 8973.419998374179 9138.083641151152 9299.83..."
},
{
"input": "4 31901",
"output": "15950.500000000000 22557.413426632053 27627.076406127377"
},
{
"input": "4 23850",
"output": "11925.000000000000 16864.496731299158 20654.705880258862"
},
{
"input": "4 72694",
"output": "36347.000000000000 51402.420351574886 62954.850702705983"
},
{
"input": "4 21538",
"output": "10769.000000000000 15229.665853195861 18652.455146709240"
},
{
"input": "4 70383",
"output": "35191.500000000000 49768.296580252774 60953.465994560145"
},
{
"input": "5 1",
"output": "0.447213595500 0.632455532034 0.774596669241 0.894427191000"
},
{
"input": "5 1",
"output": "0.447213595500 0.632455532034 0.774596669241 0.894427191000"
},
{
"input": "5 1",
"output": "0.447213595500 0.632455532034 0.774596669241 0.894427191000"
},
{
"input": "5 1",
"output": "0.447213595500 0.632455532034 0.774596669241 0.894427191000"
},
{
"input": "5 1",
"output": "0.447213595500 0.632455532034 0.774596669241 0.894427191000"
},
{
"input": "20 1",
"output": "0.223606797750 0.316227766017 0.387298334621 0.447213595500 0.500000000000 0.547722557505 0.591607978310 0.632455532034 0.670820393250 0.707106781187 0.741619848710 0.774596669241 0.806225774830 0.836660026534 0.866025403784 0.894427191000 0.921954445729 0.948683298051 0.974679434481"
},
{
"input": "775 1",
"output": "0.035921060405 0.050800050800 0.062217101684 0.071842120811 0.080321932890 0.087988269013 0.095038192662 0.101600101600 0.107763181216 0.113592366849 0.119136679436 0.124434203368 0.129515225161 0.134404301006 0.139121668728 0.143684241621 0.148106326235 0.152400152400 0.156576272252 0.160643865780 0.164610978351 0.168484707835 0.172271353843 0.175976538026 0.179605302027 0.183162187956 0.186651305051 0.190076385325 0.193440830330 0.196747750735 0.200000000000 0.203200203200 0.206350781829 0.209453975235 0..."
},
{
"input": "531 1",
"output": "0.043396303660 0.061371641193 0.075164602800 0.086792607321 0.097037084957 0.106298800691 0.114815827305 0.122743282386 0.130188910981 0.137231161599 0.143929256529 0.150329205601 0.156467598013 0.162374100149 0.168073161363 0.173585214641 0.178927543753 0.184114923580 0.189160102178 0.194074169913 0.198866846404 0.203546706606 0.208121361089 0.212597601381 0.216981518301 0.221278599182 0.225493808401 0.229631654609 0.233696247231 0.237691344271 0.241620392998 0.245486564773 0.249292785005 0.253041759057 0..."
},
{
"input": "724 1",
"output": "0.037164707312 0.052558833123 0.064371161313 0.074329414625 0.083102811914 0.091034569355 0.098328573097 0.105117666246 0.111494121937 0.117525123681 0.123261389598 0.128742322627 0.133999257852 0.139057601643 0.143938292487 0.148658829249 0.153234013794 0.157676499368 0.161997203441 0.166205623829 0.170310084440 0.174317928887 0.178235674883 0.182069138710 0.185823536562 0.189503567803 0.193113483940 0.196657146194 0.200138073886 0.203559485381 0.206924332929 0.210235332491 0.213494989396 0.216705620524 0..."
},
{
"input": "917 1",
"output": "0.033022909334 0.046701446249 0.057197356781 0.066045818668 0.073841470086 0.080889277691 0.087370405666 0.093402892499 0.099068728003 0.104427608461 0.109524599747 0.114394713561 0.119065792869 0.123560412643 0.127897177895 0.132091637337 0.136156943250 0.140104338748 0.143943524609 0.147682940172 0.151329981692 0.154891174376 0.158372309576 0.161778555382 0.165114546671 0.168384459091 0.171592070342 0.174740811332 0.177833809176 0.180873923568 0.183863777748 0.186805784998 0.189702171441 0.192554995756 0..."
},
{
"input": "458 100",
"output": "4.672693135160 6.608186004551 8.093341918275 9.345386270320 10.448459488214 11.445713905748 12.362783988552 13.216372009102 14.018079405480 14.776353114139 15.497569889795 16.186683836551 16.847634693328 17.483616785299 18.097262694412 18.690772540640 19.266007352363 19.824558013653 20.367797170339 20.896918976429 21.412969991171 21.916873521973 22.409449036367 22.891427811495 23.363465675800 23.826153477212 24.280025754826 24.725567977104 25.163222626003 25.593394344267 26.016454316384 26.432744018204 26...."
}
] | 1,494,751,718
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 77
| 307,200
|
n, H = (int(x) for x in input().split())
list_of_h = [H * ((i / n) ** 0.5) for i in range(1, n)]
print(*list_of_h)
|
Title: Cutting Carrot
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Igor the analyst has adopted *n* little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into *n* pieces of equal area.
Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to *h*. Igor wants to make *n*<=-<=1 cuts parallel to the base to cut the carrot into *n* pieces. He wants to make sure that all *n* pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area?
Input Specification:
The first and only line of input contains two space-separated integers, *n* and *h* (2<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=105).
Output Specification:
The output should contain *n*<=-<=1 real numbers *x*1,<=*x*2,<=...,<=*x**n*<=-<=1. The number *x**i* denotes that the *i*-th cut must be made *x**i* units away from the apex of the carrot. In addition, 0<=<<=*x*1<=<<=*x*2<=<<=...<=<<=*x**n*<=-<=1<=<<=*h* must hold.
Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10<=-<=6.
Formally, let your answer be *a*, and the jury's answer be *b*. Your answer is considered correct if .
Demo Input:
['3 2\n', '2 100000\n']
Demo Output:
['1.154700538379 1.632993161855\n', '70710.678118654752\n']
Note:
Definition of isosceles triangle: [https://en.wikipedia.org/wiki/Isosceles_triangle](https://en.wikipedia.org/wiki/Isosceles_triangle).
|
```python
n, H = (int(x) for x in input().split())
list_of_h = [H * ((i / n) ** 0.5) for i in range(1, n)]
print(*list_of_h)
```
| 3
|
|
454
|
A
|
Little Pony and Crystal Mine
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size *n* (*n* is odd; *n*<=><=1) is an *n*<=×<=*n* matrix with a diamond inscribed into it.
You are given an odd integer *n*. You need to draw a crystal of size *n*. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw.
|
The only line contains an integer *n* (3<=≤<=*n*<=≤<=101; *n* is odd).
|
Output a crystal of size *n*.
|
[
"3\n",
"5\n",
"7\n"
] |
[
"*D*\nDDD\n*D*\n",
"**D**\n*DDD*\nDDDDD\n*DDD*\n**D**\n",
"***D***\n**DDD**\n*DDDDD*\nDDDDDDD\n*DDDDD*\n**DDD**\n***D***\n"
] |
none
| 500
|
[
{
"input": "3",
"output": "*D*\nDDD\n*D*"
},
{
"input": "5",
"output": "**D**\n*DDD*\nDDDDD\n*DDD*\n**D**"
},
{
"input": "7",
"output": "***D***\n**DDD**\n*DDDDD*\nDDDDDDD\n*DDDDD*\n**DDD**\n***D***"
},
{
"input": "11",
"output": "*****D*****\n****DDD****\n***DDDDD***\n**DDDDDDD**\n*DDDDDDDDD*\nDDDDDDDDDDD\n*DDDDDDDDD*\n**DDDDDDD**\n***DDDDD***\n****DDD****\n*****D*****"
},
{
"input": "15",
"output": "*******D*******\n******DDD******\n*****DDDDD*****\n****DDDDDDD****\n***DDDDDDDDD***\n**DDDDDDDDDDD**\n*DDDDDDDDDDDDD*\nDDDDDDDDDDDDDDD\n*DDDDDDDDDDDDD*\n**DDDDDDDDDDD**\n***DDDDDDDDD***\n****DDDDDDD****\n*****DDDDD*****\n******DDD******\n*******D*******"
},
{
"input": "21",
"output": "**********D**********\n*********DDD*********\n********DDDDD********\n*******DDDDDDD*******\n******DDDDDDDDD******\n*****DDDDDDDDDDD*****\n****DDDDDDDDDDDDD****\n***DDDDDDDDDDDDDDD***\n**DDDDDDDDDDDDDDDDD**\n*DDDDDDDDDDDDDDDDDDD*\nDDDDDDDDDDDDDDDDDDDDD\n*DDDDDDDDDDDDDDDDDDD*\n**DDDDDDDDDDDDDDDDD**\n***DDDDDDDDDDDDDDD***\n****DDDDDDDDDDDDD****\n*****DDDDDDDDDDD*****\n******DDDDDDDDD******\n*******DDDDDDD*******\n********DDDDD********\n*********DDD*********\n**********D**********"
},
{
"input": "33",
"output": "****************D****************\n***************DDD***************\n**************DDDDD**************\n*************DDDDDDD*************\n************DDDDDDDDD************\n***********DDDDDDDDDDD***********\n**********DDDDDDDDDDDDD**********\n*********DDDDDDDDDDDDDDD*********\n********DDDDDDDDDDDDDDDDD********\n*******DDDDDDDDDDDDDDDDDDD*******\n******DDDDDDDDDDDDDDDDDDDDD******\n*****DDDDDDDDDDDDDDDDDDDDDDD*****\n****DDDDDDDDDDDDDDDDDDDDDDDDD****\n***DDDDDDDDDDDDDDDDDDDDDDDDDDD***\n**DDDDDDDDDDDDDDDDDDD..."
},
{
"input": "57",
"output": "****************************D****************************\n***************************DDD***************************\n**************************DDDDD**************************\n*************************DDDDDDD*************************\n************************DDDDDDDDD************************\n***********************DDDDDDDDDDD***********************\n**********************DDDDDDDDDDDDD**********************\n*********************DDDDDDDDDDDDDDD*********************\n********************DDDDDDDDDDDDDDDDD**..."
},
{
"input": "69",
"output": "**********************************D**********************************\n*********************************DDD*********************************\n********************************DDDDD********************************\n*******************************DDDDDDD*******************************\n******************************DDDDDDDDD******************************\n*****************************DDDDDDDDDDD*****************************\n****************************DDDDDDDDDDDDD****************************\n**************..."
},
{
"input": "81",
"output": "****************************************D****************************************\n***************************************DDD***************************************\n**************************************DDDDD**************************************\n*************************************DDDDDDD*************************************\n************************************DDDDDDDDD************************************\n***********************************DDDDDDDDDDD***********************************\n*************..."
},
{
"input": "91",
"output": "*********************************************D*********************************************\n********************************************DDD********************************************\n*******************************************DDDDD*******************************************\n******************************************DDDDDDD******************************************\n*****************************************DDDDDDDDD*****************************************\n****************************************DDDDDD..."
},
{
"input": "101",
"output": "**************************************************D**************************************************\n*************************************************DDD*************************************************\n************************************************DDDDD************************************************\n***********************************************DDDDDDD***********************************************\n**********************************************DDDDDDDDD********************************************..."
}
] | 1,597,262,697
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 12
| 108
| 0
|
n = int(input())
for i in range(int(n/2)):
print(('D'*((2*i)+1)).center(n,'*'))
print('D'*n)
for i in range(int(n/2),0,-1):
print(('D'*((2*i)-1)).center(n,'*'))
|
Title: Little Pony and Crystal Mine
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size *n* (*n* is odd; *n*<=><=1) is an *n*<=×<=*n* matrix with a diamond inscribed into it.
You are given an odd integer *n*. You need to draw a crystal of size *n*. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw.
Input Specification:
The only line contains an integer *n* (3<=≤<=*n*<=≤<=101; *n* is odd).
Output Specification:
Output a crystal of size *n*.
Demo Input:
['3\n', '5\n', '7\n']
Demo Output:
['*D*\nDDD\n*D*\n', '**D**\n*DDD*\nDDDDD\n*DDD*\n**D**\n', '***D***\n**DDD**\n*DDDDD*\nDDDDDDD\n*DDDDD*\n**DDD**\n***D***\n']
Note:
none
|
```python
n = int(input())
for i in range(int(n/2)):
print(('D'*((2*i)+1)).center(n,'*'))
print('D'*n)
for i in range(int(n/2),0,-1):
print(('D'*((2*i)-1)).center(n,'*'))
```
| 3
|
|
770
|
A
|
New Password
|
PROGRAMMING
| 800
|
[
"*special",
"implementation"
] | null | null |
Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help.
Innokentiy decides that new password should satisfy the following conditions:
- the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct.
Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions.
|
The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it.
Pay attention that a desired new password always exists.
|
Print any password which satisfies all conditions given by Innokentiy.
|
[
"4 3\n",
"6 6\n",
"5 2\n"
] |
[
"java\n",
"python\n",
"phphp\n"
] |
In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it.
In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters.
In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it.
Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests.
| 500
|
[
{
"input": "4 3",
"output": "abca"
},
{
"input": "6 6",
"output": "abcdef"
},
{
"input": "5 2",
"output": "ababa"
},
{
"input": "3 2",
"output": "aba"
},
{
"input": "10 2",
"output": "ababababab"
},
{
"input": "26 13",
"output": "abcdefghijklmabcdefghijklm"
},
{
"input": "100 2",
"output": "abababababababababababababababababababababababababababababababababababababababababababababababababab"
},
{
"input": "100 10",
"output": "abcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij"
},
{
"input": "3 3",
"output": "abc"
},
{
"input": "6 3",
"output": "abcabc"
},
{
"input": "10 3",
"output": "abcabcabca"
},
{
"input": "50 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcab"
},
{
"input": "90 2",
"output": "ababababababababababababababababababababababababababababababababababababababababababababab"
},
{
"input": "6 2",
"output": "ababab"
},
{
"input": "99 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabc"
},
{
"input": "4 2",
"output": "abab"
},
{
"input": "100 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca"
},
{
"input": "40 22",
"output": "abcdefghijklmnopqrstuvabcdefghijklmnopqr"
},
{
"input": "13 8",
"output": "abcdefghabcde"
},
{
"input": "16 15",
"output": "abcdefghijklmnoa"
},
{
"input": "17 17",
"output": "abcdefghijklmnopq"
},
{
"input": "19 4",
"output": "abcdabcdabcdabcdabc"
},
{
"input": "100 26",
"output": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuv"
},
{
"input": "100 25",
"output": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxy"
},
{
"input": "26 26",
"output": "abcdefghijklmnopqrstuvwxyz"
},
{
"input": "27 26",
"output": "abcdefghijklmnopqrstuvwxyza"
},
{
"input": "2 2",
"output": "ab"
},
{
"input": "26 25",
"output": "abcdefghijklmnopqrstuvwxya"
},
{
"input": "99 2",
"output": "abababababababababababababababababababababababababababababababababababababababababababababababababa"
},
{
"input": "99 26",
"output": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstu"
},
{
"input": "4 4",
"output": "abcd"
},
{
"input": "5 3",
"output": "abcab"
},
{
"input": "5 4",
"output": "abcda"
},
{
"input": "5 5",
"output": "abcde"
},
{
"input": "24 22",
"output": "abcdefghijklmnopqrstuvab"
},
{
"input": "26 14",
"output": "abcdefghijklmnabcdefghijkl"
},
{
"input": "26 15",
"output": "abcdefghijklmnoabcdefghijk"
},
{
"input": "30 12",
"output": "abcdefghijklabcdefghijklabcdef"
},
{
"input": "35 4",
"output": "abcdabcdabcdabcdabcdabcdabcdabcdabc"
},
{
"input": "79 3",
"output": "abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca"
},
{
"input": "79 14",
"output": "abcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghijklmnabcdefghi"
},
{
"input": "85 13",
"output": "abcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefghijklmabcdefg"
},
{
"input": "90 25",
"output": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxyabcdefghijklmno"
},
{
"input": "90 19",
"output": "abcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmnopqrsabcdefghijklmn"
},
{
"input": "26 24",
"output": "abcdefghijklmnopqrstuvwxab"
},
{
"input": "100 17",
"output": "abcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmnopqabcdefghijklmno"
},
{
"input": "26 2",
"output": "ababababababababababababab"
}
] | 1,579,100,453
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 47
| 124
| 0
|
n, k = (int(x) for x in input().split())
pwd = [chr(ord('a') + (i % k)) for i in range(n)]
print(*pwd, sep='')
|
Title: New Password
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Innokentiy decides to change the password in the social net "Contact!", but he is too lazy to invent a new password by himself. That is why he needs your help.
