message stringlengths 2 28.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 21 109k | cluster float64 7 7 | __index_level_0__ int64 42 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a grid with H rows and W columns. At first, all cells were painted white.
Snuke painted N of these cells. The i-th ( 1 \leq i \leq N ) cell he painted is the cell at the a_i-th row and b_i-th column.
Compute the following:
* For each integer j ( 0 \leq j \leq 9 ), how many subrectangles of size 3×3 of the grid contains exactly j black cells, after Snuke painted N cells?
Constraints
* 3 \leq H \leq 10^9
* 3 \leq W \leq 10^9
* 0 \leq N \leq min(10^5,H×W)
* 1 \leq a_i \leq H (1 \leq i \leq N)
* 1 \leq b_i \leq W (1 \leq i \leq N)
* (a_i, b_i) \neq (a_j, b_j) (i \neq j)
Input
The input is given from Standard Input in the following format:
H W N
a_1 b_1
:
a_N b_N
Output
Print 10 lines. The (j+1)-th ( 0 \leq j \leq 9 ) line should contain the number of the subrectangles of size 3×3 of the grid that contains exactly j black cells.
Examples
Input
4 5 8
1 1
1 4
1 5
2 3
3 1
3 2
3 4
4 4
Output
0
0
0
2
4
0
0
0
0
0
Input
10 10 20
1 1
1 4
1 9
2 5
3 10
4 2
4 7
5 9
6 4
6 6
6 7
7 1
7 3
7 7
8 1
8 5
8 10
9 2
10 4
10 9
Output
4
26
22
10
2
0
0
0
0
0
Input
1000000000 1000000000 0
Output
999999996000000004
0
0
0
0
0
0
0
0
0
Submitted Solution:
```
# coding: utf-8
# Your code here!
h,w,n = map(int,input().split())
s = [[0 for j in range(w+2)] for i in range(h+2)]
#print(s)
for i in range(0,n):
x,y = map(int,input().split())
s[x-1][y-1] += 1
s[x-1][y] += 1
s[x-1][y+1] += 1
s[x][y-1] += 1
s[x][y] += 1
s[x][y+1] += 1
s[x+1][y-1] += 1
s[x+1][y] += 1
s[x+1][y+1] += 1
#print(s)
ans = [0] * 10
for i in range(2,h):
for j in range(2,w):
ans[s[i][j]] += 1
for i in range(0,10):
print(ans[i])
``` | instruction | 0 | 91,610 | 7 | 183,220 |
No | output | 1 | 91,610 | 7 | 183,221 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of n painters who decided to organize their work as follows.
Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these n colors. Adding the j-th color to the i-th picture takes the j-th painter tij units of time.
Order is important everywhere, so the painters' work is ordered by the following rules:
* Each picture is first painted by the first painter, then by the second one, and so on. That is, after the j-th painter finishes working on the picture, it must go to the (j + 1)-th painter (if j < n);
* each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on;
* each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest;
* as soon as the j-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.
Given that the painters start working at time 0, find for each picture the time when it is ready for sale.
Input
The first line of the input contains integers m, n (1 ≤ m ≤ 50000, 1 ≤ n ≤ 5), where m is the number of pictures and n is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains n integers ti1, ti2, ..., tin (1 ≤ tij ≤ 1000), where tij is the time the j-th painter needs to work on the i-th picture.
Output
Print the sequence of m integers r1, r2, ..., rm, where ri is the moment when the n-th painter stopped working on the i-th picture.
Examples
Input
5 1
1
2
3
4
5
Output
1 3 6 10 15
Input
4 2
2 5
3 1
5 3
10 1
Output
7 8 13 21 | instruction | 0 | 92,114 | 7 | 184,228 |
Tags: brute force, dp, implementation
Correct Solution:
```
n,m=map(int,input().split())
l=[]
b=[0]*m
for i in range(n):
l.append(list(map(int,input().split())))
for i in range(n):
for j in range(m):
if j==0:
b[0]+=l[i][0]
else:
if b[j]>=b[j-1]:
b[j]+=l[i][j]
else:
b[j]=b[j-1]+l[i][j]
print(b[-1],end=" ")
``` | output | 1 | 92,114 | 7 | 184,229 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of n painters who decided to organize their work as follows.
Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these n colors. Adding the j-th color to the i-th picture takes the j-th painter tij units of time.
Order is important everywhere, so the painters' work is ordered by the following rules:
* Each picture is first painted by the first painter, then by the second one, and so on. That is, after the j-th painter finishes working on the picture, it must go to the (j + 1)-th painter (if j < n);
* each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on;
* each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest;
* as soon as the j-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.
Given that the painters start working at time 0, find for each picture the time when it is ready for sale.
Input
The first line of the input contains integers m, n (1 ≤ m ≤ 50000, 1 ≤ n ≤ 5), where m is the number of pictures and n is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains n integers ti1, ti2, ..., tin (1 ≤ tij ≤ 1000), where tij is the time the j-th painter needs to work on the i-th picture.
Output
Print the sequence of m integers r1, r2, ..., rm, where ri is the moment when the n-th painter stopped working on the i-th picture.
Examples
Input
5 1
1
2
3
4
5
Output
1 3 6 10 15
Input
4 2
2 5
3 1
5 3
10 1
Output
7 8 13 21 | instruction | 0 | 92,115 | 7 | 184,230 |
Tags: brute force, dp, implementation
Correct Solution:
```
from sys import stdin,stdout
a,b=map(int,stdin.readline().split())
z=[list(map(int,stdin.readline().split())) for _ in " "*a]
k=[[0]*b for _ in " "*a];ans=['']*a
for i in range(a):
if i==0:
for j in range(b):k[i][j]+=k[i][j-1]+z[i][j]
else:
for j in range(b):
if j==0:k[i][j]+=k[i-1][j]+z[i][j]
else:k[i][j]=max(k[i-1][j],k[i][j-1])+z[i][j]
ans[i]=str(k[i][b-1])
stdout.write(' '.join(ans))
``` | output | 1 | 92,115 | 7 | 184,231 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of n painters who decided to organize their work as follows.
Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these n colors. Adding the j-th color to the i-th picture takes the j-th painter tij units of time.
Order is important everywhere, so the painters' work is ordered by the following rules:
* Each picture is first painted by the first painter, then by the second one, and so on. That is, after the j-th painter finishes working on the picture, it must go to the (j + 1)-th painter (if j < n);
* each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on;
* each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest;
* as soon as the j-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.
Given that the painters start working at time 0, find for each picture the time when it is ready for sale.
Input
The first line of the input contains integers m, n (1 ≤ m ≤ 50000, 1 ≤ n ≤ 5), where m is the number of pictures and n is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains n integers ti1, ti2, ..., tin (1 ≤ tij ≤ 1000), where tij is the time the j-th painter needs to work on the i-th picture.
Output
Print the sequence of m integers r1, r2, ..., rm, where ri is the moment when the n-th painter stopped working on the i-th picture.
Examples
Input
5 1
1
2
3
4
5
Output
1 3 6 10 15
Input
4 2
2 5
3 1
5 3
10 1
Output
7 8 13 21 | instruction | 0 | 92,116 | 7 | 184,232 |
Tags: brute force, dp, implementation
Correct Solution:
```
m, n = map(int, input().split())
t = [[int(i) for i in input().split()] for _ in range(m)]
a = [[0 for _ in range(n)] for _ in range(m)]
for i in range(m):
for j in range(n):
x = 0
if i - 1 >= 0:
x = max(x, a[i - 1][j])
if j - 1 >= 0:
x = max(x, a[i][j - 1])
a[i][j] = x + t[i][j]
for i in range(m):
print(a[i][-1], end=' ')
print()
``` | output | 1 | 92,116 | 7 | 184,233 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of n painters who decided to organize their work as follows.
Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these n colors. Adding the j-th color to the i-th picture takes the j-th painter tij units of time.
Order is important everywhere, so the painters' work is ordered by the following rules:
* Each picture is first painted by the first painter, then by the second one, and so on. That is, after the j-th painter finishes working on the picture, it must go to the (j + 1)-th painter (if j < n);
* each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on;
* each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest;
* as soon as the j-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.
Given that the painters start working at time 0, find for each picture the time when it is ready for sale.
Input
The first line of the input contains integers m, n (1 ≤ m ≤ 50000, 1 ≤ n ≤ 5), where m is the number of pictures and n is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains n integers ti1, ti2, ..., tin (1 ≤ tij ≤ 1000), where tij is the time the j-th painter needs to work on the i-th picture.
Output
Print the sequence of m integers r1, r2, ..., rm, where ri is the moment when the n-th painter stopped working on the i-th picture.
Examples
Input
5 1
1
2
3
4
5
Output
1 3 6 10 15
Input
4 2
2 5
3 1
5 3
10 1
Output
7 8 13 21 | instruction | 0 | 92,117 | 7 | 184,234 |
Tags: brute force, dp, implementation
Correct Solution:
```
m, n = map(int, input().split())
mat = [list(map(int, input().split())) for _ in range(m)]
t = [[0] * m for _ in range(n)]
for p in range(n):
for i in range(m):
t[p][i] = max(t[p - 1][i] if p > 0 else 0, t[p][i - 1]) + mat[i][p]
print(' '.join(map(str, t[n - 1])))
``` | output | 1 | 92,117 | 7 | 184,235 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of n painters who decided to organize their work as follows.
Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these n colors. Adding the j-th color to the i-th picture takes the j-th painter tij units of time.
Order is important everywhere, so the painters' work is ordered by the following rules:
* Each picture is first painted by the first painter, then by the second one, and so on. That is, after the j-th painter finishes working on the picture, it must go to the (j + 1)-th painter (if j < n);
* each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on;
* each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest;
* as soon as the j-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.
Given that the painters start working at time 0, find for each picture the time when it is ready for sale.
Input
The first line of the input contains integers m, n (1 ≤ m ≤ 50000, 1 ≤ n ≤ 5), where m is the number of pictures and n is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains n integers ti1, ti2, ..., tin (1 ≤ tij ≤ 1000), where tij is the time the j-th painter needs to work on the i-th picture.
Output
Print the sequence of m integers r1, r2, ..., rm, where ri is the moment when the n-th painter stopped working on the i-th picture.
