message stringlengths 2 28.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 21 109k | cluster float64 7 7 | __index_level_0__ int64 42 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
Consider a circle whose perimeter is divided by N points into N arcs of equal length, and each of the arcs is painted red or blue. Such a circle is said to generate a string S from every point when the following condition is satisfied:
* We will arbitrarily choose one of the N points on the perimeter and place a piece on it.
* Then, we will perform the following move M times: move the piece clockwise or counter-clockwise to an adjacent point.
* Here, whatever point we choose initially, it is always possible to move the piece so that the color of the i-th arc the piece goes along is S_i, by properly deciding the directions of the moves.
Assume that, if S_i is `R`, it represents red; if S_i is `B`, it represents blue. Note that the directions of the moves can be decided separately for each choice of the initial point.
You are given a string S of length M consisting of `R` and `B`. Out of the 2^N ways to paint each of the arcs red or blue in a circle whose perimeter is divided into N arcs of equal length, find the number of ways resulting in a circle that generates S from every point, modulo 10^9+7.
Note that the rotations of the same coloring are also distinguished.
Constraints
* 2 \leq N \leq 2 \times 10^5
* 1 \leq M \leq 2 \times 10^5
* |S|=M
* S_i is `R` or `B`.
Input
Input is given from Standard Input in the following format:
N M
S
Output
Print the number of ways to paint each of the arcs that satisfy the condition, modulo 10^9+7.
Examples
Input
4 7
RBRRBRR
Output
2
Input
3 3
BBB
Output
4
Input
12 10
RRRRBRRRRB
Output
78 | instruction | 0 | 71,431 | 7 | 142,862 |
"Correct Solution:
```
p=int(1e9+7)
def calc(n,l,r):
f=[0]* (n+1)
f[0]=1
sum=0
for i in range (1,n+1):
if i>=l:
sum+=f[i-l]
f[i]=sum%p
if (i>=r):
sum-=f[i-r]
ans=0
for i in range (l,r+1):
ans=(ans+f[n-i]*i)%p
return ans
n,m=input().split()
n=int(n)
m=int(m)
s=input()
a=s.split(chr(ord('R')+ord('B')-ord(s[0])))
if len(a)==1:
print((calc(n,2,n)+1)%p)
elif n%2==1:
print(0)
else:
r=len(a[0])+1-len(a[0])%2
for i in range(1,len(a)-1):
if len(a[i])%2==1 :
r=min(r,len(a[i]));
n=n//2
r=(r+1)//2
print(calc(n,1,r)*2%p)
``` | output | 1 | 71,431 | 7 | 142,863 |
Provide a correct Python 3 solution for this coding contest problem.
Consider a circle whose perimeter is divided by N points into N arcs of equal length, and each of the arcs is painted red or blue. Such a circle is said to generate a string S from every point when the following condition is satisfied:
* We will arbitrarily choose one of the N points on the perimeter and place a piece on it.
* Then, we will perform the following move M times: move the piece clockwise or counter-clockwise to an adjacent point.
* Here, whatever point we choose initially, it is always possible to move the piece so that the color of the i-th arc the piece goes along is S_i, by properly deciding the directions of the moves.
Assume that, if S_i is `R`, it represents red; if S_i is `B`, it represents blue. Note that the directions of the moves can be decided separately for each choice of the initial point.
You are given a string S of length M consisting of `R` and `B`. Out of the 2^N ways to paint each of the arcs red or blue in a circle whose perimeter is divided into N arcs of equal length, find the number of ways resulting in a circle that generates S from every point, modulo 10^9+7.
Note that the rotations of the same coloring are also distinguished.
Constraints
* 2 \leq N \leq 2 \times 10^5
* 1 \leq M \leq 2 \times 10^5
* |S|=M
* S_i is `R` or `B`.
Input
Input is given from Standard Input in the following format:
N M
S
Output
Print the number of ways to paint each of the arcs that satisfy the condition, modulo 10^9+7.
Examples
Input
4 7
RBRRBRR
Output
2
Input
3 3
BBB
Output
4
Input
12 10
RRRRBRRRRB
Output
78 | instruction | 0 | 71,432 | 7 | 142,864 |
"Correct Solution:
```
n,m = map(int,input().split())
s = input().rstrip()
mod = 10**9+7
if s.count("R") == 0 or s.count("B") == 0:
dp = [[0 for i in range(2)] for j in range(n)]
dp[0][0] = 1
dp[0][1] = 1
for i in range(1,n):
dp[i][0] = (dp[i-1][0]+dp[i-1][1])%mod
dp[i][1] = dp[i-1][0]
print((dp[n-1][0]+dp[n-2][1])%mod)
exit()
if n%2:
print(0)
exit()
x = s[0]
flg = 1
cnt = 0
ncnt = 0
for i in range(m):
if s[i] == x:
cnt += 1
elif flg == 1:
flg = 0
ncnt = cnt
cnt = 0
else:
if cnt%2:
ncnt = min(ncnt,cnt)
cnt = 0
if ncnt == 1:
print(2)
exit()
y = ncnt//2
dp = [[0 for i in range(2)] for j in range(n//2)]
dp[0][0] = 1
dp[0][1] = 1
for i in range(1,n//2):
dp[i][0] = dp[i-1][0]+dp[i-1][1]
if i<y:
dp[i][1] = dp[i-1][0]+dp[i-1][1]
elif i == y:
dp[i][1] = sum([dp[j][0] for j in range(y)])
else:
dp[i][1] = dp[i-1][1]+dp[i-1][0]-dp[i-y-1][0]
dp[i][0] %= mod
dp[i][1] %= mod
ans = dp[-1][0]+dp[-1][1]
for j in range(1,y+1):
if n//2-(y+j)-2 >= 0:
ans = (ans-(dp[n//2-(y+j)-2][0])*(y+1-j))%mod
elif n//2-(y+j) == 1:
ans -= y+1-j
print(2*ans%mod)
``` | output | 1 | 71,432 | 7 | 142,865 |
Provide a correct Python 3 solution for this coding contest problem.
Consider a circle whose perimeter is divided by N points into N arcs of equal length, and each of the arcs is painted red or blue. Such a circle is said to generate a string S from every point when the following condition is satisfied:
* We will arbitrarily choose one of the N points on the perimeter and place a piece on it.
* Then, we will perform the following move M times: move the piece clockwise or counter-clockwise to an adjacent point.
* Here, whatever point we choose initially, it is always possible to move the piece so that the color of the i-th arc the piece goes along is S_i, by properly deciding the directions of the moves.
Assume that, if S_i is `R`, it represents red; if S_i is `B`, it represents blue. Note that the directions of the moves can be decided separately for each choice of the initial point.
You are given a string S of length M consisting of `R` and `B`. Out of the 2^N ways to paint each of the arcs red or blue in a circle whose perimeter is divided into N arcs of equal length, find the number of ways resulting in a circle that generates S from every point, modulo 10^9+7.
Note that the rotations of the same coloring are also distinguished.
Constraints
* 2 \leq N \leq 2 \times 10^5
* 1 \leq M \leq 2 \times 10^5
* |S|=M
* S_i is `R` or `B`.
Input
Input is given from Standard Input in the following format:
N M
S
Output
Print the number of ways to paint each of the arcs that satisfy the condition, modulo 10^9+7.
Examples
Input
4 7
RBRRBRR
Output
2
Input
3 3
BBB
Output
4
Input
12 10
RRRRBRRRRB
Output
78 | instruction | 0 | 71,433 | 7 | 142,866 |
"Correct Solution:
```
def solve(n, m, s):
f = s[0]
MOD = 10 ** 9 + 7
INF = 10 ** 6
p = None
seq = 0
min_seq = INF
for c in s:
if c == p:
seq += 1
else:
if p == f and (min_seq == INF or seq % 2 == 1):
min_seq = min(min_seq, seq)
seq = 1
p = c
if min_seq == INF:
a, b = 1, 0
for _ in range(n - 2):
a, b = (a + b) % MOD, a
return (3 * a + b) % MOD
if n % 2 == 1:
return 0
n2 = n // 2
ms = min(n2, min_seq // 2 + 1)
dp = [0] * (n2 + 1)
dp[0] = 1
acc, reject = 1, 0
for i in range(1, n2 + 1):
dp[i] = (acc - reject) % MOD
acc = (acc + dp[i]) % MOD
if i >= ms:
reject = (reject + dp[i - ms]) % MOD
ans = 0
for d in range(1, ms + 1):
ans = (ans + dp[n2 - d] * d * 2) % MOD
return ans
n, m = map(int, input().split())
s = input()
print(solve(n, m, s))
``` | output | 1 | 71,433 | 7 | 142,867 |
Provide a correct Python 3 solution for this coding contest problem.
Consider a circle whose perimeter is divided by N points into N arcs of equal length, and each of the arcs is painted red or blue. Such a circle is said to generate a string S from every point when the following condition is satisfied:
* We will arbitrarily choose one of the N points on the perimeter and place a piece on it.
* Then, we will perform the following move M times: move the piece clockwise or counter-clockwise to an adjacent point.
* Here, whatever point we choose initially, it is always possible to move the piece so that the color of the i-th arc the piece goes along is S_i, by properly deciding the directions of the moves.
Assume that, if S_i is `R`, it represents red; if S_i is `B`, it represents blue. Note that the directions of the moves can be decided separately for each choice of the initial point.
You are given a string S of length M consisting of `R` and `B`. Out of the 2^N ways to paint each of the arcs red or blue in a circle whose perimeter is divided into N arcs of equal length, find the number of ways resulting in a circle that generates S from every point, modulo 10^9+7.
Note that the rotations of the same coloring are also distinguished.
Constraints
* 2 \leq N \leq 2 \times 10^5
* 1 \leq M \leq 2 \times 10^5
* |S|=M
* S_i is `R` or `B`.
Input
Input is given from Standard Input in the following format:
N M
S
Output
Print the number of ways to paint each of the arcs that satisfy the condition, modulo 10^9+7.