Innokentiy decides that new password should satisfy the following conditions:
- the length of the password must be equal to *n*, - the password should consist only of lowercase Latin letters, - the number of distinct symbols in the password must be equal to *k*, - any two consecutive symbols in the password must be distinct.
Your task is to help Innokentiy and to invent a new password which will satisfy all given conditions.
Input Specification:
The first line contains two positive integers *n* and *k* (2<=≤<=*n*<=≤<=100, 2<=≤<=*k*<=≤<=*min*(*n*,<=26)) — the length of the password and the number of distinct symbols in it.
Pay attention that a desired new password always exists.
Output Specification:
Print any password which satisfies all conditions given by Innokentiy.
Demo Input:
['4 3\n', '6 6\n', '5 2\n']
Demo Output:
['java\n', 'python\n', 'phphp\n']
Note:
In the first test there is one of the appropriate new passwords — java, because its length is equal to 4 and 3 distinct lowercase letters a, j and v are used in it.
In the second test there is one of the appropriate new passwords — python, because its length is equal to 6 and it consists of 6 distinct lowercase letters.
In the third test there is one of the appropriate new passwords — phphp, because its length is equal to 5 and 2 distinct lowercase letters p and h are used in it.
Pay attention the condition that no two identical symbols are consecutive is correct for all appropriate passwords in tests.
|
```python
n, k = (int(x) for x in input().split())
pwd = [chr(ord('a') + (i % k)) for i in range(n)]
print(*pwd, sep='')
```
| 3
|
|
862
|
A
|
Mahmoud and Ehab and the MEX
|
PROGRAMMING
| 1,000
|
[
"greedy",
"implementation"
] | null | null |
Dr. Evil kidnapped Mahmoud and Ehab in the evil land because of their performance in the Evil Olympiad in Informatics (EOI). He decided to give them some problems to let them go.
Dr. Evil is interested in sets, He has a set of *n* integers. Dr. Evil calls a set of integers evil if the MEX of it is exactly *x*. the MEX of a set of integers is the minimum non-negative integer that doesn't exist in it. For example, the MEX of the set {0,<=2,<=4} is 1 and the MEX of the set {1,<=2,<=3} is 0 .
Dr. Evil is going to make his set evil. To do this he can perform some operations. During each operation he can add some non-negative integer to his set or erase some element from it. What is the minimal number of operations Dr. Evil has to perform to make his set evil?
|
The first line contains two integers *n* and *x* (1<=≤<=*n*<=≤<=100, 0<=≤<=*x*<=≤<=100) — the size of the set Dr. Evil owns, and the desired MEX.
The second line contains *n* distinct non-negative integers not exceeding 100 that represent the set.
|
The only line should contain one integer — the minimal number of operations Dr. Evil should perform.
|
[
"5 3\n0 4 5 6 7\n",
"1 0\n0\n",
"5 0\n1 2 3 4 5\n"
] |
[
"2\n",
"1\n",
"0\n"
] |
For the first test case Dr. Evil should add 1 and 2 to the set performing 2 operations.
For the second test case Dr. Evil should erase 0 from the set. After that, the set becomes empty, so the MEX of it is 0.
In the third test case the set is already evil.
| 500
|
[
{
"input": "5 3\n0 4 5 6 7",
"output": "2"
},
{
"input": "1 0\n0",
"output": "1"
},
{
"input": "5 0\n1 2 3 4 5",
"output": "0"
},
{
"input": "10 5\n57 1 47 9 93 37 76 70 78 15",
"output": "4"
},
{
"input": "10 5\n99 98 93 97 95 100 92 94 91 96",
"output": "5"
},
{
"input": "10 5\n1 2 3 4 59 45 0 58 51 91",
"output": "0"
},
{
"input": "100 100\n79 13 21 11 3 87 28 40 29 4 96 34 8 78 61 46 33 45 99 30 92 67 22 97 39 86 73 31 74 44 62 55 57 2 54 63 80 69 25 48 77 98 17 93 15 16 89 12 43 23 37 95 14 38 83 90 49 56 72 10 20 0 50 71 70 88 19 1 76 81 52 41 82 68 85 47 6 7 35 60 18 64 75 84 27 9 65 91 94 42 53 24 66 26 59 36 51 32 5 58",
"output": "0"
},
{
"input": "100 50\n95 78 46 92 80 18 79 58 30 72 19 89 39 29 44 65 15 100 59 8 96 9 62 67 41 42 82 14 57 32 71 77 40 5 7 51 28 53 85 23 16 35 3 91 6 11 75 61 17 66 13 47 36 56 10 22 83 60 48 24 26 97 4 33 76 86 70 0 34 64 52 43 21 49 55 74 1 73 81 25 54 63 94 84 20 68 87 12 31 88 38 93 37 90 98 69 99 45 27 2",
"output": "0"
},
{
"input": "100 33\n28 11 79 92 88 62 77 72 7 41 96 97 67 84 44 8 81 35 38 1 64 68 46 17 98 83 31 12 74 21 2 22 47 6 36 75 65 61 37 26 25 45 59 48 100 51 93 76 78 49 3 57 16 4 87 29 55 82 70 39 53 0 60 15 24 71 58 20 66 89 95 42 13 43 63 90 85 52 50 30 54 40 56 23 27 34 32 18 10 19 69 9 99 73 91 14 5 80 94 86",
"output": "0"
},
{
"input": "99 33\n25 76 41 95 55 20 47 59 58 84 87 92 16 27 35 65 72 63 93 54 36 96 15 86 5 69 24 46 67 73 48 60 40 6 61 74 97 10 100 8 52 26 77 18 7 62 37 2 14 66 11 56 68 91 0 64 75 99 30 21 53 1 89 81 3 98 12 88 39 38 29 83 22 90 9 28 45 43 78 44 32 57 4 50 70 17 13 51 80 85 71 94 82 19 34 42 23 79 49",
"output": "1"
},
{
"input": "100 100\n65 56 84 46 44 33 99 74 62 72 93 67 43 92 75 88 38 34 66 12 55 76 58 90 78 8 14 45 97 59 48 32 64 18 39 89 31 51 54 81 29 36 70 77 40 22 49 27 3 1 73 13 98 42 87 37 2 57 4 6 50 25 23 79 28 86 68 61 80 17 19 10 15 63 52 11 35 60 21 16 24 85 30 91 7 5 69 20 71 82 53 94 41 95 96 9 26 83 0 47",
"output": "0"
},
{
"input": "100 100\n58 88 12 71 22 1 40 19 73 20 67 48 57 17 69 36 100 35 33 37 72 55 52 8 89 85 47 42 78 70 81 86 11 9 68 99 6 16 21 61 53 98 23 62 32 59 51 0 87 24 50 30 65 10 80 95 7 92 25 74 60 79 91 5 13 31 75 38 90 94 46 66 93 34 14 41 28 2 76 84 43 96 3 56 49 82 27 77 64 63 4 45 18 29 54 39 15 26 83 44",
"output": "2"
},
{
"input": "89 100\n58 96 17 41 86 34 28 84 18 40 8 77 87 89 68 79 33 35 53 49 0 6 22 12 72 90 48 55 21 50 56 62 75 2 37 95 69 74 14 20 44 46 27 32 31 59 63 60 10 85 71 70 38 52 94 30 61 51 80 26 36 23 39 47 76 45 100 57 15 78 97 66 54 13 99 16 93 73 24 4 83 5 98 81 92 25 29 88 65",
"output": "13"
},
{
"input": "100 50\n7 95 24 76 81 78 60 69 83 84 100 1 65 31 48 92 73 39 18 89 38 97 10 42 8 55 98 51 21 90 62 77 16 91 0 94 4 37 19 17 67 35 45 41 56 20 15 85 75 28 59 27 12 54 61 68 36 5 79 93 66 11 70 49 50 34 30 25 96 46 64 14 32 22 47 40 58 23 43 9 87 82 26 53 80 52 3 86 13 99 33 71 6 88 57 74 2 44 72 63",
"output": "2"
},
{
"input": "77 0\n27 8 20 92 21 41 53 98 17 65 67 35 81 11 55 49 61 44 2 66 51 89 40 28 52 62 86 91 64 24 18 5 94 82 96 99 71 6 39 83 26 29 16 30 45 97 80 90 69 12 13 33 76 73 46 19 78 56 88 38 42 34 57 77 47 4 59 58 7 100 95 72 9 74 15 43 54",
"output": "0"
},
{
"input": "100 50\n55 36 0 32 81 6 17 43 24 13 30 19 8 59 71 45 15 74 3 41 99 42 86 47 2 94 35 1 66 95 38 49 4 27 96 89 34 44 92 25 51 39 54 28 80 77 20 14 48 40 68 56 31 63 33 78 69 37 18 26 83 70 23 82 91 65 67 52 61 53 7 22 60 21 12 73 72 87 75 100 90 29 64 79 98 85 5 62 93 84 50 46 97 58 57 16 9 10 76 11",
"output": "1"
},
{
"input": "77 0\n12 8 19 87 9 54 55 86 97 7 27 85 25 48 94 73 26 1 13 57 72 69 76 39 38 91 75 40 42 28 93 21 70 84 65 11 60 90 20 95 66 89 59 47 34 99 6 61 52 100 50 3 77 81 82 53 15 24 0 45 44 14 68 96 58 5 18 35 10 98 29 74 92 49 83 71 17",
"output": "1"
},
{
"input": "100 70\n25 94 66 65 10 99 89 6 70 31 7 40 20 92 64 27 21 72 77 98 17 43 47 44 48 81 38 56 100 39 90 22 88 76 3 83 86 29 33 55 82 79 49 11 2 16 12 78 85 69 32 97 26 15 53 24 23 91 51 67 34 35 52 5 62 50 95 18 71 13 75 8 30 42 93 36 45 60 63 46 57 41 87 0 84 54 74 37 4 58 28 19 96 61 80 9 1 14 73 68",
"output": "2"
},
{
"input": "89 19\n14 77 85 81 79 38 91 45 55 51 50 11 62 67 73 76 2 27 16 23 3 29 65 98 78 17 4 58 22 20 34 66 64 31 72 5 32 44 12 75 80 47 18 25 99 0 61 56 71 84 48 88 10 7 86 8 49 24 43 21 37 28 33 54 46 57 40 89 36 97 6 96 39 95 26 74 1 69 9 100 52 30 83 87 68 60 92 90 35",
"output": "2"
},
{
"input": "89 100\n69 61 56 45 11 41 42 32 28 29 0 76 7 65 13 35 36 82 10 39 26 34 38 40 92 12 17 54 24 46 88 70 66 27 100 52 85 62 22 48 86 68 21 49 53 94 67 20 1 90 77 84 31 87 58 47 95 33 4 72 93 83 8 51 91 80 99 43 71 19 44 59 98 97 64 9 81 16 79 63 25 37 3 75 2 55 50 6 18",
"output": "13"
},
{
"input": "77 0\n38 76 24 74 42 88 29 75 96 46 90 32 59 97 98 60 41 57 80 37 100 49 25 63 95 31 61 68 53 78 27 66 84 48 94 83 30 26 36 99 71 62 45 47 70 28 35 54 34 85 79 43 91 72 86 33 67 92 77 65 69 52 82 55 87 64 56 40 50 44 51 73 89 81 58 93 39",
"output": "0"
},
{
"input": "89 100\n38 90 80 64 35 44 56 11 15 89 23 12 49 70 72 60 63 85 92 10 45 83 8 88 41 33 16 6 61 76 62 71 87 13 25 77 74 0 1 37 96 93 7 94 21 82 34 78 4 73 65 20 81 95 50 32 48 17 69 55 68 5 51 27 53 43 91 67 59 46 86 84 99 24 22 3 97 98 40 36 26 58 57 9 42 30 52 2 47",
"output": "11"
},
{
"input": "77 0\n55 71 78 86 68 35 53 10 59 32 81 19 74 97 62 61 93 87 96 44 25 18 43 82 84 16 34 48 92 39 64 36 49 91 45 76 95 31 57 29 75 79 13 2 14 24 52 23 33 20 47 99 63 15 5 80 58 67 12 3 85 6 1 27 73 90 4 42 37 70 8 11 89 77 9 22 94",
"output": "0"
},
{
"input": "77 0\n12 75 31 71 44 8 3 82 21 77 50 29 57 74 40 10 15 42 84 2 100 9 28 72 92 0 49 11 90 55 17 36 19 54 68 52 4 69 97 91 5 39 59 45 89 62 53 83 16 94 76 60 95 47 30 51 7 48 20 70 67 32 58 78 63 34 56 93 99 88 24 1 66 22 25 14 13",
"output": "1"
},
{
"input": "100 70\n91 82 8 85 26 25 95 97 40 87 81 93 7 73 38 94 64 96 74 18 90 19 65 68 72 61 23 43 36 41 60 88 30 33 71 24 52 39 15 3 16 89 86 79 55 4 9 58 67 44 46 29 6 48 84 69 27 21 78 54 51 57 80 53 76 50 47 77 45 12 34 10 100 0 17 31 56 99 98 11 92 5 2 42 32 59 66 62 37 63 28 75 35 1 22 13 83 49 20 14",
"output": "0"
},
{
"input": "77 0\n51 5 81 62 30 22 11 0 83 16 79 85 52 70 69 10 8 47 58 3 24 34 44 14 82 66 99 17 28 31 64 67 23 49 94 45 4 12 27 15 21 6 43 72 87 2 63 92 35 39 59 9 90 78 93 20 65 36 60 89 50 41 61 84 77 86 76 100 38 68 53 97 96 95 7 19 88",
"output": "1"
},
{
"input": "1 100\n0",
"output": "99"
},
{
"input": "1 0\n100",
"output": "0"
},
{
"input": "1 100\n100",
"output": "101"
},
{
"input": "2 100\n0 100",
"output": "100"
},
{
"input": "5 5\n1 2 3 4 5",
"output": "2"
},
{
"input": "5 3\n0 3 4 5 6",
"output": "3"
},
{
"input": "7 10\n0 1 2 3 4 5 10",
"output": "5"
},
{
"input": "2 2\n0 2",
"output": "2"
},
{
"input": "2 1\n1 2",
"output": "2"
},
{
"input": "5 1\n1 2 3 4 5",
"output": "2"
},
{
"input": "5 2\n1 2 3 4 5",
"output": "2"
},
{
"input": "5 6\n0 1 2 3 4",
"output": "1"
},
{
"input": "3 2\n3 4 5",
"output": "2"
}
] | 1,565,294,645
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 39
| 109
| 0
|
n,x = map(int,input().split())
list1 = list(map(int,input().split()))
t = 0
for i in range(0,x):
if i in list1:
continue
else:
t = t + 1
if x in list1:
t = t + 1
print(t)
|
Title: Mahmoud and Ehab and the MEX
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dr. Evil kidnapped Mahmoud and Ehab in the evil land because of their performance in the Evil Olympiad in Informatics (EOI). He decided to give them some problems to let them go.
Dr. Evil is interested in sets, He has a set of *n* integers. Dr. Evil calls a set of integers evil if the MEX of it is exactly *x*. the MEX of a set of integers is the minimum non-negative integer that doesn't exist in it. For example, the MEX of the set {0,<=2,<=4} is 1 and the MEX of the set {1,<=2,<=3} is 0 .
Dr. Evil is going to make his set evil. To do this he can perform some operations. During each operation he can add some non-negative integer to his set or erase some element from it. What is the minimal number of operations Dr. Evil has to perform to make his set evil?