Examples
Input
5 1
1
2
3
4
5
Output
1 3 6 10 15
Input
4 2
2 5
3 1
5 3
10 1
Output
7 8 13 21 | instruction | 0 | 92,118 | 7 | 184,236 |
Tags: brute force, dp, implementation
Correct Solution:
```
n , m = map(int,input().split())
painters = [0] * (m+1)
l = []
for i in range(n):
l.append(list(map(int,input().split())))
ans = []
for i in range(n):
for j in range(m):
if i == 0 :
if j == 0 :
painters[j+1] += l[i][j]
else:
painters[j+1]= painters[j] + l[i][j]
else:
if j == 0 :
painters[j+1] += l[i][j]
else:
if painters[j] <= painters[j+1]:
painters[j+1] += l[i][j]
else:
painters[j+1] = painters[j] + l[i][j]
ans.append(painters[m])
print(*ans)
``` | output | 1 | 92,118 | 7 | 184,237 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of n painters who decided to organize their work as follows.
Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these n colors. Adding the j-th color to the i-th picture takes the j-th painter tij units of time.
Order is important everywhere, so the painters' work is ordered by the following rules:
* Each picture is first painted by the first painter, then by the second one, and so on. That is, after the j-th painter finishes working on the picture, it must go to the (j + 1)-th painter (if j < n);
* each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on;
* each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest;
* as soon as the j-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.
Given that the painters start working at time 0, find for each picture the time when it is ready for sale.
Input
The first line of the input contains integers m, n (1 ≤ m ≤ 50000, 1 ≤ n ≤ 5), where m is the number of pictures and n is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains n integers ti1, ti2, ..., tin (1 ≤ tij ≤ 1000), where tij is the time the j-th painter needs to work on the i-th picture.
Output
Print the sequence of m integers r1, r2, ..., rm, where ri is the moment when the n-th painter stopped working on the i-th picture.
Examples
Input
5 1
1
2
3
4
5
Output
1 3 6 10 15
Input
4 2
2 5
3 1
5 3
10 1
Output
7 8 13 21 | instruction | 0 | 92,119 | 7 | 184,238 |
Tags: brute force, dp, implementation
Correct Solution:
```
R = lambda:map(int, input().split())
m, n = R()
t = [0] * (n + 1)
c = []
for i in range(m):
a = list(R())
for j in range(n):
t[j + 1] = max(t[j], t[j + 1]) + a[j]
c += [str(t[n])]
print(" ".join(c))
``` | output | 1 | 92,119 | 7 | 184,239 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of n painters who decided to organize their work as follows.
Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these n colors. Adding the j-th color to the i-th picture takes the j-th painter tij units of time.
Order is important everywhere, so the painters' work is ordered by the following rules:
* Each picture is first painted by the first painter, then by the second one, and so on. That is, after the j-th painter finishes working on the picture, it must go to the (j + 1)-th painter (if j < n);
* each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on;
* each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest;
* as soon as the j-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.
Given that the painters start working at time 0, find for each picture the time when it is ready for sale.
Input
The first line of the input contains integers m, n (1 ≤ m ≤ 50000, 1 ≤ n ≤ 5), where m is the number of pictures and n is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains n integers ti1, ti2, ..., tin (1 ≤ tij ≤ 1000), where tij is the time the j-th painter needs to work on the i-th picture.
Output
Print the sequence of m integers r1, r2, ..., rm, where ri is the moment when the n-th painter stopped working on the i-th picture.
Examples
Input
5 1
1
2
3
4
5
Output
1 3 6 10 15
Input
4 2
2 5
3 1
5 3
10 1
Output
7 8 13 21 | instruction | 0 | 92,120 | 7 | 184,240 |
Tags: brute force, dp, implementation
Correct Solution:
```
m, n = map(int, input().split())
t = [[]] * m
ans = [[]] * m
for i in range(m):
t[i] = list(map(int, input().split()))
ans[i] = [0] * n
for i in range(m):
for j in range(n):
a1 = ans[i - 1][j] if i > 0 else 0
a2 = ans[i][j - 1] if j > 0 else 0
ans[i][j] = max(a1, a2) + t[i][j]
print(' '.join(map(lambda o: str(ans[o][n - 1]), range(m))))
``` | output | 1 | 92,120 | 7 | 184,241 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of n painters who decided to organize their work as follows.
Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these n colors. Adding the j-th color to the i-th picture takes the j-th painter tij units of time.
Order is important everywhere, so the painters' work is ordered by the following rules:
* Each picture is first painted by the first painter, then by the second one, and so on. That is, after the j-th painter finishes working on the picture, it must go to the (j + 1)-th painter (if j < n);
* each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on;
* each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest;
* as soon as the j-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.
Given that the painters start working at time 0, find for each picture the time when it is ready for sale.
Input
The first line of the input contains integers m, n (1 ≤ m ≤ 50000, 1 ≤ n ≤ 5), where m is the number of pictures and n is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains n integers ti1, ti2, ..., tin (1 ≤ tij ≤ 1000), where tij is the time the j-th painter needs to work on the i-th picture.
Output
Print the sequence of m integers r1, r2, ..., rm, where ri is the moment when the n-th painter stopped working on the i-th picture.
Examples
Input
5 1
1
2
3
4
5
Output
1 3 6 10 15
Input
4 2
2 5
3 1
5 3
10 1
Output
7 8 13 21 | instruction | 0 | 92,121 | 7 | 184,242 |
Tags: brute force, dp, implementation
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
##########################################################
from collections import Counter
# c=sorted((i,int(val))for i,val in enumerate(input().split()))
import heapq
# c=sorted((i,int(val))for i,val in enumerate(input().split()))
# n = int(input())
# ls = list(map(int, input().split()))
# n, k = map(int, input().split())
# n =int(input())
# e=list(map(int, input().split()))
from collections import Counter
# n =int(input())
n, k = map(int, input().split())
t=[]
for _ in range(n):
#n, k = map(int, input().split())
arr=list(map(int, input().split()))
t.append(arr)
time=[0]*n
for i in range(k):
for j in range(n):
if j==0:
time[j]+=t[j][i]
else:
if j>0 and n>1:
time[j]=max(time[j],time[j-1])+t[j][i]
else:
time[j]+=t[j][i]
#print(time)
print(*time)
``` | output | 1 | 92,121 | 7 | 184,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of n painters who decided to organize their work as follows.
Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these n colors. Adding the j-th color to the i-th picture takes the j-th painter tij units of time.
Order is important everywhere, so the painters' work is ordered by the following rules:
* Each picture is first painted by the first painter, then by the second one, and so on. That is, after the j-th painter finishes working on the picture, it must go to the (j + 1)-th painter (if j < n);
* each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on;
* each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest;
* as soon as the j-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.
Given that the painters start working at time 0, find for each picture the time when it is ready for sale.
Input
The first line of the input contains integers m, n (1 ≤ m ≤ 50000, 1 ≤ n ≤ 5), where m is the number of pictures and n is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains n integers ti1, ti2, ..., tin (1 ≤ tij ≤ 1000), where tij is the time the j-th painter needs to work on the i-th picture.
Output
Print the sequence of m integers r1, r2, ..., rm, where ri is the moment when the n-th painter stopped working on the i-th picture.
Examples
Input
5 1
1
2
3
4
5
Output
1 3 6 10 15
Input
4 2
2 5
3 1
5 3
10 1
Output
7 8 13 21
Submitted Solution:
```
m,n = list(map(int,input().split()))
t = [[0]*n] + [list(map(int,input().split())) for _ in range(m)]
for i in range(1,m+1):
t[i][0] += t[i-1][0]
for j in range(1,n):
t[i][j] += max(t[i-1][j],t[i][j-1])
print(*[t[i][-1] for i in range(1,m+1)])
``` | instruction | 0 | 92,122 | 7 | 184,244 |
Yes | output | 1 | 92,122 | 7 | 184,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of n painters who decided to organize their work as follows.
Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these n colors. Adding the j-th color to the i-th picture takes the j-th painter tij units of time.
Order is important everywhere, so the painters' work is ordered by the following rules:
* Each picture is first painted by the first painter, then by the second one, and so on. That is, after the j-th painter finishes working on the picture, it must go to the (j + 1)-th painter (if j < n);
* each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on;
* each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest;
* as soon as the j-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.
Given that the painters start working at time 0, find for each picture the time when it is ready for sale.
Input
The first line of the input contains integers m, n (1 ≤ m ≤ 50000, 1 ≤ n ≤ 5), where m is the number of pictures and n is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains n integers ti1, ti2, ..., tin (1 ≤ tij ≤ 1000), where tij is the time the j-th painter needs to work on the i-th picture.
Output
Print the sequence of m integers r1, r2, ..., rm, where ri is the moment when the n-th painter stopped working on the i-th picture.
Examples
Input
5 1
1
2
3
4
5
Output
1 3 6 10 15
Input
4 2
2 5
3 1
5 3
10 1
Output
7 8 13 21
Submitted Solution:
```
"""
Codeforces Round 241 Div 1 Problem B
Author : chaotic_iak
Language: Python 3.3.4
"""
class InputHandlerObject(object):
inputs = []
def getInput(self, n = 0):
res = ""
inputs = self.inputs
if not inputs: inputs.extend(input().split(" "))
if n == 0:
res = inputs[:]
inputs[:] = []
while n > len(inputs):
inputs.extend(input().split(" "))
if n > 0:
res = inputs[:n]
inputs[:n] = []
return res
InputHandler = InputHandlerObject()
g = InputHandler.getInput
############################## SOLUTION ##############################
m,n = [int(x) for x in g()]
q = []
for i in range(m):
q.append([int(x) for x in g()])
p = q[:]
for j in range(n):
for i in range(m):
if i == 0 and j == 0:
continue
if i == 0:
p[i][j] = p[i][j-1] + q[i][j]
continue
if j == 0:
p[i][j] = p[i-1][j] + q[i][j]
continue
p[i][j] = max(p[i-1][j], p[i][j-1]) + q[i][j]
for i in range(m):
print(p[i][n-1], end=" ")
``` | instruction | 0 | 92,123 | 7 | 184,246 |
Yes | output | 1 | 92,123 | 7 | 184,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of n painters who decided to organize their work as follows.
Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these n colors. Adding the j-th color to the i-th picture takes the j-th painter tij units of time.