Examples
Input
4 7
RBRRBRR
Output
2
Input
3 3
BBB
Output
4
Input
12 10
RRRRBRRRRB
Output
78 | instruction | 0 | 71,434 | 7 | 142,868 |
"Correct Solution:
```
"""
https://atcoder.jp/contests/agc033/tasks/agc033_e
Sの1文字目をRとしてよい
RB から始まる場合 → 全部交互以外無理(Nが奇数なら0)
RRB から始まる場合 → Rは3連で置けば可能…
R*X + B*Y + R*Z …
とつながっていく
どこからスタートしても、RだけをX回移動したときにBの隣に来なくてはいけない
Rの長さが1なら可能
R*?,B,R*?,…
でつながっていく
最初のRがX個連続の時
片方の端との距離がX-2tでなくてはならない
Xが偶数の時、Rの連続長さはX+1以下の奇数
Xが奇数の時、端からスタートした奴は反対側に抜けなくてはいけないので
Rの連続長さはX以下の奇数？
結局は奇数個連続でしか置けない
●整理(Rから始まり、Bがある場合)：
Rは奇数個連続でしか置けない
Bは1個連続でしか置けない
→つまり、Nが奇数だとアウト
Bが来た以降を考えてみる
奇数だと、各点Rの左右のどちらかにしか抜けられない→対称的な移動になるはず
→つまり、最初のRを消化した後、全てのBの両端にいる場合が存在する
Bは奇数個の時のみ意味がある
→結局、Bの両端にいた場合が交換されるだけ
→よって、BはRの区切り以上の意味はない
Rが偶数個来るた場合、交互に移動すればおk
Rが奇数個来た場合、反対側に抜ける必要がある
結論：
RBがどちらも存在する場合、Nが偶数なら0、奇数の場合は
BでsplitしたRの個数の集合を考え,
r1,r2…,r? とする
Rは奇数個連続で置けて、その個数の最大は M = min( r1+1 , r? ) #r?が奇数の物
後は、M+1個以下の偶数個単位で区切る場合の数を調べればいい
dp→あらかじめ1番目の区間の位置に、ずれた分も考慮して値を入れておけばBITでもらうdpできる
RB片方の場合、Bが連続しないような置き方をすればいい
1番目がRの場合をdp → N番目までやる
1番目がBの場合をdp → N-1番目までやる(N番目は赤なので)
最後のRは意味がないので消しておく
REの原因究明したい
K < NNの場合？
"""
from sys import stdin
import sys
#0-indexed , 半開区間[a,b)
#calc変更で演算変更
class SegTree:
def __init__(self,N,first):
self.NO = 2**(N-1).bit_length()
self.First = first
self.data = [first] * (2*self.NO)
def calc(self,l,r):
return (l+r) % mod
def update(self,ind,x):
ind += self.NO - 1
self.data[ind] = x
while ind >= 0:
ind = (ind - 1)//2
self.data[ind] = self.calc(self.data[2*ind+1],self.data[2*ind+2])
def query(self,l,r):
L = l + self.NO
R = r + self.NO
s = self.First
while L < R:
if R & 1:
R -= 1
s = self.calc(s , self.data[R-1])
if L & 1:
s = self.calc(s , self.data[L-1])
L += 1
L >>= 1
R >>= 1
return s
def get(self , ind):
ind += self.NO - 1
return self.data[ind]
N,M = map(int,stdin.readline().split())
S = list(stdin.readline()[:-1])
mod = 10**9+7
if "R" in S and "B" in S:
if N % 2 == 1:
print (0)
sys.exit()
lis = []
now = 0
while S[-1] == S[0]: #最後のRを消す
del S[-1]
for i in S:
if i == S[0]:
now += 1
elif now > 0:
lis.append(now)
now = 0
#print (lis)
nmax = 0
for i in range(len(lis)):
if i == 0:
if lis[i] % 2 == 0:
nmax = lis[i] + 1
else:
nmax = lis[i]
elif lis[i] % 2 == 1:
nmax = min(nmax , lis[i])
NN = N//2
K = (nmax+1)//2
print (NN,K,file=sys.stderr)
if NN <= K:
print ((pow(2,NN,mod)-1) * 2 % mod)
sys.exit()
ST = SegTree(NN,0)
for i in range(K):
ST.update(i,i+1)
for i in range(1,NN):
now = ST.get(i)
plus = ST.query( max(0,i-K),i )
ST.update( i , (now+plus) % mod )
print (ST.get(NN-1) * 2 % mod)
else:
ans = 0
#1番目がRの場合
dp = [1,0]
for i in range(N-1):
ndp = [0,0]
ndp[0] = (dp[0] + dp[1]) % mod
ndp[1] = dp[0]
dp = ndp
ans += sum(dp)
#1番目がBの場合
dp = [0,1]
for i in range(N-2):
ndp = [0,0]
ndp[0] = (dp[0] + dp[1]) % mod
ndp[1] = dp[0]
dp = ndp
ans += sum(dp)
print (ans % mod)
``` | output | 1 | 71,434 | 7 | 142,869 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider a circle whose perimeter is divided by N points into N arcs of equal length, and each of the arcs is painted red or blue. Such a circle is said to generate a string S from every point when the following condition is satisfied:
* We will arbitrarily choose one of the N points on the perimeter and place a piece on it.
* Then, we will perform the following move M times: move the piece clockwise or counter-clockwise to an adjacent point.
* Here, whatever point we choose initially, it is always possible to move the piece so that the color of the i-th arc the piece goes along is S_i, by properly deciding the directions of the moves.
Assume that, if S_i is `R`, it represents red; if S_i is `B`, it represents blue. Note that the directions of the moves can be decided separately for each choice of the initial point.
You are given a string S of length M consisting of `R` and `B`. Out of the 2^N ways to paint each of the arcs red or blue in a circle whose perimeter is divided into N arcs of equal length, find the number of ways resulting in a circle that generates S from every point, modulo 10^9+7.
Note that the rotations of the same coloring are also distinguished.
Constraints
* 2 \leq N \leq 2 \times 10^5
* 1 \leq M \leq 2 \times 10^5
* |S|=M
* S_i is `R` or `B`.
Input
Input is given from Standard Input in the following format:
N M
S
Output
Print the number of ways to paint each of the arcs that satisfy the condition, modulo 10^9+7.
Examples
Input
4 7
RBRRBRR
Output
2
Input
3 3
BBB
Output
4
Input
12 10
RRRRBRRRRB
Output
78
Submitted Solution:
```
n,m = map(int,input().split())
s = input().rstrip()
mod = 10**9+7
if s.count("R") == 0 or s.count("B") == 0:
dp = [[0 for i in range(2)] for j in range(n)]
dp[0][0] = 1
dp[0][1] = 1
for i in range(1,n):
dp[i][0] = (dp[i-1][0]+dp[i-1][1])%mod
dp[i][1] = dp[i-1][0]
print((dp[n-1][0]+dp[n-2][1])%mod)
exit()
if n%2:
print(0)
exit()
x = s[0]
flg = 1
cnt = 0
ncnt = 0
for i in range(m):
if s[i] == x:
cnt += 1
elif flg == 1:
flg = 0
ncnt = cnt
cnt = 0
else:
if cnt%2:
ncnt = min(ncnt,cnt)
cnt = 0
if ncnt == 1:
print(2)
exit()
y = ncnt//2
dp = [[0 for i in range(2)] for j in range(n//2)]
dp[0][0] = 1
dp[0][1] = 1
for i in range(1,n//2):
dp[i][0] = dp[i-1][0]+dp[i-1][1]
if i<y:
dp[i][1] = dp[i-1][0]+dp[i-1][1]
elif i == y:
dp[i][1] = sum([dp[j][0] for j in range(y)])
else:
dp[i][1] = dp[i-1][1]+dp[i-1][0]-dp[i-y-1][0]
dp[i][0] %= mod
dp[i][1] %= mod
ans = dp[-1][0]+dp[-1][1]
for j in range(1,y+1):
if n//2-(y+j)-2 >= 0:
ans = (ans-(dp[n//2-(y+j)-2][0])*(y+1-j))%mod
elif n//2-(y+j) >= 0:
ans -= 1
print(2*ans%mod)
``` | instruction | 0 | 71,435 | 7 | 142,870 |
No | output | 1 | 71,435 | 7 | 142,871 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider a circle whose perimeter is divided by N points into N arcs of equal length, and each of the arcs is painted red or blue. Such a circle is said to generate a string S from every point when the following condition is satisfied:
* We will arbitrarily choose one of the N points on the perimeter and place a piece on it.
* Then, we will perform the following move M times: move the piece clockwise or counter-clockwise to an adjacent point.
* Here, whatever point we choose initially, it is always possible to move the piece so that the color of the i-th arc the piece goes along is S_i, by properly deciding the directions of the moves.
Assume that, if S_i is `R`, it represents red; if S_i is `B`, it represents blue. Note that the directions of the moves can be decided separately for each choice of the initial point.
You are given a string S of length M consisting of `R` and `B`. Out of the 2^N ways to paint each of the arcs red or blue in a circle whose perimeter is divided into N arcs of equal length, find the number of ways resulting in a circle that generates S from every point, modulo 10^9+7.
Note that the rotations of the same coloring are also distinguished.
Constraints
* 2 \leq N \leq 2 \times 10^5
* 1 \leq M \leq 2 \times 10^5
* |S|=M
* S_i is `R` or `B`.
Input
Input is given from Standard Input in the following format:
N M
S
Output
Print the number of ways to paint each of the arcs that satisfy the condition, modulo 10^9+7.
Examples
Input
4 7
RBRRBRR
Output
2
Input
3 3
BBB
Output
4
Input
12 10
RRRRBRRRRB
Output
78
Submitted Solution:
```
n,m = map(int,input().split())
s = input().rstrip()
mod = 10**9+7
if s.count("R") == 0 or s.count("B") == 0:
dp = [[0 for i in range(2)] for j in range(n)]
dp[0][0] = 1
dp[0][1] = 1
for i in range(1,n):
dp[i][0] = (dp[i-1][0]+dp[i-1][1])%mod
dp[i][1] = dp[i-1][0]
print((dp[n-1][0]+dp[n-2][1])%mod)
exit()
if n%2:
print(0)
exit()
x = s[0]
flg = 1
cnt = 0
ncnt = 10**18
for i in range(m):
if s[i] == x:
cnt += 1
else:
if cnt%2:
ncnt = min(ncnt,cnt)
cnt = 0
if ncnt == 1:
print(2)
exit()
y = ncnt//2
dp = [[0 for i in range(2)] for j in range(n//2)]
dp[0][0] = 1
dp[0][1] = 1
for i in range(1,n//2):
dp[i][0] = dp[i-1][0]+dp[i-1][1]
if i<y:
dp[i][1] = dp[i-1][0]+dp[i-1][1]
elif i == y:
dp[i][1] = sum([dp[j][0] for j in range(y)])
else:
dp[i][1] = dp[i-1][1]+dp[i-1][0]-dp[i-y-1][0]
dp[i][0] %= mod
dp[i][1] %= mod
ans = dp[-1][0]+dp[-1][1]
for j in range(1,y+1):
if n//2-(y+j)-2 >= 0:
ans = (ans-(dp[n//2-(y+j)-2][0])*(y+1-j))%mod
elif n//2-(y+j) >= 0:
ans -= 1
print(2*ans%mod)
``` | instruction | 0 | 71,436 | 7 | 142,872 |
No | output | 1 | 71,436 | 7 | 142,873 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider a circle whose perimeter is divided by N points into N arcs of equal length, and each of the arcs is painted red or blue. Such a circle is said to generate a string S from every point when the following condition is satisfied:
* We will arbitrarily choose one of the N points on the perimeter and place a piece on it.
* Then, we will perform the following move M times: move the piece clockwise or counter-clockwise to an adjacent point.
* Here, whatever point we choose initially, it is always possible to move the piece so that the color of the i-th arc the piece goes along is S_i, by properly deciding the directions of the moves.
Assume that, if S_i is `R`, it represents red; if S_i is `B`, it represents blue. Note that the directions of the moves can be decided separately for each choice of the initial point.
You are given a string S of length M consisting of `R` and `B`. Out of the 2^N ways to paint each of the arcs red or blue in a circle whose perimeter is divided into N arcs of equal length, find the number of ways resulting in a circle that generates S from every point, modulo 10^9+7.
Note that the rotations of the same coloring are also distinguished.
Constraints
* 2 \leq N \leq 2 \times 10^5
* 1 \leq M \leq 2 \times 10^5
* |S|=M
* S_i is `R` or `B`.
Input
Input is given from Standard Input in the following format:
N M
S
Output
Print the number of ways to paint each of the arcs that satisfy the condition, modulo 10^9+7.
Examples
Input
4 7
RBRRBRR
Output
2
Input
3 3
BBB
Output
4
Input
12 10
RRRRBRRRRB
Output
78
Submitted Solution:
```
n,m = map(int,input().split())
s = input().rstrip()
mod = 10**9+7
if s.count("R") == 0 or s.count("B") == 0:
dp = [[0 for i in range(2)] for j in range(n)]
dp[0][0] = 1
dp[0][1] = 1
for i in range(1,n):
dp[i][0] = (dp[i-1][0]+dp[i-1][1])%mod
dp[i][1] = dp[i-1][0]
print((dp[n-1][0]+dp[n-2][1])%mod)
exit()
if n%2:
print(0)
exit()
x = s[0]
flg = 1
cnt = 0
ncnt = 0
for i in range(m):
if s[i] == x:
cnt += 1
elif flg == 1:
flg = 0
ncnt = cnt
cnt = 0
else:
if cnt%2:
flg = -1
cnt = 0
if flg == -1:
print(2)
exit()
y = ncnt//2
dp = [[0 for i in range(2)] for j in range(n//2)]
dp[0][0] = 1
dp[0][1] = 1
for i in range(1,n//2):
dp[i][0] = dp[i-1][0]+dp[i-1][1]
if i<y:
dp[i][1] = dp[i-1][0]+dp[i-1][1]
elif i == y:
dp[i][1] = sum([dp[j][0] for j in range(y)])
else:
dp[i][1] = dp[i-1][1]+dp[i-1][0]-dp[i-y-1][0]
dp[i][0] %= mod
dp[i][1] %= mod
ans = dp[-1][0]+dp[-1][1]
for j in range(1,y+1):
if n//2-(y+j)-2 >= 0:
ans = (ans-(dp[n//2-(y+j)-2][0])*(y+1-j))%mod
print(2*ans%mod)
``` | instruction | 0 | 71,437 | 7 | 142,874 |
No | output | 1 | 71,437 | 7 | 142,875 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider a circle whose perimeter is divided by N points into N arcs of equal length, and each of the arcs is painted red or blue. Such a circle is said to generate a string S from every point when the following condition is satisfied:
* We will arbitrarily choose one of the N points on the perimeter and place a piece on it.