Input Specification:
The first line contains two integers *n* and *x* (1<=≤<=*n*<=≤<=100, 0<=≤<=*x*<=≤<=100) — the size of the set Dr. Evil owns, and the desired MEX.
The second line contains *n* distinct non-negative integers not exceeding 100 that represent the set.
Output Specification:
The only line should contain one integer — the minimal number of operations Dr. Evil should perform.
Demo Input:
['5 3\n0 4 5 6 7\n', '1 0\n0\n', '5 0\n1 2 3 4 5\n']
Demo Output:
['2\n', '1\n', '0\n']
Note:
For the first test case Dr. Evil should add 1 and 2 to the set performing 2 operations.
For the second test case Dr. Evil should erase 0 from the set. After that, the set becomes empty, so the MEX of it is 0.
In the third test case the set is already evil.
|
```python
n,x = map(int,input().split())
list1 = list(map(int,input().split()))
t = 0
for i in range(0,x):
if i in list1:
continue
else:
t = t + 1
if x in list1:
t = t + 1
print(t)
```
| 3
|
|
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": "3 3\n1 2 1\n1 3 2\n2 3 1",
"output": "1 3 "
},
{
"input": "10 10\n1 5 12\n2 4 140\n2 10 149\n3 6 154\n3 7 9\n3 8 226\n3 10 132\n4 10 55\n5 8 33\n7 8 173",
"output": "1 5 8 7 3 10 "
},
{
"input": "10 10\n1 5 178\n1 8 221\n2 7 92\n2 8 159\n3 5 55\n3 6 179\n3 10 237\n4 8 205\n5 6 191\n8 10 157",
"output": "1 8 10 "
},
{
"input": "10 10\n1 4 200\n1 9 197\n3 4 79\n3 5 213\n3 6 149\n5 8 3\n5 9 189\n6 7 130\n6 9 51\n8 10 135",
"output": "1 9 5 8 10 "
},
{
"input": "10 10\n1 4 201\n2 3 238\n3 4 40\n3 6 231\n3 8 45\n4 5 227\n4 6 58\n4 9 55\n5 7 14\n6 10 242",
"output": "1 4 6 10 "
}
] | 1,666,345,106
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 3
| 46
| 0
|
# import sys
import heapq
# sys.stdin = open("input.txt")
n, m = map(int, input().split())
adjList = [[] for _ in range(n)]
for _ in range(m):
u, v, w = map(int, input().split())
adjList[u - 1].append((w, v - 1))
adjList[v - 1].append((w, u - 1))
# Initialization
dist = [float("inf") for _ in range(n)]
prev = [-1 for _ in range(n)]
dist[0] = 0
queue = [(dist[i], i) for i in range(n)]
heapq.heapify(queue)
while queue:
d, u = heapq.heappop(queue)
for w, v in adjList[u]:
if dist[v] > dist[u] + w:
dist[v] = dist[u] + w
prev[v] = u
heapq.heappush(queue, (dist[v], v))
shortest_path = []
d = n - 1
while d != -1:
shortest_path.append(d)
d = prev[d]
print(*[i + 1 for i in shortest_path[::-1]], sep=' ')
|
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
# import sys
import heapq
# sys.stdin = open("input.txt")
n, m = map(int, input().split())
adjList = [[] for _ in range(n)]
for _ in range(m):
u, v, w = map(int, input().split())
adjList[u - 1].append((w, v - 1))
adjList[v - 1].append((w, u - 1))
# Initialization
dist = [float("inf") for _ in range(n)]
prev = [-1 for _ in range(n)]
dist[0] = 0
queue = [(dist[i], i) for i in range(n)]
heapq.heapify(queue)
while queue:
d, u = heapq.heappop(queue)
for w, v in adjList[u]:
if dist[v] > dist[u] + w:
dist[v] = dist[u] + w
prev[v] = u
heapq.heappush(queue, (dist[v], v))
shortest_path = []
d = n - 1
while d != -1:
shortest_path.append(d)
d = prev[d]
print(*[i + 1 for i in shortest_path[::-1]], sep=' ')
```
| 0
|
408
|
A
|
Line to Cashier
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Little Vasya went to the supermarket to get some groceries. He walked about the supermarket for a long time and got a basket full of products. Now he needs to choose the cashier to pay for the products.
There are *n* cashiers at the exit from the supermarket. At the moment the queue for the *i*-th cashier already has *k**i* people. The *j*-th person standing in the queue to the *i*-th cashier has *m**i*,<=*j* items in the basket. Vasya knows that:
- the cashier needs 5 seconds to scan one item; - after the cashier scans each item of some customer, he needs 15 seconds to take the customer's money and give him the change.
Of course, Vasya wants to select a queue so that he can leave the supermarket as soon as possible. Help him write a program that displays the minimum number of seconds after which Vasya can get to one of the cashiers.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of cashes in the shop. The second line contains *n* space-separated integers: *k*1,<=*k*2,<=...,<=*k**n* (1<=≤<=*k**i*<=≤<=100), where *k**i* is the number of people in the queue to the *i*-th cashier.
The *i*-th of the next *n* lines contains *k**i* space-separated integers: *m**i*,<=1,<=*m**i*,<=2,<=...,<=*m**i*,<=*k**i* (1<=≤<=*m**i*,<=*j*<=≤<=100) — the number of products the *j*-th person in the queue for the *i*-th cash has.
|
Print a single integer — the minimum number of seconds Vasya needs to get to the cashier.
|
[
"1\n1\n1\n",
"4\n1 4 3 2\n100\n1 2 2 3\n1 9 1\n7 8\n"
] |
[
"20\n",
"100\n"
] |
In the second test sample, if Vasya goes to the first queue, he gets to the cashier in 100·5 + 15 = 515 seconds. But if he chooses the second queue, he will need 1·5 + 2·5 + 2·5 + 3·5 + 4·15 = 100 seconds. He will need 1·5 + 9·5 + 1·5 + 3·15 = 100 seconds for the third one and 7·5 + 8·5 + 2·15 = 105 seconds for the fourth one. Thus, Vasya gets to the cashier quicker if he chooses the second or the third queue.
| 500
|
[
{
"input": "1\n1\n1",
"output": "20"
},
{
"input": "4\n1 4 3 2\n100\n1 2 2 3\n1 9 1\n7 8",
"output": "100"
},
{
"input": "4\n5 4 5 5\n3 1 3 1 2\n3 1 1 3\n1 1 1 2 2\n2 2 1 1 3",
"output": "100"
},
{
"input": "5\n5 3 6 6 4\n7 5 3 3 9\n6 8 2\n1 10 8 5 9 2\n9 7 8 5 9 10\n9 8 3 3",
"output": "125"
},
{
"input": "5\n10 10 10 10 10\n6 7 8 6 8 5 9 8 10 5\n9 6 9 8 7 8 8 10 8 5\n8 7 7 8 7 5 6 8 9 5\n6 5 10 5 5 10 7 8 5 5\n10 9 8 7 6 9 7 9 6 5",
"output": "480"
},
{
"input": "10\n9 10 10 10 9 5 9 7 8 7\n11 6 10 4 4 15 7 15 5\n3 9 11 12 11 1 13 13 1 5\n6 15 9 12 3 2 8 12 11 10\n7 1 1 6 10 2 6 1 14 2\n8 14 2 3 6 1 14 1 12\n6 10 9 3 5\n13 12 12 7 13 4 4 8 10\n5 6 4 3 14 9 13\n8 12 1 5 7 4 13 1\n1 9 5 3 5 1 4",
"output": "240"
},
{
"input": "10\n5 5 5 5 5 5 5 5 5 5\n5 5 4 5 4\n6 5 7 7 6\n5 4 4 5 5\n4 4 5 5 5\n7 6 4 5 7\n4 6 5 4 5\n6 6 7 6 6\n4 5 4 4 7\n7 5 4 4 5\n6 6 7 4 4",
"output": "190"
},
{
"input": "1\n1\n100",
"output": "515"
},
{
"input": "1\n90\n90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90",
"output": "41850"
}
] | 1,682,161,367
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 46
| 0
|
n=int(input())
lol=list(map(int,input().split()))
ans=sum(list(map(int,input().split())))*5+15*lol[0]
for _ in range(n-1):
ans=min(ans,sum(list(map(int,input().split())))*5+15*lol[_+1])
print(ans)
|
Title: Line to Cashier
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Vasya went to the supermarket to get some groceries. He walked about the supermarket for a long time and got a basket full of products. Now he needs to choose the cashier to pay for the products.
There are *n* cashiers at the exit from the supermarket. At the moment the queue for the *i*-th cashier already has *k**i* people. The *j*-th person standing in the queue to the *i*-th cashier has *m**i*,<=*j* items in the basket. Vasya knows that:
- the cashier needs 5 seconds to scan one item; - after the cashier scans each item of some customer, he needs 15 seconds to take the customer's money and give him the change.
Of course, Vasya wants to select a queue so that he can leave the supermarket as soon as possible. Help him write a program that displays the minimum number of seconds after which Vasya can get to one of the cashiers.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of cashes in the shop. The second line contains *n* space-separated integers: *k*1,<=*k*2,<=...,<=*k**n* (1<=≤<=*k**i*<=≤<=100), where *k**i* is the number of people in the queue to the *i*-th cashier.
The *i*-th of the next *n* lines contains *k**i* space-separated integers: *m**i*,<=1,<=*m**i*,<=2,<=...,<=*m**i*,<=*k**i* (1<=≤<=*m**i*,<=*j*<=≤<=100) — the number of products the *j*-th person in the queue for the *i*-th cash has.
Output Specification:
Print a single integer — the minimum number of seconds Vasya needs to get to the cashier.
Demo Input:
['1\n1\n1\n', '4\n1 4 3 2\n100\n1 2 2 3\n1 9 1\n7 8\n']
Demo Output:
['20\n', '100\n']
Note:
In the second test sample, if Vasya goes to the first queue, he gets to the cashier in 100·5 + 15 = 515 seconds. But if he chooses the second queue, he will need 1·5 + 2·5 + 2·5 + 3·5 + 4·15 = 100 seconds. He will need 1·5 + 9·5 + 1·5 + 3·15 = 100 seconds for the third one and 7·5 + 8·5 + 2·15 = 105 seconds for the fourth one. Thus, Vasya gets to the cashier quicker if he chooses the second or the third queue.
|
```python
n=int(input())
lol=list(map(int,input().split()))
ans=sum(list(map(int,input().split())))*5+15*lol[0]
for _ in range(n-1):
ans=min(ans,sum(list(map(int,input().split())))*5+15*lol[_+1])
print(ans)
```
| 3
|
|
846
|
C
|
Four Segments
|
PROGRAMMING
| 1,800
|
[
"brute force",
"data structures",
"dp"
] | null | null |
You are given an array of *n* integer numbers. Let *sum*(*l*,<=*r*) be the sum of all numbers on positions from *l* to *r* non-inclusive (*l*-th element is counted, *r*-th element is not counted). For indices *l* and *r* holds 0<=≤<=*l*<=≤<=*r*<=≤<=*n*. Indices in array are numbered from 0.
For example, if *a*<==<=[<=-<=5,<=3,<=9,<=4], then *sum*(0,<=1)<==<=<=-<=5, *sum*(0,<=2)<==<=<=-<=2, *sum*(1,<=4)<==<=16 and *sum*(*i*,<=*i*)<==<=0 for each *i* from 0 to 4.
Choose the indices of three delimiters *delim*0, *delim*1, *delim*2 (0<=≤<=*delim*0<=≤<=*delim*1<=≤<=*delim*2<=≤<=*n*) and divide the array in such a way that the value of *res*<==<=*sum*(0,<=*delim*0) - *sum*(*delim*0,<=*delim*1) + *sum*(*delim*1,<=*delim*2) - *sum*(*delim*2,<=*n*) is maximal.
Note that some of the expressions *sum*(*l*,<=*r*) can correspond to empty segments (if *l*<==<=*r* for some segment).
|
The first line contains one integer number *n* (1<=≤<=*n*<=≤<=5000).
The second line contains *n* numbers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 (<=-<=109<=≤<=*a**i*<=≤<=109).
|
Choose three indices so that the value of *res* is maximal. If there are multiple answers, print any of them.
|
[
"3\n-1 2 3\n",
"4\n0 0 -1 0\n",
"1\n10000\n"
] |
[
"0 1 3\n",
"0 0 0\n",
"1 1 1\n"
] |
none
| 0
|
[
{
"input": "3\n-1 2 3",
"output": "0 1 3"
},
{
"input": "4\n0 0 -1 0",
"output": "0 0 0"
},
{
"input": "1\n10000",
"output": "0 0 1"
},
{
"input": "1\n-1",
"output": "0 0 0"
},
{
"input": "1\n0",
"output": "0 0 0"
},
{
"input": "10\n0 0 0 0 0 0 0 0 0 0",
"output": "0 0 0"
},
{
"input": "100\n-1 -1 0 1 -1 0 0 -1 -1 1 1 0 1 0 -1 1 0 0 -1 0 0 0 1 0 0 1 1 1 1 0 -1 -1 0 0 0 1 1 -1 0 0 1 1 1 0 -1 -1 0 -1 0 -1 -1 0 1 0 1 -1 1 -1 -1 -1 0 0 -1 0 -1 -1 0 -1 1 1 -1 1 0 1 -1 -1 1 1 -1 1 0 -1 1 -1 1 1 1 1 1 -1 0 1 0 0 -1 0 1 1 -1 -1",
"output": "43 68 98"
},
{
"input": "100\n2 1 -2 -1 -2 0 -1 -2 1 2 0 1 0 -2 -1 -2 0 2 0 1 -2 -2 2 0 -2 2 -2 0 2 0 2 0 -1 0 -2 2 -1 -1 -2 -1 0 -2 2 0 -2 -2 -2 -1 1 0 -2 -1 2 -1 -2 1 -1 1 1 2 -2 1 -2 1 2 2 -2 1 -2 0 -1 -1 -2 -2 1 0 -1 -1 2 0 2 0 -1 2 1 -1 2 0 2 1 1 1 -1 -1 1 -2 0 0 -2 0",
"output": "2 78 92"
},
{
"input": "10\n-6 -4 -7 0 7 9 8 3 8 7",
"output": "0 3 10"
},
{
"input": "4\n-1 1 -2 -2",
"output": "0 1 2"
},
{
"input": "2\n-3 -1",
"output": "0 0 0"
}
] | 1,504,854,907
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 0
|
def list_input():
return list(map(int,input().split()))
def map_input():
return map(int,input().split())
def map_string():
return input().split()
n = int(input())
a = list_input()
pref = [0]
for i in range(n): pref.append(pref[i]+a[i])
ans = -10000000000
x=y=z=0
for i in range(n+1):
# print(i,"----")
cur = pref[i]
dp = [0]*(n+1)
mx = -10000000000
ind1 = -1
ind2 = -1
sinc = -1
for j in range(max(1,i+1),n+1):
if dp[j-1] > 0:
dp[j] = dp[j-1]+a[j-1]
else:
sinc = i
dp[j] = a[j-1]
if dp[j] > mx:
mx = dp[j]
ind1 = sinc
ind2 = j
# print(dp,ind1,ind2)
if mx+cur > ans:
ans = mx+cur
x = i
y = ind1+1
z = ind2
print(x,y,z)
|
Title: Four Segments
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array of *n* integer numbers. Let *sum*(*l*,<=*r*) be the sum of all numbers on positions from *l* to *r* non-inclusive (*l*-th element is counted, *r*-th element is not counted). For indices *l* and *r* holds 0<=≤<=*l*<=≤<=*r*<=≤<=*n*. Indices in array are numbered from 0.
For example, if *a*<==<=[<=-<=5,<=3,<=9,<=4], then *sum*(0,<=1)<==<=<=-<=5, *sum*(0,<=2)<==<=<=-<=2, *sum*(1,<=4)<==<=16 and *sum*(*i*,<=*i*)<==<=0 for each *i* from 0 to 4.
Choose the indices of three delimiters *delim*0, *delim*1, *delim*2 (0<=≤<=*delim*0<=≤<=*delim*1<=≤<=*delim*2<=≤<=*n*) and divide the array in such a way that the value of *res*<==<=*sum*(0,<=*delim*0) - *sum*(*delim*0,<=*delim*1) + *sum*(*delim*1,<=*delim*2) - *sum*(*delim*2,<=*n*) is maximal.
