message stringlengths 2 65.1k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 0 108k | cluster float64 14 14 | __index_level_0__ int64 0 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A dragon symbolizes wisdom, power and wealth. On Lunar New Year's Day, people model a dragon with bamboo strips and clothes, raise them with rods, and hold the rods high and low to resemble a flying dragon.
A performer holding the rod low is represented by a 1, while one holding it high is represented by a 2. Thus, the line of performers can be represented by a sequence a1, a2, ..., an.
Little Tommy is among them. He would like to choose an interval [l, r] (1 β€ l β€ r β€ n), then reverse al, al + 1, ..., ar so that the length of the longest non-decreasing subsequence of the new sequence is maximum.
A non-decreasing subsequence is a sequence of indices p1, p2, ..., pk, such that p1 < p2 < ... < pk and ap1 β€ ap2 β€ ... β€ apk. The length of the subsequence is k.
Input
The first line contains an integer n (1 β€ n β€ 2000), denoting the length of the original sequence.
The second line contains n space-separated integers, describing the original sequence a1, a2, ..., an (1 β€ ai β€ 2, i = 1, 2, ..., n).
Output
Print a single integer, which means the maximum possible length of the longest non-decreasing subsequence of the new sequence.
Examples
Input
4
1 2 1 2
Output
4
Input
10
1 1 2 2 2 1 1 2 2 1
Output
9
Note
In the first example, after reversing [2, 3], the array will become [1, 1, 2, 2], where the length of the longest non-decreasing subsequence is 4.
In the second example, after reversing [3, 7], the array will become [1, 1, 1, 1, 2, 2, 2, 2, 2, 1], where the length of the longest non-decreasing subsequence is 9.
Submitted Solution:
```
n = int(input())
*A, = map(int, input().split())
seq = []
cur = 1; cnt = 0
for a in A:
if cur == a:
cnt += 1
else:
seq.append(cnt)
cur = a; cnt = 1
if cnt > 0:
seq.append(cnt)
ans = seq[0]
l = len(seq)
for i in range(1, l, 2):
res = seq[i-1] + seq[i]
if i + 1 < l:
res += seq[i+1]
if i + 2 < l:
res += seq[i+2]
ans = max(ans, res)
if i + 1 < l:
for j in range(i-3, -1, -2):
print(i, j)
ans = max(ans, seq[i] + seq[i+1] + seq[j])
for j in range(i+4, l, 2):
ans = max(ans, seq[i] + seq[i+1] + seq[j])
#print(seq)
print(ans)
``` | instruction | 0 | 51,889 | 14 | 103,778 |
No | output | 1 | 51,889 | 14 | 103,779 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A dragon symbolizes wisdom, power and wealth. On Lunar New Year's Day, people model a dragon with bamboo strips and clothes, raise them with rods, and hold the rods high and low to resemble a flying dragon.
A performer holding the rod low is represented by a 1, while one holding it high is represented by a 2. Thus, the line of performers can be represented by a sequence a1, a2, ..., an.
Little Tommy is among them. He would like to choose an interval [l, r] (1 β€ l β€ r β€ n), then reverse al, al + 1, ..., ar so that the length of the longest non-decreasing subsequence of the new sequence is maximum.
A non-decreasing subsequence is a sequence of indices p1, p2, ..., pk, such that p1 < p2 < ... < pk and ap1 β€ ap2 β€ ... β€ apk. The length of the subsequence is k.
Input
The first line contains an integer n (1 β€ n β€ 2000), denoting the length of the original sequence.
The second line contains n space-separated integers, describing the original sequence a1, a2, ..., an (1 β€ ai β€ 2, i = 1, 2, ..., n).
Output
Print a single integer, which means the maximum possible length of the longest non-decreasing subsequence of the new sequence.
Examples
Input
4
1 2 1 2
Output
4
Input
10
1 1 2 2 2 1 1 2 2 1
Output
9
Note
In the first example, after reversing [2, 3], the array will become [1, 1, 2, 2], where the length of the longest non-decreasing subsequence is 4.
In the second example, after reversing [3, 7], the array will become [1, 1, 1, 1, 2, 2, 2, 2, 2, 1], where the length of the longest non-decreasing subsequence is 9.
Submitted Solution:
```
n = int(input())
*A, = map(int, input().split())
seq = []
cur = 1; cnt = 0
for a in A:
if cur == a:
cnt += 1
else:
seq.append(cnt)
cur = a; cnt = 1
if cnt > 0:
seq.append(cnt)
ans = seq[0]
l = len(seq)
for i in range(1, l, 2):
res = seq[i-1] + seq[i]
if i + 1 < l:
res += seq[i+1]
if i + 2 < l:
res += seq[i+2]
ans = max(ans, res)
#print(seq)
print(ans)
``` | instruction | 0 | 51,890 | 14 | 103,780 |
No | output | 1 | 51,890 | 14 | 103,781 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
A dragon symbolizes wisdom, power and wealth. On Lunar New Year's Day, people model a dragon with bamboo strips and clothes, raise them with rods, and hold the rods high and low to resemble a flying dragon.
A performer holding the rod low is represented by a 1, while one holding it high is represented by a 2. Thus, the line of performers can be represented by a sequence a1, a2, ..., an.
Little Tommy is among them. He would like to choose an interval [l, r] (1 β€ l β€ r β€ n), then reverse al, al + 1, ..., ar so that the length of the longest non-decreasing subsequence of the new sequence is maximum.
A non-decreasing subsequence is a sequence of indices p1, p2, ..., pk, such that p1 < p2 < ... < pk and ap1 β€ ap2 β€ ... β€ apk. The length of the subsequence is k.
Input
The first line contains an integer n (1 β€ n β€ 2000), denoting the length of the original sequence.
The second line contains n space-separated integers, describing the original sequence a1, a2, ..., an (1 β€ ai β€ 2, i = 1, 2, ..., n).
Output
Print a single integer, which means the maximum possible length of the longest non-decreasing subsequence of the new sequence.
Examples
Input
4
1 2 1 2
Output
4
Input
10
1 1 2 2 2 1 1 2 2 1
Output
9
Note
In the first example, after reversing [2, 3], the array will become [1, 1, 2, 2], where the length of the longest non-decreasing subsequence is 4.
In the second example, after reversing [3, 7], the array will become [1, 1, 1, 1, 2, 2, 2, 2, 2, 1], where the length of the longest non-decreasing subsequence is 9.
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
n=in_num()
l=in_arr()
arr=[]
c=1
curr=l[0]
for i in range(1,n):
if l[i]==l[i-1]:
c+=1
else:
arr.append((curr,c))
c=1
curr=l[i]
if c:
arr.append((curr,c))
n=len(arr)
dp=[[0,0,0,0] for i in range(n)]
ans=0
for i in range(n):
if arr[i][0]==1:
dp[i][0]=arr[i][1]
for j in range(i-1,-1,-1):
dp[i][2]=max(dp[i][2],dp[j][1]+arr[i][1])
else:
for j in range(i-1,-1,-1):
dp[i][1]=max(dp[i][1],dp[j][0]+arr[i][1])
for j in range(i-1,-1,-1):
dp[i][3]=max(dp[i][3],dp[j][2]+arr[i][1])
ans=max(ans,dp[i][1],dp[i][3])
pr_num(ans)
``` | instruction | 0 | 51,891 | 14 | 103,782 |
No | output | 1 | 51,891 | 14 | 103,783 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Uniqlo Uzuki and Rin Meguro are coming to SEARIGHT LIVE FES for sale. As shown in the figure below, the product sales line consists of two lines, with W people in the horizontal direction and H people in the vertical direction, forming a U-shape. The person at the top of the line is removed from the line after shopping, and the people behind the same line are lined up for one person each.
picture of line
The two are first next to each other at the end of the product sales line, and are lined up in different lines (orange and dark blue positions in the figure below). The two are very close, so I always want to be as close as possible. However, since the way people are brushed differs depending on the row, there is a possibility that the positions of two people in different rows will be separated.
Your job is to find the number of times two people will be next to each other given how to brush the line. However, when both of them are in the corner of the U-shape (in the black state in the above figure), they are not considered to be next to each other. Also, immediately after lining up, they are lined up next to each other, but this is not counted.
Constraints
* 4 β€ W β€ 100
* 4 β€ H β€ 100
* 0 β€ N β€ 100
Input
The input is given in the following format.
W H
N
p1 p2 ... pN
The integers W and H that represent the size of the column are given in the first row, separated by blanks. W represents the number of people on the outside arranged horizontally in a U-shape, and H represents the number of people on the outside arranged vertically. The second row is given the number of times N a person can pull out of either column. In the third row, information about how to remove columns p1 ... pN is given separated by blanks. pi is '0' or '1', and when it is '0', it means that one person on the outside can get rid of it, and when it is '1', it means that one person on the inside can get rid of it. However, Uniqlo Uzuki or Rin Meguro will not leave the line.
Output
Output the number of times two people are next to each other in one line.
Examples
Input
11 11
5
0 0 0 0 0
Output
0
Input
11 11
5
0 0 1 1 1
Output
1 | instruction | 0 | 52,130 | 14 | 104,260 |
"Correct Solution:
```
pos = [[0,0], [1,0]]
way = [[0,1], [0,1]]
ans = 0
w, h = map(int, input().split())
n = int(input())
li = list(map(int, input().split()))
for p in li:
ni = pos[p][0]+way[p][0]
nj = pos[p][1]+way[p][1]
if p == 0:
if nj >= w or ni >= h:
way[p][0],way[p][1]=way[p][1],-way[p][0]
ni = pos[p][0]+way[p][0]
nj = pos[p][1]+way[p][1]
else:
if nj >= w-1 or ni >= h-1:
way[p][0],way[p][1]=way[p][1],-way[p][0]
ni = pos[p][0]+way[p][0]
nj = pos[p][1]+way[p][1]
pos[p][0] = ni
pos[p][1] = nj
if abs(pos[0][0]-pos[1][0])+abs(pos[0][1]-pos[1][1]) == 1:
ans += 1
print(ans)
``` | output | 1 | 52,130 | 14 | 104,261 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Uniqlo Uzuki and Rin Meguro are coming to SEARIGHT LIVE FES for sale. As shown in the figure below, the product sales line consists of two lines, with W people in the horizontal direction and H people in the vertical direction, forming a U-shape. The person at the top of the line is removed from the line after shopping, and the people behind the same line are lined up for one person each.
picture of line
The two are first next to each other at the end of the product sales line, and are lined up in different lines (orange and dark blue positions in the figure below). The two are very close, so I always want to be as close as possible. However, since the way people are brushed differs depending on the row, there is a possibility that the positions of two people in different rows will be separated.
Your job is to find the number of times two people will be next to each other given how to brush the line. However, when both of them are in the corner of the U-shape (in the black state in the above figure), they are not considered to be next to each other. Also, immediately after lining up, they are lined up next to each other, but this is not counted.
Constraints
* 4 β€ W β€ 100
* 4 β€ H β€ 100
* 0 β€ N β€ 100
Input
The input is given in the following format.
W H
N
p1 p2 ... pN
The integers W and H that represent the size of the column are given in the first row, separated by blanks. W represents the number of people on the outside arranged horizontally in a U-shape, and H represents the number of people on the outside arranged vertically. The second row is given the number of times N a person can pull out of either column. In the third row, information about how to remove columns p1 ... pN is given separated by blanks. pi is '0' or '1', and when it is '0', it means that one person on the outside can get rid of it, and when it is '1', it means that one person on the inside can get rid of it. However, Uniqlo Uzuki or Rin Meguro will not leave the line.
Output
Output the number of times two people are next to each other in one line.
Examples
Input
11 11
5
0 0 0 0 0
Output
0
Input
11 11
5
0 0 1 1 1
Output
1 | instruction | 0 | 52,131 | 14 | 104,262 |
"Correct Solution:
```
w,h = map(int, input().split())
n = int(input())
p = list(map(int, input().split()))
u = m = cnt = 0
for i in range(n):
if p[i] == 0:
u += 1
else:
m += 1
if u <= w-2:
if u == m:
cnt += 1
elif w <= u <= w+h-3:
if u == m+2:
cnt += 1
elif h+h-1 <= u:
if u == m+4:
cnt += 1
print(cnt)
``` | output | 1 | 52,131 | 14 | 104,263 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
Uniqlo Uzuki and Rin Meguro are coming to SEARIGHT LIVE FES for sale. As shown in the figure below, the product sales line consists of two lines, with W people in the horizontal direction and H people in the vertical direction, forming a U-shape. The person at the top of the line is removed from the line after shopping, and the people behind the same line are lined up for one person each.
picture of line
The two are first next to each other at the end of the product sales line, and are lined up in different lines (orange and dark blue positions in the figure below). The two are very close, so I always want to be as close as possible. However, since the way people are brushed differs depending on the row, there is a possibility that the positions of two people in different rows will be separated.
Your job is to find the number of times two people will be next to each other given how to brush the line. However, when both of them are in the corner of the U-shape (in the black state in the above figure), they are not considered to be next to each other. Also, immediately after lining up, they are lined up next to each other, but this is not counted.
Constraints
* 4 β€ W β€ 100
* 4 β€ H β€ 100
* 0 β€ N β€ 100
Input
The input is given in the following format.
W H
N
p1 p2 ... pN
The integers W and H that represent the size of the column are given in the first row, separated by blanks. W represents the number of people on the outside arranged horizontally in a U-shape, and H represents the number of people on the outside arranged vertically. The second row is given the number of times N a person can pull out of either column. In the third row, information about how to remove columns p1 ... pN is given separated by blanks. pi is '0' or '1', and when it is '0', it means that one person on the outside can get rid of it, and when it is '1', it means that one person on the inside can get rid of it. However, Uniqlo Uzuki or Rin Meguro will not leave the line.
Output
Output the number of times two people are next to each other in one line.
Examples
Input
11 11
5
0 0 0 0 0
Output
0
Input
11 11
5
0 0 1 1 1
Output
1 | instruction | 0 | 52,132 | 14 | 104,264 |
"Correct Solution:
```
# AOJ 1575: Product Sale Lines
# Python3 2018.7.13 bal4u
W, H = map(int, input().split())
N = int(input())
P = list(map(int, input().split()))
ans = x = y = 0
for p in P:
if p == 0: x += 1
else: y += 1
if x < W-1:
if x == y: ans += 1
elif x == W-1: pass
elif x < W+H-2:
if x == y+2: ans += 1
elif x == W+H-2: pass
else:
if x == y+4: ans += 1
print(ans)
``` | output | 1 | 52,132 | 14 | 104,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Uniqlo Uzuki and Rin Meguro are coming to SEARIGHT LIVE FES for sale. As shown in the figure below, the product sales line consists of two lines, with W people in the horizontal direction and H people in the vertical direction, forming a U-shape. The person at the top of the line is removed from the line after shopping, and the people behind the same line are lined up for one person each.
picture of line
The two are first next to each other at the end of the product sales line, and are lined up in different lines (orange and dark blue positions in the figure below). The two are very close, so I always want to be as close as possible. However, since the way people are brushed differs depending on the row, there is a possibility that the positions of two people in different rows will be separated.