* Then, we will perform the following move M times: move the piece clockwise or counter-clockwise to an adjacent point.
* Here, whatever point we choose initially, it is always possible to move the piece so that the color of the i-th arc the piece goes along is S_i, by properly deciding the directions of the moves.
Assume that, if S_i is `R`, it represents red; if S_i is `B`, it represents blue. Note that the directions of the moves can be decided separately for each choice of the initial point.
You are given a string S of length M consisting of `R` and `B`. Out of the 2^N ways to paint each of the arcs red or blue in a circle whose perimeter is divided into N arcs of equal length, find the number of ways resulting in a circle that generates S from every point, modulo 10^9+7.
Note that the rotations of the same coloring are also distinguished.
Constraints
* 2 \leq N \leq 2 \times 10^5
* 1 \leq M \leq 2 \times 10^5
* |S|=M
* S_i is `R` or `B`.
Input
Input is given from Standard Input in the following format:
N M
S
Output
Print the number of ways to paint each of the arcs that satisfy the condition, modulo 10^9+7.
Examples
Input
4 7
RBRRBRR
Output
2
Input
3 3
BBB
Output
4
Input
12 10
RRRRBRRRRB
Output
78
Submitted Solution:
```
"""
https://atcoder.jp/contests/agc033/tasks/agc033_e
Sの1文字目をRとしてよい
RB から始まる場合 → 全部交互以外無理(Nが奇数なら0)
RRB から始まる場合 → Rは3連で置けば可能…
R*X + B*Y + R*Z …
とつながっていく
どこからスタートしても、RだけをX回移動したときにBの隣に来なくてはいけない
Rの長さが1なら可能
R*?,B,R*?,…
でつながっていく
最初のRがX個連続の時
片方の端との距離がX-2tでなくてはならない
Xが偶数の時、Rの連続長さはX+1以下の奇数
Xが奇数の時、端からスタートした奴は反対側に抜けなくてはいけないので
Rの連続長さはX以下の奇数？
結局は奇数個連続でしか置けない
●整理(Rから始まり、Bがある場合)：
Rは奇数個連続でしか置けない
Bは1個連続でしか置けない
→つまり、Nが奇数だとアウト
Bが来た以降を考えてみる
奇数だと、各点Rの左右のどちらかにしか抜けられない→対称的な移動になるはず
→つまり、最初のRを消化した後、全てのBの両端にいる場合が存在する
Bは奇数個の時のみ意味がある
→結局、Bの両端にいた場合が交換されるだけ
→よって、BはRの区切り以上の意味はない
Rが偶数個来るた場合、交互に移動すればおk
Rが奇数個来た場合、反対側に抜ける必要がある
結論：
RBがどちらも存在する場合、Nが偶数なら0、奇数の場合は
BでsplitしたRの個数の集合を考え,
r1,r2…,r? とする
Rは奇数個連続で置けて、その個数の最大は M = min( r1+1 , r? ) #r?が奇数の物
後は、M+1個以下の偶数個単位で区切る場合の数を調べればいい
dp→あらかじめ1番目の区間の位置に、ずれた分も考慮して値を入れておけばBITでもらうdpできる
RB片方の場合、Bが連続しないような置き方をすればいい
1番目がRの場合をdp → N番目までやる
1番目がBの場合をdp → N-1番目までやる(N番目は赤なので)
最後のRは意味がないので消しておく
REの原因究明したい
K < NNの場合？
"""
from sys import stdin
import sys
#0-indexed , 半開区間[a,b)
#calc変更で演算変更
class SegTree:
def __init__(self,N,first):
self.NO = 2**(N-1).bit_length()
self.First = first
self.data = [first] * (2*self.NO)
def calc(self,l,r):
return (l+r) % mod
def update(self,ind,x):
ind += self.NO - 1
self.data[ind] = x
while ind >= 0:
ind = (ind - 1)//2
self.data[ind] = self.calc(self.data[2*ind+1],self.data[2*ind+2])
def query(self,l,r):
L = l + self.NO
R = r + self.NO
s = self.First
while L < R:
if R & 1:
R -= 1
s = self.calc(s , self.data[R-1])
if L & 1:
s = self.calc(s , self.data[L-1])
L += 1
L >>= 1
R >>= 1
return s
def get(self , ind):
ind += self.NO - 1
return self.data[ind]
N,M = map(int,stdin.readline().split())
S = list(stdin.readline()[:-1])
mod = 10**9+7
if "R" in S and "B" in S:
if N % 2 == 1:
print (0)
sys.exit()
lis = []
now = 0
while S[-1] == S[0]: #最後のRを消す
del S[-1]
for i in S:
if i == S[0]:
now += 1
elif now > 0:
lis.append(now)
now = 0
#print (lis)
nmax = 0
for i in range(len(lis)):
if i == 0:
if lis[i] % 2 == 0:
nmax = lis[i] + 1
else:
nmax = lis[i]
elif lis[i] % 2 == 1:
nmax = min(nmax , lis[i])
NN = N//2
K = (nmax+1)//2
#print (NN,K)
if NN <= K:
print (pow(2,NN,mod))
sys.exit()
ST = SegTree(NN,0)
for i in range(K):
ST.update(i,i+1)
for i in range(1,NN):
now = ST.get(i)
plus = ST.query( max(0,i-K),i )
ST.update( i , (now+plus) % mod )
print (ST.get(NN-1) * 2 % mod)
else:
ans = 0
#1番目がRの場合
dp = [1,0]
for i in range(N-1):
ndp = [0,0]
ndp[0] = (dp[0] + dp[1]) % mod
ndp[1] = dp[0]
dp = ndp
ans += sum(dp)
#1番目がBの場合
dp = [0,1]
for i in range(N-2):
ndp = [0,0]
ndp[0] = (dp[0] + dp[1]) % mod
ndp[1] = dp[0]
dp = ndp
ans += sum(dp)
print (ans % mod)
``` | instruction | 0 | 71,438 | 7 | 142,876 |
No | output | 1 | 71,438 | 7 | 142,877 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A Christmas party in city S. had n children. All children came in mittens. The mittens can be of different colors, but each child had the left and the right mitten of the same color. Let's say that the colors of the mittens are numbered with integers from 1 to m, and the children are numbered from 1 to n. Then the i-th child has both mittens of color ci.
The Party had Santa Claus ('Father Frost' in Russian), his granddaughter Snow Girl, the children danced around the richly decorated Christmas tree. In fact, everything was so bright and diverse that the children wanted to wear mittens of distinct colors. The children decided to swap the mittens so that each of them got one left and one right mitten in the end, and these two mittens were of distinct colors. All mittens are of the same size and fit all the children.
The children started exchanging the mittens haphazardly, but they couldn't reach the situation when each child has a pair of mittens of distinct colors. Vasily Petrov, the dad of one of the children, noted that in the general case the children's idea may turn out impossible. Besides, he is a mathematician and he came up with such scheme of distributing mittens that the number of children that have distinct-colored mittens was maximum. You task is to repeat his discovery. Note that the left and right mittens are different: each child must end up with one left and one right mitten.
Input
The first line contains two integers n and m — the number of the children and the number of possible mitten colors (1 ≤ n ≤ 5000, 1 ≤ m ≤ 100). The second line contains n integers c1, c2, ... cn, where ci is the color of the mittens of the i-th child (1 ≤ ci ≤ m).
Output
In the first line, print the maximum number of children who can end up with a distinct-colored pair of mittens. In the next n lines print the way the mittens can be distributed in this case. On the i-th of these lines print two space-separated integers: the color of the left and the color of the right mitten the i-th child will get. If there are multiple solutions, you can print any of them.
Examples
Input
6 3
1 3 2 2 1 1
Output
6
2 1
1 2
2 1
1 3
1 2
3 1
Input
4 2
1 2 1 1
Output
2
1 2
1 1
2 1
1 1 | instruction | 0 | 72,003 | 7 | 144,006 |
Tags: constructive algorithms, greedy, sortings
Correct Solution:
```
from collections import Counter
def main():
n, m = map(int, input().split())
c = list(map(int, input().split()))
sc = sorted(range(n), key=lambda x: c[x])
mc = Counter(c).most_common(1)[0][1]
print(n if mc <= n - mc else 2*(n - mc))
for i in range(n):
print(c[sc[i]], c[sc[i-mc]])
if __name__ == '__main__':
main()
# Made By Mostafa_Khaled
``` | output | 1 | 72,003 | 7 | 144,007 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A Christmas party in city S. had n children. All children came in mittens. The mittens can be of different colors, but each child had the left and the right mitten of the same color. Let's say that the colors of the mittens are numbered with integers from 1 to m, and the children are numbered from 1 to n. Then the i-th child has both mittens of color ci.
The Party had Santa Claus ('Father Frost' in Russian), his granddaughter Snow Girl, the children danced around the richly decorated Christmas tree. In fact, everything was so bright and diverse that the children wanted to wear mittens of distinct colors. The children decided to swap the mittens so that each of them got one left and one right mitten in the end, and these two mittens were of distinct colors. All mittens are of the same size and fit all the children.
The children started exchanging the mittens haphazardly, but they couldn't reach the situation when each child has a pair of mittens of distinct colors. Vasily Petrov, the dad of one of the children, noted that in the general case the children's idea may turn out impossible. Besides, he is a mathematician and he came up with such scheme of distributing mittens that the number of children that have distinct-colored mittens was maximum. You task is to repeat his discovery. Note that the left and right mittens are different: each child must end up with one left and one right mitten.
Input
The first line contains two integers n and m — the number of the children and the number of possible mitten colors (1 ≤ n ≤ 5000, 1 ≤ m ≤ 100). The second line contains n integers c1, c2, ... cn, where ci is the color of the mittens of the i-th child (1 ≤ ci ≤ m).
Output
In the first line, print the maximum number of children who can end up with a distinct-colored pair of mittens. In the next n lines print the way the mittens can be distributed in this case. On the i-th of these lines print two space-separated integers: the color of the left and the color of the right mitten the i-th child will get. If there are multiple solutions, you can print any of them.
Examples
Input
6 3
1 3 2 2 1 1
Output
6
2 1
1 2
2 1
1 3
1 2
3 1
Input
4 2
1 2 1 1
Output
2
1 2
1 1
2 1
1 1 | instruction | 0 | 72,004 | 7 | 144,008 |
Tags: constructive algorithms, greedy, sortings
Correct Solution:
```
n, m = [int(i) for i in input().split()]
c1 = [int(i) for i in input().split()]
c1.sort()
ms = 0
i = 0
while i < n - 1:
start = i
while i < n - 1 and c1[i] == c1[i + 1]:
i += 1
ms = max(ms, i - start)
i += 1
ms += 1
c2 = c1[-ms:] + c1[:-ms]
cnt = 0
for i in range(n):
if c1[i] != c2[i]:
cnt += 1
print(cnt)
for i in range(n):
print(c1[i], c2[i])
``` | output | 1 | 72,004 | 7 | 144,009 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A Christmas party in city S. had n children. All children came in mittens. The mittens can be of different colors, but each child had the left and the right mitten of the same color. Let's say that the colors of the mittens are numbered with integers from 1 to m, and the children are numbered from 1 to n. Then the i-th child has both mittens of color ci.
The Party had Santa Claus ('Father Frost' in Russian), his granddaughter Snow Girl, the children danced around the richly decorated Christmas tree. In fact, everything was so bright and diverse that the children wanted to wear mittens of distinct colors. The children decided to swap the mittens so that each of them got one left and one right mitten in the end, and these two mittens were of distinct colors. All mittens are of the same size and fit all the children.