Note that some of the expressions *sum*(*l*,<=*r*) can correspond to empty segments (if *l*<==<=*r* for some segment).
Input Specification:
The first line contains one integer number *n* (1<=≤<=*n*<=≤<=5000).
The second line contains *n* numbers *a*0,<=*a*1,<=...,<=*a**n*<=-<=1 (<=-<=109<=≤<=*a**i*<=≤<=109).
Output Specification:
Choose three indices so that the value of *res* is maximal. If there are multiple answers, print any of them.
Demo Input:
['3\n-1 2 3\n', '4\n0 0 -1 0\n', '1\n10000\n']
Demo Output:
['0 1 3\n', '0 0 0\n', '1 1 1\n']
Note:
none
|
```python
def list_input():
return list(map(int,input().split()))
def map_input():
return map(int,input().split())
def map_string():
return input().split()
n = int(input())
a = list_input()
pref = [0]
for i in range(n): pref.append(pref[i]+a[i])
ans = -10000000000
x=y=z=0
for i in range(n+1):
# print(i,"----")
cur = pref[i]
dp = [0]*(n+1)
mx = -10000000000
ind1 = -1
ind2 = -1
sinc = -1
for j in range(max(1,i+1),n+1):
if dp[j-1] > 0:
dp[j] = dp[j-1]+a[j-1]
else:
sinc = i
dp[j] = a[j-1]
if dp[j] > mx:
mx = dp[j]
ind1 = sinc
ind2 = j
# print(dp,ind1,ind2)
if mx+cur > ans:
ans = mx+cur
x = i
y = ind1+1
z = ind2
print(x,y,z)
```
| 0
|
|
755
|
A
|
PolandBall and Hypothesis
|
PROGRAMMING
| 800
|
[
"brute force",
"graphs",
"math",
"number theory"
] | null | null |
PolandBall is a young, clever Ball. He is interested in prime numbers. He has stated a following hypothesis: "There exists such a positive integer *n* that for each positive integer *m* number *n*·*m*<=+<=1 is a prime number".
Unfortunately, PolandBall is not experienced yet and doesn't know that his hypothesis is incorrect. Could you prove it wrong? Write a program that finds a counterexample for any *n*.
|
The only number in the input is *n* (1<=≤<=*n*<=≤<=1000) — number from the PolandBall's hypothesis.
|
Output such *m* that *n*·*m*<=+<=1 is not a prime number. Your answer will be considered correct if you output any suitable *m* such that 1<=≤<=*m*<=≤<=103. It is guaranteed the the answer exists.
|
[
"3\n",
"4\n"
] |
[
"1",
"2"
] |
A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.
For the first sample testcase, 3·1 + 1 = 4. We can output 1.
In the second sample testcase, 4·1 + 1 = 5. We cannot output 1 because 5 is prime. However, *m* = 2 is okay since 4·2 + 1 = 9, which is not a prime number.
| 500
|
[
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "2"
},
{
"input": "10",
"output": "2"
},
{
"input": "153",
"output": "1"
},
{
"input": "1000",
"output": "1"
},
{
"input": "1",
"output": "3"
},
{
"input": "2",
"output": "4"
},
{
"input": "5",
"output": "1"
},
{
"input": "6",
"output": "4"
},
{
"input": "7",
"output": "1"
},
{
"input": "8",
"output": "1"
},
{
"input": "9",
"output": "1"
},
{
"input": "11",
"output": "1"
},
{
"input": "998",
"output": "1"
},
{
"input": "996",
"output": "3"
},
{
"input": "36",
"output": "4"
},
{
"input": "210",
"output": "4"
},
{
"input": "270",
"output": "4"
},
{
"input": "306",
"output": "4"
},
{
"input": "330",
"output": "5"
},
{
"input": "336",
"output": "4"
},
{
"input": "600",
"output": "4"
},
{
"input": "726",
"output": "4"
},
{
"input": "988",
"output": "1"
},
{
"input": "12",
"output": "2"
},
{
"input": "987",
"output": "1"
},
{
"input": "13",
"output": "1"
},
{
"input": "986",
"output": "1"
},
{
"input": "14",
"output": "1"
},
{
"input": "985",
"output": "1"
},
{
"input": "15",
"output": "1"
},
{
"input": "984",
"output": "1"
},
{
"input": "16",
"output": "2"
},
{
"input": "983",
"output": "1"
},
{
"input": "17",
"output": "1"
},
{
"input": "982",
"output": "2"
},
{
"input": "18",
"output": "3"
},
{
"input": "981",
"output": "1"
},
{
"input": "19",
"output": "1"
},
{
"input": "980",
"output": "1"
},
{
"input": "20",
"output": "1"
},
{
"input": "979",
"output": "1"
},
{
"input": "21",
"output": "1"
},
{
"input": "978",
"output": "1"
},
{
"input": "22",
"output": "2"
},
{
"input": "977",
"output": "1"
},
{
"input": "23",
"output": "1"
},
{
"input": "976",
"output": "2"
},
{
"input": "24",
"output": "1"
},
{
"input": "975",
"output": "1"
},
{
"input": "25",
"output": "1"
},
{
"input": "2",
"output": "4"
},
{
"input": "6",
"output": "4"
},
{
"input": "996",
"output": "3"
},
{
"input": "8",
"output": "1"
},
{
"input": "998",
"output": "1"
},
{
"input": "330",
"output": "5"
},
{
"input": "18",
"output": "3"
},
{
"input": "12",
"output": "2"
},
{
"input": "24",
"output": "1"
},
{
"input": "999",
"output": "1"
},
{
"input": "726",
"output": "4"
},
{
"input": "876",
"output": "3"
}
] | 1,649,128,168
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 63
| 77
| 0
|
a = int(input())
if a%2==1:
print(3)
else:
print(min(a+2,999))
|
Title: PolandBall and Hypothesis
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
PolandBall is a young, clever Ball. He is interested in prime numbers. He has stated a following hypothesis: "There exists such a positive integer *n* that for each positive integer *m* number *n*·*m*<=+<=1 is a prime number".
Unfortunately, PolandBall is not experienced yet and doesn't know that his hypothesis is incorrect. Could you prove it wrong? Write a program that finds a counterexample for any *n*.
Input Specification:
The only number in the input is *n* (1<=≤<=*n*<=≤<=1000) — number from the PolandBall's hypothesis.
Output Specification:
Output such *m* that *n*·*m*<=+<=1 is not a prime number. Your answer will be considered correct if you output any suitable *m* such that 1<=≤<=*m*<=≤<=103. It is guaranteed the the answer exists.
Demo Input:
['3\n', '4\n']
Demo Output:
['1', '2']
Note:
A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.
For the first sample testcase, 3·1 + 1 = 4. We can output 1.
In the second sample testcase, 4·1 + 1 = 5. We cannot output 1 because 5 is prime. However, *m* = 2 is okay since 4·2 + 1 = 9, which is not a prime number.
|
```python
a = int(input())
if a%2==1:
print(3)
else:
print(min(a+2,999))
```
| 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",
"output": "1"
},
{
"input": "1000 9",
"output": "1"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "999 9",
"output": "1"
},
{
"input": "999 8",
"output": "2"
},
{
"input": "105 6",
"output": "2"
},
{
"input": "403 9",
"output": "3"
},
{
"input": "546 4",
"output": "4"
},
{
"input": "228 9",
"output": "5"
},
{
"input": "57 2",
"output": "6"
},
{
"input": "437 9",
"output": "7"
},
{
"input": "997 6",
"output": "8"
},
{
"input": "109 1",
"output": "9"
},
{
"input": "998 9",
"output": "5"
},
{
"input": "4 2",
"output": "3"
},
{
"input": "9 3",
"output": "7"
},
{
"input": "8 2",
"output": "4"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "1 7",
"output": "7"
},
{
"input": "1 8",
"output": "8"
},
{
"input": "100 3",
"output": "1"
},
{
"input": "1000 2",
"output": "1"
},
{
"input": "1000 4",
"output": "1"
},
{
"input": "1000 5",
"output": "1"
},
{
"input": "1000 6",
"output": "1"
},
{
"input": "1000 7",
"output": "1"
},
{
"input": "1000 8",
"output": "1"
},
{
"input": "23 4",
"output": "8"
},
{
"input": "33 1",
"output": "7"
},
{
"input": "33 2",
"output": "4"
},
{
"input": "666 5",
"output": "5"
},
{
"input": "2 3",
"output": "5"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "12 4",
"output": "2"
},
{
"input": "15 5",
"output": "1"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "25 5",
"output": "1"
},
{
"input": "2 9",
"output": "5"
},
{
"input": "6 7",
"output": "5"
},
{
"input": "8 9",
"output": "5"
},
{
"input": "2 7",
"output": "5"
},
{
"input": "4 7",
"output": "5"
},
{
"input": "2 1",
"output": "5"
},
{
"input": "261 1",
"output": "1"
}
] | 1,679,202,789
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 46
| 0
|
price_of_1_shovel, r = map(int, input().split())
for i in range(1, 100):
x = ((price_of_1_shovel * i) - r )%10
if x == 0 :
print(i)
break
y = (price_of_1_shovel * i)%10
if y == 0:
print(i)
break
|
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
price_of_1_shovel, r = map(int, input().split())
for i in range(1, 100):
x = ((price_of_1_shovel * i) - r )%10
if x == 0 :
print(i)
break
y = (price_of_1_shovel * i)%10
if y == 0:
print(i)
break
```
| 3
|
|
899
|
A
|
Splitting in Teams
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"greedy",
"math"
] | null | null |
There were *n* groups of students which came to write a training contest. A group is either one person who can write the contest with anyone else, or two people who want to write the contest in the same team.
The coach decided to form teams of exactly three people for this training. Determine the maximum number of teams of three people he can form. It is possible that he can't use all groups to form teams. For groups of two, either both students should write the contest, or both should not. If two students from a group of two will write the contest, they should be in the same team.
|
The first line contains single integer *n* (2<=≤<=*n*<=≤<=2·105) — the number of groups.
The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=2), where *a**i* is the number of people in group *i*.
|
Print the maximum number of teams of three people the coach can form.
|
[
"4\n1 1 2 1\n",
"2\n2 2\n",
"7\n2 2 2 1 1 1 1\n",
"3\n1 1 1\n"
] |
[
"1\n",
"0\n",
"3\n",
"1\n"
] |
In the first example the coach can form one team. For example, he can take students from the first, second and fourth groups.
In the second example he can't make a single team.
In the third example the coach can form three teams. For example, he can do this in the following way:
- The first group (of two people) and the seventh group (of one person), - The second group (of two people) and the sixth group (of one person), - The third group (of two people) and the fourth group (of one person).
| 500
|
[
{
"input": "4\n1 1 2 1",
"output": "1"
},
{
"input": "2\n2 2",
"output": "0"
},
{
"input": "7\n2 2 2 1 1 1 1",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "1"
},
{
"input": "3\n2 2 2",
"output": "0"
},
{
"input": "3\n1 2 1",
"output": "1"
},
{
"input": "5\n2 2 1 1 1",
"output": "2"
},
{
"input": "7\n1 1 2 2 1 2 1",
"output": "3"
},
{
"input": "10\n1 2 2 1 2 2 1 2 1 1",
"output": "5"
},
{
"input": "5\n2 2 2 1 2",
"output": "1"
},
{
"input": "43\n1 2 2 2 1 1 2 2 1 1 2 2 2 2 1 2 2 2 2 2 1 2 1 2 1 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2",
"output": "10"
},
{
"input": "72\n1 2 1 2 2 1 2 1 1 1 1 2 2 1 2 1 2 1 2 2 2 2 1 2 2 2 2 1 2 1 1 2 2 1 1 2 2 2 2 2 1 1 1 1 2 2 1 1 2 1 1 1 1 2 2 1 2 2 1 2 1 1 2 1 2 2 1 1 1 2 2 2",
"output": "34"
},
{
"input": "64\n2 2 1 1 1 2 1 1 1 2 2 1 2 2 2 1 2 2 2 1 1 1 1 2 1 2 1 2 1 1 2 2 1 1 2 2 1 1 1 1 2 2 1 1 1 2 1 2 2 2 2 2 2 2 1 1 2 1 1 1 2 2 1 2",
"output": "32"
},
{
"input": "20\n1 1 1 1 2 1 2 2 2 1 2 1 2 1 2 1 1 2 1 2",
"output": "9"
},
{
"input": "23\n1 1 1 1 2 1 2 1 1 1 2 2 2 2 2 2 1 2 1 2 2 1 1",
"output": "11"
},
{
"input": "201\n1 1 2 2 2 2 1 1 1 2 2 1 2 1 2 1 2 2 2 1 1 2 1 1 1 2 1 2 1 1 1 2 1 1 2 1 2 2 1 1 1 1 2 1 1 2 1 1 1 2 2 2 2 1 2 1 2 2 2 2 2 2 1 1 1 2 2 1 1 1 1 2 2 1 2 1 1 2 2 1 1 2 2 2 1 1 1 2 1 1 2 1 2 2 1 2 2 2 2 1 1 1 2 1 2 2 2 2 2 1 2 1 1 1 2 2 2 2 2 1 2 1 1 2 2 2 1 1 2 2 1 2 2 2 1 1 1 2 1 1 1 2 1 1 2 2 2 1 2 1 1 1 2 2 1 1 2 2 2 2 2 2 1 2 2 1 2 2 2 1 1 2 2 1 1 2 1 1 1 1 2 1 1 1 2 2 1 2 1 1 2 2 1 1 2 1 2 1 1 1 2",
"output": "100"
},
{
"input": "247\n2 2 1 2 1 2 2 2 2 2 2 1 1 2 2 1 2 1 1 1 2 1 1 1 1 2 1 1 2 2 1 2 1 1 1 2 2 2 1 1 2 1 1 2 1 1 1 2 1 2 1 2 2 1 1 2 1 2 2 1 2 1 2 1 1 2 1 1 1 2 2 1 1 2 2 1 1 2 1 1 1 2 2 2 2 1 2 2 2 2 2 2 1 2 2 2 2 1 1 1 1 1 1 1 1 1 2 1 2 2 1 2 1 2 2 2 1 2 2 2 1 1 2 2 1 1 1 2 1 1 1 1 2 2 1 2 2 1 1 1 2 1 2 2 1 2 1 1 1 2 2 2 2 2 1 2 2 2 1 1 1 2 1 2 1 1 2 2 2 2 1 1 2 2 2 1 2 2 2 1 2 1 1 2 2 2 2 1 2 2 1 1 1 2 1 2 1 1 1 2 2 1 1 2 1 1 2 1 2 1 1 2 1 1 1 1 2 1 1 1 1 2 2 1 2 1 1 2 1 2 2 1 2 2 2 1 2 2 1 2 2 1 1 1 2 2 2",
"output": "123"
},
{
"input": "4\n2 2 2 2",
"output": "0"
},
{
"input": "4\n1 1 1 1",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "2\n1 2",
"output": "1"
},
{
"input": "3\n1 1 2",
"output": "1"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n2 1 1",
"output": "1"
},
{
"input": "3\n2 1 2",
"output": "1"
},
{
"input": "3\n2 2 1",
"output": "1"
},
{
"input": "4\n1 1 1 2",
"output": "1"
},
{
"input": "4\n1 1 2 2",
"output": "2"
},
{
"input": "4\n1 2 2 2",
"output": "1"
},
{
"input": "4\n2 1 1 1",
"output": "1"
},
{
"input": "5\n1 1 1 1 2",
"output": "2"
},
{
"input": "14\n1 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "1"
},
{
"input": "38\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1",
"output": "1"
},
{
"input": "30\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1",
"output": "1"
},
{
"input": "20\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1",
"output": "1"
},
{
"input": "26\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2",
"output": "1"
},
{
"input": "6\n1 1 1 2 2 2",
"output": "3"
},
{
"input": "5\n2 1 1 1 1",
"output": "2"
},
{
"input": "9\n1 1 1 1 1 1 2 2 2",
"output": "4"
},
{
"input": "10\n2 2 1 1 1 1 1 1 1 1",
"output": "4"
},
{
"input": "6\n1 1 1 1 1 1",
"output": "2"
}
] | 1,561,152,918
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 1
| 140
| 409,600
|
from collections import *
input()
a = Counter(input().split())
print(min(a.values()) + (a['1'] - min(a.values())) // 3)
|
Title: Splitting in Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There were *n* groups of students which came to write a training contest. A group is either one person who can write the contest with anyone else, or two people who want to write the contest in the same team.