Order is important everywhere, so the painters' work is ordered by the following rules:
* Each picture is first painted by the first painter, then by the second one, and so on. That is, after the j-th painter finishes working on the picture, it must go to the (j + 1)-th painter (if j < n);
* each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on;
* each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest;
* as soon as the j-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.
Given that the painters start working at time 0, find for each picture the time when it is ready for sale.
Input
The first line of the input contains integers m, n (1 ≤ m ≤ 50000, 1 ≤ n ≤ 5), where m is the number of pictures and n is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains n integers ti1, ti2, ..., tin (1 ≤ tij ≤ 1000), where tij is the time the j-th painter needs to work on the i-th picture.
Output
Print the sequence of m integers r1, r2, ..., rm, where ri is the moment when the n-th painter stopped working on the i-th picture.
Examples
Input
5 1
1
2
3
4
5
Output
1 3 6 10 15
Input
4 2
2 5
3 1
5 3
10 1
Output
7 8 13 21
Submitted Solution:
```
#from dust i have come dust i will be
import sys
m,n=map(int,input().split())
a=[[]*n for i in range(m)]
for i in range(m):
a[i]=list(map(int,input().split()))
for i in range(1,m):
a[i][0]+=a[i-1][0]
for i in range(1,n):
a[0][i]+=a[0][i-1]
for i in range(1,m):
for j in range(1,n):
a[i][j]+=max(a[i-1][j],a[i][j-1])
for i in range(m):
sys.stdout.write(str(a[i][n-1])+" ")
``` | instruction | 0 | 92,124 | 7 | 184,248 |
Yes | output | 1 | 92,124 | 7 | 184,249 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of n painters who decided to organize their work as follows.
Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these n colors. Adding the j-th color to the i-th picture takes the j-th painter tij units of time.
Order is important everywhere, so the painters' work is ordered by the following rules:
* Each picture is first painted by the first painter, then by the second one, and so on. That is, after the j-th painter finishes working on the picture, it must go to the (j + 1)-th painter (if j < n);
* each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on;
* each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest;
* as soon as the j-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.
Given that the painters start working at time 0, find for each picture the time when it is ready for sale.
Input
The first line of the input contains integers m, n (1 ≤ m ≤ 50000, 1 ≤ n ≤ 5), where m is the number of pictures and n is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains n integers ti1, ti2, ..., tin (1 ≤ tij ≤ 1000), where tij is the time the j-th painter needs to work on the i-th picture.
Output
Print the sequence of m integers r1, r2, ..., rm, where ri is the moment when the n-th painter stopped working on the i-th picture.
Examples
Input
5 1
1
2
3
4
5
Output
1 3 6 10 15
Input
4 2
2 5
3 1
5 3
10 1
Output
7 8 13 21
Submitted Solution:
```
n , m = map(int, input().split())
mat = [[0 for i in range(m+1)]for j in range(n+1)]
a = []
for i in range(n):
a.append(list(map(int , input().split())))
for i in range(1 , n+1):
for j in range(1 , m+1):
mat[i][j] = a[i-1][j-1] + max(mat[i-1][j], mat[i][j-1])
#print(mat)
for i in range(1 , n+1):
print(mat[i][m] , end = " ")
``` | instruction | 0 | 92,125 | 7 | 184,250 |
Yes | output | 1 | 92,125 | 7 | 184,251 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of n painters who decided to organize their work as follows.
Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these n colors. Adding the j-th color to the i-th picture takes the j-th painter tij units of time.
Order is important everywhere, so the painters' work is ordered by the following rules:
* Each picture is first painted by the first painter, then by the second one, and so on. That is, after the j-th painter finishes working on the picture, it must go to the (j + 1)-th painter (if j < n);
* each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on;
* each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest;
* as soon as the j-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.
Given that the painters start working at time 0, find for each picture the time when it is ready for sale.
Input
The first line of the input contains integers m, n (1 ≤ m ≤ 50000, 1 ≤ n ≤ 5), where m is the number of pictures and n is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains n integers ti1, ti2, ..., tin (1 ≤ tij ≤ 1000), where tij is the time the j-th painter needs to work on the i-th picture.
Output
Print the sequence of m integers r1, r2, ..., rm, where ri is the moment when the n-th painter stopped working on the i-th picture.
Examples
Input
5 1
1
2
3
4
5
Output
1 3 6 10 15
Input
4 2
2 5
3 1
5 3
10 1
Output
7 8 13 21
Submitted Solution:
```
m,n=map(int, input().split())
a=[0]*8
for i in range(m):
x=list(map(int, input().split()))
for j in range(1, n+1):
a[j]=a[j-1]+x[j-1]
print(a[n], end=' ')
``` | instruction | 0 | 92,126 | 7 | 184,252 |
No | output | 1 | 92,126 | 7 | 184,253 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of n painters who decided to organize their work as follows.
Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these n colors. Adding the j-th color to the i-th picture takes the j-th painter tij units of time.
Order is important everywhere, so the painters' work is ordered by the following rules:
* Each picture is first painted by the first painter, then by the second one, and so on. That is, after the j-th painter finishes working on the picture, it must go to the (j + 1)-th painter (if j < n);
* each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on;
* each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest;
* as soon as the j-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.
Given that the painters start working at time 0, find for each picture the time when it is ready for sale.
Input
The first line of the input contains integers m, n (1 ≤ m ≤ 50000, 1 ≤ n ≤ 5), where m is the number of pictures and n is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains n integers ti1, ti2, ..., tin (1 ≤ tij ≤ 1000), where tij is the time the j-th painter needs to work on the i-th picture.
Output
Print the sequence of m integers r1, r2, ..., rm, where ri is the moment when the n-th painter stopped working on the i-th picture.
Examples
Input
5 1
1
2
3
4
5
Output
1 3 6 10 15
Input
4 2
2 5
3 1
5 3
10 1
Output
7 8 13 21
Submitted Solution:
```
n,m=map(int,input().split())
L=[list(map(int,input().split())) for i in range(n)]
L1=[[0 for j in range(m)] for i in range(n)]
g=0
p=sum(L[0])
out=[p]
for i in range(1,n) :
d=sum(L[i-1])
d-=L[i-1][0]
d+=g
e=d
g=max(0,d-L[i][0])
for j in range(m-1) :
f=d-L[i][j]
if f>0 :
L1[i][j]=0
e-=L[i-1][j+1]
d-=L[i][j]
d=min(d,e)
else :
L1[i][j]=abs(d-L[i][j])
d=0
L1[i][m-1]=L[i][m-1]
for i in range(1,n) :
p+=sum(L1[i])
out.append(p)
print(*out)
``` | instruction | 0 | 92,127 | 7 | 184,254 |
No | output | 1 | 92,127 | 7 | 184,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of n painters who decided to organize their work as follows.
Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these n colors. Adding the j-th color to the i-th picture takes the j-th painter tij units of time.
Order is important everywhere, so the painters' work is ordered by the following rules:
* Each picture is first painted by the first painter, then by the second one, and so on. That is, after the j-th painter finishes working on the picture, it must go to the (j + 1)-th painter (if j < n);
* each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on;
* each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest;
* as soon as the j-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.
Given that the painters start working at time 0, find for each picture the time when it is ready for sale.
Input
The first line of the input contains integers m, n (1 ≤ m ≤ 50000, 1 ≤ n ≤ 5), where m is the number of pictures and n is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains n integers ti1, ti2, ..., tin (1 ≤ tij ≤ 1000), where tij is the time the j-th painter needs to work on the i-th picture.
Output
Print the sequence of m integers r1, r2, ..., rm, where ri is the moment when the n-th painter stopped working on the i-th picture.
Examples
Input
5 1
1
2
3
4
5
Output
1 3 6 10 15
Input
4 2
2 5
3 1
5 3
10 1
Output
7 8 13 21
Submitted Solution:
```
m,n=map(int, input().split())
a=[0]*8;
for i in range(m):
for j in range(1,n):
x=int(input())
a[j]=max(a[j], a[j-1])+x
print(a[n], end=' ')
``` | instruction | 0 | 92,128 | 7 | 184,256 |
No | output | 1 | 92,128 | 7 | 184,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of n painters who decided to organize their work as follows.
Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these n colors. Adding the j-th color to the i-th picture takes the j-th painter tij units of time.
Order is important everywhere, so the painters' work is ordered by the following rules:
* Each picture is first painted by the first painter, then by the second one, and so on. That is, after the j-th painter finishes working on the picture, it must go to the (j + 1)-th painter (if j < n);
* each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on;
* each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest;
* as soon as the j-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.
Given that the painters start working at time 0, find for each picture the time when it is ready for sale.
Input
The first line of the input contains integers m, n (1 ≤ m ≤ 50000, 1 ≤ n ≤ 5), where m is the number of pictures and n is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains n integers ti1, ti2, ..., tin (1 ≤ tij ≤ 1000), where tij is the time the j-th painter needs to work on the i-th picture.
Output
Print the sequence of m integers r1, r2, ..., rm, where ri is the moment when the n-th painter stopped working on the i-th picture.
Examples
Input
5 1
1
2
3
4
5
Output
1 3 6 10 15
Input
4 2
2 5
3 1
5 3
10 1
Output
7 8 13 21
Submitted Solution:
```
m,n=[int(x) for x in input().split()]
ans=[[0 for i in range(n)] for i in range(m)]
a=[]
for i in range(m):
a.append([int(x) for x in input().split()])
ans[0][0]=a[0][0]
for i in range(1,m):
ans[i][0]=ans[i-1][0]+a[i][0]
for i in range(1,n):
ans[0][i]=ans[0][i-1]+a[0][i]
for i in range(1,n):
key=ans[0][i]
for j in range(1,m):
if ans[j][i-1]<=key:
ans[j][i]=ans[j-1][i]+a[j][i]
else:
ans[j][i]=ans[j][i-1]+a[j][i]
key=a[j][i]
# print(ans[n-1])
for i in range(m):
print(ans[i][n-1],end=" ")
``` | instruction | 0 | 92,129 | 7 | 184,258 |
No | output | 1 | 92,129 | 7 | 184,259 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is unrest in the Galactic Senate. Several thousand solar systems have declared their intentions to leave the Republic. Master Heidi needs to select the Jedi Knights who will go on peacekeeping missions throughout the galaxy. It is well-known that the success of any peacekeeping mission depends on the colors of the lightsabers of the Jedi who will go on that mission.