The children started exchanging the mittens haphazardly, but they couldn't reach the situation when each child has a pair of mittens of distinct colors. Vasily Petrov, the dad of one of the children, noted that in the general case the children's idea may turn out impossible. Besides, he is a mathematician and he came up with such scheme of distributing mittens that the number of children that have distinct-colored mittens was maximum. You task is to repeat his discovery. Note that the left and right mittens are different: each child must end up with one left and one right mitten.
Input
The first line contains two integers n and m — the number of the children and the number of possible mitten colors (1 ≤ n ≤ 5000, 1 ≤ m ≤ 100). The second line contains n integers c1, c2, ... cn, where ci is the color of the mittens of the i-th child (1 ≤ ci ≤ m).
Output
In the first line, print the maximum number of children who can end up with a distinct-colored pair of mittens. In the next n lines print the way the mittens can be distributed in this case. On the i-th of these lines print two space-separated integers: the color of the left and the color of the right mitten the i-th child will get. If there are multiple solutions, you can print any of them.
Examples
Input
6 3
1 3 2 2 1 1
Output
6
2 1
1 2
2 1
1 3
1 2
3 1
Input
4 2
1 2 1 1
Output
2
1 2
1 1
2 1
1 1 | instruction | 0 | 72,005 | 7 | 144,010 |
Tags: constructive algorithms, greedy, sortings
Correct Solution:
```
import math
import sys
import collections
def In():
return map(int, sys.stdin.readline().split())
def mittens():
n, m = In()
l = list(collections.Counter(map(int, input().split())).items())
for i in range(len(l)):
l[i] = [l[i][0], l[i][1] * 2]
l.sort(key=lambda x: x[1], reverse=True)
m = len(l)
ans = []
while m > 1:
for i in range(len(l)):
for i1 in range(len(l) - 1, i, -1):
if l[i][1] and l[i1][1]:
l[i][1] -= 2
if not l[i][1]:
m -= 1
l[i1][1] -= 2
if not l[i1][1]:
m -= 1
ans.append([l[i][0], l[i1][0]])
ans.append([l[i1][0], l[i][0]])
print(len(ans))
for i in ans:
print(*i)
if m:
for i in l:
while i[1]:
print(i[0], i[0])
i[1] -= 2
# mittens()
def mittens2():
n, m = In()
l = list(map(int, input().split()))
ml = [0] * (m + 1)
mx = 0
for i in l:
ml[i] += 1
mx = max(mx, ml[i])
for i in range(n):
l[i] = [l[i], ml[l[i]]]
l.sort(key=lambda x: (x[1], x[0]),reverse=True)
pos = mx
ans = []
n1 = 0
for i in range(n):
num,num1 = l[pos%n][0],l[pos-mx][0]
if num != num1:
n1 += 1
ans.append([num,num1])
pos += 1
print(n1)
for i in ans:
print(*i)
mittens2()
``` | output | 1 | 72,005 | 7 | 144,011 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A Christmas party in city S. had n children. All children came in mittens. The mittens can be of different colors, but each child had the left and the right mitten of the same color. Let's say that the colors of the mittens are numbered with integers from 1 to m, and the children are numbered from 1 to n. Then the i-th child has both mittens of color ci.
The Party had Santa Claus ('Father Frost' in Russian), his granddaughter Snow Girl, the children danced around the richly decorated Christmas tree. In fact, everything was so bright and diverse that the children wanted to wear mittens of distinct colors. The children decided to swap the mittens so that each of them got one left and one right mitten in the end, and these two mittens were of distinct colors. All mittens are of the same size and fit all the children.
The children started exchanging the mittens haphazardly, but they couldn't reach the situation when each child has a pair of mittens of distinct colors. Vasily Petrov, the dad of one of the children, noted that in the general case the children's idea may turn out impossible. Besides, he is a mathematician and he came up with such scheme of distributing mittens that the number of children that have distinct-colored mittens was maximum. You task is to repeat his discovery. Note that the left and right mittens are different: each child must end up with one left and one right mitten.
Input
The first line contains two integers n and m — the number of the children and the number of possible mitten colors (1 ≤ n ≤ 5000, 1 ≤ m ≤ 100). The second line contains n integers c1, c2, ... cn, where ci is the color of the mittens of the i-th child (1 ≤ ci ≤ m).
Output
In the first line, print the maximum number of children who can end up with a distinct-colored pair of mittens. In the next n lines print the way the mittens can be distributed in this case. On the i-th of these lines print two space-separated integers: the color of the left and the color of the right mitten the i-th child will get. If there are multiple solutions, you can print any of them.
Examples
Input
6 3
1 3 2 2 1 1
Output
6
2 1
1 2
2 1
1 3
1 2
3 1
Input
4 2
1 2 1 1
Output
2
1 2
1 1
2 1
1 1 | instruction | 0 | 72,006 | 7 | 144,012 |
Tags: constructive algorithms, greedy, sortings
Correct Solution:
```
#!/usr/bin/python3
def readln(): return tuple(map(int, input().split()))
n, m = readln()
cnt = [0] * (m + 1)
for c in readln():
cnt[c] += 1
ans = [0] * n
j = 0
for _ in range(1, m + 1):
v = max(cnt)
i = cnt.index(v)
while cnt[i]:
ans[j] = i
cnt[i] -= 1
j += 2
if j >= n:
j = 1
print(len([1 for i in range(n) if ans[i] != ans[(i + 1) % n]]))
for i in range(n):
print(ans[i], ans[(i + 1) % n])
``` | output | 1 | 72,006 | 7 | 144,013 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A Christmas party in city S. had n children. All children came in mittens. The mittens can be of different colors, but each child had the left and the right mitten of the same color. Let's say that the colors of the mittens are numbered with integers from 1 to m, and the children are numbered from 1 to n. Then the i-th child has both mittens of color ci.
The Party had Santa Claus ('Father Frost' in Russian), his granddaughter Snow Girl, the children danced around the richly decorated Christmas tree. In fact, everything was so bright and diverse that the children wanted to wear mittens of distinct colors. The children decided to swap the mittens so that each of them got one left and one right mitten in the end, and these two mittens were of distinct colors. All mittens are of the same size and fit all the children.
The children started exchanging the mittens haphazardly, but they couldn't reach the situation when each child has a pair of mittens of distinct colors. Vasily Petrov, the dad of one of the children, noted that in the general case the children's idea may turn out impossible. Besides, he is a mathematician and he came up with such scheme of distributing mittens that the number of children that have distinct-colored mittens was maximum. You task is to repeat his discovery. Note that the left and right mittens are different: each child must end up with one left and one right mitten.
Input
The first line contains two integers n and m — the number of the children and the number of possible mitten colors (1 ≤ n ≤ 5000, 1 ≤ m ≤ 100). The second line contains n integers c1, c2, ... cn, where ci is the color of the mittens of the i-th child (1 ≤ ci ≤ m).
Output
In the first line, print the maximum number of children who can end up with a distinct-colored pair of mittens. In the next n lines print the way the mittens can be distributed in this case. On the i-th of these lines print two space-separated integers: the color of the left and the color of the right mitten the i-th child will get. If there are multiple solutions, you can print any of them.
Examples
Input
6 3
1 3 2 2 1 1
Output
6
2 1
1 2
2 1
1 3
1 2
3 1
Input
4 2
1 2 1 1
Output
2
1 2
1 1
2 1
1 1 | instruction | 0 | 72,007 | 7 | 144,014 |
Tags: constructive algorithms, greedy, sortings
Correct Solution:
```
def takefirst(eml):
return eml[0]
n,m=map(int,input().split())
a=list(map(int,input().split()))
a.sort()
b=[]
a.append(0)
max1=0
c=0
for i in range(len(a)-1):
if a[i]!=a[i+1]:
c+=1
max1=max(max1,c)
c=0
else:
c+=1
a.pop()
ans=[]
total=0
for i in range(len(a)):
ans.append((a[i],a[(i+max1)%len(a)]))
if a[i]!=a[(i+max1)%len(a)]:
total+=1
print(total)
for i in range(len(ans)):
print('{0} {1}'.format(ans[i][0],ans[i][1]))
``` | output | 1 | 72,007 | 7 | 144,015 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A Christmas party in city S. had n children. All children came in mittens. The mittens can be of different colors, but each child had the left and the right mitten of the same color. Let's say that the colors of the mittens are numbered with integers from 1 to m, and the children are numbered from 1 to n. Then the i-th child has both mittens of color ci.
The Party had Santa Claus ('Father Frost' in Russian), his granddaughter Snow Girl, the children danced around the richly decorated Christmas tree. In fact, everything was so bright and diverse that the children wanted to wear mittens of distinct colors. The children decided to swap the mittens so that each of them got one left and one right mitten in the end, and these two mittens were of distinct colors. All mittens are of the same size and fit all the children.
The children started exchanging the mittens haphazardly, but they couldn't reach the situation when each child has a pair of mittens of distinct colors. Vasily Petrov, the dad of one of the children, noted that in the general case the children's idea may turn out impossible. Besides, he is a mathematician and he came up with such scheme of distributing mittens that the number of children that have distinct-colored mittens was maximum. You task is to repeat his discovery. Note that the left and right mittens are different: each child must end up with one left and one right mitten.
Input
The first line contains two integers n and m — the number of the children and the number of possible mitten colors (1 ≤ n ≤ 5000, 1 ≤ m ≤ 100). The second line contains n integers c1, c2, ... cn, where ci is the color of the mittens of the i-th child (1 ≤ ci ≤ m).
Output
In the first line, print the maximum number of children who can end up with a distinct-colored pair of mittens. In the next n lines print the way the mittens can be distributed in this case. On the i-th of these lines print two space-separated integers: the color of the left and the color of the right mitten the i-th child will get. If there are multiple solutions, you can print any of them.
Examples
Input
6 3
1 3 2 2 1 1
Output
6
2 1
1 2
2 1
1 3
1 2
3 1
Input
4 2
1 2 1 1
Output
2
1 2
1 1
2 1
1 1 | instruction | 0 | 72,008 | 7 | 144,016 |
Tags: constructive algorithms, greedy, sortings
Correct Solution:
```
n, m = map(int, input().split())
arr = sorted(list(map(int, input().split())))
mit, col = 1, 1
for i in range(1, m+1):
t = arr.count(i)
if t > col:
mit = i
col = t
print(n - max(0, 2*col - n))
for i in range(n):
print(arr[(i+col) % n], arr[i])
``` | output | 1 | 72,008 | 7 | 144,017 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A Christmas party in city S. had n children. All children came in mittens. The mittens can be of different colors, but each child had the left and the right mitten of the same color. Let's say that the colors of the mittens are numbered with integers from 1 to m, and the children are numbered from 1 to n. Then the i-th child has both mittens of color ci.
The Party had Santa Claus ('Father Frost' in Russian), his granddaughter Snow Girl, the children danced around the richly decorated Christmas tree. In fact, everything was so bright and diverse that the children wanted to wear mittens of distinct colors. The children decided to swap the mittens so that each of them got one left and one right mitten in the end, and these two mittens were of distinct colors. All mittens are of the same size and fit all the children.
The children started exchanging the mittens haphazardly, but they couldn't reach the situation when each child has a pair of mittens of distinct colors. Vasily Petrov, the dad of one of the children, noted that in the general case the children's idea may turn out impossible. Besides, he is a mathematician and he came up with such scheme of distributing mittens that the number of children that have distinct-colored mittens was maximum. You task is to repeat his discovery. Note that the left and right mittens are different: each child must end up with one left and one right mitten.
Input
The first line contains two integers n and m — the number of the children and the number of possible mitten colors (1 ≤ n ≤ 5000, 1 ≤ m ≤ 100). The second line contains n integers c1, c2, ... cn, where ci is the color of the mittens of the i-th child (1 ≤ ci ≤ m).
Output
In the first line, print the maximum number of children who can end up with a distinct-colored pair of mittens. In the next n lines print the way the mittens can be distributed in this case. On the i-th of these lines print two space-separated integers: the color of the left and the color of the right mitten the i-th child will get. If there are multiple solutions, you can print any of them.