The coach decided to form teams of exactly three people for this training. Determine the maximum number of teams of three people he can form. It is possible that he can't use all groups to form teams. For groups of two, either both students should write the contest, or both should not. If two students from a group of two will write the contest, they should be in the same team.
Input Specification:
The first line contains single integer *n* (2<=≤<=*n*<=≤<=2·105) — the number of groups.
The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=2), where *a**i* is the number of people in group *i*.
Output Specification:
Print the maximum number of teams of three people the coach can form.
Demo Input:
['4\n1 1 2 1\n', '2\n2 2\n', '7\n2 2 2 1 1 1 1\n', '3\n1 1 1\n']
Demo Output:
['1\n', '0\n', '3\n', '1\n']
Note:
In the first example the coach can form one team. For example, he can take students from the first, second and fourth groups.
In the second example he can't make a single team.
In the third example the coach can form three teams. For example, he can do this in the following way:
- The first group (of two people) and the seventh group (of one person), - The second group (of two people) and the sixth group (of one person), - The third group (of two people) and the fourth group (of one person).
|
```python
from collections import *
input()
a = Counter(input().split())
print(min(a.values()) + (a['1'] - min(a.values())) // 3)
```
| 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"
},
{
"input": "8\n68931246",
"output": "77553333332222222"
},
{
"input": "7\n4424368",
"output": "75333332222222222"
},
{
"input": "6\n576825",
"output": "7755532222"
},
{
"input": "5\n97715",
"output": "7775332"
},
{
"input": "3\n915",
"output": "75332"
},
{
"input": "2\n26",
"output": "532"
},
{
"input": "1\n4",
"output": "322"
},
{
"input": "15\n028745260720699",
"output": "7777755533333332222222222"
},
{
"input": "13\n5761790121605",
"output": "7775555333322"
},
{
"input": "10\n3312667105",
"output": "755533332"
},
{
"input": "1\n7",
"output": "7"
},
{
"input": "15\n989898989898989",
"output": "777777777777777333333333333333322222222222222222222222222222"
},
{
"input": "15\n000000000000007",
"output": "7"
},
{
"input": "15\n999999999999990",
"output": "77777777777777333333333333333333333333333322222222222222"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "1\n3",
"output": "3"
},
{
"input": "1\n4",
"output": "322"
},
{
"input": "1\n5",
"output": "5"
},
{
"input": "1\n6",
"output": "53"
},
{
"input": "1\n7",
"output": "7"
},
{
"input": "1\n9",
"output": "7332"
},
{
"input": "2\n09",
"output": "7332"
},
{
"input": "13\n1337251172966",
"output": "777555333333222"
},
{
"input": "15\n987654329876543",
"output": "777777555533333333332222222222222"
},
{
"input": "9\n234567899",
"output": "777755333333322222222"
},
{
"input": "2\n99",
"output": "77333322"
},
{
"input": "2\n66",
"output": "5533"
},
{
"input": "3\n999",
"output": "777333333222"
},
{
"input": "5\n99999",
"output": "77777333333333322222"
},
{
"input": "9\n123456789",
"output": "77755333332222222"
},
{
"input": "9\n987654321",
"output": "77755333332222222"
},
{
"input": "3\n666",
"output": "555333"
},
{
"input": "2\n26",
"output": "532"
},
{
"input": "6\n555777",
"output": "777555"
},
{
"input": "10\n1234567899",
"output": "777755333333322222222"
},
{
"input": "4\n6666",
"output": "55553333"
},
{
"input": "4\n9754",
"output": "775333222"
},
{
"input": "2\n95",
"output": "75332"
},
{
"input": "14\n11122233344455",
"output": "55333333222222222"
},
{
"input": "12\n836544897832",
"output": "77777553333333222222222222222"
}
] | 1,650,371,342
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 4
| 46
| 0
|
li={'4':['2','2','3'],'6':['5','3','2'],'8':['2','2','2','7'],'9':['2','2','2','7','3','3']}
n=int(input())
s=list(input())
j=[]
for i in s:
if i!='0' and i!='1':
if i not in li:
j+=i
else:
j+=li[i]
j.sort(reverse=True)
print(''.join(j))
|
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
li={'4':['2','2','3'],'6':['5','3','2'],'8':['2','2','2','7'],'9':['2','2','2','7','3','3']}
n=int(input())
s=list(input())
j=[]
for i in s:
if i!='0' and i!='1':
if i not in li:
j+=i
else:
j+=li[i]
j.sort(reverse=True)
print(''.join(j))
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly.
Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of actions Valentin did.
The next *n* lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types:
1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and *w* is the word that Valentin said. 1. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and *w* is the word that Valentin said. 1. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and *s* is the guess — a lowercase English letter.
All words consist only of lowercase English letters. The total length of all words does not exceed 105.
It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter.
|
Output a single integer — the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined.
|
[
"5\n! abc\n. ad\n. b\n! cd\n? c\n",
"8\n! hello\n! codeforces\n? c\n. o\n? d\n? h\n. l\n? e\n",
"7\n! ababahalamaha\n? a\n? b\n? a\n? b\n? a\n? h\n"
] |
[
"1\n",
"2\n",
"0\n"
] |
In the first test case after the first action it becomes clear that the selected letter is one of the following: *a*, *b*, *c*. After the second action we can note that the selected letter is not *a*. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is *c*, but Valentin pronounces the word *cd* and gets an excessive electric shock.
In the second test case after the first two electric shocks we understand that the selected letter is *e* or *o*. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is *e*, but Valentin makes 3 more actions before he makes a correct hypothesis.
In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks.
| 0
|
[
{
"input": "5\n! abc\n. ad\n. b\n! cd\n? c",
"output": "1"
},
{
"input": "8\n! hello\n! codeforces\n? c\n. o\n? d\n? h\n. l\n? e",
"output": "2"
},
{
"input": "7\n! ababahalamaha\n? a\n? b\n? a\n? b\n? a\n? h",
"output": "0"
},
{
"input": "4\n! abcd\n! cdef\n? d\n? c",
"output": "0"
},
{
"input": "1\n? q",
"output": "0"
},
{
"input": "15\n. r\n? e\n. s\n. rw\n? y\n. fj\n. zftyd\n? r\n! wq\n? w\n? p\n. ours\n. dto\n. lbyfru\n? q",
"output": "2"
},
{
"input": "3\n. abcdefghijklmnopqrstuvwxy\n? a\n? z",
"output": "1"
},
{
"input": "3\n. abcdefghijklmnopqrstuvwxy\n! z\n? z",
"output": "1"
}
] | 1,514,042,188
| 4,288
|
Python 3
|
OK
|
TESTS
| 38
| 218
| 5,939,200
|
n=int(input())
alf='abcdefghijklmnopqrstuvwxyz'
so='';
j=n
for i in range(n):
s=input()
j-=1
if s[0]=='!':
so = '';
for e in alf:
if s.find(e)>-1: so+=e;
alf=so
elif s[0]=='.':
so = '';
for e in alf:
if s.find(e)==-1: so+=e;
alf=so
else:
alf=alf.replace(s[2],'')
if len(alf)==1: break
lol=0;
p=j;
for i in range(j-1):
s = input()
if s[0]=='!' or s[0]=='?': lol+=1
if not(p==0):s = input()
print(lol)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly.
Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of actions Valentin did.
The next *n* lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types:
1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and *w* is the word that Valentin said. 1. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and *w* is the word that Valentin said. 1. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and *s* is the guess — a lowercase English letter.
All words consist only of lowercase English letters. The total length of all words does not exceed 105.
It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter.
Output Specification:
Output a single integer — the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined.
Demo Input:
['5\n! abc\n. ad\n. b\n! cd\n? c\n', '8\n! hello\n! codeforces\n? c\n. o\n? d\n? h\n. l\n? e\n', '7\n! ababahalamaha\n? a\n? b\n? a\n? b\n? a\n? h\n']
Demo Output:
['1\n', '2\n', '0\n']
Note:
In the first test case after the first action it becomes clear that the selected letter is one of the following: *a*, *b*, *c*. After the second action we can note that the selected letter is not *a*. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is *c*, but Valentin pronounces the word *cd* and gets an excessive electric shock.
In the second test case after the first two electric shocks we understand that the selected letter is *e* or *o*. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is *e*, but Valentin makes 3 more actions before he makes a correct hypothesis.
In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks.
|
```python
n=int(input())
alf='abcdefghijklmnopqrstuvwxyz'
so='';
j=n
for i in range(n):
s=input()
j-=1
if s[0]=='!':
so = '';
for e in alf:
if s.find(e)>-1: so+=e;
alf=so
elif s[0]=='.':
so = '';
for e in alf:
if s.find(e)==-1: so+=e;
alf=so
else:
alf=alf.replace(s[2],'')
if len(alf)==1: break
lol=0;
p=j;
for i in range(j-1):
s = input()
if s[0]=='!' or s[0]=='?': lol+=1
if not(p==0):s = input()
print(lol)
```
| 3
|
|
116
|
A
|
Tram
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
|
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
|
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
|
[
"4\n0 3\n2 5\n4 2\n4 0\n"
] |
[
"6\n"
] |
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.
| 500
|
[
{
"input": "4\n0 3\n2 5\n4 2\n4 0",
"output": "6"
},
{
"input": "5\n0 4\n4 6\n6 5\n5 4\n4 0",
"output": "6"
},
{
"input": "10\n0 5\n1 7\n10 8\n5 3\n0 5\n3 3\n8 8\n0 6\n10 1\n9 0",
"output": "18"
},
{
"input": "3\n0 1\n1 1\n1 0",
"output": "1"
},
{
"input": "4\n0 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "3\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "5\n0 73\n73 189\n189 766\n766 0\n0 0",
"output": "766"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 1\n1 0",
"output": "1"
},
{
"input": "5\n0 917\n917 923\n904 992\n1000 0\n11 0",
"output": "1011"
},
{
"input": "5\n0 1\n1 2\n2 1\n1 2\n2 0",
"output": "2"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "20\n0 7\n2 1\n2 2\n5 7\n2 6\n6 10\n2 4\n0 4\n7 4\n8 0\n10 6\n2 1\n6 1\n1 7\n0 3\n8 7\n6 3\n6 3\n1 1\n3 0",
"output": "22"
},
{
"input": "5\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "10\n0 592\n258 598\n389 203\n249 836\n196 635\n478 482\n994 987\n1000 0\n769 0\n0 0",
"output": "1776"
},
{
"input": "10\n0 1\n1 0\n0 0\n0 0\n0 0\n0 1\n1 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "10\n0 926\n926 938\n938 931\n931 964\n937 989\n983 936\n908 949\n997 932\n945 988\n988 0",
"output": "1016"
},
{
"input": "10\n0 1\n1 2\n1 2\n2 2\n2 2\n2 2\n1 1\n1 1\n2 1\n2 0",
"output": "3"
},
{
"input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "10\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "50\n0 332\n332 268\n268 56\n56 711\n420 180\n160 834\n149 341\n373 777\n763 93\n994 407\n86 803\n700 132\n471 608\n429 467\n75 5\n638 305\n405 853\n316 478\n643 163\n18 131\n648 241\n241 766\n316 847\n640 380\n923 759\n789 41\n125 421\n421 9\n9 388\n388 829\n408 108\n462 856\n816 411\n518 688\n290 7\n405 912\n397 772\n396 652\n394 146\n27 648\n462 617\n514 433\n780 35\n710 705\n460 390\n194 508\n643 56\n172 469\n1000 0\n194 0",
"output": "2071"
},
{
"input": "50\n0 0\n0 1\n1 1\n0 1\n0 0\n1 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 1\n1 0\n0 1\n0 0\n1 1\n1 0\n0 1\n0 0\n1 1\n0 1\n1 0\n1 1\n1 0\n0 0\n1 1\n1 0\n0 1\n0 0\n0 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 0\n0 1\n1 0\n0 0\n0 1\n1 1\n1 1\n0 1\n0 0\n1 0\n1 0",
"output": "3"
},
{
"input": "50\n0 926\n926 971\n915 980\n920 965\n954 944\n928 952\n955 980\n916 980\n906 935\n944 913\n905 923\n912 922\n965 934\n912 900\n946 930\n931 983\n979 905\n925 969\n924 926\n910 914\n921 977\n934 979\n962 986\n942 909\n976 903\n982 982\n991 941\n954 929\n902 980\n947 983\n919 924\n917 943\n916 905\n907 913\n964 977\n984 904\n905 999\n950 970\n986 906\n993 970\n960 994\n963 983\n918 986\n980 900\n931 986\n993 997\n941 909\n907 909\n1000 0\n278 0",
"output": "1329"
},
{
"input": "2\n0 863\n863 0",
"output": "863"
},
{
"input": "50\n0 1\n1 2\n2 2\n1 1\n1 1\n1 2\n1 2\n1 1\n1 2\n1 1\n1 1\n1 2\n1 2\n1 1\n2 1\n2 2\n1 2\n2 2\n1 2\n2 1\n2 1\n2 2\n2 1\n1 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n2 2\n2 1\n1 2\n2 2\n1 2\n1 1\n1 1\n2 1\n2 1\n2 2\n2 1\n2 1\n1 2\n1 2\n1 2\n1 2\n2 0\n2 0\n2 0\n0 0",
"output": "8"
},
{
"input": "50\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "100\n0 1\n0 0\n0 0\n1 0\n0 0\n0 1\n0 1\n1 1\n0 0\n0 0\n1 1\n0 0\n1 1\n0 1\n1 1\n0 1\n1 1\n1 0\n1 0\n0 0\n1 0\n0 1\n1 0\n0 0\n0 0\n1 1\n1 1\n0 1\n0 0\n1 0\n1 1\n0 1\n1 0\n1 1\n0 1\n1 1\n1 0\n0 0\n0 0\n0 1\n0 0\n0 1\n1 1\n0 0\n1 1\n1 1\n0 0\n0 1\n1 0\n0 1\n0 0\n0 1\n0 1\n1 1\n1 1\n1 1\n0 0\n0 0\n1 1\n0 1\n0 1\n1 0\n0 0\n0 0\n1 1\n0 1\n0 1\n1 1\n1 1\n0 1\n1 1\n1 1\n0 0\n1 0\n0 1\n0 0\n0 0\n1 1\n1 1\n1 1\n1 1\n0 1\n1 0\n1 0\n1 0\n1 0\n1 0\n0 0\n1 0\n1 0\n0 0\n1 0\n0 0\n0 1\n1 0\n0 1\n1 0\n1 0\n1 0\n1 0",
"output": "11"
},
{
"input": "100\n0 2\n1 2\n2 1\n1 2\n1 2\n2 1\n2 2\n1 1\n1 1\n2 1\n1 2\n2 1\n1 2\n2 2\n2 2\n2 2\n1 2\n2 2\n2 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 2\n1 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n1 1\n2 2\n2 1\n1 2\n1 1\n1 2\n2 1\n2 2\n1 1\n2 1\n1 1\n2 1\n1 1\n1 2\n2 2\n2 2\n1 1\n2 2\n1 2\n2 1\n2 1\n1 1\n1 1\n1 2\n1 2\n1 1\n1 1\n2 1\n1 2\n1 2\n2 1\n2 2\n2 2\n2 2\n2 1\n2 2\n1 1\n1 2\n1 2\n1 1\n2 2\n2 2\n1 1\n2 1\n1 1\n1 2\n1 2\n1 2\n1 1\n1 1\n2 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 1\n2 2\n1 2\n2 0\n2 0\n2 0\n1 0",
"output": "7"
},
{
"input": "100\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 900\n900 0",
"output": "900"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1000\n1000 0",
"output": "1000"
},
{
"input": "3\n0 802\n175 188\n815 0",
"output": "815"
},
{
"input": "3\n0 910\n910 976\n976 0",
"output": "976"
},
{
"input": "3\n0 2\n2 1\n1 0",
"output": "2"
}
] | 1,688,039,035
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 71
| 92
| 0
|
n = int(input())
currentCapacity = 0
minCapacity = 0
for _ in range(n):
a, b = map(int, input().split())
currentCapacity -= a
currentCapacity += b
minCapacity = max(minCapacity, currentCapacity)
print(minCapacity)
|
Title: Tram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
Input Specification:
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
Output Specification:
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
Demo Input:
['4\n0 3\n2 5\n4 2\n4 0\n']
Demo Output:
['6\n']
Note:
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.
|
```python
n = int(input())
currentCapacity = 0
minCapacity = 0
for _ in range(n):
a, b = map(int, input().split())
currentCapacity -= a
currentCapacity += b
minCapacity = max(minCapacity, currentCapacity)
print(minCapacity)
```
| 3
|
|
59
|
A
|
Word
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Word
|
2
|
256
|
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
|
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
|
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
|
[
"HoUse\n",
"ViP\n",
"maTRIx\n"
] |
[
"house\n",
"VIP\n",
"matrix\n"
] |
none
| 500
|
[
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,592,974,388
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 218
| 6,656,000
|
letters = input()
countLwr = 0
countUpr = 0
for i in letters:
if i.islower():
countLwr += 1
else:
countUpr += 1
if countLwr == countUpr or countLwr > countUpr:
print(letters.lower())
else:
print(letters.upper())
|
Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none
|
```python
letters = input()
countLwr = 0
countUpr = 0
for i in letters:
if i.islower():
countLwr += 1
else:
countUpr += 1
if countLwr == countUpr or countLwr > countUpr:
print(letters.lower())
else:
print(letters.upper())
```
| 3.933102
|
602
|
B
|
Approximating a Constant Range
|
PROGRAMMING
| 1,400
|
[
"dp",
"implementation",
"two pointers"
] | null | null |
When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?