Your job is to find the number of times two people will be next to each other given how to brush the line. However, when both of them are in the corner of the U-shape (in the black state in the above figure), they are not considered to be next to each other. Also, immediately after lining up, they are lined up next to each other, but this is not counted.
Constraints
* 4 β€ W β€ 100
* 4 β€ H β€ 100
* 0 β€ N β€ 100
Input
The input is given in the following format.
W H
N
p1 p2 ... pN
The integers W and H that represent the size of the column are given in the first row, separated by blanks. W represents the number of people on the outside arranged horizontally in a U-shape, and H represents the number of people on the outside arranged vertically. The second row is given the number of times N a person can pull out of either column. In the third row, information about how to remove columns p1 ... pN is given separated by blanks. pi is '0' or '1', and when it is '0', it means that one person on the outside can get rid of it, and when it is '1', it means that one person on the inside can get rid of it. However, Uniqlo Uzuki or Rin Meguro will not leave the line.
Output
Output the number of times two people are next to each other in one line.
Examples
Input
11 11
5
0 0 0 0 0
Output
0
Input
11 11
5
0 0 1 1 1
Output
1
Submitted Solution:
```
h, w = map(int, input().split())
n = int(input())
a = list(map(int, input().split()))
u, s, c = 0, 0, 0
for i in range(len(a)):
u += a[i]
s += 1- a[i]
if u < w - 2:
if u == s:
c += 1
elif u == w - 2:
if u == s or u + 2 == s:
c += 1
elif u < w + h - 5:
if u + 2 == s:
c += 1
elif u == w + h - 5:
if u + 2 == s or u + 4 == s:
c += 1
elif u + 4 == s:c += 1
print(c)
``` | instruction | 0 | 52,133 | 14 | 104,266 |
No | output | 1 | 52,133 | 14 | 104,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Uniqlo Uzuki and Rin Meguro are coming to SEARIGHT LIVE FES for sale. As shown in the figure below, the product sales line consists of two lines, with W people in the horizontal direction and H people in the vertical direction, forming a U-shape. The person at the top of the line is removed from the line after shopping, and the people behind the same line are lined up for one person each.
picture of line
The two are first next to each other at the end of the product sales line, and are lined up in different lines (orange and dark blue positions in the figure below). The two are very close, so I always want to be as close as possible. However, since the way people are brushed differs depending on the row, there is a possibility that the positions of two people in different rows will be separated.
Your job is to find the number of times two people will be next to each other given how to brush the line. However, when both of them are in the corner of the U-shape (in the black state in the above figure), they are not considered to be next to each other. Also, immediately after lining up, they are lined up next to each other, but this is not counted.
Constraints
* 4 β€ W β€ 100
* 4 β€ H β€ 100
* 0 β€ N β€ 100
Input
The input is given in the following format.
W H
N
p1 p2 ... pN
The integers W and H that represent the size of the column are given in the first row, separated by blanks. W represents the number of people on the outside arranged horizontally in a U-shape, and H represents the number of people on the outside arranged vertically. The second row is given the number of times N a person can pull out of either column. In the third row, information about how to remove columns p1 ... pN is given separated by blanks. pi is '0' or '1', and when it is '0', it means that one person on the outside can get rid of it, and when it is '1', it means that one person on the inside can get rid of it. However, Uniqlo Uzuki or Rin Meguro will not leave the line.
Output
Output the number of times two people are next to each other in one line.
Examples
Input
11 11
5
0 0 0 0 0
Output
0
Input
11 11
5
0 0 1 1 1
Output
1
Submitted Solution:
```
h, w = map(int, input().split())
n = int(input())
a = list(map(int, input().split()))
u, s, c = 0, 0, 0
for i in range(n):
u += a[i]
s += 1- a[i]
if u <= w - 2 and u == s:c += 1
elif w - 2 <= u and u <= w + h - 5 and u + 2 == s:c += 1
elif w + h - 5 <= u and u + 4 == s:c += 1
print(c)
``` | instruction | 0 | 52,134 | 14 | 104,268 |
No | output | 1 | 52,134 | 14 | 104,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Uniqlo Uzuki and Rin Meguro are coming to SEARIGHT LIVE FES for sale. As shown in the figure below, the product sales line consists of two lines, with W people in the horizontal direction and H people in the vertical direction, forming a U-shape. The person at the top of the line is removed from the line after shopping, and the people behind the same line are lined up for one person each.
picture of line
The two are first next to each other at the end of the product sales line, and are lined up in different lines (orange and dark blue positions in the figure below). The two are very close, so I always want to be as close as possible. However, since the way people are brushed differs depending on the row, there is a possibility that the positions of two people in different rows will be separated.
Your job is to find the number of times two people will be next to each other given how to brush the line. However, when both of them are in the corner of the U-shape (in the black state in the above figure), they are not considered to be next to each other. Also, immediately after lining up, they are lined up next to each other, but this is not counted.
Constraints
* 4 β€ W β€ 100
* 4 β€ H β€ 100
* 0 β€ N β€ 100
Input
The input is given in the following format.
W H
N
p1 p2 ... pN
The integers W and H that represent the size of the column are given in the first row, separated by blanks. W represents the number of people on the outside arranged horizontally in a U-shape, and H represents the number of people on the outside arranged vertically. The second row is given the number of times N a person can pull out of either column. In the third row, information about how to remove columns p1 ... pN is given separated by blanks. pi is '0' or '1', and when it is '0', it means that one person on the outside can get rid of it, and when it is '1', it means that one person on the inside can get rid of it. However, Uniqlo Uzuki or Rin Meguro will not leave the line.
Output
Output the number of times two people are next to each other in one line.
Examples
Input
11 11
5
0 0 0 0 0
Output
0
Input
11 11
5
0 0 1 1 1
Output
1
Submitted Solution:
```
# AOJ 1575: Product Sale Lines
# Python3 2018.7.13 bal4u
W, H = map(int, input().split())
N = int(input())
p = list(map(int, input().split()))
ans = x = y = 0
for x in p:
if x == 0: x += 1
else: y += 1
if x < W-1:
if x == y: ans += 1
elif x == W-1: pass
elif x < W+H-2:
if x == y+2: ans += 1
elif x == W+H-2: pass
else:
if x == y+4: ans += 1
print(ans)
``` | instruction | 0 | 52,135 | 14 | 104,270 |
No | output | 1 | 52,135 | 14 | 104,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Uniqlo Uzuki and Rin Meguro are coming to SEARIGHT LIVE FES for sale. As shown in the figure below, the product sales line consists of two lines, with W people in the horizontal direction and H people in the vertical direction, forming a U-shape. The person at the top of the line is removed from the line after shopping, and the people behind the same line are lined up for one person each.
picture of line
The two are first next to each other at the end of the product sales line, and are lined up in different lines (orange and dark blue positions in the figure below). The two are very close, so I always want to be as close as possible. However, since the way people are brushed differs depending on the row, there is a possibility that the positions of two people in different rows will be separated.
Your job is to find the number of times two people will be next to each other given how to brush the line. However, when both of them are in the corner of the U-shape (in the black state in the above figure), they are not considered to be next to each other. Also, immediately after lining up, they are lined up next to each other, but this is not counted.
Constraints
* 4 β€ W β€ 100
* 4 β€ H β€ 100
* 0 β€ N β€ 100
Input
The input is given in the following format.
W H
N
p1 p2 ... pN
The integers W and H that represent the size of the column are given in the first row, separated by blanks. W represents the number of people on the outside arranged horizontally in a U-shape, and H represents the number of people on the outside arranged vertically. The second row is given the number of times N a person can pull out of either column. In the third row, information about how to remove columns p1 ... pN is given separated by blanks. pi is '0' or '1', and when it is '0', it means that one person on the outside can get rid of it, and when it is '1', it means that one person on the inside can get rid of it. However, Uniqlo Uzuki or Rin Meguro will not leave the line.
Output
Output the number of times two people are next to each other in one line.
Examples
Input
11 11
5
0 0 0 0 0
Output
0
Input
11 11
5
0 0 1 1 1
Output
1
Submitted Solution:
```
h, w = map(int, input().split())
n = int(input())
a = list(map(int, input().split()))
u, s, c = 0, 0, 0
for i in range(n):
u += a[i]
s += 1- a[i]
if u <= w - 2 and u == s:c += 1
elif w - 2 <= u and u <= w + h - 5 and u + 2 == s:c += 1
elif w + h - 5 <= u and u + 3 == s:c += 1
print(c)
``` | instruction | 0 | 52,136 | 14 | 104,272 |
No | output | 1 | 52,136 | 14 | 104,273 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The three friends, Kuro, Shiro, and Katie, met up again! It's time for a party...
What the cats do when they unite? Right, they have a party. Since they wanted to have as much fun as possible, they invited all their friends. Now n cats are at the party, sitting in a circle and eating soup. The rules are simple: anyone having finished their soup leaves the circle.
Katie suddenly notices that whenever a cat leaves, the place where she was sitting becomes an empty space, which means the circle is divided into smaller continuous groups of cats sitting next to each other. At the moment Katie observes, there are m cats who left the circle. This raises a question for Katie: what is the maximum possible number of groups the circle is divided into at the moment?
Could you help her with this curiosity?
You can see the examples and their descriptions with pictures in the "Note" section.
Input
The only line contains two integers n and m (2 β€ n β€ 1000, 0 β€ m β€ n) β the initial number of cats at the party and the number of cats who left the circle at the moment Katie observes, respectively.
Output
Print a single integer β the maximum number of groups of cats at the moment Katie observes.
Examples
Input
7 4
Output
3
Input
6 2
Output
2
Input
3 0
Output
1
Input
2 2
Output
0
Note
In the first example, originally there are 7 cats sitting as shown below, creating a single group:
<image>
At the observed moment, 4 cats have left the table. Suppose the cats 2, 3, 5 and 7 have left, then there are 3 groups remaining. It is possible to show that it is the maximum possible number of groups remaining.
<image>
In the second example, there are 6 cats sitting as shown below:
<image>
At the observed moment, 2 cats have left the table. Suppose the cats numbered 3 and 6 left, then there will be 2 groups remaining (\{1, 2\} and \{4, 5\}). It is impossible to have more than 2 groups of cats remaining.
<image>
In the third example, no cats have left, so there is 1 group consisting of all cats.
In the fourth example, all cats have left the circle, so there are 0 groups. | instruction | 0 | 52,266 | 14 | 104,532 |
Tags: greedy, math
Correct Solution:
```
n,m=map(int,input().split())
print(min(max(m,1),n-m))
``` | output | 1 | 52,266 | 14 | 104,533 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The three friends, Kuro, Shiro, and Katie, met up again! It's time for a party...
What the cats do when they unite? Right, they have a party. Since they wanted to have as much fun as possible, they invited all their friends. Now n cats are at the party, sitting in a circle and eating soup. The rules are simple: anyone having finished their soup leaves the circle.
Katie suddenly notices that whenever a cat leaves, the place where she was sitting becomes an empty space, which means the circle is divided into smaller continuous groups of cats sitting next to each other. At the moment Katie observes, there are m cats who left the circle. This raises a question for Katie: what is the maximum possible number of groups the circle is divided into at the moment?
Could you help her with this curiosity?
You can see the examples and their descriptions with pictures in the "Note" section.
Input
The only line contains two integers n and m (2 β€ n β€ 1000, 0 β€ m β€ n) β the initial number of cats at the party and the number of cats who left the circle at the moment Katie observes, respectively.
Output
Print a single integer β the maximum number of groups of cats at the moment Katie observes.
Examples
Input
7 4
Output
3
Input
6 2
Output
2
Input
3 0
Output
1
Input
2 2
Output
0
Note
In the first example, originally there are 7 cats sitting as shown below, creating a single group:
<image>
At the observed moment, 4 cats have left the table. Suppose the cats 2, 3, 5 and 7 have left, then there are 3 groups remaining. It is possible to show that it is the maximum possible number of groups remaining.
<image>
In the second example, there are 6 cats sitting as shown below:
<image>
At the observed moment, 2 cats have left the table. Suppose the cats numbered 3 and 6 left, then there will be 2 groups remaining (\{1, 2\} and \{4, 5\}). It is impossible to have more than 2 groups of cats remaining.
<image>
In the third example, no cats have left, so there is 1 group consisting of all cats.
In the fourth example, all cats have left the circle, so there are 0 groups. | instruction | 0 | 52,267 | 14 | 104,534 |
Tags: greedy, math
Correct Solution:
```
K=list(map(int, input().strip().split()))
n=K[0]
m=K[1]
if m>=n/2:
print(n-m)
elif m==0:
print(1)
else:
print(m)
``` | output | 1 | 52,267 | 14 | 104,535 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The three friends, Kuro, Shiro, and Katie, met up again! It's time for a party...
What the cats do when they unite? Right, they have a party. Since they wanted to have as much fun as possible, they invited all their friends. Now n cats are at the party, sitting in a circle and eating soup. The rules are simple: anyone having finished their soup leaves the circle.
Katie suddenly notices that whenever a cat leaves, the place where she was sitting becomes an empty space, which means the circle is divided into smaller continuous groups of cats sitting next to each other. At the moment Katie observes, there are m cats who left the circle. This raises a question for Katie: what is the maximum possible number of groups the circle is divided into at the moment?
Could you help her with this curiosity?
You can see the examples and their descriptions with pictures in the "Note" section.
Input
The only line contains two integers n and m (2 β€ n β€ 1000, 0 β€ m β€ n) β the initial number of cats at the party and the number of cats who left the circle at the moment Katie observes, respectively.
Output
Print a single integer β the maximum number of groups of cats at the moment Katie observes.
Examples
Input
7 4
Output
3
Input
6 2
Output
2
Input
3 0
Output
1
Input
2 2
Output
0
Note
In the first example, originally there are 7 cats sitting as shown below, creating a single group:
<image>
At the observed moment, 4 cats have left the table. Suppose the cats 2, 3, 5 and 7 have left, then there are 3 groups remaining. It is possible to show that it is the maximum possible number of groups remaining.
<image>
In the second example, there are 6 cats sitting as shown below:
<image>
At the observed moment, 2 cats have left the table. Suppose the cats numbered 3 and 6 left, then there will be 2 groups remaining (\{1, 2\} and \{4, 5\}). It is impossible to have more than 2 groups of cats remaining.