Examples
Input
6 3
1 3 2 2 1 1
Output
6
2 1
1 2
2 1
1 3
1 2
3 1
Input
4 2
1 2 1 1
Output
2
1 2
1 1
2 1
1 1 | instruction | 0 | 72,009 | 7 | 144,018 |
Tags: constructive algorithms, greedy, sortings
Correct Solution:
```
n, m = map(int, input().split())
p = [0] * (m + 1)
for i in map(int, input().split()): p[i] += 1
r, q = [], sorted((p[i], i) for i in range(1, m + 1))
k = q[m - 1][0]
print(n - max(2 * k - n, 0))
for i in q: r.extend([str(i[1])] * i[0])
print('\n'.join(i + ' ' + j for i, j in zip(*(r, r[k: ] + r[: k]))))
``` | output | 1 | 72,009 | 7 | 144,019 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A Christmas party in city S. had n children. All children came in mittens. The mittens can be of different colors, but each child had the left and the right mitten of the same color. Let's say that the colors of the mittens are numbered with integers from 1 to m, and the children are numbered from 1 to n. Then the i-th child has both mittens of color ci.
The Party had Santa Claus ('Father Frost' in Russian), his granddaughter Snow Girl, the children danced around the richly decorated Christmas tree. In fact, everything was so bright and diverse that the children wanted to wear mittens of distinct colors. The children decided to swap the mittens so that each of them got one left and one right mitten in the end, and these two mittens were of distinct colors. All mittens are of the same size and fit all the children.
The children started exchanging the mittens haphazardly, but they couldn't reach the situation when each child has a pair of mittens of distinct colors. Vasily Petrov, the dad of one of the children, noted that in the general case the children's idea may turn out impossible. Besides, he is a mathematician and he came up with such scheme of distributing mittens that the number of children that have distinct-colored mittens was maximum. You task is to repeat his discovery. Note that the left and right mittens are different: each child must end up with one left and one right mitten.
Input
The first line contains two integers n and m — the number of the children and the number of possible mitten colors (1 ≤ n ≤ 5000, 1 ≤ m ≤ 100). The second line contains n integers c1, c2, ... cn, where ci is the color of the mittens of the i-th child (1 ≤ ci ≤ m).
Output
In the first line, print the maximum number of children who can end up with a distinct-colored pair of mittens. In the next n lines print the way the mittens can be distributed in this case. On the i-th of these lines print two space-separated integers: the color of the left and the color of the right mitten the i-th child will get. If there are multiple solutions, you can print any of them.
Examples
Input
6 3
1 3 2 2 1 1
Output
6
2 1
1 2
2 1
1 3
1 2
3 1
Input
4 2
1 2 1 1
Output
2
1 2
1 1
2 1
1 1 | instruction | 0 | 72,010 | 7 | 144,020 |
Tags: constructive algorithms, greedy, sortings
Correct Solution:
```
n, m = map(int, input().split())
arr = list(map(int, input().split()))
arr.sort()
mit = 1
col = 1
for i in range(1, m+1):
t =arr.count(i)
if t > col:
mit = i
col = t
print(n-max(0, col-(n-col)))
for i in range(n):
print(arr[(i+col) % n], arr[i])
``` | output | 1 | 72,010 | 7 | 144,021 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A Christmas party in city S. had n children. All children came in mittens. The mittens can be of different colors, but each child had the left and the right mitten of the same color. Let's say that the colors of the mittens are numbered with integers from 1 to m, and the children are numbered from 1 to n. Then the i-th child has both mittens of color ci.
The Party had Santa Claus ('Father Frost' in Russian), his granddaughter Snow Girl, the children danced around the richly decorated Christmas tree. In fact, everything was so bright and diverse that the children wanted to wear mittens of distinct colors. The children decided to swap the mittens so that each of them got one left and one right mitten in the end, and these two mittens were of distinct colors. All mittens are of the same size and fit all the children.
The children started exchanging the mittens haphazardly, but they couldn't reach the situation when each child has a pair of mittens of distinct colors. Vasily Petrov, the dad of one of the children, noted that in the general case the children's idea may turn out impossible. Besides, he is a mathematician and he came up with such scheme of distributing mittens that the number of children that have distinct-colored mittens was maximum. You task is to repeat his discovery. Note that the left and right mittens are different: each child must end up with one left and one right mitten.
Input
The first line contains two integers n and m — the number of the children and the number of possible mitten colors (1 ≤ n ≤ 5000, 1 ≤ m ≤ 100). The second line contains n integers c1, c2, ... cn, where ci is the color of the mittens of the i-th child (1 ≤ ci ≤ m).
Output
In the first line, print the maximum number of children who can end up with a distinct-colored pair of mittens. In the next n lines print the way the mittens can be distributed in this case. On the i-th of these lines print two space-separated integers: the color of the left and the color of the right mitten the i-th child will get. If there are multiple solutions, you can print any of them.
Examples
Input
6 3
1 3 2 2 1 1
Output
6
2 1
1 2
2 1
1 3
1 2
3 1
Input
4 2
1 2 1 1
Output
2
1 2
1 1
2 1
1 1
Submitted Solution:
```
n, m = map(int, input().split())
arr = sorted(list(map(int, input().split())))
max_cnt = 1
for i in range(1, m+1):
max_cnt = max(max_cnt, arr.count(i))
print(n - max(0, 2*max_cnt - n))
for i in range(n):
print(arr[(i+max_cnt) % n], arr[i])
``` | instruction | 0 | 72,011 | 7 | 144,022 |
Yes | output | 1 | 72,011 | 7 | 144,023 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A Christmas party in city S. had n children. All children came in mittens. The mittens can be of different colors, but each child had the left and the right mitten of the same color. Let's say that the colors of the mittens are numbered with integers from 1 to m, and the children are numbered from 1 to n. Then the i-th child has both mittens of color ci.
The Party had Santa Claus ('Father Frost' in Russian), his granddaughter Snow Girl, the children danced around the richly decorated Christmas tree. In fact, everything was so bright and diverse that the children wanted to wear mittens of distinct colors. The children decided to swap the mittens so that each of them got one left and one right mitten in the end, and these two mittens were of distinct colors. All mittens are of the same size and fit all the children.
The children started exchanging the mittens haphazardly, but they couldn't reach the situation when each child has a pair of mittens of distinct colors. Vasily Petrov, the dad of one of the children, noted that in the general case the children's idea may turn out impossible. Besides, he is a mathematician and he came up with such scheme of distributing mittens that the number of children that have distinct-colored mittens was maximum. You task is to repeat his discovery. Note that the left and right mittens are different: each child must end up with one left and one right mitten.
Input
The first line contains two integers n and m — the number of the children and the number of possible mitten colors (1 ≤ n ≤ 5000, 1 ≤ m ≤ 100). The second line contains n integers c1, c2, ... cn, where ci is the color of the mittens of the i-th child (1 ≤ ci ≤ m).
Output
In the first line, print the maximum number of children who can end up with a distinct-colored pair of mittens. In the next n lines print the way the mittens can be distributed in this case. On the i-th of these lines print two space-separated integers: the color of the left and the color of the right mitten the i-th child will get. If there are multiple solutions, you can print any of them.
Examples
Input
6 3
1 3 2 2 1 1
Output
6
2 1
1 2
2 1
1 3
1 2
3 1
Input
4 2
1 2 1 1
Output
2
1 2
1 1
2 1
1 1
Submitted Solution:
```
from collections import Counter
def main():
n, m = map(int, input().split())
c = list(map(int, input().split()))
sc = sorted(range(n), key=lambda x: c[x])
mc = Counter(c).most_common(1)[0][1]
print(n if mc <= n - mc else 2*(n - mc))
for i in range(n):
print(c[sc[i]], c[sc[i-mc]])
if __name__ == '__main__':
main()
``` | instruction | 0 | 72,012 | 7 | 144,024 |
Yes | output | 1 | 72,012 | 7 | 144,025 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A Christmas party in city S. had n children. All children came in mittens. The mittens can be of different colors, but each child had the left and the right mitten of the same color. Let's say that the colors of the mittens are numbered with integers from 1 to m, and the children are numbered from 1 to n. Then the i-th child has both mittens of color ci.
The Party had Santa Claus ('Father Frost' in Russian), his granddaughter Snow Girl, the children danced around the richly decorated Christmas tree. In fact, everything was so bright and diverse that the children wanted to wear mittens of distinct colors. The children decided to swap the mittens so that each of them got one left and one right mitten in the end, and these two mittens were of distinct colors. All mittens are of the same size and fit all the children.
The children started exchanging the mittens haphazardly, but they couldn't reach the situation when each child has a pair of mittens of distinct colors. Vasily Petrov, the dad of one of the children, noted that in the general case the children's idea may turn out impossible. Besides, he is a mathematician and he came up with such scheme of distributing mittens that the number of children that have distinct-colored mittens was maximum. You task is to repeat his discovery. Note that the left and right mittens are different: each child must end up with one left and one right mitten.
Input
The first line contains two integers n and m — the number of the children and the number of possible mitten colors (1 ≤ n ≤ 5000, 1 ≤ m ≤ 100). The second line contains n integers c1, c2, ... cn, where ci is the color of the mittens of the i-th child (1 ≤ ci ≤ m).
Output
In the first line, print the maximum number of children who can end up with a distinct-colored pair of mittens. In the next n lines print the way the mittens can be distributed in this case. On the i-th of these lines print two space-separated integers: the color of the left and the color of the right mitten the i-th child will get. If there are multiple solutions, you can print any of them.
Examples
Input
6 3
1 3 2 2 1 1
Output
6
2 1
1 2
2 1
1 3
1 2
3 1
Input
4 2
1 2 1 1
Output
2
1 2
1 1
2 1
1 1
Submitted Solution:
```
line = input().split()
n, nc = int(line[0]), int(line[1])
line = input().split()
sum = 0
colors = []
pairs = []
dif_pairs = 0
for i in range(nc):
colors.append([i + 1 , 0 ])
for i in range(n):
colors[int(line[i]) - 1][1] += 1
pairs.append([0,0])
colorss = sorted(colors, key = lambda t:t[1])
i = 0
while i < nc and sum < n :
sum += colorss[i][1] * 2
if sum <= n:
for j in range(dif_pairs,sum):
pairs[j][0] = colorss[i][0]
dif_pairs = sum
else:
for j in range(dif_pairs,n):
pairs[j][0] = colorss[i][0]
for j in range(sum - n):
pairs[j][1] = colorss[i][0]
if i < nc - 1:
dif_pairs = n
i += 1
k = sum - n
while i < nc:
for j in range(colorss[i][1] * 2):
pairs[k][1] = colorss[i][0]
k += 1
i += 1
print(dif_pairs)
i = 0
while i < n:
k = colors[pairs[i][0] - 1][1]
for j in range(k) :
if i < n:
pairs[i][0], pairs[i][1]= pairs[i][1], pairs[i][0]
else:
break
i += 1
for j in range(k):
i +=1
for p in pairs:
print(p[0],p[1])
``` | instruction | 0 | 72,013 | 7 | 144,026 |
No | output | 1 | 72,013 | 7 | 144,027 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A Christmas party in city S. had n children. All children came in mittens. The mittens can be of different colors, but each child had the left and the right mitten of the same color. Let's say that the colors of the mittens are numbered with integers from 1 to m, and the children are numbered from 1 to n. Then the i-th child has both mittens of color ci.
The Party had Santa Claus ('Father Frost' in Russian), his granddaughter Snow Girl, the children danced around the richly decorated Christmas tree. In fact, everything was so bright and diverse that the children wanted to wear mittens of distinct colors. The children decided to swap the mittens so that each of them got one left and one right mitten in the end, and these two mittens were of distinct colors. All mittens are of the same size and fit all the children.
The children started exchanging the mittens haphazardly, but they couldn't reach the situation when each child has a pair of mittens of distinct colors. Vasily Petrov, the dad of one of the children, noted that in the general case the children's idea may turn out impossible. Besides, he is a mathematician and he came up with such scheme of distributing mittens that the number of children that have distinct-colored mittens was maximum. You task is to repeat his discovery. Note that the left and right mittens are different: each child must end up with one left and one right mitten.
Input
The first line contains two integers n and m — the number of the children and the number of possible mitten colors (1 ≤ n ≤ 5000, 1 ≤ m ≤ 100). The second line contains n integers c1, c2, ... cn, where ci is the color of the mittens of the i-th child (1 ≤ ci ≤ m).