Heidi has n Jedi Knights standing in front of her, each one with a lightsaber of one of m possible colors. She knows that for the mission to be the most effective, she needs to select some contiguous interval of knights such that there are exactly k1 knights with lightsabers of the first color, k2 knights with lightsabers of the second color, ..., km knights with lightsabers of the m-th color. Help her find out if this is possible.
Input
The first line of the input contains n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ n). The second line contains n integers in the range {1, 2, ..., m} representing colors of the lightsabers of the subsequent Jedi Knights. The third line contains m integers k1, k2, ..., km (with <image>) – the desired counts of lightsabers of each color from 1 to m.
Output
Output YES if an interval with prescribed color counts exists, or output NO if there is none.
Example
Input
5 2
1 1 2 2 1
1 2
Output
YES | instruction | 0 | 92,353 | 7 | 184,706 |
Tags: implementation
Correct Solution:
```
INF = 1 << 60
n, m = map(int, input().split())
a = list(map(int, input().split()))
Constraint = [0] + list(map(int, input().split()))
pos = 0
satisfied_color = 0
for i in range(1,m + 1):
if Constraint[i] == 0:
satisfied_color += 1
GETCOLOR = [0]*(n + 1)
ans = INF
waste = 0
for i in range(n):
while satisfied_color < m and pos < n:
now_color = a[pos]
GETCOLOR[now_color] += 1
if GETCOLOR[now_color] == Constraint[now_color]:
satisfied_color += 1
elif GETCOLOR[now_color] > Constraint[now_color]:
waste += 1
pos += 1
if satisfied_color == m:
ans = min(ans, waste)
removed_color = a[i]
if GETCOLOR[removed_color] > Constraint[removed_color]:
GETCOLOR[removed_color] -= 1
waste -= 1
continue
elif GETCOLOR[removed_color] == Constraint[removed_color]:
GETCOLOR[removed_color] -= 1
satisfied_color -= 1
if ans < INF:
print(ans)
else:
print(-1)
``` | output | 1 | 92,353 | 7 | 184,707 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is unrest in the Galactic Senate. Several thousand solar systems have declared their intentions to leave the Republic. Master Heidi needs to select the Jedi Knights who will go on peacekeeping missions throughout the galaxy. It is well-known that the success of any peacekeeping mission depends on the colors of the lightsabers of the Jedi who will go on that mission.
Heidi has n Jedi Knights standing in front of her, each one with a lightsaber of one of m possible colors. She knows that for the mission to be the most effective, she needs to select some contiguous interval of knights such that there are exactly k1 knights with lightsabers of the first color, k2 knights with lightsabers of the second color, ..., km knights with lightsabers of the m-th color. Help her find out if this is possible.
Input
The first line of the input contains n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ n). The second line contains n integers in the range {1, 2, ..., m} representing colors of the lightsabers of the subsequent Jedi Knights. The third line contains m integers k1, k2, ..., km (with <image>) – the desired counts of lightsabers of each color from 1 to m.
Output
Output YES if an interval with prescribed color counts exists, or output NO if there is none.
Example
Input
5 2
1 1 2 2 1
1 2
Output
YES | instruction | 0 | 92,354 | 7 | 184,708 |
Tags: implementation
Correct Solution:
```
n, m = map(int, input().split())
c = list(map(int, input().split()))
k = list(map(int, input().split()))
p = [0] * m
c = [o - 1 for o in c]
s = sum(k)
ok = k.count(0)
i1 = 0
i2 = 0
while i2 < n and ok < m:
p[c[i2]] += 1
if p[c[i2]] == k[c[i2]]:
ok += 1
i2 += 1
if ok != m:
print(-1)
exit(0)
ans = i2 - i1
while i1 < n:
p[c[i1]] -= 1
while i2 < n and p[c[i1]] < k[c[i1]]:
p[c[i2]] += 1
i2 += 1
if p[c[i1]] >= k[c[i1]]:
ans = min(ans, i2 - i1 - 1)
elif i2 == n:
break
i1 += 1
print(ans - s)
``` | output | 1 | 92,354 | 7 | 184,709 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is unrest in the Galactic Senate. Several thousand solar systems have declared their intentions to leave the Republic. Master Heidi needs to select the Jedi Knights who will go on peacekeeping missions throughout the galaxy. It is well-known that the success of any peacekeeping mission depends on the colors of the lightsabers of the Jedi who will go on that mission.
Heidi has n Jedi Knights standing in front of her, each one with a lightsaber of one of m possible colors. She knows that for the mission to be the most effective, she needs to select some contiguous interval of knights such that there are exactly k1 knights with lightsabers of the first color, k2 knights with lightsabers of the second color, ..., km knights with lightsabers of the m-th color. Help her find out if this is possible.
Input
The first line of the input contains n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ n). The second line contains n integers in the range {1, 2, ..., m} representing colors of the lightsabers of the subsequent Jedi Knights. The third line contains m integers k1, k2, ..., km (with <image>) – the desired counts of lightsabers of each color from 1 to m.
Output
Output YES if an interval with prescribed color counts exists, or output NO if there is none.
Example
Input
5 2
1 1 2 2 1
1 2
Output
YES | instruction | 0 | 92,355 | 7 | 184,710 |
Tags: implementation
Correct Solution:
```
n,m=map(int,input().split())
a=list(map(int,input().split()))
t=list(map(int,input().split()))
x=sum(t)
for i in range(n-x+1):
freq=[0]*m
for j in range(i,i+x):
freq[a[j]-1]+=1
flag=True
for j in range(m):
if freq[j]!=t[j]:
flag=False
break
if flag:
break
if flag:
print("YES")
else:
print("NO")
``` | output | 1 | 92,355 | 7 | 184,711 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is unrest in the Galactic Senate. Several thousand solar systems have declared their intentions to leave the Republic. Master Heidi needs to select the Jedi Knights who will go on peacekeeping missions throughout the galaxy. It is well-known that the success of any peacekeeping mission depends on the colors of the lightsabers of the Jedi who will go on that mission.
Heidi has n Jedi Knights standing in front of her, each one with a lightsaber of one of m possible colors. She knows that for the mission to be the most effective, she needs to select some contiguous interval of knights such that there are exactly k1 knights with lightsabers of the first color, k2 knights with lightsabers of the second color, ..., km knights with lightsabers of the m-th color. Help her find out if this is possible.
Input
The first line of the input contains n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ n). The second line contains n integers in the range {1, 2, ..., m} representing colors of the lightsabers of the subsequent Jedi Knights. The third line contains m integers k1, k2, ..., km (with <image>) – the desired counts of lightsabers of each color from 1 to m.
Output
Output YES if an interval with prescribed color counts exists, or output NO if there is none.
Example
Input
5 2
1 1 2 2 1
1 2
Output
YES | instruction | 0 | 92,356 | 7 | 184,712 |
Tags: implementation
Correct Solution:
```
import sys
n, m = [int(x) for x in input().split(' ')]
arr = [int(x)-1 for x in input().split(' ')]
k = [int(x) for x in input().split(' ')]
summ = 0
for i in range(m):
summ += k[i]
for i in range(n):
if i >= summ:
k[arr[i-summ]] += 1
k[arr[i]] -= 1
done = True
for j in range(m):
if k[j] != 0:
done = False
if done:
print('YES')
sys.exit()
print('NO')
``` | output | 1 | 92,356 | 7 | 184,713 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is unrest in the Galactic Senate. Several thousand solar systems have declared their intentions to leave the Republic. Master Heidi needs to select the Jedi Knights who will go on peacekeeping missions throughout the galaxy. It is well-known that the success of any peacekeeping mission depends on the colors of the lightsabers of the Jedi who will go on that mission.
Heidi has n Jedi Knights standing in front of her, each one with a lightsaber of one of m possible colors. She knows that for the mission to be the most effective, she needs to select some contiguous interval of knights such that there are exactly k1 knights with lightsabers of the first color, k2 knights with lightsabers of the second color, ..., km knights with lightsabers of the m-th color. Help her find out if this is possible.
Input
The first line of the input contains n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ n). The second line contains n integers in the range {1, 2, ..., m} representing colors of the lightsabers of the subsequent Jedi Knights. The third line contains m integers k1, k2, ..., km (with <image>) – the desired counts of lightsabers of each color from 1 to m.
Output
Output YES if an interval with prescribed color counts exists, or output NO if there is none.
Example
Input
5 2
1 1 2 2 1
1 2
Output
YES | instruction | 0 | 92,357 | 7 | 184,714 |
Tags: implementation
Correct Solution:
```
n, m = map(int, input().split())
c = list(map(int, input().split()))
k = list(map(int, input().split()))
p = [0] * m
c = [i -1 for i in c]
s = sum(k)
cnt = k.count(0)
l = 0
r = 0
while r < n and cnt < m:
p[c[r]] += 1
if p[c[r]] == k[c[r]]:
cnt += 1
r += 1
if cnt != m:
print(-1)
exit(0)
ans = r-l
while l < n:
p[c[l]] -= 1
while r < n and p[c[l]] < k[c[l]]:
p[c[r]] += 1
r += 1
if p[c[l]] >= k[c[l]]:
ans = min(ans, r-l-1)
elif r == n:
break
l += 1
print(ans-s)
``` | output | 1 | 92,357 | 7 | 184,715 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is unrest in the Galactic Senate. Several thousand solar systems have declared their intentions to leave the Republic. Master Heidi needs to select the Jedi Knights who will go on peacekeeping missions throughout the galaxy. It is well-known that the success of any peacekeeping mission depends on the colors of the lightsabers of the Jedi who will go on that mission.
Heidi has n Jedi Knights standing in front of her, each one with a lightsaber of one of m possible colors. She knows that for the mission to be the most effective, she needs to select some contiguous interval of knights such that there are exactly k1 knights with lightsabers of the first color, k2 knights with lightsabers of the second color, ..., km knights with lightsabers of the m-th color. Help her find out if this is possible.
Input
The first line of the input contains n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ n). The second line contains n integers in the range {1, 2, ..., m} representing colors of the lightsabers of the subsequent Jedi Knights. The third line contains m integers k1, k2, ..., km (with <image>) – the desired counts of lightsabers of each color from 1 to m.