You're given a sequence of *n* data points *a*1,<=...,<=*a**n*. There aren't any big jumps between consecutive data points — for each 1<=≤<=*i*<=<<=*n*, it's guaranteed that |*a**i*<=+<=1<=-<=*a**i*|<=≤<=1.
A range [*l*,<=*r*] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let *M* be the maximum and *m* the minimum value of *a**i* for *l*<=≤<=*i*<=≤<=*r*; the range [*l*,<=*r*] is almost constant if *M*<=-<=*m*<=≤<=1.
Find the length of the longest almost constant range.
|
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000) — the number of data points.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000).
|
Print a single number — the maximum length of an almost constant range of the given sequence.
|
[
"5\n1 2 3 3 2\n",
"11\n5 4 5 5 6 7 8 8 8 7 6\n"
] |
[
"4\n",
"5\n"
] |
In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.
In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].
| 1,000
|
[
{
"input": "5\n1 2 3 3 2",
"output": "4"
},
{
"input": "11\n5 4 5 5 6 7 8 8 8 7 6",
"output": "5"
},
{
"input": "2\n3 2",
"output": "2"
},
{
"input": "4\n1001 1000 1000 1001",
"output": "4"
},
{
"input": "4\n1 1 2 3",
"output": "3"
},
{
"input": "3\n1 2 1",
"output": "3"
},
{
"input": "3\n1 2 3",
"output": "2"
},
{
"input": "18\n10 9 8 7 6 5 4 3 2 1 2 3 4 5 6 7 8 9",
"output": "3"
},
{
"input": "3\n1 2 2",
"output": "3"
},
{
"input": "4\n10 9 10 9",
"output": "4"
},
{
"input": "4\n4 3 2 3",
"output": "3"
},
{
"input": "4\n8 8 7 7",
"output": "4"
},
{
"input": "3\n99998 99999 100000",
"output": "2"
},
{
"input": "3\n100000 99999 99998",
"output": "2"
},
{
"input": "3\n1 1 1",
"output": "3"
},
{
"input": "2\n99999 100000",
"output": "2"
},
{
"input": "2\n100000 100000",
"output": "2"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "15\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "15"
}
] | 1,602,671,525
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 107
| 218
| 10,649,600
|
n = int(input())
arr = list(map(int,input().split()))
prev = - 1
cur = arr[0]
curcount = 1
prevcount = 0
ans = 1
next = 0
for i in range(1 , n ):
next = arr[i]
if (next == cur):
curcount += 1
elif (next == prev):
prevcount += curcount
prev = cur
cur = next
curcount = 1
else:
ans = max(ans , curcount + prevcount)
prev = cur
prevcount = curcount
cur = next
curcount = 1
#print(ans)
print(max(ans , prevcount + curcount))
|
Title: Approximating a Constant Range
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?
You're given a sequence of *n* data points *a*1,<=...,<=*a**n*. There aren't any big jumps between consecutive data points — for each 1<=≤<=*i*<=<<=*n*, it's guaranteed that |*a**i*<=+<=1<=-<=*a**i*|<=≤<=1.
A range [*l*,<=*r*] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let *M* be the maximum and *m* the minimum value of *a**i* for *l*<=≤<=*i*<=≤<=*r*; the range [*l*,<=*r*] is almost constant if *M*<=-<=*m*<=≤<=1.
Find the length of the longest almost constant range.
Input Specification:
The first line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000) — the number of data points.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000).
Output Specification:
Print a single number — the maximum length of an almost constant range of the given sequence.
Demo Input:
['5\n1 2 3 3 2\n', '11\n5 4 5 5 6 7 8 8 8 7 6\n']
Demo Output:
['4\n', '5\n']
Note:
In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.
In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].
|
```python
n = int(input())
arr = list(map(int,input().split()))
prev = - 1
cur = arr[0]
curcount = 1
prevcount = 0
ans = 1
next = 0
for i in range(1 , n ):
next = arr[i]
if (next == cur):
curcount += 1
elif (next == prev):
prevcount += curcount
prev = cur
cur = next
curcount = 1
else:
ans = max(ans , curcount + prevcount)
prev = cur
prevcount = curcount
cur = next
curcount = 1
#print(ans)
print(max(ans , prevcount + curcount))
```
| 3
|
|
41
|
A
|
Translation
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Translation
|
2
|
256
|
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
|
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
|
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
|
[
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,575,795,813
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 40
| 280
| 0
|
s = list(input())
t = list(input())
n = len(s)
N = len(t)
v = 0
if n != N:
print('NO')
else:
for i in range(n):
if s[i] != t[-i-1]:
v += 1
if v == 0:
print('YES')
else:
print('NO')
|
Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
s = list(input())
t = list(input())
n = len(s)
N = len(t)
v = 0
if n != N:
print('NO')
else:
for i in range(n):
if s[i] != t[-i-1]:
v += 1
if v == 0:
print('YES')
else:
print('NO')
```
| 3.93
|
906
|
A
|
Shockers
|
PROGRAMMING
| 1,600
|
[
"implementation",
"strings"
] | null | null |
Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly.
Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of actions Valentin did.
The next *n* lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types:
1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and *w* is the word that Valentin said. 1. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and *w* is the word that Valentin said. 1. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and *s* is the guess — a lowercase English letter.
All words consist only of lowercase English letters. The total length of all words does not exceed 105.
It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter.
|
Output a single integer — the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined.
|
[
"5\n! abc\n. ad\n. b\n! cd\n? c\n",
"8\n! hello\n! codeforces\n? c\n. o\n? d\n? h\n. l\n? e\n",
"7\n! ababahalamaha\n? a\n? b\n? a\n? b\n? a\n? h\n"
] |
[
"1\n",
"2\n",
"0\n"
] |
In the first test case after the first action it becomes clear that the selected letter is one of the following: *a*, *b*, *c*. After the second action we can note that the selected letter is not *a*. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is *c*, but Valentin pronounces the word *cd* and gets an excessive electric shock.
In the second test case after the first two electric shocks we understand that the selected letter is *e* or *o*. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is *e*, but Valentin makes 3 more actions before he makes a correct hypothesis.
In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks.
| 500
|
[
{
"input": "5\n! abc\n. ad\n. b\n! cd\n? c",
"output": "1"
},
{
"input": "8\n! hello\n! codeforces\n? c\n. o\n? d\n? h\n. l\n? e",
"output": "2"
},
{
"input": "7\n! ababahalamaha\n? a\n? b\n? a\n? b\n? a\n? h",
"output": "0"
},
{
"input": "4\n! abcd\n! cdef\n? d\n? c",
"output": "0"
},
{
"input": "1\n? q",
"output": "0"
},
{
"input": "15\n. r\n? e\n. s\n. rw\n? y\n. fj\n. zftyd\n? r\n! wq\n? w\n? p\n. ours\n. dto\n. lbyfru\n? q",
"output": "2"
},
{
"input": "3\n. abcdefghijklmnopqrstuvwxy\n? a\n? z",
"output": "1"
},
{
"input": "3\n. abcdefghijklmnopqrstuvwxy\n! z\n? z",
"output": "1"
}
] | 1,593,020,334
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 38
| 265
| 25,395,200
|
import math,sys
from sys import stdin, stdout
from collections import Counter, defaultdict, deque
input = stdin.readline
I = lambda:int(input())
li = lambda:list(map(int,input().split()))
def solve():
n=I()
vis=[1]*26
c=0
flag=0
ans=0
for i in range(n-1):
a,b=input().split(" ")
b=b.strip()
if(flag and a!='.'):
ans+=1
continue
if(a!='!'):
for i in range(len(b)):
vis[ord(b[i])-97]=0
elif(a=='!'):
c=Counter(b)
for i in range(26):
if(chr(97+i) not in c and vis[i]==1):
vis[i]=0
c=0
#print(*vis)
for i in range(26):
if(vis[i]==1):
c+=1
if(c==1):
flag=1
a,b=input().split(" ")
print(ans)
for _ in range(1):
solve()
|
Title: Shockers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly.
Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of actions Valentin did.
The next *n* lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types:
1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and *w* is the word that Valentin said. 1. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and *w* is the word that Valentin said. 1. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and *s* is the guess — a lowercase English letter.
All words consist only of lowercase English letters. The total length of all words does not exceed 105.
It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter.
Output Specification:
Output a single integer — the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined.
Demo Input:
['5\n! abc\n. ad\n. b\n! cd\n? c\n', '8\n! hello\n! codeforces\n? c\n. o\n? d\n? h\n. l\n? e\n', '7\n! ababahalamaha\n? a\n? b\n? a\n? b\n? a\n? h\n']
Demo Output:
['1\n', '2\n', '0\n']
Note:
In the first test case after the first action it becomes clear that the selected letter is one of the following: *a*, *b*, *c*. After the second action we can note that the selected letter is not *a*. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is *c*, but Valentin pronounces the word *cd* and gets an excessive electric shock.
In the second test case after the first two electric shocks we understand that the selected letter is *e* or *o*. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is *e*, but Valentin makes 3 more actions before he makes a correct hypothesis.
In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks.
|
```python
import math,sys
from sys import stdin, stdout
from collections import Counter, defaultdict, deque
input = stdin.readline
I = lambda:int(input())
li = lambda:list(map(int,input().split()))
def solve():
n=I()
vis=[1]*26
c=0
flag=0
ans=0
for i in range(n-1):
a,b=input().split(" ")
b=b.strip()
if(flag and a!='.'):
ans+=1
continue
if(a!='!'):
for i in range(len(b)):
vis[ord(b[i])-97]=0
elif(a=='!'):
c=Counter(b)
for i in range(26):
if(chr(97+i) not in c and vis[i]==1):
vis[i]=0
c=0
#print(*vis)
for i in range(26):
if(vis[i]==1):
c+=1
if(c==1):
flag=1
a,b=input().split(" ")
print(ans)
for _ in range(1):
solve()
```
| 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\n00000100101011",
"output": "10"
},
{
"input": "14\n01111101111111",
"output": "2"
},
{
"input": "18\n110010101101111111",
"output": "10"
},
{
"input": "11\n00010000011",
"output": "4"
},
{
"input": "10\n1000010110",
"output": "6"
},
{
"input": "15\n100000100000011",
"output": "4"
},
{
"input": "18\n011010101110111101",
"output": "8"
},
{
"input": "10\n0011011111",
"output": "6"
},
{
"input": "3\n011",
"output": "2"
},
{
"input": "14\n11111111111111",
"output": "0"
},
{
"input": "65\n11001000110001001011011110111100000000101001001010101111000100010",
"output": "48"
},
{
"input": "10\n0011111000",
"output": "10"
},
{
"input": "13\n1110000001110",
"output": "12"
},
{
"input": "3\n110",
"output": "2"
},
{
"input": "4\n1110",
"output": "2"
},
{
"input": "9\n001011001",
"output": "8"
},
{
"input": "2\n10",
"output": "2"
},
{
"input": "2\n01",
"output": "2"
},
{
"input": "12\n110000110000",
"output": "8"
},
{
"input": "3\n001",
"output": "2"
},
{
"input": "14\n11000011000000",
"output": "8"
},
{
"input": "19\n0000011111111100000",
"output": "18"
},
{
"input": "45\n011010001100001111110001011100000001101100111",
"output": "44"
},
{
"input": "18\n000001111111100000",
"output": "16"
},
{
"input": "4\n0101",
"output": "4"
},
{
"input": "12\n000101011001",
"output": "10"
}
] | 1,691,082,865
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 53
| 109
| 11,059,200
|
n = int(input())
s = input()
balance_indices = {0: -1}
balance = 0
max_length = 0
for i in range(n):
if s[i] == '0':
balance -= 1
else:
balance += 1
if balance in balance_indices:
max_length = max(max_length, i - balance_indices[balance])
else:
balance_indices[balance] = i
print(max_length)# 1691082864.9898152
|
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
n = int(input())
s = input()
balance_indices = {0: -1}
balance = 0
max_length = 0
for i in range(n):
if s[i] == '0':
balance -= 1
else:
balance += 1
if balance in balance_indices:
max_length = max(max_length, i - balance_indices[balance])
else:
balance_indices[balance] = i
print(max_length)# 1691082864.9898152
```
| 3
|
|
1,009
|
B
|
Minimum Ternary String
|
PROGRAMMING
| 1,400
|
[
"greedy",
"implementation"
] | null | null |
You are given a ternary string (it is a string which consists only of characters '0', '1' and '2').
You can swap any two adjacent (consecutive) characters '0' and '1' (i.e. replace "01" with "10" or vice versa) or any two adjacent (consecutive) characters '1' and '2' (i.e. replace "12" with "21" or vice versa).
For example, for string "010210" we can perform the following moves:
- "010210" $\rightarrow$ "100210"; - "010210" $\rightarrow$ "001210"; - "010210" $\rightarrow$ "010120"; - "010210" $\rightarrow$ "010201".
Note than you cannot swap "02" $\rightarrow$ "20" and vice versa. You cannot perform any other operations with the given string excluding described above.
You task is to obtain the minimum possible (lexicographically) string by using these swaps arbitrary number of times (possibly, zero).
String $a$ is lexicographically less than string $b$ (if strings $a$ and $b$ have the same length) if there exists some position $i$ ($1 \le i \le |a|$, where $|s|$ is the length of the string $s$) such that for every $j < i$ holds $a_j = b_j$, and $a_i < b_i$.
|
The first line of the input contains the string $s$ consisting only of characters '0', '1' and '2', its length is between $1$ and $10^5$ (inclusive).
|
Print a single string — the minimum possible (lexicographically) string you can obtain by using the swaps described above arbitrary number of times (possibly, zero).
|
[
"100210\n",
"11222121\n",
"20\n"
] |
[
"001120\n",
"11112222\n",
"20\n"
] |
none
| 0
|
[
{
"input": "100210",
"output": "001120"
},
{
"input": "11222121",
"output": "11112222"
},
{
"input": "20",
"output": "20"
},
{
"input": "1002",
"output": "0012"
},
{
"input": "10",
"output": "01"
},
{
"input": "000021",
"output": "000012"
},
{
"input": "021",
"output": "012"
},
{
"input": "2",
"output": "2"
},
{
"input": "201",
"output": "120"
},
{
"input": "2112120",
"output": "1112220"
},
{
"input": "102",
"output": "012"
},
{
"input": "202",
"output": "202"
},
{
"input": "220201",
"output": "122020"
},
{
"input": "12",
"output": "12"
},
{
"input": "100022202",
"output": "000122202"
},
{
"input": "01",
"output": "01"
},
{
"input": "1",
"output": "1"
}
] | 1,689,430,598
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 31
| 0
|
print("_RANDOM_GUESS_1689430598.602514")# 1689430598.6025329
|
Title: Minimum Ternary String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a ternary string (it is a string which consists only of characters '0', '1' and '2').