<image>
In the third example, no cats have left, so there is 1 group consisting of all cats.
In the fourth example, all cats have left the circle, so there are 0 groups. | instruction | 0 | 52,268 | 14 | 104,536 |
Tags: greedy, math
Correct Solution:
```
n, m = map(int, input().split())
k = n//2
if m == 0:
print(1)
elif m <= k:
print(m)
else:
print(n-m)
``` | output | 1 | 52,268 | 14 | 104,537 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The three friends, Kuro, Shiro, and Katie, met up again! It's time for a party...
What the cats do when they unite? Right, they have a party. Since they wanted to have as much fun as possible, they invited all their friends. Now n cats are at the party, sitting in a circle and eating soup. The rules are simple: anyone having finished their soup leaves the circle.
Katie suddenly notices that whenever a cat leaves, the place where she was sitting becomes an empty space, which means the circle is divided into smaller continuous groups of cats sitting next to each other. At the moment Katie observes, there are m cats who left the circle. This raises a question for Katie: what is the maximum possible number of groups the circle is divided into at the moment?
Could you help her with this curiosity?
You can see the examples and their descriptions with pictures in the "Note" section.
Input
The only line contains two integers n and m (2 β€ n β€ 1000, 0 β€ m β€ n) β the initial number of cats at the party and the number of cats who left the circle at the moment Katie observes, respectively.
Output
Print a single integer β the maximum number of groups of cats at the moment Katie observes.
Examples
Input
7 4
Output
3
Input
6 2
Output
2
Input
3 0
Output
1
Input
2 2
Output
0
Note
In the first example, originally there are 7 cats sitting as shown below, creating a single group:
<image>
At the observed moment, 4 cats have left the table. Suppose the cats 2, 3, 5 and 7 have left, then there are 3 groups remaining. It is possible to show that it is the maximum possible number of groups remaining.
<image>
In the second example, there are 6 cats sitting as shown below:
<image>
At the observed moment, 2 cats have left the table. Suppose the cats numbered 3 and 6 left, then there will be 2 groups remaining (\{1, 2\} and \{4, 5\}). It is impossible to have more than 2 groups of cats remaining.
<image>
In the third example, no cats have left, so there is 1 group consisting of all cats.
In the fourth example, all cats have left the circle, so there are 0 groups. | instruction | 0 | 52,269 | 14 | 104,538 |
Tags: greedy, math
Correct Solution:
```
from sys import stdin
n,m=map(int,stdin.readline().split())
if m==0:
print(1)
elif n==m:
print(0)
else:
if m>n//2:
print(n-m)
else:
print(m)
``` | output | 1 | 52,269 | 14 | 104,539 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The three friends, Kuro, Shiro, and Katie, met up again! It's time for a party...
What the cats do when they unite? Right, they have a party. Since they wanted to have as much fun as possible, they invited all their friends. Now n cats are at the party, sitting in a circle and eating soup. The rules are simple: anyone having finished their soup leaves the circle.
Katie suddenly notices that whenever a cat leaves, the place where she was sitting becomes an empty space, which means the circle is divided into smaller continuous groups of cats sitting next to each other. At the moment Katie observes, there are m cats who left the circle. This raises a question for Katie: what is the maximum possible number of groups the circle is divided into at the moment?
Could you help her with this curiosity?
You can see the examples and their descriptions with pictures in the "Note" section.
Input
The only line contains two integers n and m (2 β€ n β€ 1000, 0 β€ m β€ n) β the initial number of cats at the party and the number of cats who left the circle at the moment Katie observes, respectively.
Output
Print a single integer β the maximum number of groups of cats at the moment Katie observes.
Examples
Input
7 4
Output
3
Input
6 2
Output
2
Input
3 0
Output
1
Input
2 2
Output
0
Note
In the first example, originally there are 7 cats sitting as shown below, creating a single group:
<image>
At the observed moment, 4 cats have left the table. Suppose the cats 2, 3, 5 and 7 have left, then there are 3 groups remaining. It is possible to show that it is the maximum possible number of groups remaining.
<image>
In the second example, there are 6 cats sitting as shown below:
<image>
At the observed moment, 2 cats have left the table. Suppose the cats numbered 3 and 6 left, then there will be 2 groups remaining (\{1, 2\} and \{4, 5\}). It is impossible to have more than 2 groups of cats remaining.
<image>
In the third example, no cats have left, so there is 1 group consisting of all cats.
In the fourth example, all cats have left the circle, so there are 0 groups. | instruction | 0 | 52,270 | 14 | 104,540 |
Tags: greedy, math
Correct Solution:
```
import sys
def res(n, m):
if m == 0:
print("1")
return
if m <= n/2:
print(m)
else:
print(n-m)
n, m = map(int, input().split())
res(n, m)
``` | output | 1 | 52,270 | 14 | 104,541 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The three friends, Kuro, Shiro, and Katie, met up again! It's time for a party...
What the cats do when they unite? Right, they have a party. Since they wanted to have as much fun as possible, they invited all their friends. Now n cats are at the party, sitting in a circle and eating soup. The rules are simple: anyone having finished their soup leaves the circle.
Katie suddenly notices that whenever a cat leaves, the place where she was sitting becomes an empty space, which means the circle is divided into smaller continuous groups of cats sitting next to each other. At the moment Katie observes, there are m cats who left the circle. This raises a question for Katie: what is the maximum possible number of groups the circle is divided into at the moment?
Could you help her with this curiosity?
You can see the examples and their descriptions with pictures in the "Note" section.
Input
The only line contains two integers n and m (2 β€ n β€ 1000, 0 β€ m β€ n) β the initial number of cats at the party and the number of cats who left the circle at the moment Katie observes, respectively.
Output
Print a single integer β the maximum number of groups of cats at the moment Katie observes.
Examples
Input
7 4
Output
3
Input
6 2
Output
2
Input
3 0
Output
1
Input
2 2
Output
0
Note
In the first example, originally there are 7 cats sitting as shown below, creating a single group:
<image>
At the observed moment, 4 cats have left the table. Suppose the cats 2, 3, 5 and 7 have left, then there are 3 groups remaining. It is possible to show that it is the maximum possible number of groups remaining.
<image>
In the second example, there are 6 cats sitting as shown below:
<image>
At the observed moment, 2 cats have left the table. Suppose the cats numbered 3 and 6 left, then there will be 2 groups remaining (\{1, 2\} and \{4, 5\}). It is impossible to have more than 2 groups of cats remaining.
<image>
In the third example, no cats have left, so there is 1 group consisting of all cats.
In the fourth example, all cats have left the circle, so there are 0 groups. | instruction | 0 | 52,271 | 14 | 104,542 |
Tags: greedy, math
Correct Solution:
```
n, m = map(int, input().split(' '))
pairs = n // 2
if n == m:
print(0)
elif m == 0 or m == 1:
print(1)
elif m <= pairs:
print(m)
elif n % 2 and m == pairs + 1:
print(pairs)
else:
print(n - m)
``` | output | 1 | 52,271 | 14 | 104,543 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The three friends, Kuro, Shiro, and Katie, met up again! It's time for a party...
What the cats do when they unite? Right, they have a party. Since they wanted to have as much fun as possible, they invited all their friends. Now n cats are at the party, sitting in a circle and eating soup. The rules are simple: anyone having finished their soup leaves the circle.
Katie suddenly notices that whenever a cat leaves, the place where she was sitting becomes an empty space, which means the circle is divided into smaller continuous groups of cats sitting next to each other. At the moment Katie observes, there are m cats who left the circle. This raises a question for Katie: what is the maximum possible number of groups the circle is divided into at the moment?
Could you help her with this curiosity?
You can see the examples and their descriptions with pictures in the "Note" section.
Input
The only line contains two integers n and m (2 β€ n β€ 1000, 0 β€ m β€ n) β the initial number of cats at the party and the number of cats who left the circle at the moment Katie observes, respectively.
Output
Print a single integer β the maximum number of groups of cats at the moment Katie observes.
Examples
Input
7 4
Output
3
Input
6 2
Output
2
Input
3 0
Output
1
Input
2 2
Output
0
Note
In the first example, originally there are 7 cats sitting as shown below, creating a single group:
<image>
At the observed moment, 4 cats have left the table. Suppose the cats 2, 3, 5 and 7 have left, then there are 3 groups remaining. It is possible to show that it is the maximum possible number of groups remaining.
<image>
In the second example, there are 6 cats sitting as shown below:
<image>
At the observed moment, 2 cats have left the table. Suppose the cats numbered 3 and 6 left, then there will be 2 groups remaining (\{1, 2\} and \{4, 5\}). It is impossible to have more than 2 groups of cats remaining.
<image>
In the third example, no cats have left, so there is 1 group consisting of all cats.
In the fourth example, all cats have left the circle, so there are 0 groups. | instruction | 0 | 52,272 | 14 | 104,544 |
Tags: greedy, math
Correct Solution:
```
a = [int(i) for i in input().split()]
n =a[0]
m=a[1]
if(m==0):
print(1)
elif(n%2==0):
if(m==1):
print(1)
elif(m <= n//2):
print(m)
else:
print(n-m)
else:
if(m==1):
print(1)
elif(m <= n//2):
print(m)
else:
print(n-m)
``` | output | 1 | 52,272 | 14 | 104,545 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The three friends, Kuro, Shiro, and Katie, met up again! It's time for a party...
What the cats do when they unite? Right, they have a party. Since they wanted to have as much fun as possible, they invited all their friends. Now n cats are at the party, sitting in a circle and eating soup. The rules are simple: anyone having finished their soup leaves the circle.
Katie suddenly notices that whenever a cat leaves, the place where she was sitting becomes an empty space, which means the circle is divided into smaller continuous groups of cats sitting next to each other. At the moment Katie observes, there are m cats who left the circle. This raises a question for Katie: what is the maximum possible number of groups the circle is divided into at the moment?
Could you help her with this curiosity?
You can see the examples and their descriptions with pictures in the "Note" section.
Input
The only line contains two integers n and m (2 β€ n β€ 1000, 0 β€ m β€ n) β the initial number of cats at the party and the number of cats who left the circle at the moment Katie observes, respectively.
Output
Print a single integer β the maximum number of groups of cats at the moment Katie observes.
Examples
Input
7 4
Output
3
Input
6 2
Output
2
Input
3 0
Output
1
Input
2 2
Output
0
Note
In the first example, originally there are 7 cats sitting as shown below, creating a single group:
<image>
At the observed moment, 4 cats have left the table. Suppose the cats 2, 3, 5 and 7 have left, then there are 3 groups remaining. It is possible to show that it is the maximum possible number of groups remaining.
<image>
In the second example, there are 6 cats sitting as shown below:
<image>
At the observed moment, 2 cats have left the table. Suppose the cats numbered 3 and 6 left, then there will be 2 groups remaining (\{1, 2\} and \{4, 5\}). It is impossible to have more than 2 groups of cats remaining.
<image>
In the third example, no cats have left, so there is 1 group consisting of all cats.
In the fourth example, all cats have left the circle, so there are 0 groups. | instruction | 0 | 52,273 | 14 | 104,546 |
Tags: greedy, math
Correct Solution:
```
n,m = map(int, input().strip().split(' '))
if n==m:
print('0')
elif m==0 or m==n-1:
print('1')
#n is even
elif n%2==0:
p=n//2
if p==m:
print(p)
elif m<p:
print(m)
else:
print(2*p-m)
#n is odd
elif n%2!=0:
p=(n//2)
if m==p or m==p+1:
print(p)
elif m>p+1:
print(2*p-m+1)
elif m<p:
print(m)
``` | output | 1 | 52,273 | 14 | 104,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The three friends, Kuro, Shiro, and Katie, met up again! It's time for a party...
What the cats do when they unite? Right, they have a party. Since they wanted to have as much fun as possible, they invited all their friends. Now n cats are at the party, sitting in a circle and eating soup. The rules are simple: anyone having finished their soup leaves the circle.
Katie suddenly notices that whenever a cat leaves, the place where she was sitting becomes an empty space, which means the circle is divided into smaller continuous groups of cats sitting next to each other. At the moment Katie observes, there are m cats who left the circle. This raises a question for Katie: what is the maximum possible number of groups the circle is divided into at the moment?
Could you help her with this curiosity?
You can see the examples and their descriptions with pictures in the "Note" section.
Input
The only line contains two integers n and m (2 β€ n β€ 1000, 0 β€ m β€ n) β the initial number of cats at the party and the number of cats who left the circle at the moment Katie observes, respectively.
Output
Print a single integer β the maximum number of groups of cats at the moment Katie observes.
Examples
Input
7 4
Output
3
Input
6 2
Output
2
Input
3 0
Output
1
Input
2 2
Output
0
Note
In the first example, originally there are 7 cats sitting as shown below, creating a single group:
<image>
At the observed moment, 4 cats have left the table. Suppose the cats 2, 3, 5 and 7 have left, then there are 3 groups remaining. It is possible to show that it is the maximum possible number of groups remaining.
<image>
In the second example, there are 6 cats sitting as shown below:
<image>
At the observed moment, 2 cats have left the table. Suppose the cats numbered 3 and 6 left, then there will be 2 groups remaining (\{1, 2\} and \{4, 5\}). It is impossible to have more than 2 groups of cats remaining.
<image>
In the third example, no cats have left, so there is 1 group consisting of all cats.
In the fourth example, all cats have left the circle, so there are 0 groups.
Submitted Solution:
```
[a,b] = list(map(int,input().split(" ")))
if(b==0):
print(1)
elif(a-b > b):
print(b)
else:
print(a-b)
``` | instruction | 0 | 52,274 | 14 | 104,548 |
Yes | output | 1 | 52,274 | 14 | 104,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The three friends, Kuro, Shiro, and Katie, met up again! It's time for a party...
What the cats do when they unite? Right, they have a party. Since they wanted to have as much fun as possible, they invited all their friends. Now n cats are at the party, sitting in a circle and eating soup. The rules are simple: anyone having finished their soup leaves the circle.
Katie suddenly notices that whenever a cat leaves, the place where she was sitting becomes an empty space, which means the circle is divided into smaller continuous groups of cats sitting next to each other. At the moment Katie observes, there are m cats who left the circle. This raises a question for Katie: what is the maximum possible number of groups the circle is divided into at the moment?
Could you help her with this curiosity?
You can see the examples and their descriptions with pictures in the "Note" section.
Input
The only line contains two integers n and m (2 β€ n β€ 1000, 0 β€ m β€ n) β the initial number of cats at the party and the number of cats who left the circle at the moment Katie observes, respectively.
Output
Print a single integer β the maximum number of groups of cats at the moment Katie observes.