Output
In the first line, print the maximum number of children who can end up with a distinct-colored pair of mittens. In the next n lines print the way the mittens can be distributed in this case. On the i-th of these lines print two space-separated integers: the color of the left and the color of the right mitten the i-th child will get. If there are multiple solutions, you can print any of them.
Examples
Input
6 3
1 3 2 2 1 1
Output
6
2 1
1 2
2 1
1 3
1 2
3 1
Input
4 2
1 2 1 1
Output
2
1 2
1 1
2 1
1 1
Submitted Solution:
```
def takefirst(eml):
return eml[0]
n,m=map(int,input().split())
a=list(map(int,input().split()))
a.sort()
b=[]
c=1
for i in range(1,len(a)):
if a[i]!=a[i-1]:
tmp=[c,a[i-1]]
b.append(tmp)
c=1
if i==len(a)-1:
cmp=[c,a[i]]
b.append(cmp)
elif i==len(a)-1:
tmp=[c,a[i]]
b.append(tmp)
else:
c+=1
b.sort(key=takefirst)
ans=[]
total=0
while len(b)>1:
for i in range(b[len(b)-2][0]):
ans.append((b[len(b)-2][1],b[len(b)-1][1]))
ans.append((b[len(b) - 1][1], b[len(b) - 2][1]))
total+=b[len(b)-2][0]*2
b[len(b) - 2][0]=b[len(b)-1][0]-b[len(b) - 2][0]
b[len(b) - 2][1]=b[len(b)-1][1]
b.pop()
print(total)
for i in range(len(ans)):
print('{0} {1}'.format(ans[i][0],ans[i][1]))
if b:
for i in range(b[0][0]):
print(('{0} {1}'.format(b[0][1],b[0][1])))
``` | instruction | 0 | 72,014 | 7 | 144,028 |
No | output | 1 | 72,014 | 7 | 144,029 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A Christmas party in city S. had n children. All children came in mittens. The mittens can be of different colors, but each child had the left and the right mitten of the same color. Let's say that the colors of the mittens are numbered with integers from 1 to m, and the children are numbered from 1 to n. Then the i-th child has both mittens of color ci.
The Party had Santa Claus ('Father Frost' in Russian), his granddaughter Snow Girl, the children danced around the richly decorated Christmas tree. In fact, everything was so bright and diverse that the children wanted to wear mittens of distinct colors. The children decided to swap the mittens so that each of them got one left and one right mitten in the end, and these two mittens were of distinct colors. All mittens are of the same size and fit all the children.
The children started exchanging the mittens haphazardly, but they couldn't reach the situation when each child has a pair of mittens of distinct colors. Vasily Petrov, the dad of one of the children, noted that in the general case the children's idea may turn out impossible. Besides, he is a mathematician and he came up with such scheme of distributing mittens that the number of children that have distinct-colored mittens was maximum. You task is to repeat his discovery. Note that the left and right mittens are different: each child must end up with one left and one right mitten.
Input
The first line contains two integers n and m — the number of the children and the number of possible mitten colors (1 ≤ n ≤ 5000, 1 ≤ m ≤ 100). The second line contains n integers c1, c2, ... cn, where ci is the color of the mittens of the i-th child (1 ≤ ci ≤ m).
Output
In the first line, print the maximum number of children who can end up with a distinct-colored pair of mittens. In the next n lines print the way the mittens can be distributed in this case. On the i-th of these lines print two space-separated integers: the color of the left and the color of the right mitten the i-th child will get. If there are multiple solutions, you can print any of them.
Examples
Input
6 3
1 3 2 2 1 1
Output
6
2 1
1 2
2 1
1 3
1 2
3 1
Input
4 2
1 2 1 1
Output
2
1 2
1 1
2 1
1 1
Submitted Solution:
```
import sys
sys.setrecursionlimit(10000000)
def solve():
n,m, = rv()
colors, = rl(1)
count = [0] * m
for val in colors:
count[val - 1] += 2
left = 0
right = 0
lastleft = -1
res = list()
while True:
if count[left] > 0:
lastleft = left
right = left + 1
right %= m
while count[right] == 0:
right += 1
right %= m
if left == right:
break #it is all over
res.append((left + 1, right + 1))
count[left] -= 1
count[right] -= 1
else:
left += 1
left %= m
if left == lastleft: break
# if sum(count) == 0: break
good = len(res)
for i in range(m):
while count[i] > 0:
res.append((i + 1, i + 1))
count[i] -= 2
print(good)
print('\n'.join(' '.join(map(str, row)) for row in res))
def prt(l): return print(' '.join(l))
def rv(): return map(int, input().split())
def rl(n): return [list(map(int, input().split())) for _ in range(n)]
if sys.hexversion == 50594544 : sys.stdin = open("test.txt")
solve()
``` | instruction | 0 | 72,015 | 7 | 144,030 |
No | output | 1 | 72,015 | 7 | 144,031 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A Christmas party in city S. had n children. All children came in mittens. The mittens can be of different colors, but each child had the left and the right mitten of the same color. Let's say that the colors of the mittens are numbered with integers from 1 to m, and the children are numbered from 1 to n. Then the i-th child has both mittens of color ci.
The Party had Santa Claus ('Father Frost' in Russian), his granddaughter Snow Girl, the children danced around the richly decorated Christmas tree. In fact, everything was so bright and diverse that the children wanted to wear mittens of distinct colors. The children decided to swap the mittens so that each of them got one left and one right mitten in the end, and these two mittens were of distinct colors. All mittens are of the same size and fit all the children.
The children started exchanging the mittens haphazardly, but they couldn't reach the situation when each child has a pair of mittens of distinct colors. Vasily Petrov, the dad of one of the children, noted that in the general case the children's idea may turn out impossible. Besides, he is a mathematician and he came up with such scheme of distributing mittens that the number of children that have distinct-colored mittens was maximum. You task is to repeat his discovery. Note that the left and right mittens are different: each child must end up with one left and one right mitten.
Input
The first line contains two integers n and m — the number of the children and the number of possible mitten colors (1 ≤ n ≤ 5000, 1 ≤ m ≤ 100). The second line contains n integers c1, c2, ... cn, where ci is the color of the mittens of the i-th child (1 ≤ ci ≤ m).
Output
In the first line, print the maximum number of children who can end up with a distinct-colored pair of mittens. In the next n lines print the way the mittens can be distributed in this case. On the i-th of these lines print two space-separated integers: the color of the left and the color of the right mitten the i-th child will get. If there are multiple solutions, you can print any of them.
Examples
Input
6 3
1 3 2 2 1 1
Output
6
2 1
1 2
2 1
1 3
1 2
3 1
Input
4 2
1 2 1 1
Output
2
1 2
1 1
2 1
1 1
Submitted Solution:
```
import math
import sys
import collections
def In():
return map(int, sys.stdin.readline().split())
def mittens():
n, m = In()
l = list(collections.Counter(map(int, input().split())).items())
for i in range(len(l)):
l[i] = [l[i][0], l[i][1]*2]
c = 0
ans = []
for i in range(len(l)):
for i1 in range(i+1,len(l)):
if l[i][1] and l[i1][1]:
l[i][1] -= 2
l[i1][1] -= 2
ans.append([l[i][0],l[i1][0]])
ans.append([l[i1][0],l[i][0]])
print(len(ans))
for i in ans:
print(*i)
for i in l:
while i[1]:
print(i[0],i[0])
i[1] -=2
mittens()
``` | instruction | 0 | 72,016 | 7 | 144,032 |
No | output | 1 | 72,016 | 7 | 144,033 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Mobiles are widely known as moving works of art. The IOI Japan Committee has decided to create mobiles to publicize JOI. JOI public relations mobiles are sticks, strings, and weights. It is constructed as follows using the three types of elements of.
* One end of the bar is painted blue and the other end is painted red.
* The rod is hung with a string with one point other than both ends as a fulcrum.
* Both the length from the fulcrum to the red end and the length from the fulcrum to the blue end are positive integers.
* At both ends of the rod, hang a weight or another rod with a string.
* The weight is hung on one end of one of the rods using a string.
* Nothing hangs on the weight.
* The weight of the weight is a positive integer.
* Only one of the strings is tied at one end to the fulcrum of that bar to hang one bar, the other end is not tied to any other component. Meet one or the other.
* Connect the end of a bar to the fulcrum of a bar.
* Connect the end of a rod to a weight.
However, all rods need to be balanced. The weights of the rods and strings are negligibly light, so consider that the weights of the rods and strings are all 0. About the stick,
(Total weight of weights suspended below the red end of the rod) × (Length from the fulcrum of the rod to the red end) = (Hung below the blue end of the rod) Total weight of the weight) × (length from the fulcrum of the rod to the end of the blue)
If, then the bar is balanced.
<image>
The length and how to tie the rods to make up the mobile has already been decided, but the weight of the weight has not been decided yet. The lighter the mobile, the easier it is to hang, so I want to make the mobile as light as possible. As mentioned above, while balancing all the rods, find a way to attach a weight that minimizes the total weight of the mobile, and create a program that outputs the total weight of the mobile at that time. Information about the configuration is given.
* Number of bars n
* Information for each bar (bar numbers 1 to n)
* Ratio of the length from the fulcrum to the red end to the length from the fulcrum to the blue end
* Number of rods to hang on the red end (0 for hanging weights)
* Number of rods to hang on the blue edge (0 for hanging weights)
<image>
input
The input consists of multiple datasets. Each dataset is given in the following format. The input ends on a line containing one zero.
The first line contains the number of bars n used in the mobile. The following n lines (1 ≤ n ≤ 100) contain the data for each bar. I + 1st line In (1 ≤ i ≤ n), four integers p, q, r, b are written with a blank as a delimiter, and on the bar i, the length from the fulcrum to the red end and the length from the fulcrum to the blue end. The ratio is p: q, the number of the bar hanging at the red end is r, and the number of the bar hanging at the blue end is b, where bar number 0 is the weight. In any input, if the minimum value of the weight of the mobile is w and the maximum value of the positive integer used to express the ratio in the input is L, then wL <231 Meet.
The number of datasets does not exceed 15.
output
The weight of the mobile is output to one line for each data set.
Examples
Input
1
6 9 0 0
4
3 2 0 4
1 3 0 0
4 4 2 1
2 2 0 0
0
Output
5
40
Input
None
Output
None | instruction | 0 | 72,432 | 7 | 144,864 |
"Correct Solution:
```
# AOJ0520
import sys
def gcd(x, y):
if x < y:
x, y = y, x
while x % y != 0:
x, y = y, x % y
return y
class frac:
def __init__(self, n, d):
self.num = n
self.den = d
def mul(self, yn, yd):
xn, xd = self.num, self.den
xn *= yn
xd *= yd
if xn != 0 and xd != 0:
g = gcd(xn, xd)
xn /= g
xd /= g
return frac(xn, xd)
def is_zero(self):
return self.num == 0
def output(self):
return "{}/{}".format(self.num, self.den)
class LightestMobile:
def __init__(self, n):
self.n = n + 1
self.mobile = [[0,0,0,0]]
self.weight = [[frac(0,1), frac(0,1)]]
self.denom = []
def read(self, p, q, r, b):
self.mobile.append([p, q, r, b])
self.weight.append([frac(0,1), frac(0,1)])
def root(self):
for i in range(1, self.n):
m = self.mobile[i]
if m[2] == 0 and m[3] == 0:
return self.parent(i)
def parent(self, c):
for i in range(1, self.n):
m = self.mobile[i]
if i != c and (m[2] == c or m[3] == c):
return self.parent(i)
return c
def solve(self, i):
self.solve2(i, frac(0,1))
lcm = 1
for d in self.denom:
lcm = lcm * d / gcd(lcm, d)
answer = self.weight[i][0].mul(lcm, 1).num + self.weight[i][1].mul(lcm, 1).num
return int(answer)
def solve2(self, i, wlr):
p, q, r, b = self.mobile[i]
# print("p,q,r,b=", self.mobile[i])
# print("i, wlr=", i, wlr.output())
if wlr.is_zero():
g = gcd(p, q)
p /= g
q /= g
self.weight[i] = [frac(q, 1), frac(p, 1)]
else:
self.weight[i] = [wlr.mul(q, p + q), wlr.mul(p, p + q)]
if r != 0:
self.solve2(r, self.weight[i][0])
if b != 0:
self.solve2(b, self.weight[i][1])
if r == 0 and b == 0:
if not self.weight[i][0].is_zero() and not self.weight[i][1].is_zero():
self.denom.append(self.weight[i][0].den)
self.denom.append(self.weight[i][1].den)
return
if __name__ == '__main__':
while True:
line = input()
n = int(line)
if n == 0:
break
lm = LightestMobile(n)
for _ in range(0, n):
p, q, r, b = map(int, list(input().split()))
lm.read(p, q, r, b)
rt = lm.root()
print(lm.solve(rt))
del lm
``` | output | 1 | 72,432 | 7 | 144,865 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Mobiles are widely known as moving works of art. The IOI Japan Committee has decided to create mobiles to publicize JOI. JOI public relations mobiles are sticks, strings, and weights. It is constructed as follows using the three types of elements of.