Output
Output YES if an interval with prescribed color counts exists, or output NO if there is none.
Example
Input
5 2
1 1 2 2 1
1 2
Output
YES | instruction | 0 | 92,358 | 7 | 184,716 |
Tags: implementation
Correct Solution:
```
s = input().split()
n, m = int(s[0]), int(s[1])
cl = list(map(int, input().split()))
com = list(map(int, input().split()))
res = False
for i in range(n):
for j in range(i, n):
e = True
t = cl[i:j+1]
for k in range(1, m+1):
e = t.count(k)==com[k-1] and e
if e:
res = True
break
if res: print('YES')
else: print('NO')
``` | output | 1 | 92,358 | 7 | 184,717 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is unrest in the Galactic Senate. Several thousand solar systems have declared their intentions to leave the Republic. Master Heidi needs to select the Jedi Knights who will go on peacekeeping missions throughout the galaxy. It is well-known that the success of any peacekeeping mission depends on the colors of the lightsabers of the Jedi who will go on that mission.
Heidi has n Jedi Knights standing in front of her, each one with a lightsaber of one of m possible colors. She knows that for the mission to be the most effective, she needs to select some contiguous interval of knights such that there are exactly k1 knights with lightsabers of the first color, k2 knights with lightsabers of the second color, ..., km knights with lightsabers of the m-th color. Help her find out if this is possible.
Input
The first line of the input contains n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ n). The second line contains n integers in the range {1, 2, ..., m} representing colors of the lightsabers of the subsequent Jedi Knights. The third line contains m integers k1, k2, ..., km (with <image>) – the desired counts of lightsabers of each color from 1 to m.
Output
Output YES if an interval with prescribed color counts exists, or output NO if there is none.
Example
Input
5 2
1 1 2 2 1
1 2
Output
YES | instruction | 0 | 92,359 | 7 | 184,718 |
Tags: implementation
Correct Solution:
```
from collections import defaultdict, deque, Counter
from sys import stdin, stdout
from heapq import heappush, heappop
import math
import io
import os
import math
import bisect
#?############################################################
def isPrime(x):
for i in range(2, x):
if i*i > x:
break
if (x % i == 0):
return False
return True
#?############################################################
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den, p - 2, p)) % p
#?############################################################
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3, int(math.sqrt(n))+1, 2):
while n % i == 0:
l.append(int(i))
n = n / i
if n > 2:
l.append(n)
return list(set(l))
#?############################################################
def power(x, y, p):
res = 1
x = x % p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1):
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
#?############################################################
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#?############################################################
def digits(n):
a = 0
while (n > 0):
n //= 10
a += 1
return a
#?############################################################
def ceil(n, x):
if (n % x == 0):
return n//x
return n//x+1
#?############################################################
def mapin():
return map(int, input().split())
#?############################################################
def solve():
n, m = mapin()
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]
s = sum(b)
for i in range(n-s+1):
cc = [0] * m
for j in range(i, i+s):
cc[a[j]-1]+=1
if cc == b:
return 1
return 0
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# python3 15.py<in>op
aa = solve()
if(aa):
print("YES")
else:
print("NO")
``` | output | 1 | 92,359 | 7 | 184,719 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is unrest in the Galactic Senate. Several thousand solar systems have declared their intentions to leave the Republic. Master Heidi needs to select the Jedi Knights who will go on peacekeeping missions throughout the galaxy. It is well-known that the success of any peacekeeping mission depends on the colors of the lightsabers of the Jedi who will go on that mission.
Heidi has n Jedi Knights standing in front of her, each one with a lightsaber of one of m possible colors. She knows that for the mission to be the most effective, she needs to select some contiguous interval of knights such that there are exactly k1 knights with lightsabers of the first color, k2 knights with lightsabers of the second color, ..., km knights with lightsabers of the m-th color. Help her find out if this is possible.
Input
The first line of the input contains n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ n). The second line contains n integers in the range {1, 2, ..., m} representing colors of the lightsabers of the subsequent Jedi Knights. The third line contains m integers k1, k2, ..., km (with <image>) – the desired counts of lightsabers of each color from 1 to m.
Output
Output YES if an interval with prescribed color counts exists, or output NO if there is none.
Example
Input
5 2
1 1 2 2 1
1 2
Output
YES | instruction | 0 | 92,360 | 7 | 184,720 |
Tags: implementation
Correct Solution:
```
n,m=list(map(int,input().split()))
arr=list(map(int,input().split()))
marr=list(map(int,input().split()))
f=[0]*(m+1)
cnt,i=0,0
valid=sum(marr) #m
while(i<n):
f[arr[i]]+=1
if f[arr[i]]<=marr[arr[i]-1]:
cnt+=1
if cnt==valid:
break
i+=1
if i==n:
print(-1)
else:
ans=(i+1)-valid
s,e=0,i
while(e<n):
while(f[arr[s]]>marr[arr[s]-1]):
f[arr[s]]-=1
s+=1
ans=min((e-s+1)-valid,ans)
e+=1
if e<n:
f[arr[e]]+=1
print(ans)
``` | output | 1 | 92,360 | 7 | 184,721 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is unrest in the Galactic Senate. Several thousand solar systems have declared their intentions to leave the Republic. Master Heidi needs to select the Jedi Knights who will go on peacekeeping missions throughout the galaxy. It is well-known that the success of any peacekeeping mission depends on the colors of the lightsabers of the Jedi who will go on that mission.
Heidi has n Jedi Knights standing in front of her, each one with a lightsaber of one of m possible colors. She knows that for the mission to be the most effective, she needs to select some contiguous interval of knights such that there are exactly k1 knights with lightsabers of the first color, k2 knights with lightsabers of the second color, ..., km knights with lightsabers of the m-th color. Help her find out if this is possible.
Input
The first line of the input contains n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ n). The second line contains n integers in the range {1, 2, ..., m} representing colors of the lightsabers of the subsequent Jedi Knights. The third line contains m integers k1, k2, ..., km (with <image>) – the desired counts of lightsabers of each color from 1 to m.
Output
Output YES if an interval with prescribed color counts exists, or output NO if there is none.
Example
Input
5 2
1 1 2 2 1
1 2
Output
YES
Submitted Solution:
```
n, m = map(int,input().split())
a = list(map(int,input().split()))
z = list(map(int,input().split()))
ch = 0
if n == sum(z):
t = 0
for i in a:
z[i-1] -= 1
if z.count(0) == m:
print('YES')
else:
print('NO')
else:
for i in range(n-sum(z)+1):
t = 0
for j in range(1,m+1):
if a[i:i+sum(z)].count(j) == z[j-1]:
t+=1
if t == m:
print('YES')
ch = 1
break
if ch == 0:
print('NO')
``` | instruction | 0 | 92,363 | 7 | 184,726 |
Yes | output | 1 | 92,363 | 7 | 184,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is unrest in the Galactic Senate. Several thousand solar systems have declared their intentions to leave the Republic. Master Heidi needs to select the Jedi Knights who will go on peacekeeping missions throughout the galaxy. It is well-known that the success of any peacekeeping mission depends on the colors of the lightsabers of the Jedi who will go on that mission.
Heidi has n Jedi Knights standing in front of her, each one with a lightsaber of one of m possible colors. She knows that for the mission to be the most effective, she needs to select some contiguous interval of knights such that there are exactly k1 knights with lightsabers of the first color, k2 knights with lightsabers of the second color, ..., km knights with lightsabers of the m-th color. Help her find out if this is possible.
Input
The first line of the input contains n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ n). The second line contains n integers in the range {1, 2, ..., m} representing colors of the lightsabers of the subsequent Jedi Knights. The third line contains m integers k1, k2, ..., km (with <image>) – the desired counts of lightsabers of each color from 1 to m.
Output
Output YES if an interval with prescribed color counts exists, or output NO if there is none.
Example
Input
5 2
1 1 2 2 1
1 2
Output
YES
Submitted Solution:
```
n,m=list(map(int,input().split()))
a=list(map(int,input().split()))
b=list(map(int,input().split()))
c=sum(b)
d=[0]*m
for i in range(c):
d[a[i]-1]+=1
e=0
for i in range(n-c+1):
if i!=0:
d[a[i-1]-1]-=1
d[a[i+c-1]-1]+=1
if d==b:
e=1
break
if e==0:
print("NO")
else:
print("YES")
``` | instruction | 0 | 92,364 | 7 | 184,728 |
Yes | output | 1 | 92,364 | 7 | 184,729 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is unrest in the Galactic Senate. Several thousand solar systems have declared their intentions to leave the Republic. Master Heidi needs to select the Jedi Knights who will go on peacekeeping missions throughout the galaxy. It is well-known that the success of any peacekeeping mission depends on the colors of the lightsabers of the Jedi who will go on that mission.
Heidi has n Jedi Knights standing in front of her, each one with a lightsaber of one of m possible colors. She knows that for the mission to be the most effective, she needs to select some contiguous interval of knights such that there are exactly k1 knights with lightsabers of the first color, k2 knights with lightsabers of the second color, ..., km knights with lightsabers of the m-th color. Help her find out if this is possible.
Input
The first line of the input contains n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ n). The second line contains n integers in the range {1, 2, ..., m} representing colors of the lightsabers of the subsequent Jedi Knights. The third line contains m integers k1, k2, ..., km (with <image>) – the desired counts of lightsabers of each color from 1 to m.
Output
Output YES if an interval with prescribed color counts exists, or output NO if there is none.
Example
Input
5 2
1 1 2 2 1
1 2
Output
YES
Submitted Solution:
```
from collections import Counter
def check():
for i in range(n):
for j in range(i + 1, n + 1):
if target == Counter(ns[i:j]):
return True
return False
n, m = map(int, input().split())
ns = list(map(int, input().split()))
ms = list(map(int, input().split()))
target = Counter({k: v for k, v in enumerate(ms, 1)})
print(['NO', 'YES'][check()])
``` | instruction | 0 | 92,365 | 7 | 184,730 |
No | output | 1 | 92,365 | 7 | 184,731 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is unrest in the Galactic Senate. Several thousand solar systems have declared their intentions to leave the Republic. Master Heidi needs to select the Jedi Knights who will go on peacekeeping missions throughout the galaxy. It is well-known that the success of any peacekeeping mission depends on the colors of the lightsabers of the Jedi who will go on that mission.