Examples
Input
7 4
Output
3
Input
6 2
Output
2
Input
3 0
Output
1
Input
2 2
Output
0
Note
In the first example, originally there are 7 cats sitting as shown below, creating a single group:
<image>
At the observed moment, 4 cats have left the table. Suppose the cats 2, 3, 5 and 7 have left, then there are 3 groups remaining. It is possible to show that it is the maximum possible number of groups remaining.
<image>
In the second example, there are 6 cats sitting as shown below:
<image>
At the observed moment, 2 cats have left the table. Suppose the cats numbered 3 and 6 left, then there will be 2 groups remaining (\{1, 2\} and \{4, 5\}). It is impossible to have more than 2 groups of cats remaining.
<image>
In the third example, no cats have left, so there is 1 group consisting of all cats.
In the fourth example, all cats have left the circle, so there are 0 groups.
Submitted Solution:
```
N, M = map(int, input().split())
M = max(M, 1)
print(min(M, N-M))
``` | instruction | 0 | 52,275 | 14 | 104,550 |
Yes | output | 1 | 52,275 | 14 | 104,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The three friends, Kuro, Shiro, and Katie, met up again! It's time for a party...
What the cats do when they unite? Right, they have a party. Since they wanted to have as much fun as possible, they invited all their friends. Now n cats are at the party, sitting in a circle and eating soup. The rules are simple: anyone having finished their soup leaves the circle.
Katie suddenly notices that whenever a cat leaves, the place where she was sitting becomes an empty space, which means the circle is divided into smaller continuous groups of cats sitting next to each other. At the moment Katie observes, there are m cats who left the circle. This raises a question for Katie: what is the maximum possible number of groups the circle is divided into at the moment?
Could you help her with this curiosity?
You can see the examples and their descriptions with pictures in the "Note" section.
Input
The only line contains two integers n and m (2 β€ n β€ 1000, 0 β€ m β€ n) β the initial number of cats at the party and the number of cats who left the circle at the moment Katie observes, respectively.
Output
Print a single integer β the maximum number of groups of cats at the moment Katie observes.
Examples
Input
7 4
Output
3
Input
6 2
Output
2
Input
3 0
Output
1
Input
2 2
Output
0
Note
In the first example, originally there are 7 cats sitting as shown below, creating a single group:
<image>
At the observed moment, 4 cats have left the table. Suppose the cats 2, 3, 5 and 7 have left, then there are 3 groups remaining. It is possible to show that it is the maximum possible number of groups remaining.
<image>
In the second example, there are 6 cats sitting as shown below:
<image>
At the observed moment, 2 cats have left the table. Suppose the cats numbered 3 and 6 left, then there will be 2 groups remaining (\{1, 2\} and \{4, 5\}). It is impossible to have more than 2 groups of cats remaining.
<image>
In the third example, no cats have left, so there is 1 group consisting of all cats.
In the fourth example, all cats have left the circle, so there are 0 groups.
Submitted Solution:
```
# cook your dish here
n,m=[int(i) for i in input().split()]
if(m==0):
print(1)
else:
if(n%2==0):
if(m<=n//2):
print(m)
else:
print(n-m)
else:
if(m<=n//2):
print(m)
elif(m==n//2+1):
print(m-1)
else:
print(n-m)
``` | instruction | 0 | 52,276 | 14 | 104,552 |
Yes | output | 1 | 52,276 | 14 | 104,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The three friends, Kuro, Shiro, and Katie, met up again! It's time for a party...
What the cats do when they unite? Right, they have a party. Since they wanted to have as much fun as possible, they invited all their friends. Now n cats are at the party, sitting in a circle and eating soup. The rules are simple: anyone having finished their soup leaves the circle.
Katie suddenly notices that whenever a cat leaves, the place where she was sitting becomes an empty space, which means the circle is divided into smaller continuous groups of cats sitting next to each other. At the moment Katie observes, there are m cats who left the circle. This raises a question for Katie: what is the maximum possible number of groups the circle is divided into at the moment?
Could you help her with this curiosity?
You can see the examples and their descriptions with pictures in the "Note" section.
Input
The only line contains two integers n and m (2 β€ n β€ 1000, 0 β€ m β€ n) β the initial number of cats at the party and the number of cats who left the circle at the moment Katie observes, respectively.
Output
Print a single integer β the maximum number of groups of cats at the moment Katie observes.
Examples
Input
7 4
Output
3
Input
6 2
Output
2
Input
3 0
Output
1
Input
2 2
Output
0
Note
In the first example, originally there are 7 cats sitting as shown below, creating a single group:
<image>
At the observed moment, 4 cats have left the table. Suppose the cats 2, 3, 5 and 7 have left, then there are 3 groups remaining. It is possible to show that it is the maximum possible number of groups remaining.
<image>
In the second example, there are 6 cats sitting as shown below:
<image>
At the observed moment, 2 cats have left the table. Suppose the cats numbered 3 and 6 left, then there will be 2 groups remaining (\{1, 2\} and \{4, 5\}). It is impossible to have more than 2 groups of cats remaining.
<image>
In the third example, no cats have left, so there is 1 group consisting of all cats.
In the fourth example, all cats have left the circle, so there are 0 groups.
Submitted Solution:
```
n, m = map(int, input().split())
print(min([max([m, 1]), n - m]))
``` | instruction | 0 | 52,277 | 14 | 104,554 |
Yes | output | 1 | 52,277 | 14 | 104,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The three friends, Kuro, Shiro, and Katie, met up again! It's time for a party...
What the cats do when they unite? Right, they have a party. Since they wanted to have as much fun as possible, they invited all their friends. Now n cats are at the party, sitting in a circle and eating soup. The rules are simple: anyone having finished their soup leaves the circle.
Katie suddenly notices that whenever a cat leaves, the place where she was sitting becomes an empty space, which means the circle is divided into smaller continuous groups of cats sitting next to each other. At the moment Katie observes, there are m cats who left the circle. This raises a question for Katie: what is the maximum possible number of groups the circle is divided into at the moment?
Could you help her with this curiosity?
You can see the examples and their descriptions with pictures in the "Note" section.
Input
The only line contains two integers n and m (2 β€ n β€ 1000, 0 β€ m β€ n) β the initial number of cats at the party and the number of cats who left the circle at the moment Katie observes, respectively.
Output
Print a single integer β the maximum number of groups of cats at the moment Katie observes.
Examples
Input
7 4
Output
3
Input
6 2
Output
2
Input
3 0
Output
1
Input
2 2
Output
0
Note
In the first example, originally there are 7 cats sitting as shown below, creating a single group:
<image>
At the observed moment, 4 cats have left the table. Suppose the cats 2, 3, 5 and 7 have left, then there are 3 groups remaining. It is possible to show that it is the maximum possible number of groups remaining.
<image>
In the second example, there are 6 cats sitting as shown below:
<image>
At the observed moment, 2 cats have left the table. Suppose the cats numbered 3 and 6 left, then there will be 2 groups remaining (\{1, 2\} and \{4, 5\}). It is impossible to have more than 2 groups of cats remaining.
<image>
In the third example, no cats have left, so there is 1 group consisting of all cats.
In the fourth example, all cats have left the circle, so there are 0 groups.
Submitted Solution:
```
n, m = map(int, input().split())
if m != 0:
calc = 0
n1 = n
for i in range(m):
if n != 1 and i < (n1 + 1) // 2:
n = (n + 1) // 2
calc += 1
elif i >= (n1 + 1) // 2:
calc -= 1
print(calc)
else:
print(1)
``` | instruction | 0 | 52,278 | 14 | 104,556 |
No | output | 1 | 52,278 | 14 | 104,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The three friends, Kuro, Shiro, and Katie, met up again! It's time for a party...
What the cats do when they unite? Right, they have a party. Since they wanted to have as much fun as possible, they invited all their friends. Now n cats are at the party, sitting in a circle and eating soup. The rules are simple: anyone having finished their soup leaves the circle.
Katie suddenly notices that whenever a cat leaves, the place where she was sitting becomes an empty space, which means the circle is divided into smaller continuous groups of cats sitting next to each other. At the moment Katie observes, there are m cats who left the circle. This raises a question for Katie: what is the maximum possible number of groups the circle is divided into at the moment?
Could you help her with this curiosity?
You can see the examples and their descriptions with pictures in the "Note" section.
Input
The only line contains two integers n and m (2 β€ n β€ 1000, 0 β€ m β€ n) β the initial number of cats at the party and the number of cats who left the circle at the moment Katie observes, respectively.
Output
Print a single integer β the maximum number of groups of cats at the moment Katie observes.
Examples
Input
7 4
Output
3
Input
6 2
Output
2
Input
3 0
Output
1
Input
2 2
Output
0
Note
In the first example, originally there are 7 cats sitting as shown below, creating a single group:
<image>
At the observed moment, 4 cats have left the table. Suppose the cats 2, 3, 5 and 7 have left, then there are 3 groups remaining. It is possible to show that it is the maximum possible number of groups remaining.
<image>
In the second example, there are 6 cats sitting as shown below:
<image>
At the observed moment, 2 cats have left the table. Suppose the cats numbered 3 and 6 left, then there will be 2 groups remaining (\{1, 2\} and \{4, 5\}). It is impossible to have more than 2 groups of cats remaining.
<image>
In the third example, no cats have left, so there is 1 group consisting of all cats.
In the fourth example, all cats have left the circle, so there are 0 groups.
Submitted Solution:
```
no_of_cats,no_of_cats_left = map(int,input().split())
l = []
if(no_of_cats%2 == 0):
if(no_of_cats//2 == no_of_cats_left):
print(no_of_cats_left)
elif(no_of_cats_left < no_of_cats//2):
print(no_of_cats_left)
else:
print(no_of_cats - no_of_cats_left)
elif(no_of_cats_left==0):
print(1)
else:
if(no_of_cats_left <= no_of_cats//2):
print(no_of_cats_left)
else:
print(no_of_cats - no_of_cats_left)
``` | instruction | 0 | 52,279 | 14 | 104,558 |
No | output | 1 | 52,279 | 14 | 104,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The three friends, Kuro, Shiro, and Katie, met up again! It's time for a party...
What the cats do when they unite? Right, they have a party. Since they wanted to have as much fun as possible, they invited all their friends. Now n cats are at the party, sitting in a circle and eating soup. The rules are simple: anyone having finished their soup leaves the circle.
Katie suddenly notices that whenever a cat leaves, the place where she was sitting becomes an empty space, which means the circle is divided into smaller continuous groups of cats sitting next to each other. At the moment Katie observes, there are m cats who left the circle. This raises a question for Katie: what is the maximum possible number of groups the circle is divided into at the moment?
Could you help her with this curiosity?
You can see the examples and their descriptions with pictures in the "Note" section.
Input
The only line contains two integers n and m (2 β€ n β€ 1000, 0 β€ m β€ n) β the initial number of cats at the party and the number of cats who left the circle at the moment Katie observes, respectively.
Output
Print a single integer β the maximum number of groups of cats at the moment Katie observes.
Examples
Input
7 4
Output
3
Input
6 2
Output
2
Input
3 0
Output
1
Input
2 2
Output
0
Note
In the first example, originally there are 7 cats sitting as shown below, creating a single group:
<image>
At the observed moment, 4 cats have left the table. Suppose the cats 2, 3, 5 and 7 have left, then there are 3 groups remaining. It is possible to show that it is the maximum possible number of groups remaining.
<image>
In the second example, there are 6 cats sitting as shown below:
<image>
At the observed moment, 2 cats have left the table. Suppose the cats numbered 3 and 6 left, then there will be 2 groups remaining (\{1, 2\} and \{4, 5\}). It is impossible to have more than 2 groups of cats remaining.
<image>
In the third example, no cats have left, so there is 1 group consisting of all cats.
In the fourth example, all cats have left the circle, so there are 0 groups.
Submitted Solution:
```
data = [int(i) for i in input().split()]
cats_plots_num = []
cats_leave = 0
for cats in range(1, data[0] + 1, 2):
if (cats_leave == data[0] - data[1] or data[1] == 0):
break
cats_plots_num += [cats]
cats_leave += 1
print(len(cats_plots_num))
``` | instruction | 0 | 52,280 | 14 | 104,560 |
No | output | 1 | 52,280 | 14 | 104,561 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The three friends, Kuro, Shiro, and Katie, met up again! It's time for a party...
What the cats do when they unite? Right, they have a party. Since they wanted to have as much fun as possible, they invited all their friends. Now n cats are at the party, sitting in a circle and eating soup. The rules are simple: anyone having finished their soup leaves the circle.
Katie suddenly notices that whenever a cat leaves, the place where she was sitting becomes an empty space, which means the circle is divided into smaller continuous groups of cats sitting next to each other. At the moment Katie observes, there are m cats who left the circle. This raises a question for Katie: what is the maximum possible number of groups the circle is divided into at the moment?
Could you help her with this curiosity?
You can see the examples and their descriptions with pictures in the "Note" section.
Input
The only line contains two integers n and m (2 β€ n β€ 1000, 0 β€ m β€ n) β the initial number of cats at the party and the number of cats who left the circle at the moment Katie observes, respectively.
Output
Print a single integer β the maximum number of groups of cats at the moment Katie observes.
Examples
Input
7 4
Output
3
Input
6 2
Output
2
Input
3 0
Output
1
Input
2 2
Output
0
Note
In the first example, originally there are 7 cats sitting as shown below, creating a single group:
<image>
At the observed moment, 4 cats have left the table. Suppose the cats 2, 3, 5 and 7 have left, then there are 3 groups remaining. It is possible to show that it is the maximum possible number of groups remaining.
<image>
In the second example, there are 6 cats sitting as shown below:
<image>
At the observed moment, 2 cats have left the table. Suppose the cats numbered 3 and 6 left, then there will be 2 groups remaining (\{1, 2\} and \{4, 5\}). It is impossible to have more than 2 groups of cats remaining.
<image>
In the third example, no cats have left, so there is 1 group consisting of all cats.
In the fourth example, all cats have left the circle, so there are 0 groups.
Submitted Solution:
```
n, m = [int(x) for x in input().split()]
cuts = min(n // 2, m)
grps = cuts
rem = max(0, m - cuts - n % 2)
print(max(0, grps - rem))
``` | instruction | 0 | 52,281 | 14 | 104,562 |
No | output | 1 | 52,281 | 14 | 104,563 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel is in the Amusement Park. And now she is in a queue in front of the Ferris wheel. There are n people (or foxes more precisely) in the queue: we use first people to refer one at the head of the queue, and n-th people to refer the last one in the queue.
There will be k gondolas, and the way we allocate gondolas looks like this:
* When the first gondolas come, the q1 people in head of the queue go into the gondolas.