* One end of the bar is painted blue and the other end is painted red.
* The rod is hung with a string with one point other than both ends as a fulcrum.
* Both the length from the fulcrum to the red end and the length from the fulcrum to the blue end are positive integers.
* At both ends of the rod, hang a weight or another rod with a string.
* The weight is hung on one end of one of the rods using a string.
* Nothing hangs on the weight.
* The weight of the weight is a positive integer.
* Only one of the strings is tied at one end to the fulcrum of that bar to hang one bar, the other end is not tied to any other component. Meet one or the other.
* Connect the end of a bar to the fulcrum of a bar.
* Connect the end of a rod to a weight.
However, all rods need to be balanced. The weights of the rods and strings are negligibly light, so consider that the weights of the rods and strings are all 0. About the stick,
(Total weight of weights suspended below the red end of the rod) × (Length from the fulcrum of the rod to the red end) = (Hung below the blue end of the rod) Total weight of the weight) × (length from the fulcrum of the rod to the end of the blue)
If, then the bar is balanced.
<image>
The length and how to tie the rods to make up the mobile has already been decided, but the weight of the weight has not been decided yet. The lighter the mobile, the easier it is to hang, so I want to make the mobile as light as possible. As mentioned above, while balancing all the rods, find a way to attach a weight that minimizes the total weight of the mobile, and create a program that outputs the total weight of the mobile at that time. Information about the configuration is given.
* Number of bars n
* Information for each bar (bar numbers 1 to n)
* Ratio of the length from the fulcrum to the red end to the length from the fulcrum to the blue end
* Number of rods to hang on the red end (0 for hanging weights)
* Number of rods to hang on the blue edge (0 for hanging weights)
<image>
input
The input consists of multiple datasets. Each dataset is given in the following format. The input ends on a line containing one zero.
The first line contains the number of bars n used in the mobile. The following n lines (1 ≤ n ≤ 100) contain the data for each bar. I + 1st line In (1 ≤ i ≤ n), four integers p, q, r, b are written with a blank as a delimiter, and on the bar i, the length from the fulcrum to the red end and the length from the fulcrum to the blue end. The ratio is p: q, the number of the bar hanging at the red end is r, and the number of the bar hanging at the blue end is b, where bar number 0 is the weight. In any input, if the minimum value of the weight of the mobile is w and the maximum value of the positive integer used to express the ratio in the input is L, then wL <231 Meet.
The number of datasets does not exceed 15.
output
The weight of the mobile is output to one line for each data set.
Examples
Input
1
6 9 0 0
4
3 2 0 4
1 3 0 0
4 4 2 1
2 2 0 0
0
Output
5
40
Input
None
Output
None | instruction | 0 | 72,433 | 7 | 144,866 |
"Correct Solution:
```
def gcd(x, y):
while y > 0:
x, y = y, x%y
return x
while True:
N = int(input())
if not N:
break
a = [list(map(int, input().split())) for _ in [0]*N]
root = (set(range(1, N+1)) - set(n for bar in a for n in bar[2:])).pop()
def solve(ratio_r, ratio_l, left, right):
d = gcd(ratio_r, ratio_l)
ratio_r //= d
ratio_l //= d
w_left = solve(*a[left-1]) if left else 0
w_right = solve(*a[right-1]) if right else 0
if w_left == w_right == 0:
w_left, w_right = ratio_l, ratio_r
elif w_left == 0:
w_right = (w_right * ratio_r) // gcd(w_right, ratio_r)
w_left = ratio_l * (w_right // ratio_r)
elif w_right == 0:
w_left = (w_left * ratio_l) // gcd(w_left, ratio_l)
w_right = ratio_r * (w_left // ratio_l)
else:
_r, _l = w_left*ratio_r, w_right*ratio_l
d = gcd(_l, _r)
w_right, w_left = w_right*(_r // d), w_left*(_l // d)
return w_right + w_left
print(solve(*a[root-1]))
``` | output | 1 | 72,433 | 7 | 144,867 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Mobiles are widely known as moving works of art. The IOI Japan Committee has decided to create mobiles to publicize JOI. JOI public relations mobiles are sticks, strings, and weights. It is constructed as follows using the three types of elements of.
* One end of the bar is painted blue and the other end is painted red.
* The rod is hung with a string with one point other than both ends as a fulcrum.
* Both the length from the fulcrum to the red end and the length from the fulcrum to the blue end are positive integers.
* At both ends of the rod, hang a weight or another rod with a string.
* The weight is hung on one end of one of the rods using a string.
* Nothing hangs on the weight.
* The weight of the weight is a positive integer.
* Only one of the strings is tied at one end to the fulcrum of that bar to hang one bar, the other end is not tied to any other component. Meet one or the other.
* Connect the end of a bar to the fulcrum of a bar.
* Connect the end of a rod to a weight.
However, all rods need to be balanced. The weights of the rods and strings are negligibly light, so consider that the weights of the rods and strings are all 0. About the stick,
(Total weight of weights suspended below the red end of the rod) × (Length from the fulcrum of the rod to the red end) = (Hung below the blue end of the rod) Total weight of the weight) × (length from the fulcrum of the rod to the end of the blue)
If, then the bar is balanced.
<image>
The length and how to tie the rods to make up the mobile has already been decided, but the weight of the weight has not been decided yet. The lighter the mobile, the easier it is to hang, so I want to make the mobile as light as possible. As mentioned above, while balancing all the rods, find a way to attach a weight that minimizes the total weight of the mobile, and create a program that outputs the total weight of the mobile at that time. Information about the configuration is given.
* Number of bars n
* Information for each bar (bar numbers 1 to n)
* Ratio of the length from the fulcrum to the red end to the length from the fulcrum to the blue end
* Number of rods to hang on the red end (0 for hanging weights)
* Number of rods to hang on the blue edge (0 for hanging weights)
<image>
input
The input consists of multiple datasets. Each dataset is given in the following format. The input ends on a line containing one zero.
The first line contains the number of bars n used in the mobile. The following n lines (1 ≤ n ≤ 100) contain the data for each bar. I + 1st line In (1 ≤ i ≤ n), four integers p, q, r, b are written with a blank as a delimiter, and on the bar i, the length from the fulcrum to the red end and the length from the fulcrum to the blue end. The ratio is p: q, the number of the bar hanging at the red end is r, and the number of the bar hanging at the blue end is b, where bar number 0 is the weight. In any input, if the minimum value of the weight of the mobile is w and the maximum value of the positive integer used to express the ratio in the input is L, then wL <231 Meet.
The number of datasets does not exceed 15.
output
The weight of the mobile is output to one line for each data set.
Examples
Input
1
6 9 0 0
4
3 2 0 4
1 3 0 0
4 4 2 1
2 2 0 0
0
Output
5
40
Input
None
Output
None | instruction | 0 | 72,434 | 7 | 144,868 |
"Correct Solution:
```
def gcd(a, b):
r = a % b
if r:
return gcd(b, r)
return b
def lcm(a, b):
if a < b:
a, b = b, a
return a * b // gcd(a, b)
def min_weight(m):
p, q, r, b = bars[m - 1]
red = p * (min_weight(r) if r else 1)
blue = q * (min_weight(b) if b else 1)
equilibrium = lcm(red, blue)
return equilibrium // p + equilibrium // q
while True:
n = int(input())
if not n:
break
bars = [tuple(map(int, input().split())) for _ in range(n)]
root_bar = set(range(1, n + 1))
for _, _, r, b in bars:
if r: root_bar.remove(r)
if b: root_bar.remove(b)
print(min_weight(root_bar.pop()))
``` | output | 1 | 72,434 | 7 | 144,869 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Mobiles are widely known as moving works of art. The IOI Japan Committee has decided to create mobiles to publicize JOI. JOI public relations mobiles are sticks, strings, and weights. It is constructed as follows using the three types of elements of.
* One end of the bar is painted blue and the other end is painted red.
* The rod is hung with a string with one point other than both ends as a fulcrum.
* Both the length from the fulcrum to the red end and the length from the fulcrum to the blue end are positive integers.
* At both ends of the rod, hang a weight or another rod with a string.
* The weight is hung on one end of one of the rods using a string.
* Nothing hangs on the weight.
* The weight of the weight is a positive integer.
* Only one of the strings is tied at one end to the fulcrum of that bar to hang one bar, the other end is not tied to any other component. Meet one or the other.
* Connect the end of a bar to the fulcrum of a bar.
* Connect the end of a rod to a weight.
However, all rods need to be balanced. The weights of the rods and strings are negligibly light, so consider that the weights of the rods and strings are all 0. About the stick,
(Total weight of weights suspended below the red end of the rod) × (Length from the fulcrum of the rod to the red end) = (Hung below the blue end of the rod) Total weight of the weight) × (length from the fulcrum of the rod to the end of the blue)
If, then the bar is balanced.
<image>
The length and how to tie the rods to make up the mobile has already been decided, but the weight of the weight has not been decided yet. The lighter the mobile, the easier it is to hang, so I want to make the mobile as light as possible. As mentioned above, while balancing all the rods, find a way to attach a weight that minimizes the total weight of the mobile, and create a program that outputs the total weight of the mobile at that time. Information about the configuration is given.
* Number of bars n
* Information for each bar (bar numbers 1 to n)
* Ratio of the length from the fulcrum to the red end to the length from the fulcrum to the blue end
* Number of rods to hang on the red end (0 for hanging weights)
* Number of rods to hang on the blue edge (0 for hanging weights)
<image>
input
The input consists of multiple datasets. Each dataset is given in the following format. The input ends on a line containing one zero.
The first line contains the number of bars n used in the mobile. The following n lines (1 ≤ n ≤ 100) contain the data for each bar. I + 1st line In (1 ≤ i ≤ n), four integers p, q, r, b are written with a blank as a delimiter, and on the bar i, the length from the fulcrum to the red end and the length from the fulcrum to the blue end. The ratio is p: q, the number of the bar hanging at the red end is r, and the number of the bar hanging at the blue end is b, where bar number 0 is the weight. In any input, if the minimum value of the weight of the mobile is w and the maximum value of the positive integer used to express the ratio in the input is L, then wL <231 Meet.
The number of datasets does not exceed 15.
output
The weight of the mobile is output to one line for each data set.
Examples
Input
1
6 9 0 0
4
3 2 0 4
1 3 0 0
4 4 2 1
2 2 0 0
0
Output
5
40
Input
None
Output
None | instruction | 0 | 72,435 | 7 | 144,870 |
"Correct Solution:
```
'''
Created on 2017/03/09
@author: 03key
'''
def gcd(x,y):
if x<y:
tmp = x
x = y
y = tmp
ans = y
if x%y != 0:
ans = gcd(y,x%y)
return ans
class Mobile:
def __init__(self,leftratio=0,rightratio=0,left=-1,right=-1):
self.weight = 0
self.leftratio = leftratio
self.rightratio = rightratio
self.left = left
self.leftweight = 0
self.right = right
self.rightweight = 0
def calc(self):
if self.weight != 0 : return self.weight
if self.left == -1:
self.leftweight = 1
else:
self.leftweight = Mobile.calc(self.left)
if self.right == -1:
self.rightweight = 1
else:
self.rightweight = Mobile.calc(self.right)
lmoment = self.leftratio*self.leftweight
rmoment = self.rightratio*self.rightweight
lcs = lmoment * rmoment // gcd(lmoment,rmoment)
self.weight = (lcs//self.leftratio) + (lcs//self.rightratio)
return self.weight
def stdout(self):
print(self.left, self.right)
while True:
n = int(input())
if n==0: break
barlist = [Mobile() for i in range(n)]
for i in range(n):
p,q,r,b = map(int, input().split())
barlist[i].leftratio = p
barlist[i].rightratio = q
if r != 0 : barlist[i].left = barlist[r-1]
if b != 0 : barlist[i].right = barlist[b-1]
maxi = 0
for i in range(n):
maxi = max(barlist[i].calc(),maxi)
print(maxi)
``` | output | 1 | 72,435 | 7 | 144,871 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Mobiles are widely known as moving works of art. The IOI Japan Committee has decided to create mobiles to publicize JOI. JOI public relations mobiles are sticks, strings, and weights. It is constructed as follows using the three types of elements of.