Heidi has n Jedi Knights standing in front of her, each one with a lightsaber of one of m possible colors. She knows that for the mission to be the most effective, she needs to select some contiguous interval of knights such that there are exactly k1 knights with lightsabers of the first color, k2 knights with lightsabers of the second color, ..., km knights with lightsabers of the m-th color. Help her find out if this is possible.
Input
The first line of the input contains n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ n). The second line contains n integers in the range {1, 2, ..., m} representing colors of the lightsabers of the subsequent Jedi Knights. The third line contains m integers k1, k2, ..., km (with <image>) – the desired counts of lightsabers of each color from 1 to m.
Output
Output YES if an interval with prescribed color counts exists, or output NO if there is none.
Example
Input
5 2
1 1 2 2 1
1 2
Output
YES
Submitted Solution:
```
from collections import Counter
n, m = map(int, input().split())
c = list(map(int, input().split()))
k = list(map(int, input().split()))
s = sum(k)
C_orig = {i + 1: k[i] for i in range(m)}
C = Counter(c[:s])
if C == C_orig:
print('YES')
exit()
for i in range(n - s):
C[c[i]] -= 1
C[c[i + s]] += 1
if C == C_orig:
print('YES')
exit()
print('NO')
``` | instruction | 0 | 92,366 | 7 | 184,732 |
No | output | 1 | 92,366 | 7 | 184,733 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a sequence with N integers: A = \\{ A_1, A_2, \cdots, A_N \\}. For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:
* If A_i and A_j (i < j) are painted with the same color, A_i < A_j.
Find the minimum number of colors required to satisfy the condition.
Constraints
* 1 \leq N \leq 10^5
* 0 \leq A_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
A_1
:
A_N
Output
Print the minimum number of colors required to satisfy the condition.
Examples
Input
5
2
1
4
5
3
Output
2
Input
4
0
0
0
0
Output
4 | instruction | 0 | 92,433 | 7 | 184,866 |
"Correct Solution:
```
import bisect
N = int(input())
LIS = [-1 * int(input())]
for _ in range(N-1):
A = -1 * int(input())
if A >= LIS[-1]:
LIS.append(A)
else:
LIS[bisect.bisect_right(LIS, A)] = A
print(len(LIS))
``` | output | 1 | 92,433 | 7 | 184,867 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a sequence with N integers: A = \\{ A_1, A_2, \cdots, A_N \\}. For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:
* If A_i and A_j (i < j) are painted with the same color, A_i < A_j.
Find the minimum number of colors required to satisfy the condition.
Constraints
* 1 \leq N \leq 10^5
* 0 \leq A_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
A_1
:
A_N
Output
Print the minimum number of colors required to satisfy the condition.
Examples
Input
5
2
1
4
5
3
Output
2
Input
4
0
0
0
0
Output
4 | instruction | 0 | 92,434 | 7 | 184,868 |
"Correct Solution:
```
from bisect import bisect as br
n=int(input())
x=[-1]*n
for _ in range(n):
a=int(input())
i=br(x,a-1)-1
x[i]=a
print(n-x.count(-1))
``` | output | 1 | 92,434 | 7 | 184,869 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a sequence with N integers: A = \\{ A_1, A_2, \cdots, A_N \\}. For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:
* If A_i and A_j (i < j) are painted with the same color, A_i < A_j.
Find the minimum number of colors required to satisfy the condition.
Constraints
* 1 \leq N \leq 10^5
* 0 \leq A_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
A_1
:
A_N
Output
Print the minimum number of colors required to satisfy the condition.
Examples
Input
5
2
1
4
5
3
Output
2
Input
4
0
0
0
0
Output
4 | instruction | 0 | 92,435 | 7 | 184,870 |
"Correct Solution:
```
N = int(input())
A = [int(input()) for i in range(N)]
from bisect import *
h = []
for a in A:
x = bisect_right(h, -a)
if len(h) <= x:
h.append(-a)
else:
h[x] = -a
result = len(h)
print(result)
``` | output | 1 | 92,435 | 7 | 184,871 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a sequence with N integers: A = \\{ A_1, A_2, \cdots, A_N \\}. For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:
* If A_i and A_j (i < j) are painted with the same color, A_i < A_j.
Find the minimum number of colors required to satisfy the condition.
Constraints
* 1 \leq N \leq 10^5
* 0 \leq A_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
A_1
:
A_N
Output
Print the minimum number of colors required to satisfy the condition.
Examples
Input
5
2
1
4
5
3
Output
2
Input
4
0
0
0
0
Output
4 | instruction | 0 | 92,436 | 7 | 184,872 |
"Correct Solution:
```
n=int(input())
a=[int(input())*(-1) for _ in range(n)]
inf=10**10
l=[inf]*n
import bisect
for i in range(n):
r=bisect.bisect_right(l,a[i])
l[r]=a[i]
cnt=0
for i in range(n):
if l[i]!=inf:
cnt+=1
else:
break
print(cnt)
``` | output | 1 | 92,436 | 7 | 184,873 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a sequence with N integers: A = \\{ A_1, A_2, \cdots, A_N \\}. For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:
* If A_i and A_j (i < j) are painted with the same color, A_i < A_j.
Find the minimum number of colors required to satisfy the condition.
Constraints
* 1 \leq N \leq 10^5
* 0 \leq A_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
A_1
:
A_N
Output
Print the minimum number of colors required to satisfy the condition.
Examples
Input
5
2
1
4
5
3
Output
2
Input
4
0
0
0
0
Output
4 | instruction | 0 | 92,437 | 7 | 184,874 |
"Correct Solution:
```
N = int(input())
A = [int(input()) for i in range(N)]
from bisect import bisect
arr = []
for a in A:
i = bisect(arr, -a)
if i==len(arr):
arr.append(-a)
else:
arr[i] = -a
print(len(arr))
``` | output | 1 | 92,437 | 7 | 184,875 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a sequence with N integers: A = \\{ A_1, A_2, \cdots, A_N \\}. For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:
* If A_i and A_j (i < j) are painted with the same color, A_i < A_j.
Find the minimum number of colors required to satisfy the condition.
Constraints
* 1 \leq N \leq 10^5
* 0 \leq A_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
A_1
:
A_N
Output
Print the minimum number of colors required to satisfy the condition.
Examples
Input
5
2
1
4
5
3
Output
2
Input
4
0
0
0
0
Output
4 | instruction | 0 | 92,438 | 7 | 184,876 |
"Correct Solution:
```
from bisect import bisect_right
N = int(input())
A = [int(input()) for i in range(N)]
A = [-a for a in A]
X = []
for a in A:
i = bisect_right(X, a)
if i == len(X):
X.append(a)
else:
X[i] = a
print(len(X))
``` | output | 1 | 92,438 | 7 | 184,877 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a sequence with N integers: A = \\{ A_1, A_2, \cdots, A_N \\}. For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:
* If A_i and A_j (i < j) are painted with the same color, A_i < A_j.
Find the minimum number of colors required to satisfy the condition.
Constraints
* 1 \leq N \leq 10^5
* 0 \leq A_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
A_1
:
A_N
Output
Print the minimum number of colors required to satisfy the condition.
Examples
Input
5
2
1
4
5
3
Output
2
Input
4
0
0
0
0
Output
4 | instruction | 0 | 92,439 | 7 | 184,878 |
"Correct Solution:
```
from bisect import bisect_right
n = int(input())
color = [0] * (n+1)
for i in range(n):
a = int(input())
j = bisect_right(color, a)
color[j-1] = a+1
ans = 0
for t in color:
if t != 0:
ans += 1
print(ans)
``` | output | 1 | 92,439 | 7 | 184,879 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a sequence with N integers: A = \\{ A_1, A_2, \cdots, A_N \\}. For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:
* If A_i and A_j (i < j) are painted with the same color, A_i < A_j.
Find the minimum number of colors required to satisfy the condition.
Constraints
* 1 \leq N \leq 10^5
* 0 \leq A_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
A_1
:
A_N
Output
Print the minimum number of colors required to satisfy the condition.
Examples
Input
5
2
1
4
5
3
Output
2
Input
4
0
0
0
0
Output
4 | instruction | 0 | 92,440 | 7 | 184,880 |
"Correct Solution:
```
from bisect import bisect
n = int(input())
a = [int(input()) for _ in range(n)]
dp = [1 for _ in range(n+2)]
for i in range(n):
b = bisect(dp, -a[i])
dp[b] = -a[i]
ans = 1
while dp[ans] <= 0:
ans += 1
print(ans)
``` | output | 1 | 92,440 | 7 | 184,881 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence with N integers: A = \\{ A_1, A_2, \cdots, A_N \\}. For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:
* If A_i and A_j (i < j) are painted with the same color, A_i < A_j.
Find the minimum number of colors required to satisfy the condition.
Constraints
* 1 \leq N \leq 10^5
* 0 \leq A_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
A_1
:
A_N
Output
Print the minimum number of colors required to satisfy the condition.
Examples
Input
5
2
1
4
5
3
Output
2
Input
4
0
0
0
0
Output
4
Submitted Solution:
```
from bisect import bisect
N, *A = map(int, open(0).read().split())
INF = 10 ** 9 + 7
dp = [INF] * (N + 1)
for i in range(N):
dp[bisect(dp, -A[i])] = -A[i]
print(sum(dp[i] < INF for i in range(N)))
``` | instruction | 0 | 92,441 | 7 | 184,882 |
Yes | output | 1 | 92,441 | 7 | 184,883 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence with N integers: A = \\{ A_1, A_2, \cdots, A_N \\}. For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:
* If A_i and A_j (i < j) are painted with the same color, A_i < A_j.
Find the minimum number of colors required to satisfy the condition.
Constraints
* 1 \leq N \leq 10^5
* 0 \leq A_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
A_1
:
A_N
Output
Print the minimum number of colors required to satisfy the condition.
Examples
Input
5
2
1
4
5
3
Output
2
Input
4
0
0
0
0
Output
4
Submitted Solution:
```
from collections import deque
import bisect
N = int(input())
A = deque()
for _ in range(N):
x = int(input())
i = bisect.bisect_left(A,x)
if i == 0:
A.appendleft(x)
else:
A[i-1] = x
print(len(A))
``` | instruction | 0 | 92,442 | 7 | 184,884 |
Yes | output | 1 | 92,442 | 7 | 184,885 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence with N integers: A = \\{ A_1, A_2, \cdots, A_N \\}. For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:
* If A_i and A_j (i < j) are painted with the same color, A_i < A_j.