* Then when the second gondolas come, the q2 people in head of the remain queue go into the gondolas.
...
* The remain qk people go into the last (k-th) gondolas.
Note that q1, q2, ..., qk must be positive. You can get from the statement that <image> and qi > 0.
You know, people don't want to stay with strangers in the gondolas, so your task is to find an optimal allocation way (that is find an optimal sequence q) to make people happy. For every pair of people i and j, there exists a value uij denotes a level of unfamiliar. You can assume uij = uji for all i, j (1 β€ i, j β€ n) and uii = 0 for all i (1 β€ i β€ n). Then an unfamiliar value of a gondolas is the sum of the levels of unfamiliar between any pair of people that is into the gondolas.
A total unfamiliar value is the sum of unfamiliar values for all gondolas. Help Fox Ciel to find the minimal possible total unfamiliar value for some optimal allocation.
Input
The first line contains two integers n and k (1 β€ n β€ 4000 and 1 β€ k β€ min(n, 800)) β the number of people in the queue and the number of gondolas. Each of the following n lines contains n integers β matrix u, (0 β€ uij β€ 9, uij = uji and uii = 0).
Please, use fast input methods (for example, please use BufferedReader instead of Scanner for Java).
Output
Print an integer β the minimal possible total unfamiliar value.
Examples
Input
5 2
0 0 1 1 1
0 0 1 1 1
1 1 0 0 0
1 1 0 0 0
1 1 0 0 0
Output
0
Input
8 3
0 1 1 1 1 1 1 1
1 0 1 1 1 1 1 1
1 1 0 1 1 1 1 1
1 1 1 0 1 1 1 1
1 1 1 1 0 1 1 1
1 1 1 1 1 0 1 1
1 1 1 1 1 1 0 1
1 1 1 1 1 1 1 0
Output
7
Input
3 2
0 2 0
2 0 3
0 3 0
Output
2
Note
In the first example, we can allocate people like this: {1, 2} goes into a gondolas, {3, 4, 5} goes into another gondolas.
In the second example, an optimal solution is : {1, 2, 3} | {4, 5, 6} | {7, 8}.
Submitted Solution:
```
def main():
N, K = map(int, input().split())
M = [list(map(int, input().split())) for _ in range(N)]
if K >= N :
print(0)
return
for s in range(2, N):
for i in range(N-s):
M[i][i+s] += M[i][i+s-1]+M[i+1][i+s]-M[i+1][i+s-1]
if K == 1:
print(M[0][N-1])
#for row in M : print(*row)
last, pivot = M[0][:], [0]*N
for g in range(1,K-1): #g+1 gondols
stack = [(g,N-1)]
pivot[N-1] = min(range(N-1), key=lambda i:last[i]+M[i+1][N-1])
pivot[g] = g-1
while stack:
a,b = stack.pop()
m = (a+b)//2
pivot[m] = min(range(pivot[a],pivot[b]+1), key=lambda i:last[i]+M[i+1][m])
if m>a+1 : stack.append((a,m))
if m<b-1 : stack.append((m,b))
new = [0]*N
for i in range(g, N):
new[i]=last[pivot[i]]+M[pivot[i]+1][i]
last, pivot = new, [0]*N
print(min(last[i]+M[i+1][N-1] for i in range(N-1)))
main()
``` | instruction | 0 | 52,540 | 14 | 105,080 |
No | output | 1 | 52,540 | 14 | 105,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel is in the Amusement Park. And now she is in a queue in front of the Ferris wheel. There are n people (or foxes more precisely) in the queue: we use first people to refer one at the head of the queue, and n-th people to refer the last one in the queue.
There will be k gondolas, and the way we allocate gondolas looks like this:
* When the first gondolas come, the q1 people in head of the queue go into the gondolas.
* Then when the second gondolas come, the q2 people in head of the remain queue go into the gondolas.
...
* The remain qk people go into the last (k-th) gondolas.
Note that q1, q2, ..., qk must be positive. You can get from the statement that <image> and qi > 0.
You know, people don't want to stay with strangers in the gondolas, so your task is to find an optimal allocation way (that is find an optimal sequence q) to make people happy. For every pair of people i and j, there exists a value uij denotes a level of unfamiliar. You can assume uij = uji for all i, j (1 β€ i, j β€ n) and uii = 0 for all i (1 β€ i β€ n). Then an unfamiliar value of a gondolas is the sum of the levels of unfamiliar between any pair of people that is into the gondolas.
A total unfamiliar value is the sum of unfamiliar values for all gondolas. Help Fox Ciel to find the minimal possible total unfamiliar value for some optimal allocation.
Input
The first line contains two integers n and k (1 β€ n β€ 4000 and 1 β€ k β€ min(n, 800)) β the number of people in the queue and the number of gondolas. Each of the following n lines contains n integers β matrix u, (0 β€ uij β€ 9, uij = uji and uii = 0).
Please, use fast input methods (for example, please use BufferedReader instead of Scanner for Java).
Output
Print an integer β the minimal possible total unfamiliar value.
Examples
Input
5 2
0 0 1 1 1
0 0 1 1 1
1 1 0 0 0
1 1 0 0 0
1 1 0 0 0
Output
0
Input
8 3
0 1 1 1 1 1 1 1
1 0 1 1 1 1 1 1
1 1 0 1 1 1 1 1
1 1 1 0 1 1 1 1
1 1 1 1 0 1 1 1
1 1 1 1 1 0 1 1
1 1 1 1 1 1 0 1
1 1 1 1 1 1 1 0
Output
7
Input
3 2
0 2 0
2 0 3
0 3 0
Output
2
Note
In the first example, we can allocate people like this: {1, 2} goes into a gondolas, {3, 4, 5} goes into another gondolas.
In the second example, an optimal solution is : {1, 2, 3} | {4, 5, 6} | {7, 8}.
Submitted Solution:
```
def main():
N, K = map(int, input().split())
C = ord('0')
M = []
for _ in range(N):
line = input()
M.append([ord(line[2*i])-C for i in range(N)])
if K >= N:
print(0)
return
for s in range(2, N):
for i in range(N-s):
M[i][i+s] += M[i][i+s-1]+M[i+1][i+s]-M[i+1][i+s-1]
if K == 1:
print(M[0][N-1])
return
#for row in M : print(*row)
last, pivot, new = M[0], [0]*N, [0]*N
for g in range(1, K-1): # g+1 gondols
stack = [(g, N-1)]
pivot[N-1] = min(range(N-1), key=lambda i: last[i]+M[i+1][N-1])
pivot[g] = g-1
for m in (g, N-1):
new[m] = last[pivot[m]]+M[pivot[m]+1][m]
for a, b in stack:
if pivot[a]==pivot[b]:
for m in range(a+1,b):
new[m] = last[pivot[m]]+M[pivot[m]+1][m]
else:
m = (a+b)//2
pivot[m] = min(range(pivot[a], pivot[b]+1), key=lambda i: last[i]+M[i+1][m])
new[m] = last[pivot[m]]+M[pivot[m]+1][m]
if m > a+1:
stack.append((a, m))
if m < b-1:
stack.append((m, b))
last, pivot, new = new, [0]*N, [0]*N
print(min(last[i]+M[i+1][N-1] for i in range(N-1)))
main()
``` | instruction | 0 | 52,541 | 14 | 105,082 |
No | output | 1 | 52,541 | 14 | 105,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel is in the Amusement Park. And now she is in a queue in front of the Ferris wheel. There are n people (or foxes more precisely) in the queue: we use first people to refer one at the head of the queue, and n-th people to refer the last one in the queue.
There will be k gondolas, and the way we allocate gondolas looks like this:
* When the first gondolas come, the q1 people in head of the queue go into the gondolas.
* Then when the second gondolas come, the q2 people in head of the remain queue go into the gondolas.
...
* The remain qk people go into the last (k-th) gondolas.
Note that q1, q2, ..., qk must be positive. You can get from the statement that <image> and qi > 0.
You know, people don't want to stay with strangers in the gondolas, so your task is to find an optimal allocation way (that is find an optimal sequence q) to make people happy. For every pair of people i and j, there exists a value uij denotes a level of unfamiliar. You can assume uij = uji for all i, j (1 β€ i, j β€ n) and uii = 0 for all i (1 β€ i β€ n). Then an unfamiliar value of a gondolas is the sum of the levels of unfamiliar between any pair of people that is into the gondolas.
A total unfamiliar value is the sum of unfamiliar values for all gondolas. Help Fox Ciel to find the minimal possible total unfamiliar value for some optimal allocation.
Input
The first line contains two integers n and k (1 β€ n β€ 4000 and 1 β€ k β€ min(n, 800)) β the number of people in the queue and the number of gondolas. Each of the following n lines contains n integers β matrix u, (0 β€ uij β€ 9, uij = uji and uii = 0).
Please, use fast input methods (for example, please use BufferedReader instead of Scanner for Java).
Output
Print an integer β the minimal possible total unfamiliar value.
Examples
Input
5 2
0 0 1 1 1
0 0 1 1 1
1 1 0 0 0
1 1 0 0 0
1 1 0 0 0
Output
0
Input
8 3
0 1 1 1 1 1 1 1
1 0 1 1 1 1 1 1
1 1 0 1 1 1 1 1
1 1 1 0 1 1 1 1
1 1 1 1 0 1 1 1
1 1 1 1 1 0 1 1
1 1 1 1 1 1 0 1
1 1 1 1 1 1 1 0
Output
7
Input
3 2
0 2 0
2 0 3
0 3 0
Output
2
Note
In the first example, we can allocate people like this: {1, 2} goes into a gondolas, {3, 4, 5} goes into another gondolas.
In the second example, an optimal solution is : {1, 2, 3} | {4, 5, 6} | {7, 8}.
Submitted Solution:
```
from sys import stdin
def main():
INPUT = stdin.buffer.read()
C = ord('0')
N = K = 0
second = False
for i in range(len(INPUT)):
if INPUT[i]==ord(' '):
second = True
elif INPUT[i]==ord('\n'):
i+=1
break
elif second:
K=10*K+INPUT[i]-C
else:
N=10*N+INPUT[i]-C
offset = i
M = []
for i in range(N):
M.append([INPUT[i*(2*N)+2*j+offset]-C for j in range(N)])
if K >= N:
print(0)
return
for s in range(2, N):
for i in range(N-s):
M[i][i+s] += M[i][i+s-1]+M[i+1][i+s]-M[i+1][i+s-1]
if K == 1:
print(M[0][N-1])
return
#for row in M : print(*row)
last, pivot, new = M[0], [0]*N, [0]*N
for g in range(1, K-1): # g+1 gondols
stack = [(g, N-1)]
pivot[N-1] = min(range(N-1), key=lambda i: last[i]+M[i+1][N-1])
pivot[g] = g-1
for m in (g, N-1):
new[m] = last[pivot[m]]+M[pivot[m]+1][m]
for a, b in stack:
if pivot[a] == pivot[b]:
for m in range(a+1, b):
new[m] = last[pivot[a]]+M[pivot[a]+1][m]
else:
m = (a+b)//2
pivot[m] = min(range(pivot[a], pivot[b]+1), key=lambda i: last[i]+M[i+1][m])
new[m] = last[pivot[m]]+M[pivot[m]+1][m]
if m > a+1:
stack.append((a, m))
if m < b-1:
stack.append((m, b))
last, pivot, new = new, [0]*N, [0]*N
print(min(last[i]+M[i+1][N-1] for i in range(N-1)))
main()
``` | instruction | 0 | 52,542 | 14 | 105,084 |
No | output | 1 | 52,542 | 14 | 105,085 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fox Ciel is in the Amusement Park. And now she is in a queue in front of the Ferris wheel. There are n people (or foxes more precisely) in the queue: we use first people to refer one at the head of the queue, and n-th people to refer the last one in the queue.
There will be k gondolas, and the way we allocate gondolas looks like this:
* When the first gondolas come, the q1 people in head of the queue go into the gondolas.
* Then when the second gondolas come, the q2 people in head of the remain queue go into the gondolas.
...
* The remain qk people go into the last (k-th) gondolas.
Note that q1, q2, ..., qk must be positive. You can get from the statement that <image> and qi > 0.
You know, people don't want to stay with strangers in the gondolas, so your task is to find an optimal allocation way (that is find an optimal sequence q) to make people happy. For every pair of people i and j, there exists a value uij denotes a level of unfamiliar. You can assume uij = uji for all i, j (1 β€ i, j β€ n) and uii = 0 for all i (1 β€ i β€ n). Then an unfamiliar value of a gondolas is the sum of the levels of unfamiliar between any pair of people that is into the gondolas.
A total unfamiliar value is the sum of unfamiliar values for all gondolas. Help Fox Ciel to find the minimal possible total unfamiliar value for some optimal allocation.
Input
The first line contains two integers n and k (1 β€ n β€ 4000 and 1 β€ k β€ min(n, 800)) β the number of people in the queue and the number of gondolas. Each of the following n lines contains n integers β matrix u, (0 β€ uij β€ 9, uij = uji and uii = 0).
Please, use fast input methods (for example, please use BufferedReader instead of Scanner for Java).
Output
Print an integer β the minimal possible total unfamiliar value.
Examples
Input
5 2
0 0 1 1 1
0 0 1 1 1
1 1 0 0 0
1 1 0 0 0
1 1 0 0 0
Output
0
Input
8 3
0 1 1 1 1 1 1 1
1 0 1 1 1 1 1 1
1 1 0 1 1 1 1 1
1 1 1 0 1 1 1 1
1 1 1 1 0 1 1 1
1 1 1 1 1 0 1 1
1 1 1 1 1 1 0 1
1 1 1 1 1 1 1 0
Output
7
Input
3 2
0 2 0
2 0 3
0 3 0
Output
2
Note
In the first example, we can allocate people like this: {1, 2} goes into a gondolas, {3, 4, 5} goes into another gondolas.
In the second example, an optimal solution is : {1, 2, 3} | {4, 5, 6} | {7, 8}.
Submitted Solution:
```
def cost(l,r):
# global unfamil,sumCost
if l==r:return 0
S = 0
if sumCost[l][r]>-1:return sumCost[l][r]
for i in range(l,r):
for j in range(i+1,r+1):
S+=unfamil[i][j]
sumCost[l][r]=S
return S
def calculate(l,p_left,g_left):
global unfamil,sumCost,n,k
# print(l,p_left,g_left)
if g_left==0:return float('inf')
if g_left>p_left: return float('inf')
if g_left==p_left:return 0
if g_left==1: return cost(l,n)
best = float('inf')
for i in range(l+1,p_left-g_left+3):
best = min(best, calculate(i,p_left-(i-l)+1,g_left-1)+cost(l,i-1))
return best
n,k = map(int,input().split())
unfamil = [[]]
sumCost = [[-1 for i in range(n+1)] for j in range(n+1)]
for i in range(n):
unfamil.append([0]+list(map(int,input().split())))
if k==n:print(0)
if k==1:print(cost(1,n))
else:
p_left,g_left=n,k
best = calculate(1,p_left,g_left)
print(best)
``` | instruction | 0 | 52,543 | 14 | 105,086 |
No | output | 1 | 52,543 | 14 | 105,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A flea is sitting at one of the n hassocks, arranged in a circle, at the moment. After minute number k the flea jumps through k - 1 hassoΡks (clockwise). For example, after the first minute the flea jumps to the neighboring hassock. You should answer: will the flea visit all the hassocks or not. We assume that flea has infinitely much time for this jumping.