You can swap any two adjacent (consecutive) characters '0' and '1' (i.e. replace "01" with "10" or vice versa) or any two adjacent (consecutive) characters '1' and '2' (i.e. replace "12" with "21" or vice versa).
For example, for string "010210" we can perform the following moves:
- "010210" $\rightarrow$ "100210"; - "010210" $\rightarrow$ "001210"; - "010210" $\rightarrow$ "010120"; - "010210" $\rightarrow$ "010201".
Note than you cannot swap "02" $\rightarrow$ "20" and vice versa. You cannot perform any other operations with the given string excluding described above.
You task is to obtain the minimum possible (lexicographically) string by using these swaps arbitrary number of times (possibly, zero).
String $a$ is lexicographically less than string $b$ (if strings $a$ and $b$ have the same length) if there exists some position $i$ ($1 \le i \le |a|$, where $|s|$ is the length of the string $s$) such that for every $j < i$ holds $a_j = b_j$, and $a_i < b_i$.
Input Specification:
The first line of the input contains the string $s$ consisting only of characters '0', '1' and '2', its length is between $1$ and $10^5$ (inclusive).
Output Specification:
Print a single string — the minimum possible (lexicographically) string you can obtain by using the swaps described above arbitrary number of times (possibly, zero).
Demo Input:
['100210\n', '11222121\n', '20\n']
Demo Output:
['001120\n', '11112222\n', '20\n']
Note:
none
|
```python
print("_RANDOM_GUESS_1689430598.602514")# 1689430598.6025329
```
| 0
|
|
507
|
A
|
Amr and Music
|
PROGRAMMING
| 1,000
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
Amr is a young coder who likes music a lot. He always wanted to learn how to play music but he was busy coding so he got an idea.
Amr has *n* instruments, it takes *a**i* days to learn *i*-th instrument. Being busy, Amr dedicated *k* days to learn how to play the maximum possible number of instruments.
Amr asked for your help to distribute his free days between instruments so that he can achieve his goal.
|
The first line contains two numbers *n*, *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=10<=000), the number of instruments and number of days respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100), representing number of days required to learn the *i*-th instrument.
|
In the first line output one integer *m* representing the maximum number of instruments Amr can learn.
In the second line output *m* space-separated integers: the indices of instruments to be learnt. You may output indices in any order.
if there are multiple optimal solutions output any. It is not necessary to use all days for studying.
|
[
"4 10\n4 3 1 2\n",
"5 6\n4 3 1 1 2\n",
"1 3\n4\n"
] |
[
"4\n1 2 3 4",
"3\n1 3 4",
"0\n"
] |
In the first test Amr can learn all 4 instruments.
In the second test other possible solutions are: {2, 3, 5} or {3, 4, 5}.
In the third test Amr doesn't have enough time to learn the only presented instrument.
| 500
|
[
{
"input": "4 10\n4 3 1 2",
"output": "4\n1 2 3 4"
},
{
"input": "5 6\n4 3 1 1 2",
"output": "3\n3 4 5"
},
{
"input": "1 3\n4",
"output": "0"
},
{
"input": "2 100\n100 100",
"output": "1\n1"
},
{
"input": "3 150\n50 50 50",
"output": "3\n1 2 3"
},
{
"input": "4 0\n100 100 100 100",
"output": "0"
},
{
"input": "100 7567\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "75\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75"
},
{
"input": "68 3250\n95 84 67 7 82 75 100 39 31 45 69 100 8 97 13 58 74 40 88 69 35 91 94 28 62 85 51 97 37 15 87 51 24 96 89 49 53 54 35 17 23 54 51 91 94 18 26 92 79 63 23 37 98 43 16 44 82 25 100 59 97 3 60 92 76 58 56 50",
"output": "60\n1 2 3 4 5 6 8 9 10 11 13 15 16 17 18 19 20 21 22 23 24 25 26 27 29 30 31 32 33 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 54 55 56 57 58 60 62 63 64 65 66 67 68"
},
{
"input": "100 10000\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100"
},
{
"input": "25 1293\n96 13 7 2 81 72 39 45 5 88 47 23 60 81 54 46 63 52 41 57 2 87 90 28 93",
"output": "25\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25"
},
{
"input": "98 7454\n71 57 94 76 52 90 76 81 67 60 99 88 98 61 73 61 80 91 88 93 53 55 88 64 71 55 81 76 52 63 87 99 84 66 65 52 83 99 92 62 95 81 90 67 64 57 80 80 67 75 77 58 71 85 97 50 97 55 52 59 55 96 57 53 85 100 95 95 74 51 78 88 66 98 97 86 94 81 56 64 61 57 67 95 85 82 85 60 76 95 69 95 76 91 74 100 69 76",
"output": "98\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98"
},
{
"input": "5 249\n96 13 7 2 81",
"output": "5\n1 2 3 4 5"
},
{
"input": "61 3331\n12 63 99 56 57 70 53 21 41 82 97 63 42 91 18 84 99 78 85 89 6 63 76 28 33 78 100 46 78 78 32 13 11 12 73 50 34 60 12 73 9 19 88 100 28 51 50 45 51 10 78 38 25 22 8 40 71 55 56 83 44",
"output": "61\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61"
},
{
"input": "99 10000\n42 88 21 63 59 38 23 100 86 37 57 86 11 22 19 89 6 19 15 64 18 77 83 29 14 26 80 73 8 51 14 19 9 98 81 96 47 77 22 19 86 71 91 61 84 8 80 28 6 25 33 95 96 21 57 92 96 57 31 88 38 32 70 19 25 67 29 78 18 90 37 50 62 33 49 16 47 39 9 33 88 69 69 29 14 66 75 76 41 98 40 52 65 25 33 47 39 24 80",
"output": "99\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99"
},
{
"input": "89 4910\n44 9 31 70 85 72 55 9 85 84 63 43 92 85 10 34 83 28 73 45 62 7 34 52 89 58 24 10 28 6 72 45 57 36 71 34 26 24 38 59 5 15 48 82 58 99 8 77 49 84 14 58 29 46 88 50 13 7 58 23 40 63 96 23 46 31 17 8 59 93 12 76 69 20 43 44 91 78 68 94 37 27 100 65 40 25 52 30 97",
"output": "89\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89"
},
{
"input": "40 2110\n91 18 52 22 26 67 59 10 55 43 97 78 20 81 99 36 33 12 86 32 82 87 70 63 48 48 45 94 78 23 77 15 68 17 71 54 44 98 54 8",
"output": "39\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40"
},
{
"input": "27 1480\n38 95 9 36 21 70 19 89 35 46 7 31 88 25 10 72 81 32 65 83 68 57 50 20 73 42 12",
"output": "27\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27"
},
{
"input": "57 2937\n84 73 23 62 93 64 23 17 53 100 47 67 52 53 90 58 19 84 33 69 46 47 50 28 73 74 40 42 92 70 32 29 57 52 23 82 42 32 46 83 45 87 40 58 50 51 48 37 57 52 78 26 21 54 16 66 93",
"output": "55\n1 2 3 4 5 6 7 8 9 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56"
},
{
"input": "6 41\n6 8 9 8 9 8",
"output": "5\n1 2 3 4 6"
},
{
"input": "9 95\n9 11 12 11 12 11 8 11 10",
"output": "9\n1 2 3 4 5 6 7 8 9"
},
{
"input": "89 6512\n80 87 61 91 85 51 58 69 79 57 81 67 74 55 88 70 77 61 55 81 56 76 79 67 92 52 54 73 67 72 81 54 72 81 65 88 83 57 83 92 62 66 63 58 61 66 92 77 73 66 71 85 92 73 82 65 76 64 58 62 64 51 90 59 79 70 86 89 86 51 72 61 60 71 52 74 58 72 77 91 91 60 76 56 64 55 61 81 52",
"output": "89\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89"
},
{
"input": "5 29\n6 3 7 2 1",
"output": "5\n1 2 3 4 5"
},
{
"input": "5 49\n16 13 7 2 1",
"output": "5\n1 2 3 4 5"
},
{
"input": "6 84\n16 21 25 6 17 16",
"output": "5\n1 2 4 5 6"
},
{
"input": "4 9\n7 4 2 1",
"output": "3\n2 3 4"
},
{
"input": "50 2500\n50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50",
"output": "50\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50"
},
{
"input": "100 10000\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100"
},
{
"input": "100 100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100"
},
{
"input": "96 514\n6 3 7 2 1 2 9 5 5 8 7 3 10 1 4 6 3 2 1 7 2 7 10 8 3 8 10 4 8 8 2 5 3 2 1 4 4 8 4 3 3 7 4 4 2 7 8 3 9 2 2 6 3 4 8 6 7 5 4 3 10 7 6 5 10 1 7 10 7 7 8 2 1 2 3 10 9 8 8 2 7 1 2 7 10 1 2 2 3 8 6 2 9 6 9 6",
"output": "96\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96"
},
{
"input": "47 350\n6 1 9 12 8 8 11 4 4 8 8 3 3 2 12 7 7 7 12 2 9 1 5 10 6 1 5 2 6 3 9 13 8 3 10 10 10 10 6 9 10 10 8 5 12 11 3",
"output": "47\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47"
},
{
"input": "100 200\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100"
},
{
"input": "2 10000\n1 1",
"output": "2\n1 2"
},
{
"input": "1 2\n1",
"output": "1\n1"
},
{
"input": "1 3\n2",
"output": "1\n1"
},
{
"input": "34 4964\n37 27 90 83 36 59 80 7 28 41 97 72 64 8 40 30 76 4 92 51 52 44 42 13 38 64 60 66 47 93 30 35 71 71",
"output": "34\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34"
},
{
"input": "2 2\n1 10",
"output": "1\n1"
},
{
"input": "2 5\n1 1",
"output": "2\n1 2"
},
{
"input": "1 4\n3",
"output": "1\n1"
},
{
"input": "4 384\n1 2 3 4",
"output": "4\n1 2 3 4"
}
] | 1,567,882,538
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 109
| 0
|
nk=input().split()
n=int(nk[0])
k=int(nk[1])
a=list(map(int,input().split()))
a1=sorted(a)
i=0
l=[]
while k>0:
if a1[i]<=k:
if a.index(a1[i])+1 in l:
l.append(a.index(a1[i])+2)
else:
l.append(a.index(a1[i])+1)
k-=a1[i]
i+=1
else:
break
l.sort()
if len(l)==0:
print(0)
else:
for i in l:
print(i,end=' ')
|
Title: Amr and Music
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Amr is a young coder who likes music a lot. He always wanted to learn how to play music but he was busy coding so he got an idea.
Amr has *n* instruments, it takes *a**i* days to learn *i*-th instrument. Being busy, Amr dedicated *k* days to learn how to play the maximum possible number of instruments.
Amr asked for your help to distribute his free days between instruments so that he can achieve his goal.
Input Specification:
The first line contains two numbers *n*, *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=10<=000), the number of instruments and number of days respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100), representing number of days required to learn the *i*-th instrument.
Output Specification:
In the first line output one integer *m* representing the maximum number of instruments Amr can learn.
In the second line output *m* space-separated integers: the indices of instruments to be learnt. You may output indices in any order.
if there are multiple optimal solutions output any. It is not necessary to use all days for studying.
Demo Input:
['4 10\n4 3 1 2\n', '5 6\n4 3 1 1 2\n', '1 3\n4\n']
Demo Output:
['4\n1 2 3 4', '3\n1 3 4', '0\n']
Note:
In the first test Amr can learn all 4 instruments.
In the second test other possible solutions are: {2, 3, 5} or {3, 4, 5}.
In the third test Amr doesn't have enough time to learn the only presented instrument.
|
```python
nk=input().split()
n=int(nk[0])
k=int(nk[1])
a=list(map(int,input().split()))
a1=sorted(a)
i=0
l=[]
while k>0:
if a1[i]<=k:
if a.index(a1[i])+1 in l:
l.append(a.index(a1[i])+2)
else:
l.append(a.index(a1[i])+1)
k-=a1[i]
i+=1
else:
break
l.sort()
if len(l)==0:
print(0)
else:
for i in l:
print(i,end=' ')
```
| 0
|
|
952
|
A
|
Quirky Quantifiers
|
PROGRAMMING
| 800
|
[
"math"
] | null | null |
The input contains a single integer *a* (10<=≤<=*a*<=≤<=999).
Output 0 or 1.
|
The input contains a single integer *a* (10<=≤<=*a*<=≤<=999).
|
Output 0 or 1.
|
[
"13\n",
"927\n",
"48\n"
] |
[
"1\n",
"1\n",
"0\n"
] |
none
| 0
|
[
{
"input": "13",
"output": "1"
},
{
"input": "927",
"output": "1"
},
{
"input": "48",
"output": "0"
},
{
"input": "10",
"output": "0"
},
{
"input": "999",
"output": "1"
},
{
"input": "142",
"output": "0"
},
{
"input": "309",
"output": "1"
},
{
"input": "572",
"output": "0"
},
{
"input": "835",
"output": "1"
},
{
"input": "990",
"output": "0"
},
{
"input": "168",
"output": "0"
},
{
"input": "431",
"output": "1"
},
{
"input": "694",
"output": "0"
},
{
"input": "957",
"output": "1"
},
{
"input": "932",
"output": "0"
}
] | 1,683,701,560
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 0
|
n = int(input())
if n%6==0 or n%6==1 or n%6==3:
print(1)
else:
print(0)
|
Title: Quirky Quantifiers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The input contains a single integer *a* (10<=≤<=*a*<=≤<=999).
Output 0 or 1.
Input Specification:
The input contains a single integer *a* (10<=≤<=*a*<=≤<=999).
Output Specification:
Output 0 or 1.
Demo Input:
['13\n', '927\n', '48\n']
Demo Output:
['1\n', '1\n', '0\n']
Note:
none
|
```python
n = int(input())
if n%6==0 or n%6==1 or n%6==3:
print(1)
else:
print(0)
```
| 0
|
|
816
|
B
|
Karen and Coffee
|
PROGRAMMING
| 1,400
|
[
"binary search",
"data structures",
"implementation"
] | null | null |
To stay woke and attentive during classes, Karen needs some coffee!
Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally acclaimed "The Art of the Covfefe".
She knows *n* coffee recipes. The *i*-th recipe suggests that coffee should be brewed between *l**i* and *r**i* degrees, inclusive, to achieve the optimal taste.
Karen thinks that a temperature is admissible if at least *k* recipes recommend it.
Karen has a rather fickle mind, and so she asks *q* questions. In each question, given that she only wants to prepare coffee with a temperature between *a* and *b*, inclusive, can you tell her how many admissible integer temperatures fall within the range?
|
The first line of input contains three integers, *n*, *k* (1<=≤<=*k*<=≤<=*n*<=≤<=200000), and *q* (1<=≤<=*q*<=≤<=200000), the number of recipes, the minimum number of recipes a certain temperature must be recommended by to be admissible, and the number of questions Karen has, respectively.
The next *n* lines describe the recipes. Specifically, the *i*-th line among these contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=200000), describing that the *i*-th recipe suggests that the coffee be brewed between *l**i* and *r**i* degrees, inclusive.