Find the minimum number of colors required to satisfy the condition.
Constraints
* 1 \leq N \leq 10^5
* 0 \leq A_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
A_1
:
A_N
Output
Print the minimum number of colors required to satisfy the condition.
Examples
Input
5
2
1
4
5
3
Output
2
Input
4
0
0
0
0
Output
4
Submitted Solution:
```
from bisect import bisect_right
N = int(input())
A = []
for i in range(N):
A.append(-int(input()))
INF = float('inf')
D = [INF] * N
for i in range(len(A)):
D[bisect_right(D, A[i])] = A[i]
print(N - D.count(INF))
``` | instruction | 0 | 92,443 | 7 | 184,886 |
Yes | output | 1 | 92,443 | 7 | 184,887 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence with N integers: A = \\{ A_1, A_2, \cdots, A_N \\}. For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:
* If A_i and A_j (i < j) are painted with the same color, A_i < A_j.
Find the minimum number of colors required to satisfy the condition.
Constraints
* 1 \leq N \leq 10^5
* 0 \leq A_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
A_1
:
A_N
Output
Print the minimum number of colors required to satisfy the condition.
Examples
Input
5
2
1
4
5
3
Output
2
Input
4
0
0
0
0
Output
4
Submitted Solution:
```
from bisect import bisect_right
N = int(input())
S = [1]
for _ in range(N):
a = int(input())
a *= -1
i = bisect_right(S, a)
if i == len(S):
S.append(a)
else:
S[i] = a
print(len(S))
``` | instruction | 0 | 92,444 | 7 | 184,888 |
Yes | output | 1 | 92,444 | 7 | 184,889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence with N integers: A = \\{ A_1, A_2, \cdots, A_N \\}. For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:
* If A_i and A_j (i < j) are painted with the same color, A_i < A_j.
Find the minimum number of colors required to satisfy the condition.
Constraints
* 1 \leq N \leq 10^5
* 0 \leq A_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
A_1
:
A_N
Output
Print the minimum number of colors required to satisfy the condition.
Examples
Input
5
2
1
4
5
3
Output
2
Input
4
0
0
0
0
Output
4
Submitted Solution:
```
N = int(input())
A = []
for _ in range(N):
a = int(input())
A.append(a)
nums = A[:]
colors = []
from bisect import bisect_left, bisect_right, insort_right
cnt = 0
for n in nums:
idx = bisect_right(colors, n)
if idx == 0:
insort_right(colors, n)
else:
if idx == len(colors):
colors[idx-1] = max(n, colors[idx-1])
else:
colors[idx] = max(n, colors[idx])
#print(colors)
print(len(colors))
``` | instruction | 0 | 92,445 | 7 | 184,890 |
No | output | 1 | 92,445 | 7 | 184,891 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence with N integers: A = \\{ A_1, A_2, \cdots, A_N \\}. For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:
* If A_i and A_j (i < j) are painted with the same color, A_i < A_j.
Find the minimum number of colors required to satisfy the condition.
Constraints
* 1 \leq N \leq 10^5
* 0 \leq A_i \leq 10^9
Input
Input is given from Standard Input in the following format:
N
A_1
:
A_N
Output
Print the minimum number of colors required to satisfy the condition.
Examples
Input
5
2
1
4
5
3
Output
2
Input
4
0
0
0
0
Output
4
Submitted Solution:
```
# ABC134E - Sequence Decomposing
def lis(A: "Array[int]") -> int:
from bisect import bisect_left
# dp := length of LIS (dp table itself is not LIS)
dp = [A[0]]
for i in A[1:]:
if i >= dp[-1]:
dp.append(i)
else:
dp[bisect_left(dp, i)] = i
return len(dp)
def main():
N, *A = map(int, open(0))
ans = lis([-i for i in A]) # need the length of lds -> -i
print(ans)
if __name__ == "__main__":
main()
``` | instruction | 0 | 92,447 | 7 | 184,894 |
No | output | 1 | 92,447 | 7 | 184,895 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Panic is rising in the committee for doggo standardization — the puppies of the new brood have been born multi-colored! In total there are 26 possible colors of puppies in the nature and they are denoted by letters from 'a' to 'z' inclusive.
The committee rules strictly prohibit even the smallest diversity between doggos and hence all the puppies should be of the same color. Thus Slava, the committee employee, has been assigned the task to recolor some puppies into other colors in order to eliminate the difference and make all the puppies have one common color.
Unfortunately, due to bureaucratic reasons and restricted budget, there's only one operation Slava can perform: he can choose a color x such that there are currently at least two puppies of color x and recolor all puppies of the color x into some arbitrary color y. Luckily, this operation can be applied multiple times (including zero).
For example, if the number of puppies is 7 and their colors are represented as the string "abababc", then in one operation Slava can get the results "zbzbzbc", "bbbbbbc", "aaaaaac", "acacacc" and others. However, if the current color sequence is "abababc", then he can't choose x='c' right now, because currently only one puppy has the color 'c'.
Help Slava and the committee determine whether it is possible to standardize all the puppies, i.e. after Slava's operations all the puppies should have the same color.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of puppies.
The second line contains a string s of length n consisting of lowercase Latin letters, where the i-th symbol denotes the i-th puppy's color.
Output
If it's possible to recolor all puppies into one color, print "Yes".
Otherwise print "No".
Output the answer without quotation signs.
Examples
Input
6
aabddc
Output
Yes
Input
3
abc
Output
No
Input
3
jjj
Output
Yes
Note
In the first example Slava can perform the following steps:
1. take all puppies of color 'a' (a total of two) and recolor them into 'b';
2. take all puppies of color 'd' (a total of two) and recolor them into 'c';
3. take all puppies of color 'b' (three puppies for now) and recolor them into 'c'.
In the second example it's impossible to recolor any of the puppies.
In the third example all the puppies' colors are the same; thus there's no need to recolor anything. | instruction | 0 | 92,613 | 7 | 185,226 |
Tags: implementation, sortings
Correct Solution:
```
n=int(input())
s=input()
print("Yes" if n==1or n>26or len(set(s))!=n else "No")
``` | output | 1 | 92,613 | 7 | 185,227 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Panic is rising in the committee for doggo standardization — the puppies of the new brood have been born multi-colored! In total there are 26 possible colors of puppies in the nature and they are denoted by letters from 'a' to 'z' inclusive.
The committee rules strictly prohibit even the smallest diversity between doggos and hence all the puppies should be of the same color. Thus Slava, the committee employee, has been assigned the task to recolor some puppies into other colors in order to eliminate the difference and make all the puppies have one common color.
Unfortunately, due to bureaucratic reasons and restricted budget, there's only one operation Slava can perform: he can choose a color x such that there are currently at least two puppies of color x and recolor all puppies of the color x into some arbitrary color y. Luckily, this operation can be applied multiple times (including zero).
For example, if the number of puppies is 7 and their colors are represented as the string "abababc", then in one operation Slava can get the results "zbzbzbc", "bbbbbbc", "aaaaaac", "acacacc" and others. However, if the current color sequence is "abababc", then he can't choose x='c' right now, because currently only one puppy has the color 'c'.
Help Slava and the committee determine whether it is possible to standardize all the puppies, i.e. after Slava's operations all the puppies should have the same color.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of puppies.
The second line contains a string s of length n consisting of lowercase Latin letters, where the i-th symbol denotes the i-th puppy's color.
Output
If it's possible to recolor all puppies into one color, print "Yes".
Otherwise print "No".
Output the answer without quotation signs.
Examples
Input
6
aabddc
Output
Yes
Input
3
abc
Output
No
Input
3
jjj
Output
Yes
Note
In the first example Slava can perform the following steps:
1. take all puppies of color 'a' (a total of two) and recolor them into 'b';
2. take all puppies of color 'd' (a total of two) and recolor them into 'c';
3. take all puppies of color 'b' (three puppies for now) and recolor them into 'c'.
In the second example it's impossible to recolor any of the puppies.
In the third example all the puppies' colors are the same; thus there's no need to recolor anything. | instruction | 0 | 92,614 | 7 | 185,228 |
Tags: implementation, sortings
Correct Solution:
```
n=int(input())
s=input().strip()
arr=[]
for i in range(26):
arr.append(0)
for i in s:
arr[ord(i)-97]+=1
flag=1
if (n==1):
print ("Yes")
else:
for i in range(26):
if (arr[i]!=0):
if (arr[i]!=1):
print ("Yes")
flag=0
break
if (flag==1):
print ("No")
``` | output | 1 | 92,614 | 7 | 185,229 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Panic is rising in the committee for doggo standardization — the puppies of the new brood have been born multi-colored! In total there are 26 possible colors of puppies in the nature and they are denoted by letters from 'a' to 'z' inclusive.
The committee rules strictly prohibit even the smallest diversity between doggos and hence all the puppies should be of the same color. Thus Slava, the committee employee, has been assigned the task to recolor some puppies into other colors in order to eliminate the difference and make all the puppies have one common color.
Unfortunately, due to bureaucratic reasons and restricted budget, there's only one operation Slava can perform: he can choose a color x such that there are currently at least two puppies of color x and recolor all puppies of the color x into some arbitrary color y. Luckily, this operation can be applied multiple times (including zero).
For example, if the number of puppies is 7 and their colors are represented as the string "abababc", then in one operation Slava can get the results "zbzbzbc", "bbbbbbc", "aaaaaac", "acacacc" and others. However, if the current color sequence is "abababc", then he can't choose x='c' right now, because currently only one puppy has the color 'c'.
Help Slava and the committee determine whether it is possible to standardize all the puppies, i.e. after Slava's operations all the puppies should have the same color.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of puppies.
The second line contains a string s of length n consisting of lowercase Latin letters, where the i-th symbol denotes the i-th puppy's color.
Output
If it's possible to recolor all puppies into one color, print "Yes".
Otherwise print "No".
Output the answer without quotation signs.