Input
The only line contains single integer: 1 β€ n β€ 1000 β number of hassocks.
Output
Output "YES" if all the hassocks will be visited and "NO" otherwise.
Examples
Input
1
Output
YES
Input
3
Output
NO | instruction | 0 | 52,603 | 14 | 105,206 |
Tags: implementation, math
Correct Solution:
```
n = int(input())
if n & n-1:
print("NO")
else:
print("YES")
``` | output | 1 | 52,603 | 14 | 105,207 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A flea is sitting at one of the n hassocks, arranged in a circle, at the moment. After minute number k the flea jumps through k - 1 hassoΡks (clockwise). For example, after the first minute the flea jumps to the neighboring hassock. You should answer: will the flea visit all the hassocks or not. We assume that flea has infinitely much time for this jumping.
Input
The only line contains single integer: 1 β€ n β€ 1000 β number of hassocks.
Output
Output "YES" if all the hassocks will be visited and "NO" otherwise.
Examples
Input
1
Output
YES
Input
3
Output
NO | instruction | 0 | 52,604 | 14 | 105,208 |
Tags: implementation, math
Correct Solution:
```
n=int(input())
tri=[]
for i in range(2,2009):
tri.append(i*(i+1)//2)
if n==1:
print('YES')
exit()
check=[0]*n
for i in range(2*n+5):
check[tri[i]%n]=1
if n&(n-1)==0:
print('YES')
else:
print('NO')
``` | output | 1 | 52,604 | 14 | 105,209 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A flea is sitting at one of the n hassocks, arranged in a circle, at the moment. After minute number k the flea jumps through k - 1 hassoΡks (clockwise). For example, after the first minute the flea jumps to the neighboring hassock. You should answer: will the flea visit all the hassocks or not. We assume that flea has infinitely much time for this jumping.
Input
The only line contains single integer: 1 β€ n β€ 1000 β number of hassocks.
Output
Output "YES" if all the hassocks will be visited and "NO" otherwise.
Examples
Input
1
Output
YES
Input
3
Output
NO | instruction | 0 | 52,605 | 14 | 105,210 |
Tags: implementation, math
Correct Solution:
```
n, c, k = int(input()), 0, 0
v = [[False] * n for i in range(n)]
while not v[c][k]:
v[c][k] = True
k = (k + 1) % n
c = (c + k) % n
print('YES' if all(any(x) for x in v) else 'NO')
``` | output | 1 | 52,605 | 14 | 105,211 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A flea is sitting at one of the n hassocks, arranged in a circle, at the moment. After minute number k the flea jumps through k - 1 hassoΡks (clockwise). For example, after the first minute the flea jumps to the neighboring hassock. You should answer: will the flea visit all the hassocks or not. We assume that flea has infinitely much time for this jumping.
Input
The only line contains single integer: 1 β€ n β€ 1000 β number of hassocks.
Output
Output "YES" if all the hassocks will be visited and "NO" otherwise.
Examples
Input
1
Output
YES
Input
3
Output
NO | instruction | 0 | 52,606 | 14 | 105,212 |
Tags: implementation, math
Correct Solution:
```
from sys import stdin, stdout
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from collections import defaultdict as dd, deque
from heapq import merge, heapify, heappop, heappush, nsmallest
from bisect import bisect_left as bl, bisect_right as br, bisect
mod = pow(10, 9) + 7
mod2 = 998244353
def inp(): return stdin.readline().strip()
def out(var, end="\n"): stdout.write(str(var)+"\n")
def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
def lmp(): return list(mp())
def mp(): return map(int, inp().split())
def smp(): return map(str, inp().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
def remadd(x, y): return 1 if x%y else 0
def ceil(a,b): return (a+b-1)//b
def isprime(x):
if x<=1: return False
if x in (2, 3): return True
if x%2 == 0: return False
for i in range(3, int(sqrt(x))+1, 2):
if x%i == 0: return False
return True
n = int(inp())
print("YES" if not n&(n-1) else "NO")
``` | output | 1 | 52,606 | 14 | 105,213 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A flea is sitting at one of the n hassocks, arranged in a circle, at the moment. After minute number k the flea jumps through k - 1 hassoΡks (clockwise). For example, after the first minute the flea jumps to the neighboring hassock. You should answer: will the flea visit all the hassocks or not. We assume that flea has infinitely much time for this jumping.
Input
The only line contains single integer: 1 β€ n β€ 1000 β number of hassocks.
Output
Output "YES" if all the hassocks will be visited and "NO" otherwise.
Examples
Input
1
Output
YES
Input
3
Output
NO | instruction | 0 | 52,607 | 14 | 105,214 |
Tags: implementation, math
Correct Solution:
```
class CodeforcesTask55ASolution:
def __init__(self):
self.result = ''
self.n = 0
def read_input(self):
self.n = int(input())
def process_task(self):
hassocks = [x + 1 for x in range(self.n)]
visits = set()
visiting = 1
for x in range(1000 * self.n):
visiting += x
visiting = visiting % self.n
if not visiting:
visiting = self.n
visits.add(visiting)
hassocks.sort()
v = list(visits)
v.sort()
if v == hassocks:
self.result = "YES"
else:
self.result = "NO"
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask55ASolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | output | 1 | 52,607 | 14 | 105,215 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A flea is sitting at one of the n hassocks, arranged in a circle, at the moment. After minute number k the flea jumps through k - 1 hassoΡks (clockwise). For example, after the first minute the flea jumps to the neighboring hassock. You should answer: will the flea visit all the hassocks or not. We assume that flea has infinitely much time for this jumping.
Input
The only line contains single integer: 1 β€ n β€ 1000 β number of hassocks.
Output
Output "YES" if all the hassocks will be visited and "NO" otherwise.
Examples
Input
1
Output
YES
Input
3
Output
NO | instruction | 0 | 52,608 | 14 | 105,216 |
Tags: implementation, math
Correct Solution:
```
i = int(input())
mods = [0] * i
k = 0
for j in range(i):
k += j
mods[k % i] = 1
if all(m == 1 for m in mods):
print("YES")
else:
print("NO")
``` | output | 1 | 52,608 | 14 | 105,217 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A flea is sitting at one of the n hassocks, arranged in a circle, at the moment. After minute number k the flea jumps through k - 1 hassoΡks (clockwise). For example, after the first minute the flea jumps to the neighboring hassock. You should answer: will the flea visit all the hassocks or not. We assume that flea has infinitely much time for this jumping.
Input
The only line contains single integer: 1 β€ n β€ 1000 β number of hassocks.
Output
Output "YES" if all the hassocks will be visited and "NO" otherwise.
Examples
Input
1
Output
YES
Input
3
Output
NO | instruction | 0 | 52,609 | 14 | 105,218 |
Tags: implementation, math
Correct Solution:
```
n = int(input())
visited = [False] * n
k = 0
for i in range(n ** 2 + 1):
k += i + 1
visited[k % n] = True
for i in visited:
if not i:
print("NO")
exit()
print("YES")
``` | output | 1 | 52,609 | 14 | 105,219 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A flea is sitting at one of the n hassocks, arranged in a circle, at the moment. After minute number k the flea jumps through k - 1 hassoΡks (clockwise). For example, after the first minute the flea jumps to the neighboring hassock. You should answer: will the flea visit all the hassocks or not. We assume that flea has infinitely much time for this jumping.
Input
The only line contains single integer: 1 β€ n β€ 1000 β number of hassocks.
Output
Output "YES" if all the hassocks will be visited and "NO" otherwise.
Examples
Input
1
Output
YES
Input
3
Output
NO | instruction | 0 | 52,610 | 14 | 105,220 |
Tags: implementation, math
Correct Solution:
```
n = int(input())
if n & (n-1) > 0:
print('NO')
else:
print('YES')
``` | output | 1 | 52,610 | 14 | 105,221 |
Provide a correct Python 3 solution for this coding contest problem.
A serious incident happened at Taro's house, which loves steamed buns. One of the three steamed buns offered at the Buddhist altar in the Japanese-style room was gone. When Taro, who was aiming for a snack someday, started an investigation to find the criminal, it turned out that there were many people who entered the Japanese-style room that day. So I decided to ask everyone to give the following form of testimony to find out the order in which these suspects entered the room.
Suspect A's testimony "I entered the room before Suspect B."
One of the suspects (?) Is a calico cat, so I can't testify, but fortunately Taro saw the last time he entered the room.
From these testimonies, Taro decided to guess the order in which he entered the room and use it for the investigation.
For example, if there are 6 suspects and Tama is suspect 2, the following testimony is obtained.
Testimony of Suspect 5 "I entered the room before 2."
Suspect 1's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 5."
Suspect 4's testimony "I entered the room before 2."
Suspect 1's testimony "I entered the room before 6."
Suspect 6's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 4."
Putting this testimony together, the order in which they entered the room
* 3 β 5 β 1 β 6 β 4 β 2
* 1 β 6 β 3 β 4 β 5 β 2
* 3 β 1 β 6 β 5 β 4 β 2
You can narrow down to several possibilities.
From the testimony of all suspects except Tama (Suspect 2), estimate the order in which they entered the room and write a program that outputs one of the possible orders. However, although some suspects may give multiple testimonies, all testimonies shall be true and consistent.
Input
The input is given in the following format:
m
n
x1 y1
x2 y2
::
xn yn
The first line gives the number of suspects m (m β€ 20), and the second line gives the number of testimonies n (n β€ 100). The following n lines are given the contents of the i-th testimony, xi and yi, respectively, on the first line. xi yi represents the testimony that "suspect xi (I) entered before suspect yi".
Output
Print the number of the suspect who entered the room first, the number of the suspect who entered the room second, ..., and the number of the suspect who entered the room second, in order on one line.
Example
Input
6
7
5 2
1 4
3 5
4 2
1 6
6 4
3 4
Output
3
5
1
6
4
2 | instruction | 0 | 52,937 | 14 | 105,874 |
"Correct Solution:
```
visited = []
def dfs(v):
for i in edges[v]:
if i not in visited:
dfs(i)
visited.append(v)
if __name__ == '__main__':
m = int(input())
n = int(input())
evidences = []
edges = [[] for i in range(m)]
for i in range(n):
x, y = map(int, input().split())
evidences.append((x, y))
edges[x - 1].append(y - 1)
for i in range(m):
if i not in visited:
dfs(i)
ans = reversed([i + 1 for i in visited])
print('\n'.join(map(str, ans)))
``` | output | 1 | 52,937 | 14 | 105,875 |
Provide a correct Python 3 solution for this coding contest problem.
A serious incident happened at Taro's house, which loves steamed buns. One of the three steamed buns offered at the Buddhist altar in the Japanese-style room was gone. When Taro, who was aiming for a snack someday, started an investigation to find the criminal, it turned out that there were many people who entered the Japanese-style room that day. So I decided to ask everyone to give the following form of testimony to find out the order in which these suspects entered the room.
Suspect A's testimony "I entered the room before Suspect B."
One of the suspects (?) Is a calico cat, so I can't testify, but fortunately Taro saw the last time he entered the room.
From these testimonies, Taro decided to guess the order in which he entered the room and use it for the investigation.
For example, if there are 6 suspects and Tama is suspect 2, the following testimony is obtained.
Testimony of Suspect 5 "I entered the room before 2."
Suspect 1's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 5."
Suspect 4's testimony "I entered the room before 2."
Suspect 1's testimony "I entered the room before 6."
Suspect 6's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 4."
Putting this testimony together, the order in which they entered the room
* 3 β 5 β 1 β 6 β 4 β 2
* 1 β 6 β 3 β 4 β 5 β 2
* 3 β 1 β 6 β 5 β 4 β 2
You can narrow down to several possibilities.
From the testimony of all suspects except Tama (Suspect 2), estimate the order in which they entered the room and write a program that outputs one of the possible orders. However, although some suspects may give multiple testimonies, all testimonies shall be true and consistent.
Input
The input is given in the following format:
m
n
x1 y1
x2 y2
::
xn yn
The first line gives the number of suspects m (m β€ 20), and the second line gives the number of testimonies n (n β€ 100). The following n lines are given the contents of the i-th testimony, xi and yi, respectively, on the first line. xi yi represents the testimony that "suspect xi (I) entered before suspect yi".
Output
Print the number of the suspect who entered the room first, the number of the suspect who entered the room second, ..., and the number of the suspect who entered the room second, in order on one line.
Example
Input
6
7
5 2
1 4
3 5
4 2
1 6
6 4
3 4
Output
3
5
1
6
4
2 | instruction | 0 | 52,938 | 14 | 105,876 |
"Correct Solution:
```
#γγγγΈγ«γ«γ½γΌγ
m = int(input())
n = int(input())
edges = [[] for _ in range(m)]
for _ in range(n):
x, y = map(int, input().split())
edges[x - 1].append(y - 1)
visited = [False for _ in range(m)]
sorted_lst = []
def visit(i):
if not visited[i]:
visited[i] = True
for to in edges[i]:
visit(to)
sorted_lst.append(i)
for i in range(m):
visit(i)
sorted_lst.reverse()
for i in sorted_lst:
print(i + 1)
``` | output | 1 | 52,938 | 14 | 105,877 |
Provide a correct Python 3 solution for this coding contest problem.
A serious incident happened at Taro's house, which loves steamed buns. One of the three steamed buns offered at the Buddhist altar in the Japanese-style room was gone. When Taro, who was aiming for a snack someday, started an investigation to find the criminal, it turned out that there were many people who entered the Japanese-style room that day. So I decided to ask everyone to give the following form of testimony to find out the order in which these suspects entered the room.
Suspect A's testimony "I entered the room before Suspect B."
One of the suspects (?) Is a calico cat, so I can't testify, but fortunately Taro saw the last time he entered the room.
From these testimonies, Taro decided to guess the order in which he entered the room and use it for the investigation.
For example, if there are 6 suspects and Tama is suspect 2, the following testimony is obtained.
Testimony of Suspect 5 "I entered the room before 2."
Suspect 1's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 5."