The next *q* lines describe the questions. Each of these lines contains *a* and *b*, (1<=≤<=*a*<=≤<=*b*<=≤<=200000), describing that she wants to know the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
|
For each question, output a single integer on a line by itself, the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
|
[
"3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100\n",
"2 1 1\n1 1\n200000 200000\n90 100\n"
] |
[
"3\n3\n0\n4\n",
"0\n"
] |
In the first test case, Karen knows 3 recipes.
1. The first one recommends brewing the coffee between 91 and 94 degrees, inclusive. 1. The second one recommends brewing the coffee between 92 and 97 degrees, inclusive. 1. The third one recommends brewing the coffee between 97 and 99 degrees, inclusive.
A temperature is admissible if at least 2 recipes recommend it.
She asks 4 questions.
In her first question, she wants to know the number of admissible integer temperatures between 92 and 94 degrees, inclusive. There are 3: 92, 93 and 94 degrees are all admissible.
In her second question, she wants to know the number of admissible integer temperatures between 93 and 97 degrees, inclusive. There are 3: 93, 94 and 97 degrees are all admissible.
In her third question, she wants to know the number of admissible integer temperatures between 95 and 96 degrees, inclusive. There are none.
In her final question, she wants to know the number of admissible integer temperatures between 90 and 100 degrees, inclusive. There are 4: 92, 93, 94 and 97 degrees are all admissible.
In the second test case, Karen knows 2 recipes.
1. The first one, "wikiHow to make Cold Brew Coffee", recommends brewing the coffee at exactly 1 degree. 1. The second one, "What good is coffee that isn't brewed at at least 36.3306 times the temperature of the surface of the sun?", recommends brewing the coffee at exactly 200000 degrees.
A temperature is admissible if at least 1 recipe recommends it.
In her first and only question, she wants to know the number of admissible integer temperatures that are actually reasonable. There are none.
| 1,000
|
[
{
"input": "3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100",
"output": "3\n3\n0\n4"
},
{
"input": "2 1 1\n1 1\n200000 200000\n90 100",
"output": "0"
},
{
"input": "1 1 1\n1 1\n1 1",
"output": "1"
},
{
"input": "1 1 1\n200000 200000\n200000 200000",
"output": "1"
}
] | 1,564,761,138
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 93
| 0
|
from itertools import groupby
n,k,q = map(int, input().split())
permissble_temp = []
while(n):
l,r = map(int, input().split())
for i in range(l, r+1):
permissble_temp.append(i)
n -= 1
sorted_permissble_temp = sorted(permissble_temp)
temp_count = dict([(key, len(list(group))) for key, group in groupby(sorted_permissble_temp)])
print(temp_count)
while(q):
a, b = map(int, input().split())
num = 0
for i in range(a, b+1):
if temp_count[i] >= k:
num += 1
q -= 1
print(num)
|
Title: Karen and Coffee
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
To stay woke and attentive during classes, Karen needs some coffee!
Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally acclaimed "The Art of the Covfefe".
She knows *n* coffee recipes. The *i*-th recipe suggests that coffee should be brewed between *l**i* and *r**i* degrees, inclusive, to achieve the optimal taste.
Karen thinks that a temperature is admissible if at least *k* recipes recommend it.
Karen has a rather fickle mind, and so she asks *q* questions. In each question, given that she only wants to prepare coffee with a temperature between *a* and *b*, inclusive, can you tell her how many admissible integer temperatures fall within the range?
Input Specification:
The first line of input contains three integers, *n*, *k* (1<=≤<=*k*<=≤<=*n*<=≤<=200000), and *q* (1<=≤<=*q*<=≤<=200000), the number of recipes, the minimum number of recipes a certain temperature must be recommended by to be admissible, and the number of questions Karen has, respectively.
The next *n* lines describe the recipes. Specifically, the *i*-th line among these contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=200000), describing that the *i*-th recipe suggests that the coffee be brewed between *l**i* and *r**i* degrees, inclusive.
The next *q* lines describe the questions. Each of these lines contains *a* and *b*, (1<=≤<=*a*<=≤<=*b*<=≤<=200000), describing that she wants to know the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
Output Specification:
For each question, output a single integer on a line by itself, the number of admissible integer temperatures between *a* and *b* degrees, inclusive.
Demo Input:
['3 2 4\n91 94\n92 97\n97 99\n92 94\n93 97\n95 96\n90 100\n', '2 1 1\n1 1\n200000 200000\n90 100\n']
Demo Output:
['3\n3\n0\n4\n', '0\n']
Note:
In the first test case, Karen knows 3 recipes.
1. The first one recommends brewing the coffee between 91 and 94 degrees, inclusive. 1. The second one recommends brewing the coffee between 92 and 97 degrees, inclusive. 1. The third one recommends brewing the coffee between 97 and 99 degrees, inclusive.
A temperature is admissible if at least 2 recipes recommend it.
She asks 4 questions.
In her first question, she wants to know the number of admissible integer temperatures between 92 and 94 degrees, inclusive. There are 3: 92, 93 and 94 degrees are all admissible.
In her second question, she wants to know the number of admissible integer temperatures between 93 and 97 degrees, inclusive. There are 3: 93, 94 and 97 degrees are all admissible.
In her third question, she wants to know the number of admissible integer temperatures between 95 and 96 degrees, inclusive. There are none.
In her final question, she wants to know the number of admissible integer temperatures between 90 and 100 degrees, inclusive. There are 4: 92, 93, 94 and 97 degrees are all admissible.
In the second test case, Karen knows 2 recipes.
1. The first one, "wikiHow to make Cold Brew Coffee", recommends brewing the coffee at exactly 1 degree. 1. The second one, "What good is coffee that isn't brewed at at least 36.3306 times the temperature of the surface of the sun?", recommends brewing the coffee at exactly 200000 degrees.
A temperature is admissible if at least 1 recipe recommends it.
In her first and only question, she wants to know the number of admissible integer temperatures that are actually reasonable. There are none.
|
```python
from itertools import groupby
n,k,q = map(int, input().split())
permissble_temp = []
while(n):
l,r = map(int, input().split())
for i in range(l, r+1):
permissble_temp.append(i)
n -= 1
sorted_permissble_temp = sorted(permissble_temp)
temp_count = dict([(key, len(list(group))) for key, group in groupby(sorted_permissble_temp)])
print(temp_count)
while(q):
a, b = map(int, input().split())
num = 0
for i in range(a, b+1):
if temp_count[i] >= k:
num += 1
q -= 1
print(num)
```
| -1
|
|
545
|
D
|
Queue
|
PROGRAMMING
| 1,300
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
Little girl Susie went shopping with her mom and she wondered how to improve service quality.
There are *n* people in the queue. For each person we know time *t**i* needed to serve him. A person will be disappointed if the time he waits is more than the time needed to serve him. The time a person waits is the total time when all the people who stand in the queue in front of him are served. Susie thought that if we swap some people in the queue, then we can decrease the number of people who are disappointed.
Help Susie find out what is the maximum number of not disappointed people can be achieved by swapping people in the queue.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105).
The next line contains *n* integers *t**i* (1<=≤<=*t**i*<=≤<=109), separated by spaces.
|
Print a single number — the maximum number of not disappointed people in the queue.
|
[
"5\n15 2 1 5 3\n"
] |
[
"4\n"
] |
Value 4 is achieved at such an arrangement, for example: 1, 2, 3, 5, 15. Thus, you can make everything feel not disappointed except for the person with time 5.
| 1,750
|
[
{
"input": "5\n15 2 1 5 3",
"output": "4"
},
{
"input": "15\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "10\n13 2 5 55 21 34 1 8 1 3",
"output": "6"
},
{
"input": "10\n8 256 16 1 2 1 64 4 128 32",
"output": "10"
},
{
"input": "10\n10000 40000 10000 50000 20000 100000 10000 100 30000 500",
"output": "6"
},
{
"input": "1\n1000000000",
"output": "1"
},
{
"input": "15\n9 11 45 86 52 65 35 3 93 7 21 45 15 11 39",
"output": "6"
},
{
"input": "20\n16839799 17525904 91276752 42650694 60106463 12243176 54892123 25142243 16015971 41250998 11150057 6994983 67700784 16562412 82163675 46178521 33914268 91966607 93976858 84100064",
"output": "5"
},
{
"input": "26\n1000 4110030 64221 131521030 942374833 1003 2055015 32110 513757 16440130 263042057 32880256 128439 557559573 16051 8220066 469240078 65760513 256878 790176315 4012 2005 1027508 928528684 8030 805074697",
"output": "21"
}
] | 1,698,232,746
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
n = int(input())
queue = list(map(int,input().split()))
wait = 0
length = 0
for i in range(n):
if queue[i] >= wait:
wait += queue[i]
length += 1
print(length)
|
Title: Queue
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little girl Susie went shopping with her mom and she wondered how to improve service quality.
There are *n* people in the queue. For each person we know time *t**i* needed to serve him. A person will be disappointed if the time he waits is more than the time needed to serve him. The time a person waits is the total time when all the people who stand in the queue in front of him are served. Susie thought that if we swap some people in the queue, then we can decrease the number of people who are disappointed.
Help Susie find out what is the maximum number of not disappointed people can be achieved by swapping people in the queue.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105).
The next line contains *n* integers *t**i* (1<=≤<=*t**i*<=≤<=109), separated by spaces.
Output Specification:
Print a single number — the maximum number of not disappointed people in the queue.
Demo Input:
['5\n15 2 1 5 3\n']
Demo Output:
['4\n']
Note:
Value 4 is achieved at such an arrangement, for example: 1, 2, 3, 5, 15. Thus, you can make everything feel not disappointed except for the person with time 5.
|
```python
n = int(input())
queue = list(map(int,input().split()))
wait = 0
length = 0
for i in range(n):
if queue[i] >= wait:
wait += queue[i]
length += 1
print(length)
```
| 0
|
|
29
|
C
|
Mail Stamps
|
PROGRAMMING
| 1,700
|
[
"data structures",
"dfs and similar",
"graphs",
"implementation"
] |
C. Mail Stamps
|
2
|
256
|
One day Bob got a letter in an envelope. Bob knows that when Berland's post officers send a letter directly from city «A» to city «B», they stamp it with «A B», or «B A». Unfortunately, often it is impossible to send a letter directly from the city of the sender to the city of the receiver, that's why the letter is sent via some intermediate cities. Post officers never send a letter in such a way that the route of this letter contains some city more than once. Bob is sure that the post officers stamp the letters accurately.
There are *n* stamps on the envelope of Bob's letter. He understands that the possible routes of this letter are only two. But the stamps are numerous, and Bob can't determine himself none of these routes. That's why he asks you to help him. Find one of the possible routes of the letter.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of mail stamps on the envelope. Then there follow *n* lines with two integers each — description of the stamps. Each stamp is described with indexes of the cities between which a letter is sent. The indexes of cities are integers from 1 to 109. Indexes of all the cities are different. Every time the letter is sent from one city to another, exactly one stamp is put on the envelope. It is guaranteed that the given stamps correspond to some valid route from some city to some other city.
|
Output *n*<=+<=1 numbers — indexes of cities in one of the two possible routes of the letter.
|
[
"2\n1 100\n100 2\n",
"3\n3 1\n100 2\n3 2\n"
] |
[
"2 100 1 ",
"100 2 3 1 "
] |
none
| 1,500
|
[
{
"input": "2\n1 100\n100 2",
"output": "2 100 1 "
},
{
"input": "3\n3 1\n100 2\n3 2",
"output": "100 2 3 1 "
},
{
"input": "3\n458744979 589655889\n248228386 824699605\n458744979 824699605",
"output": "589655889 458744979 824699605 248228386 "
},
{
"input": "4\n90104473 221011623\n18773664 221011623\n90104473 74427905\n74427905 186329050",
"output": "186329050 74427905 90104473 221011623 18773664 "
},
{
"input": "5\n695442143 421284135\n641835294 542627184\n852367357 120042890\n641835294 852367357\n542627184 421284135",
"output": "695442143 421284135 542627184 641835294 852367357 120042890 "
},
{
"input": "6\n264896923 2497658\n57071588 447086061\n2497658 483723090\n57071588 264896923\n158310110 483723090\n158310110 72866107",
"output": "447086061 57071588 264896923 2497658 483723090 158310110 72866107 "
},
{
"input": "1\n1 1000000000",
"output": "1000000000 1 "
},
{
"input": "1\n1000000000 999999999",
"output": "1000000000 999999999 "
},
{
"input": "10\n661239801 721746596\n225324231 116454751\n687002568 865423160\n799202882 865423160\n661239801 116454751\n387882517 687002568\n748798833 721746596\n179630172 225324231\n945958362 387882517\n179630172 945958362",
"output": "799202882 865423160 687002568 387882517 945958362 179630172 225324231 116454751 661239801 721746596 748798833 "
},
{
"input": "21\n280810160 291988863\n760364563 140163983\n16417017 364832782\n400253359 677358550\n597688496 794948223\n400253359 603304541\n589408417 603304541\n385039298 307729574\n293170375 805849550\n140163983 219301181\n732214548 760364563\n307729574 280810160\n131915938 219301181\n4615652 347722938\n396478457 805849550\n16417017 732214548\n4615652 677358550\n131915938 589408417\n291988863 364832782\n396478457 794948223\n385039298 597688496",
"output": "347722938 4615652 677358550 400253359 603304541 589408417 131915938 219301181 140163983 760364563 732214548 16417017 364832782 291988863 280810160 307729574 385039298 597688496 794948223 396478457 805849550 293170375 "
},
{
"input": "1\n2105127 227379126",
"output": "227379126 2105127 "
}
] | 1,591,046,516
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 13
| 2,000
| 22,118,400
|
n=int(input())
d=dict()
a=[]
for j in range(n):
x,y=map(int,input().split())
if x in d.keys():
d[x].append(y)
else:
a.append(x)
d[x]=[y]
if y in d.keys():
d[y].append(x)
else:
a.append(y)
d[y] = [x]
r=len(a)
for j in range(r):
if len(d[a[j]])==1:
k=a[j]
break
c=[]
s=[k]
v=dict()
while(s):
p=s.pop()
v[p]=1
c.append(p)
if p!=k and len(d[p])==1:
break
if d[p][0] in v.keys():
s.append(d[p][1])
else:
s.append(d[p][0])
print(*c)
|
Title: Mail Stamps
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Bob got a letter in an envelope. Bob knows that when Berland's post officers send a letter directly from city «A» to city «B», they stamp it with «A B», or «B A». Unfortunately, often it is impossible to send a letter directly from the city of the sender to the city of the receiver, that's why the letter is sent via some intermediate cities. Post officers never send a letter in such a way that the route of this letter contains some city more than once. Bob is sure that the post officers stamp the letters accurately.
There are *n* stamps on the envelope of Bob's letter. He understands that the possible routes of this letter are only two. But the stamps are numerous, and Bob can't determine himself none of these routes. That's why he asks you to help him. Find one of the possible routes of the letter.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of mail stamps on the envelope. Then there follow *n* lines with two integers each — description of the stamps. Each stamp is described with indexes of the cities between which a letter is sent. The indexes of cities are integers from 1 to 109. Indexes of all the cities are different. Every time the letter is sent from one city to another, exactly one stamp is put on the envelope. It is guaranteed that the given stamps correspond to some valid route from some city to some other city.
Output Specification:
Output *n*<=+<=1 numbers — indexes of cities in one of the two possible routes of the letter.
Demo Input:
['2\n1 100\n100 2\n', '3\n3 1\n100 2\n3 2\n']
Demo Output:
['2 100 1 ', '100 2 3 1 ']
Note:
none
|
```python
n=int(input())
d=dict()
a=[]
for j in range(n):
x,y=map(int,input().split())
if x in d.keys():
d[x].append(y)
else:
a.append(x)
d[x]=[y]
if y in d.keys():
d[y].append(x)
else:
a.append(y)
d[y] = [x]
r=len(a)
for j in range(r):
if len(d[a[j]])==1:
k=a[j]
break
c=[]
s=[k]
v=dict()
while(s):
p=s.pop()
v[p]=1
c.append(p)
if p!=k and len(d[p])==1:
break
if d[p][0] in v.keys():
s.append(d[p][1])
else:
s.append(d[p][0])
print(*c)
```
| 0
|
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