Examples
Input
6
aabddc
Output
Yes
Input
3
abc
Output
No
Input
3
jjj
Output
Yes
Note
In the first example Slava can perform the following steps:
1. take all puppies of color 'a' (a total of two) and recolor them into 'b';
2. take all puppies of color 'd' (a total of two) and recolor them into 'c';
3. take all puppies of color 'b' (three puppies for now) and recolor them into 'c'.
In the second example it's impossible to recolor any of the puppies.
In the third example all the puppies' colors are the same; thus there's no need to recolor anything. | instruction | 0 | 92,615 | 7 | 185,230 |
Tags: implementation, sortings
Correct Solution:
```
n = int(input())
s = input()
arr = [0] * 26
for i in s:
arr[ord(i)-97] += 1
if n == 1:
print('YES')
exit()
else:
for i in arr:
if i >= 2:
print('YES')
exit()
print('NO')
``` | output | 1 | 92,615 | 7 | 185,231 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Panic is rising in the committee for doggo standardization — the puppies of the new brood have been born multi-colored! In total there are 26 possible colors of puppies in the nature and they are denoted by letters from 'a' to 'z' inclusive.
The committee rules strictly prohibit even the smallest diversity between doggos and hence all the puppies should be of the same color. Thus Slava, the committee employee, has been assigned the task to recolor some puppies into other colors in order to eliminate the difference and make all the puppies have one common color.
Unfortunately, due to bureaucratic reasons and restricted budget, there's only one operation Slava can perform: he can choose a color x such that there are currently at least two puppies of color x and recolor all puppies of the color x into some arbitrary color y. Luckily, this operation can be applied multiple times (including zero).
For example, if the number of puppies is 7 and their colors are represented as the string "abababc", then in one operation Slava can get the results "zbzbzbc", "bbbbbbc", "aaaaaac", "acacacc" and others. However, if the current color sequence is "abababc", then he can't choose x='c' right now, because currently only one puppy has the color 'c'.
Help Slava and the committee determine whether it is possible to standardize all the puppies, i.e. after Slava's operations all the puppies should have the same color.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of puppies.
The second line contains a string s of length n consisting of lowercase Latin letters, where the i-th symbol denotes the i-th puppy's color.
Output
If it's possible to recolor all puppies into one color, print "Yes".
Otherwise print "No".
Output the answer without quotation signs.
Examples
Input
6
aabddc
Output
Yes
Input
3
abc
Output
No
Input
3
jjj
Output
Yes
Note
In the first example Slava can perform the following steps:
1. take all puppies of color 'a' (a total of two) and recolor them into 'b';
2. take all puppies of color 'd' (a total of two) and recolor them into 'c';
3. take all puppies of color 'b' (three puppies for now) and recolor them into 'c'.
In the second example it's impossible to recolor any of the puppies.
In the third example all the puppies' colors are the same; thus there's no need to recolor anything. | instruction | 0 | 92,616 | 7 | 185,232 |
Tags: implementation, sortings
Correct Solution:
```
import sys
#sys.stdin = open("py_in.txt","r")
freq = [0] * 27
n = int(input())
p = input()
if(n==1):
print("Yes")
else:
flag = False
for i in range(n):
id = ord(p[i]) - ord('a')
freq[id] += 1
if(freq[id]>1):
flag = True
break
if(flag==True):
print("Yes")
else:
print("No")
``` | output | 1 | 92,616 | 7 | 185,233 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Panic is rising in the committee for doggo standardization — the puppies of the new brood have been born multi-colored! In total there are 26 possible colors of puppies in the nature and they are denoted by letters from 'a' to 'z' inclusive.
The committee rules strictly prohibit even the smallest diversity between doggos and hence all the puppies should be of the same color. Thus Slava, the committee employee, has been assigned the task to recolor some puppies into other colors in order to eliminate the difference and make all the puppies have one common color.
Unfortunately, due to bureaucratic reasons and restricted budget, there's only one operation Slava can perform: he can choose a color x such that there are currently at least two puppies of color x and recolor all puppies of the color x into some arbitrary color y. Luckily, this operation can be applied multiple times (including zero).
For example, if the number of puppies is 7 and their colors are represented as the string "abababc", then in one operation Slava can get the results "zbzbzbc", "bbbbbbc", "aaaaaac", "acacacc" and others. However, if the current color sequence is "abababc", then he can't choose x='c' right now, because currently only one puppy has the color 'c'.
Help Slava and the committee determine whether it is possible to standardize all the puppies, i.e. after Slava's operations all the puppies should have the same color.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of puppies.
The second line contains a string s of length n consisting of lowercase Latin letters, where the i-th symbol denotes the i-th puppy's color.
Output
If it's possible to recolor all puppies into one color, print "Yes".
Otherwise print "No".
Output the answer without quotation signs.
Examples
Input
6
aabddc
Output
Yes
Input
3
abc
Output
No
Input
3
jjj
Output
Yes
Note
In the first example Slava can perform the following steps:
1. take all puppies of color 'a' (a total of two) and recolor them into 'b';
2. take all puppies of color 'd' (a total of two) and recolor them into 'c';
3. take all puppies of color 'b' (three puppies for now) and recolor them into 'c'.
In the second example it's impossible to recolor any of the puppies.
In the third example all the puppies' colors are the same; thus there's no need to recolor anything. | instruction | 0 | 92,617 | 7 | 185,234 |
Tags: implementation, sortings
Correct Solution:
```
#-------------Program--------------
#----Kuzlyaev-Nikita-Codeforces----
#-------------Training-------------
#----------------------------------
n=int(input())
s=list(str(input()))
if n>26:print("Yes")
else:
l=list(set(s))
if len(l)==len(s) and n!=1:
print("No")
else:
print("Yes")
``` | output | 1 | 92,617 | 7 | 185,235 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Panic is rising in the committee for doggo standardization — the puppies of the new brood have been born multi-colored! In total there are 26 possible colors of puppies in the nature and they are denoted by letters from 'a' to 'z' inclusive.
The committee rules strictly prohibit even the smallest diversity between doggos and hence all the puppies should be of the same color. Thus Slava, the committee employee, has been assigned the task to recolor some puppies into other colors in order to eliminate the difference and make all the puppies have one common color.
Unfortunately, due to bureaucratic reasons and restricted budget, there's only one operation Slava can perform: he can choose a color x such that there are currently at least two puppies of color x and recolor all puppies of the color x into some arbitrary color y. Luckily, this operation can be applied multiple times (including zero).
For example, if the number of puppies is 7 and their colors are represented as the string "abababc", then in one operation Slava can get the results "zbzbzbc", "bbbbbbc", "aaaaaac", "acacacc" and others. However, if the current color sequence is "abababc", then he can't choose x='c' right now, because currently only one puppy has the color 'c'.
Help Slava and the committee determine whether it is possible to standardize all the puppies, i.e. after Slava's operations all the puppies should have the same color.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of puppies.
The second line contains a string s of length n consisting of lowercase Latin letters, where the i-th symbol denotes the i-th puppy's color.
Output
If it's possible to recolor all puppies into one color, print "Yes".
Otherwise print "No".
Output the answer without quotation signs.
Examples
Input
6
aabddc
Output
Yes
Input
3
abc
Output
No
Input
3
jjj
Output
Yes
Note
In the first example Slava can perform the following steps:
1. take all puppies of color 'a' (a total of two) and recolor them into 'b';
2. take all puppies of color 'd' (a total of two) and recolor them into 'c';
3. take all puppies of color 'b' (three puppies for now) and recolor them into 'c'.
In the second example it's impossible to recolor any of the puppies.
In the third example all the puppies' colors are the same; thus there's no need to recolor anything. | instruction | 0 | 92,618 | 7 | 185,236 |
Tags: implementation, sortings
Correct Solution:
```
import functools
x = input()
sequence = input()
sequence = ''.join(sorted(sequence))
previous_char = None
is_possible = False
if len(sequence) == 1:
print("YES")
else:
for char in sequence:
if char == previous_char:
is_possible = True
break
previous_char = char
if is_possible:
print("YES")
else:
print("NO")
``` | output | 1 | 92,618 | 7 | 185,237 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Panic is rising in the committee for doggo standardization — the puppies of the new brood have been born multi-colored! In total there are 26 possible colors of puppies in the nature and they are denoted by letters from 'a' to 'z' inclusive.
The committee rules strictly prohibit even the smallest diversity between doggos and hence all the puppies should be of the same color. Thus Slava, the committee employee, has been assigned the task to recolor some puppies into other colors in order to eliminate the difference and make all the puppies have one common color.
Unfortunately, due to bureaucratic reasons and restricted budget, there's only one operation Slava can perform: he can choose a color x such that there are currently at least two puppies of color x and recolor all puppies of the color x into some arbitrary color y. Luckily, this operation can be applied multiple times (including zero).
For example, if the number of puppies is 7 and their colors are represented as the string "abababc", then in one operation Slava can get the results "zbzbzbc", "bbbbbbc", "aaaaaac", "acacacc" and others. However, if the current color sequence is "abababc", then he can't choose x='c' right now, because currently only one puppy has the color 'c'.
Help Slava and the committee determine whether it is possible to standardize all the puppies, i.e. after Slava's operations all the puppies should have the same color.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of puppies.
The second line contains a string s of length n consisting of lowercase Latin letters, where the i-th symbol denotes the i-th puppy's color.
Output
If it's possible to recolor all puppies into one color, print "Yes".
Otherwise print "No".
Output the answer without quotation signs.
Examples
Input
6
aabddc
Output
Yes
Input
3
abc
Output
No
Input
3
jjj
Output
Yes
Note
In the first example Slava can perform the following steps:
1. take all puppies of color 'a' (a total of two) and recolor them into 'b';
2. take all puppies of color 'd' (a total of two) and recolor them into 'c';
3. take all puppies of color 'b' (three puppies for now) and recolor them into 'c'.
In the second example it's impossible to recolor any of the puppies.
In the third example all the puppies' colors are the same; thus there's no need to recolor anything. | instruction | 0 | 92,619 | 7 | 185,238 |
Tags: implementation, sortings
Correct Solution:
```
#!/usr/bin/python
import sys
a = input()
a = input()
count = [0] * 26
for i in a:
count[ord(i) - ord('a')] += 1
if (max(count) >= 2) or (len(a) == 1):
print('Yes')
else:
print('No')
``` | output | 1 | 92,619 | 7 | 185,239 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.