Suspect 4's testimony "I entered the room before 2."
Suspect 1's testimony "I entered the room before 6."
Suspect 6's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 4."
Putting this testimony together, the order in which they entered the room
* 3 β 5 β 1 β 6 β 4 β 2
* 1 β 6 β 3 β 4 β 5 β 2
* 3 β 1 β 6 β 5 β 4 β 2
You can narrow down to several possibilities.
From the testimony of all suspects except Tama (Suspect 2), estimate the order in which they entered the room and write a program that outputs one of the possible orders. However, although some suspects may give multiple testimonies, all testimonies shall be true and consistent.
Input
The input is given in the following format:
m
n
x1 y1
x2 y2
::
xn yn
The first line gives the number of suspects m (m β€ 20), and the second line gives the number of testimonies n (n β€ 100). The following n lines are given the contents of the i-th testimony, xi and yi, respectively, on the first line. xi yi represents the testimony that "suspect xi (I) entered before suspect yi".
Output
Print the number of the suspect who entered the room first, the number of the suspect who entered the room second, ..., and the number of the suspect who entered the room second, in order on one line.
Example
Input
6
7
5 2
1 4
3 5
4 2
1 6
6 4
3 4
Output
3
5
1
6
4
2 | instruction | 0 | 52,939 | 14 | 105,878 |
"Correct Solution:
```
from collections import deque
def solve():
M = int(input())
G = [[] for i in range(M)]
deg = [0]*M
N = int(input())
for i in range(N):
x, y = map(int, input().split())
G[x-1].append(y-1)
deg[y-1] += 1
que = deque()
ans = []
for i in range(M):
if deg[i] == 0:
que.append(i)
while que:
v = que.popleft()
ans.append(v+1)
for w in G[v]:
deg[w] -= 1
if deg[w] == 0:
que.append(w)
print(*ans, sep='\n')
solve()
``` | output | 1 | 52,939 | 14 | 105,879 |
Provide a correct Python 3 solution for this coding contest problem.
A serious incident happened at Taro's house, which loves steamed buns. One of the three steamed buns offered at the Buddhist altar in the Japanese-style room was gone. When Taro, who was aiming for a snack someday, started an investigation to find the criminal, it turned out that there were many people who entered the Japanese-style room that day. So I decided to ask everyone to give the following form of testimony to find out the order in which these suspects entered the room.
Suspect A's testimony "I entered the room before Suspect B."
One of the suspects (?) Is a calico cat, so I can't testify, but fortunately Taro saw the last time he entered the room.
From these testimonies, Taro decided to guess the order in which he entered the room and use it for the investigation.
For example, if there are 6 suspects and Tama is suspect 2, the following testimony is obtained.
Testimony of Suspect 5 "I entered the room before 2."
Suspect 1's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 5."
Suspect 4's testimony "I entered the room before 2."
Suspect 1's testimony "I entered the room before 6."
Suspect 6's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 4."
Putting this testimony together, the order in which they entered the room
* 3 β 5 β 1 β 6 β 4 β 2
* 1 β 6 β 3 β 4 β 5 β 2
* 3 β 1 β 6 β 5 β 4 β 2
You can narrow down to several possibilities.
From the testimony of all suspects except Tama (Suspect 2), estimate the order in which they entered the room and write a program that outputs one of the possible orders. However, although some suspects may give multiple testimonies, all testimonies shall be true and consistent.
Input
The input is given in the following format:
m
n
x1 y1
x2 y2
::
xn yn
The first line gives the number of suspects m (m β€ 20), and the second line gives the number of testimonies n (n β€ 100). The following n lines are given the contents of the i-th testimony, xi and yi, respectively, on the first line. xi yi represents the testimony that "suspect xi (I) entered before suspect yi".
Output
Print the number of the suspect who entered the room first, the number of the suspect who entered the room second, ..., and the number of the suspect who entered the room second, in order on one line.
Example
Input
6
7
5 2
1 4
3 5
4 2
1 6
6 4
3 4
Output
3
5
1
6
4
2 | instruction | 0 | 52,940 | 14 | 105,880 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
import os
import math
m = int(input())
n = int(input())
# Manage people who entered earlier than me in a list
G = [[] for i in range(m)]
for i in range(n):
x, y = map(int, input().split())
x -= 1
y -= 1
G[y].append(x)
printed = [False] * m
while printed.count(False) > 1:
# Delete node with degree 0
for i, node in enumerate(G):
if not printed[i] and len(node) == 0:
print(i+1)
printed[i] = True
# Erase his existence
for node in G:
if i in node:
node.remove(i)
# cat
if False in printed:
i = printed.index(False)
print(i+1)
``` | output | 1 | 52,940 | 14 | 105,881 |
Provide a correct Python 3 solution for this coding contest problem.
A serious incident happened at Taro's house, which loves steamed buns. One of the three steamed buns offered at the Buddhist altar in the Japanese-style room was gone. When Taro, who was aiming for a snack someday, started an investigation to find the criminal, it turned out that there were many people who entered the Japanese-style room that day. So I decided to ask everyone to give the following form of testimony to find out the order in which these suspects entered the room.
Suspect A's testimony "I entered the room before Suspect B."
One of the suspects (?) Is a calico cat, so I can't testify, but fortunately Taro saw the last time he entered the room.
From these testimonies, Taro decided to guess the order in which he entered the room and use it for the investigation.
For example, if there are 6 suspects and Tama is suspect 2, the following testimony is obtained.
Testimony of Suspect 5 "I entered the room before 2."
Suspect 1's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 5."
Suspect 4's testimony "I entered the room before 2."
Suspect 1's testimony "I entered the room before 6."
Suspect 6's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 4."
Putting this testimony together, the order in which they entered the room
* 3 β 5 β 1 β 6 β 4 β 2
* 1 β 6 β 3 β 4 β 5 β 2
* 3 β 1 β 6 β 5 β 4 β 2
You can narrow down to several possibilities.
From the testimony of all suspects except Tama (Suspect 2), estimate the order in which they entered the room and write a program that outputs one of the possible orders. However, although some suspects may give multiple testimonies, all testimonies shall be true and consistent.
Input
The input is given in the following format:
m
n
x1 y1
x2 y2
::
xn yn
The first line gives the number of suspects m (m β€ 20), and the second line gives the number of testimonies n (n β€ 100). The following n lines are given the contents of the i-th testimony, xi and yi, respectively, on the first line. xi yi represents the testimony that "suspect xi (I) entered before suspect yi".
Output
Print the number of the suspect who entered the room first, the number of the suspect who entered the room second, ..., and the number of the suspect who entered the room second, in order on one line.
Example
Input
6
7
5 2
1 4
3 5
4 2
1 6
6 4
3 4
Output
3
5
1
6
4
2 | instruction | 0 | 52,941 | 14 | 105,882 |
"Correct Solution:
```
m = int(input())
n = int(input())
X = [0 for i in range(n)]
Y = [0 for i in range(n)]
for l in range(n):
X[l],Y[l] = [int(i) for i in input().split()]
ans = []
while True:
flag = True
for i in range(1, m+1):
if not (i in Y) and not (i in ans):
ans.append(i)
for j in range(len(X))[::-1]:
if X[j] == i:
X.pop(j)
Y.pop(j)
flag = False
break
if flag:
break
for i in range(len(ans)):
print(ans[i])
``` | output | 1 | 52,941 | 14 | 105,883 |
Provide a correct Python 3 solution for this coding contest problem.
A serious incident happened at Taro's house, which loves steamed buns. One of the three steamed buns offered at the Buddhist altar in the Japanese-style room was gone. When Taro, who was aiming for a snack someday, started an investigation to find the criminal, it turned out that there were many people who entered the Japanese-style room that day. So I decided to ask everyone to give the following form of testimony to find out the order in which these suspects entered the room.
Suspect A's testimony "I entered the room before Suspect B."
One of the suspects (?) Is a calico cat, so I can't testify, but fortunately Taro saw the last time he entered the room.
From these testimonies, Taro decided to guess the order in which he entered the room and use it for the investigation.
For example, if there are 6 suspects and Tama is suspect 2, the following testimony is obtained.
Testimony of Suspect 5 "I entered the room before 2."
Suspect 1's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 5."
Suspect 4's testimony "I entered the room before 2."
Suspect 1's testimony "I entered the room before 6."
Suspect 6's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 4."
Putting this testimony together, the order in which they entered the room
* 3 β 5 β 1 β 6 β 4 β 2
* 1 β 6 β 3 β 4 β 5 β 2
* 3 β 1 β 6 β 5 β 4 β 2
You can narrow down to several possibilities.
From the testimony of all suspects except Tama (Suspect 2), estimate the order in which they entered the room and write a program that outputs one of the possible orders. However, although some suspects may give multiple testimonies, all testimonies shall be true and consistent.
Input
The input is given in the following format:
m
n
x1 y1
x2 y2
::
xn yn
The first line gives the number of suspects m (m β€ 20), and the second line gives the number of testimonies n (n β€ 100). The following n lines are given the contents of the i-th testimony, xi and yi, respectively, on the first line. xi yi represents the testimony that "suspect xi (I) entered before suspect yi".
Output
Print the number of the suspect who entered the room first, the number of the suspect who entered the room second, ..., and the number of the suspect who entered the room second, in order on one line.
Example
Input
6
7
5 2
1 4
3 5
4 2
1 6
6 4
3 4
Output
3
5
1
6
4
2 | instruction | 0 | 52,942 | 14 | 105,884 |
"Correct Solution:
```
visited = []
def dfs(v):
for i in edges[v]:
if i not in visited:
dfs(i)
visited.append(v)
if __name__ == '__main__':
m = int(input())
n = int(input())
edges = [[] for i in range(m)]
for i in range(n):
x, y = map(int, input().split())
edges[x - 1].append(y - 1)
for i in range(m):
if i not in visited:
dfs(i)
ans = reversed([i + 1 for i in visited])
print('\n'.join(map(str, ans)))
``` | output | 1 | 52,942 | 14 | 105,885 |
Provide a correct Python 3 solution for this coding contest problem.
A serious incident happened at Taro's house, which loves steamed buns. One of the three steamed buns offered at the Buddhist altar in the Japanese-style room was gone. When Taro, who was aiming for a snack someday, started an investigation to find the criminal, it turned out that there were many people who entered the Japanese-style room that day. So I decided to ask everyone to give the following form of testimony to find out the order in which these suspects entered the room.
Suspect A's testimony "I entered the room before Suspect B."
One of the suspects (?) Is a calico cat, so I can't testify, but fortunately Taro saw the last time he entered the room.
From these testimonies, Taro decided to guess the order in which he entered the room and use it for the investigation.
For example, if there are 6 suspects and Tama is suspect 2, the following testimony is obtained.
Testimony of Suspect 5 "I entered the room before 2."
Suspect 1's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 5."
Suspect 4's testimony "I entered the room before 2."
Suspect 1's testimony "I entered the room before 6."
Suspect 6's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 4."
Putting this testimony together, the order in which they entered the room
* 3 β 5 β 1 β 6 β 4 β 2
* 1 β 6 β 3 β 4 β 5 β 2
* 3 β 1 β 6 β 5 β 4 β 2
You can narrow down to several possibilities.
From the testimony of all suspects except Tama (Suspect 2), estimate the order in which they entered the room and write a program that outputs one of the possible orders. However, although some suspects may give multiple testimonies, all testimonies shall be true and consistent.
Input
The input is given in the following format:
m
n
x1 y1
x2 y2
::
xn yn
The first line gives the number of suspects m (m β€ 20), and the second line gives the number of testimonies n (n β€ 100). The following n lines are given the contents of the i-th testimony, xi and yi, respectively, on the first line. xi yi represents the testimony that "suspect xi (I) entered before suspect yi".
Output
Print the number of the suspect who entered the room first, the number of the suspect who entered the room second, ..., and the number of the suspect who entered the room second, in order on one line.
Example
Input
6
7
5 2
1 4
3 5
4 2
1 6
6 4
3 4
Output
3
5
1
6
4
2 | instruction | 0 | 52,943 | 14 | 105,886 |
"Correct Solution:
```
m = int(input())
n = int(input())
edges = [[] for _ in range(m)]
for _ in range(n):
x,y = map(int,input().split())
edges[x-1].append(y-1)
L = []
visited = [False]*m
def visit(node):
if not visited[node]:
visited[node] = True
for nn in edges[node]:
visit(nn)
L.append(node)
for node in range(m):
visit(node)
for e in L[::-1]:
print(e+1)
``` | output | 1 | 52,943 | 14 | 105,887 |
Provide a correct Python 3 solution for this coding contest problem.
A serious incident happened at Taro's house, which loves steamed buns. One of the three steamed buns offered at the Buddhist altar in the Japanese-style room was gone. When Taro, who was aiming for a snack someday, started an investigation to find the criminal, it turned out that there were many people who entered the Japanese-style room that day. So I decided to ask everyone to give the following form of testimony to find out the order in which these suspects entered the room.
Suspect A's testimony "I entered the room before Suspect B."
One of the suspects (?) Is a calico cat, so I can't testify, but fortunately Taro saw the last time he entered the room.
From these testimonies, Taro decided to guess the order in which he entered the room and use it for the investigation.
For example, if there are 6 suspects and Tama is suspect 2, the following testimony is obtained.
Testimony of Suspect 5 "I entered the room before 2."
Suspect 1's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 5."
Suspect 4's testimony "I entered the room before 2."
Suspect 1's testimony "I entered the room before 6."
Suspect 6's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 4."
Putting this testimony together, the order in which they entered the room
* 3 β 5 β 1 β 6 β 4 β 2
* 1 β 6 β 3 β 4 β 5 β 2
* 3 β 1 β 6 β 5 β 4 β 2
You can narrow down to several possibilities.
From the testimony of all suspects except Tama (Suspect 2), estimate the order in which they entered the room and write a program that outputs one of the possible orders. However, although some suspects may give multiple testimonies, all testimonies shall be true and consistent.
Input
The input is given in the following format:
m
n
x1 y1
x2 y2
::
xn yn
The first line gives the number of suspects m (m β€ 20), and the second line gives the number of testimonies n (n β€ 100). The following n lines are given the contents of the i-th testimony, xi and yi, respectively, on the first line. xi yi represents the testimony that "suspect xi (I) entered before suspect yi".
Output
Print the number of the suspect who entered the room first, the number of the suspect who entered the room second, ..., and the number of the suspect who entered the room second, in order on one line.
Example
Input
6
7
5 2
1 4
3 5
4 2
1 6
6 4
3 4
Output
3
5
1
6
4
2 | instruction | 0 | 52,944 | 14 | 105,888 |
"Correct Solution:
```
# Aizu Problem 00119: Taro's Obsession
#
import sys, math, os
# read input:
PYDEV = os.environ.get('PYDEV')
if PYDEV=="True":
sys.stdin = open("sample-input.txt", "rt")
class DirectedGraph():
def __init__(self):
self.edges = {}
def AddNode(self, node):
if node not in self.edges:
self.edges[node] = []
def GetNodes(self):
return list(self.edges.keys())
def GetEdges(self):
return self.edges
def AddEdge(self, node1, node2):
if node1 not in self.GetNodes():
self.AddNode(node1)
if node2 not in self.edges[node1]:
self.edges[node1].append(node2)
def RemoveEdge(self, node1, node2):
if node1 not in self.edges or node2 not in self.edges[node1]:
print("ERROR: can't remove edge", node1, "->", node2)
print(self.edges)
return
self.edges[node1].remove(node2)
def OutDegree(self, node):
return len(self.edges[node])
def InDegree(self, node):
deg = 0
for node2 in self.GetNodes():
if node in self.GetNeighbors(node2):
deg += 1
return deg
def GetNeighbors(self, node):
return self.edges.get(node, [])
def GetPath(self, source, target):
# find path by BFS:
predecessor = {node: -1 for node in self.GetNodes()}
predecessor[source] = 0
visited = {node: False for node in self.GetNodes()}
queue = [source]
while len(queue) > 0:
current = queue.pop(0)
visited[current] = True
for neighbor in self.GetNeighbors(current):
if not visited[neighbor] and neighbor not in queue:
queue.append(neighbor)
predecessor[neighbor] = current
if neighbor == target:
queue = []
break
# build path from predecessor info:
if predecessor[target] == -1:
return []
path = [target]
while path[0] != source:
path = [predecessor[path[0]]] + path[:]
return path
def TopologicalSort(self):
# determine topological sorting of graph by Kahn's algorithm;
# edges are removed by this algorithm
L = []
S = [node for node in self.GetNodes() if self.InDegree(node) == 0]
while len(S) > 0:
n = S.pop(0)
L.append(n)
for m in self.GetNeighbors(n)[:]:
self.RemoveEdge(n, m)
if self.InDegree(m) == 0:
S.append(m)
for node in self.GetNodes():
if self.edges[node] != []:
print("ERROR: graph has cycle!", self.edges)
return
return L
m = int(input())
n = int(input())
G = DirectedGraph()
for __ in range(n):
a, b = [int(_) for _ in input().split()]
G.AddEdge(a, b)
tsort = G.TopologicalSort()
for k in tsort:
print(k)
``` | output | 1 | 52,944 | 14 | 105,889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A serious incident happened at Taro's house, which loves steamed buns. One of the three steamed buns offered at the Buddhist altar in the Japanese-style room was gone. When Taro, who was aiming for a snack someday, started an investigation to find the criminal, it turned out that there were many people who entered the Japanese-style room that day. So I decided to ask everyone to give the following form of testimony to find out the order in which these suspects entered the room.
Suspect A's testimony "I entered the room before Suspect B."
One of the suspects (?) Is a calico cat, so I can't testify, but fortunately Taro saw the last time he entered the room.
From these testimonies, Taro decided to guess the order in which he entered the room and use it for the investigation.
For example, if there are 6 suspects and Tama is suspect 2, the following testimony is obtained.
Testimony of Suspect 5 "I entered the room before 2."
Suspect 1's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 5."
Suspect 4's testimony "I entered the room before 2."
Suspect 1's testimony "I entered the room before 6."
Suspect 6's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 4."
Putting this testimony together, the order in which they entered the room
* 3 β 5 β 1 β 6 β 4 β 2
* 1 β 6 β 3 β 4 β 5 β 2
* 3 β 1 β 6 β 5 β 4 β 2
You can narrow down to several possibilities.
From the testimony of all suspects except Tama (Suspect 2), estimate the order in which they entered the room and write a program that outputs one of the possible orders. However, although some suspects may give multiple testimonies, all testimonies shall be true and consistent.
Input
The input is given in the following format:
m
n
x1 y1
x2 y2
::
xn yn
The first line gives the number of suspects m (m β€ 20), and the second line gives the number of testimonies n (n β€ 100). The following n lines are given the contents of the i-th testimony, xi and yi, respectively, on the first line. xi yi represents the testimony that "suspect xi (I) entered before suspect yi".
Output
Print the number of the suspect who entered the room first, the number of the suspect who entered the room second, ..., and the number of the suspect who entered the room second, in order on one line.
Example
Input
6
7
5 2
1 4
3 5
4 2
1 6
6 4
3 4
Output
3
5
1
6
4
2
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0119
"""
import sys
from collections import deque
def bfs(u, result, out_edge, in_degree, processed):
q = deque()
q.append(u)
processed[u] = True
while q:
u = q.popleft()
result.append(u)
for e in out_edge[u]:
in_degree[e] -= 1 # ?????Β¨?????????????????????(u)?????????????????Β¨???u?????\?ΒΆ????(e)????????\????Β¬???Β°???1???????????Β¨?????????
if in_degree[e] == 0 and processed[e] == False: # ??\????Β¬???Β°???1??????????????????0???????????Β°??????????????????????Β¬??????????????Β£??????????
processed[e] = True
q.append(e)
def solve(m, in_degree, out_edge):
result = []
processed = [False for _ in range(m+1)] # ???????????????????????????????????????????????Β°
for i, u in enumerate(in_degree[1:], start=1):
if u == 0 and processed[i] == False: # ??\????Β¬???Β°???0???????????????
bfs(i, result, out_edge, in_degree, processed)
return result
def main(args):
m = int(input())
n = int(input())
in_degree = [0 for _ in range(m+1)] # ????????????????????\????Β¬???Β°
out_edge = [[] for _ in range(m+1)] # ???????????????????????????(??\?ΒΆ?)?????????????????????
for _ in range(n):
s, t = [int(x) for x in input().split(' ')]
out_edge[s].append(t)
in_degree[t] += 1
result = solve(m, in_degree, out_edge)
print('\n'.join(map(str, result)))
if __name__ == '__main__':
main(sys.argv[1:])
``` | instruction | 0 | 52,945 | 14 | 105,890 |
Yes | output | 1 | 52,945 | 14 | 105,891 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A serious incident happened at Taro's house, which loves steamed buns. One of the three steamed buns offered at the Buddhist altar in the Japanese-style room was gone. When Taro, who was aiming for a snack someday, started an investigation to find the criminal, it turned out that there were many people who entered the Japanese-style room that day. So I decided to ask everyone to give the following form of testimony to find out the order in which these suspects entered the room.
Suspect A's testimony "I entered the room before Suspect B."
One of the suspects (?) Is a calico cat, so I can't testify, but fortunately Taro saw the last time he entered the room.
From these testimonies, Taro decided to guess the order in which he entered the room and use it for the investigation.
For example, if there are 6 suspects and Tama is suspect 2, the following testimony is obtained.
Testimony of Suspect 5 "I entered the room before 2."
Suspect 1's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 5."
Suspect 4's testimony "I entered the room before 2."
Suspect 1's testimony "I entered the room before 6."
Suspect 6's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 4."
Putting this testimony together, the order in which they entered the room
* 3 β 5 β 1 β 6 β 4 β 2
* 1 β 6 β 3 β 4 β 5 β 2
* 3 β 1 β 6 β 5 β 4 β 2
You can narrow down to several possibilities.
From the testimony of all suspects except Tama (Suspect 2), estimate the order in which they entered the room and write a program that outputs one of the possible orders. However, although some suspects may give multiple testimonies, all testimonies shall be true and consistent.
Input
The input is given in the following format:
m
n
x1 y1
x2 y2
::
xn yn
The first line gives the number of suspects m (m β€ 20), and the second line gives the number of testimonies n (n β€ 100). The following n lines are given the contents of the i-th testimony, xi and yi, respectively, on the first line. xi yi represents the testimony that "suspect xi (I) entered before suspect yi".
Output
Print the number of the suspect who entered the room first, the number of the suspect who entered the room second, ..., and the number of the suspect who entered the room second, in order on one line.
Example
Input
6
7
5 2
1 4
3 5
4 2
1 6
6 4
3 4
Output
3
5
1
6
4
2
Submitted Solution:
```
# AOJ 0119: Taro's Obsession
# Python3 2018.6.20 bal4u
def topological_sort(V, to):
cnt = [0]*V
for i in range(V):
for j in to[i]: cnt[j] += 1
Q = []
for i in range(V):
if cnt[i] == 0:
Q.append(i)
while len(Q) > 0:
i = Q[0]
del Q[0]
print(i+1)
for k in to[i]:
cnt[k] -= 1
if cnt[k] == 0: Q.append(k)
m = int(input())
to = [[] for i in range(m)]
n = int(input())
for i in range(n):
x, y = list(map(int, input().split()))
to[x-1].append(y-1)
topological_sort(m, to)
``` | instruction | 0 | 52,946 | 14 | 105,892 |
Yes | output | 1 | 52,946 | 14 | 105,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A serious incident happened at Taro's house, which loves steamed buns. One of the three steamed buns offered at the Buddhist altar in the Japanese-style room was gone. When Taro, who was aiming for a snack someday, started an investigation to find the criminal, it turned out that there were many people who entered the Japanese-style room that day. So I decided to ask everyone to give the following form of testimony to find out the order in which these suspects entered the room.
Suspect A's testimony "I entered the room before Suspect B."
One of the suspects (?) Is a calico cat, so I can't testify, but fortunately Taro saw the last time he entered the room.
From these testimonies, Taro decided to guess the order in which he entered the room and use it for the investigation.
For example, if there are 6 suspects and Tama is suspect 2, the following testimony is obtained.
Testimony of Suspect 5 "I entered the room before 2."
Suspect 1's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 5."
Suspect 4's testimony "I entered the room before 2."
Suspect 1's testimony "I entered the room before 6."
Suspect 6's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 4."
Putting this testimony together, the order in which they entered the room
* 3 β 5 β 1 β 6 β 4 β 2
* 1 β 6 β 3 β 4 β 5 β 2
* 3 β 1 β 6 β 5 β 4 β 2
You can narrow down to several possibilities.
From the testimony of all suspects except Tama (Suspect 2), estimate the order in which they entered the room and write a program that outputs one of the possible orders. However, although some suspects may give multiple testimonies, all testimonies shall be true and consistent.
Input
The input is given in the following format:
m
n
x1 y1
x2 y2
::
xn yn
The first line gives the number of suspects m (m β€ 20), and the second line gives the number of testimonies n (n β€ 100). The following n lines are given the contents of the i-th testimony, xi and yi, respectively, on the first line. xi yi represents the testimony that "suspect xi (I) entered before suspect yi".
Output
Print the number of the suspect who entered the room first, the number of the suspect who entered the room second, ..., and the number of the suspect who entered the room second, in order on one line.
Example
Input
6
7
5 2
1 4
3 5
4 2
1 6
6 4
3 4
Output
3
5
1
6
4
2
Submitted Solution:
```
# -*- coding: utf-8 -*-
import sys
import os
import math
m = int(input())
n = int(input())
# Manage people who entered earlier than me in a list
G = [[] for i in range(m)]
for i in range(n):
x, y = map(int, input().split())
x -= 1
y -= 1
G[y].append(x)
printed = [False] * m
while printed.count(False) > 1:
# Delete node with degree 0
for i, node in enumerate(G):
if not printed[i] and len(node) == 0:
print(i+1)
printed[i] = True
# Erase his existence
for node in G:
if i in node:
node.remove(i)
# cat
i = printed.index(False)
print(i+1)
``` | instruction | 0 | 52,947 | 14 | 105,894 |
No | output | 1 | 52,947 | 14 | 105,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A serious incident happened at Taro's house, which loves steamed buns. One of the three steamed buns offered at the Buddhist altar in the Japanese-style room was gone. When Taro, who was aiming for a snack someday, started an investigation to find the criminal, it turned out that there were many people who entered the Japanese-style room that day. So I decided to ask everyone to give the following form of testimony to find out the order in which these suspects entered the room.
Suspect A's testimony "I entered the room before Suspect B."
One of the suspects (?) Is a calico cat, so I can't testify, but fortunately Taro saw the last time he entered the room.
From these testimonies, Taro decided to guess the order in which he entered the room and use it for the investigation.
For example, if there are 6 suspects and Tama is suspect 2, the following testimony is obtained.
Testimony of Suspect 5 "I entered the room before 2."
Suspect 1's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 5."
Suspect 4's testimony "I entered the room before 2."
Suspect 1's testimony "I entered the room before 6."
Suspect 6's testimony "I entered the room before 4."
Testimony of Suspect 3 "I entered the room before 4."
Putting this testimony together, the order in which they entered the room
* 3 β 5 β 1 β 6 β 4 β 2
* 1 β 6 β 3 β 4 β 5 β 2
* 3 β 1 β 6 β 5 β 4 β 2
You can narrow down to several possibilities.
From the testimony of all suspects except Tama (Suspect 2), estimate the order in which they entered the room and write a program that outputs one of the possible orders. However, although some suspects may give multiple testimonies, all testimonies shall be true and consistent.
Input
The input is given in the following format:
m
n
x1 y1
x2 y2
::
xn yn
The first line gives the number of suspects m (m β€ 20), and the second line gives the number of testimonies n (n β€ 100). The following n lines are given the contents of the i-th testimony, xi and yi, respectively, on the first line. xi yi represents the testimony that "suspect xi (I) entered before suspect yi".
Output
Print the number of the suspect who entered the room first, the number of the suspect who entered the room second, ..., and the number of the suspect who entered the room second, in order on one line.
Example
Input
6
7
5 2
1 4
3 5
4 2
1 6
6 4
3 4
Output
3
5
1
6
4
2
Submitted Solution:
```
count = int(input())
comment_count = int(input())
input_data = [input().split(" ") for _ in range(comment_count)]
input_data = [(int(item1), int(item2)) for item1, item2 in input_data]
people_list = [2]
search_list = [False] * (count + 1)
search_list[2] = True
index_set = {0: 2}
while 0 < comment_count:
data_list = [item for item in input_data if search_list[item[1]]]
search_list = [False] * (count + 1)
comment_count -= len(data_list)
for i, you in data_list:
if i not in index_set:
people_list.append(i)
index_set[i] = len(people_list) - 1
else:
index1 = index_set[i]
index2 = index_set[you]
if index1 < index2:
people_list.insert(index2 + 1, i)
people_list.pop(index1)
index_set[i] = index2 + 1
search_list[i] = True
people_list = [str(item) for item in people_list]
print("\n".join(people_list))
``` | instruction | 0 | 52,948 | 14 | 105,896 |
No | output | 1 | 52,948 | 14 | 105,897 |
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