* One end of the bar is painted blue and the other end is painted red.
* The rod is hung with a string with one point other than both ends as a fulcrum.
* Both the length from the fulcrum to the red end and the length from the fulcrum to the blue end are positive integers.
* At both ends of the rod, hang a weight or another rod with a string.
* The weight is hung on one end of one of the rods using a string.
* Nothing hangs on the weight.
* The weight of the weight is a positive integer.
* Only one of the strings is tied at one end to the fulcrum of that bar to hang one bar, the other end is not tied to any other component. Meet one or the other.
* Connect the end of a bar to the fulcrum of a bar.
* Connect the end of a rod to a weight.
However, all rods need to be balanced. The weights of the rods and strings are negligibly light, so consider that the weights of the rods and strings are all 0. About the stick,
(Total weight of weights suspended below the red end of the rod) × (Length from the fulcrum of the rod to the red end) = (Hung below the blue end of the rod) Total weight of the weight) × (length from the fulcrum of the rod to the end of the blue)
If, then the bar is balanced.
<image>
The length and how to tie the rods to make up the mobile has already been decided, but the weight of the weight has not been decided yet. The lighter the mobile, the easier it is to hang, so I want to make the mobile as light as possible. As mentioned above, while balancing all the rods, find a way to attach a weight that minimizes the total weight of the mobile, and create a program that outputs the total weight of the mobile at that time. Information about the configuration is given.
* Number of bars n
* Information for each bar (bar numbers 1 to n)
* Ratio of the length from the fulcrum to the red end to the length from the fulcrum to the blue end
* Number of rods to hang on the red end (0 for hanging weights)
* Number of rods to hang on the blue edge (0 for hanging weights)
<image>
input
The input consists of multiple datasets. Each dataset is given in the following format. The input ends on a line containing one zero.
The first line contains the number of bars n used in the mobile. The following n lines (1 ≤ n ≤ 100) contain the data for each bar. I + 1st line In (1 ≤ i ≤ n), four integers p, q, r, b are written with a blank as a delimiter, and on the bar i, the length from the fulcrum to the red end and the length from the fulcrum to the blue end. The ratio is p: q, the number of the bar hanging at the red end is r, and the number of the bar hanging at the blue end is b, where bar number 0 is the weight. In any input, if the minimum value of the weight of the mobile is w and the maximum value of the positive integer used to express the ratio in the input is L, then wL <231 Meet.
The number of datasets does not exceed 15.
output
The weight of the mobile is output to one line for each data set.
Examples
Input
1
6 9 0 0
4
3 2 0 4
1 3 0 0
4 4 2 1
2 2 0 0
0
Output
5
40
Input
None
Output
None | instruction | 0 | 72,436 | 7 | 144,872 |
"Correct Solution:
```
from math import gcd
while True:
try:
n = int(input())
if not n: break
lst = [[]]
weight = [-1 for i in range(n + 1)]
def setWeight(n):
p,q,r,b = lst[n]
if weight[n] != -1:pass
elif r == 0 and b == 0:
weight[n] = (p + q) // gcd(p,q)
elif r == 0:
setWeight(b)
wb = weight[b]
g = gcd(wb * q, p)
lcm = wb * q * p // g
weight[n] = lcm // q + lcm // p
# print("wr:",str(lcm//p),"wb:",str(wb),"g:",g,"p:",p,"lcm",lcm)
elif b == 0:
setWeight(r)
wr = weight[r]
g = gcd(wr * p, q)
lcm = wr * p * q // g
weight[n] = lcm // p + lcm // q
else:
if weight[r] == -1: setWeight(r)
if weight[b] == -1: setWeight(b)
wr = weight[r]
wb = weight[b]
g = gcd(wr * p, wb * q)
lcm = wr * p * wb * q // g
weight[n] = lcm // p + lcm // q
for i in range(n):
lst.append(list(map(int,input().split())))
for i in range(0,n):
if weight[i + 1] == -1:
setWeight(i+1)
print(max(weight))
# print(weight)
except EOFError:
break
``` | output | 1 | 72,436 | 7 | 144,873 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Mobiles are widely known as moving works of art. The IOI Japan Committee has decided to create mobiles to publicize JOI. JOI public relations mobiles are sticks, strings, and weights. It is constructed as follows using the three types of elements of.
* One end of the bar is painted blue and the other end is painted red.
* The rod is hung with a string with one point other than both ends as a fulcrum.
* Both the length from the fulcrum to the red end and the length from the fulcrum to the blue end are positive integers.
* At both ends of the rod, hang a weight or another rod with a string.
* The weight is hung on one end of one of the rods using a string.
* Nothing hangs on the weight.
* The weight of the weight is a positive integer.
* Only one of the strings is tied at one end to the fulcrum of that bar to hang one bar, the other end is not tied to any other component. Meet one or the other.
* Connect the end of a bar to the fulcrum of a bar.
* Connect the end of a rod to a weight.
However, all rods need to be balanced. The weights of the rods and strings are negligibly light, so consider that the weights of the rods and strings are all 0. About the stick,
(Total weight of weights suspended below the red end of the rod) × (Length from the fulcrum of the rod to the red end) = (Hung below the blue end of the rod) Total weight of the weight) × (length from the fulcrum of the rod to the end of the blue)
If, then the bar is balanced.
<image>
The length and how to tie the rods to make up the mobile has already been decided, but the weight of the weight has not been decided yet. The lighter the mobile, the easier it is to hang, so I want to make the mobile as light as possible. As mentioned above, while balancing all the rods, find a way to attach a weight that minimizes the total weight of the mobile, and create a program that outputs the total weight of the mobile at that time. Information about the configuration is given.
* Number of bars n
* Information for each bar (bar numbers 1 to n)
* Ratio of the length from the fulcrum to the red end to the length from the fulcrum to the blue end
* Number of rods to hang on the red end (0 for hanging weights)
* Number of rods to hang on the blue edge (0 for hanging weights)
<image>
input
The input consists of multiple datasets. Each dataset is given in the following format. The input ends on a line containing one zero.
The first line contains the number of bars n used in the mobile. The following n lines (1 ≤ n ≤ 100) contain the data for each bar. I + 1st line In (1 ≤ i ≤ n), four integers p, q, r, b are written with a blank as a delimiter, and on the bar i, the length from the fulcrum to the red end and the length from the fulcrum to the blue end. The ratio is p: q, the number of the bar hanging at the red end is r, and the number of the bar hanging at the blue end is b, where bar number 0 is the weight. In any input, if the minimum value of the weight of the mobile is w and the maximum value of the positive integer used to express the ratio in the input is L, then wL <231 Meet.
The number of datasets does not exceed 15.
output
The weight of the mobile is output to one line for each data set.
Examples
Input
1
6 9 0 0
4
3 2 0 4
1 3 0 0
4 4 2 1
2 2 0 0
0
Output
5
40
Input
None
Output
None | instruction | 0 | 72,437 | 7 | 144,874 |
"Correct Solution:
```
# AOJ 0520: Lightest Mobile
# Python3 2018.6.30 bal4u
def lcm(a, b):
return a // gcd(a, b) * b
def gcd(a, b):
while b != 0:
r = a % b
a, b = b, r
return a
def calc(i):
wr = calc(t[i][2]) if t[i][2] > 0 else 1
wb = calc(t[i][3]) if t[i][3] > 0 else 1
w = lcm(t[i][0] * wr, t[i][1] * wb)
return w//t[i][0] + w//t[i][1]
while True:
n = int(input())
if n == 0: break
f = [0]*(n+1)
t = [(0,0,0,0)]
for i in range(1, n+1):
p, q, r, b = map(int, input().split())
t.append((p, q, r, b))
if r > 0: f[r] = i
if b > 0: f[b] = i
for i in range(1, n+1):
if f[i] == 0: break
print(calc(i))
``` | output | 1 | 72,437 | 7 | 144,875 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Mobiles are widely known as moving works of art. The IOI Japan Committee has decided to create mobiles to publicize JOI. JOI public relations mobiles are sticks, strings, and weights. It is constructed as follows using the three types of elements of.
* One end of the bar is painted blue and the other end is painted red.
* The rod is hung with a string with one point other than both ends as a fulcrum.
* Both the length from the fulcrum to the red end and the length from the fulcrum to the blue end are positive integers.
* At both ends of the rod, hang a weight or another rod with a string.
* The weight is hung on one end of one of the rods using a string.
* Nothing hangs on the weight.
* The weight of the weight is a positive integer.
* Only one of the strings is tied at one end to the fulcrum of that bar to hang one bar, the other end is not tied to any other component. Meet one or the other.
* Connect the end of a bar to the fulcrum of a bar.
* Connect the end of a rod to a weight.
However, all rods need to be balanced. The weights of the rods and strings are negligibly light, so consider that the weights of the rods and strings are all 0. About the stick,
(Total weight of weights suspended below the red end of the rod) × (Length from the fulcrum of the rod to the red end) = (Hung below the blue end of the rod) Total weight of the weight) × (length from the fulcrum of the rod to the end of the blue)
If, then the bar is balanced.
<image>
The length and how to tie the rods to make up the mobile has already been decided, but the weight of the weight has not been decided yet. The lighter the mobile, the easier it is to hang, so I want to make the mobile as light as possible. As mentioned above, while balancing all the rods, find a way to attach a weight that minimizes the total weight of the mobile, and create a program that outputs the total weight of the mobile at that time. Information about the configuration is given.
* Number of bars n
* Information for each bar (bar numbers 1 to n)
* Ratio of the length from the fulcrum to the red end to the length from the fulcrum to the blue end
* Number of rods to hang on the red end (0 for hanging weights)
* Number of rods to hang on the blue edge (0 for hanging weights)
<image>
input
The input consists of multiple datasets. Each dataset is given in the following format. The input ends on a line containing one zero.
The first line contains the number of bars n used in the mobile. The following n lines (1 ≤ n ≤ 100) contain the data for each bar. I + 1st line In (1 ≤ i ≤ n), four integers p, q, r, b are written with a blank as a delimiter, and on the bar i, the length from the fulcrum to the red end and the length from the fulcrum to the blue end. The ratio is p: q, the number of the bar hanging at the red end is r, and the number of the bar hanging at the blue end is b, where bar number 0 is the weight. In any input, if the minimum value of the weight of the mobile is w and the maximum value of the positive integer used to express the ratio in the input is L, then wL <231 Meet.
The number of datasets does not exceed 15.
output
The weight of the mobile is output to one line for each data set.
Examples
Input
1
6 9 0 0
4
3 2 0 4
1 3 0 0
4 4 2 1
2 2 0 0
0
Output
5
40
Input
None
Output
None | instruction | 0 | 72,438 | 7 | 144,876 |
"Correct Solution:
```
def solve():
from math import gcd
# Post-order tree traversal
def traversal(bar_id):
if bar_id == 0:
return 1
p, q, r, b = nodes[bar_id]
r_w = traversal(r)
b_w = traversal(b)
cf = gcd(p * r_w, q * b_w)
return r_w * b_w * (p + q) // cf
from sys import stdin
file_input = stdin
while True:
n = int(file_input.readline())
if n == 0:
break
root_id = n * (n + 1) // 2
nodes = [None]
for i in range(1, n + 1):
p, q, r, b = map(int, file_input.readline().split())
g = gcd(p, q)
nodes.append((p // g, q // g, r, b))
root_id -= (r + b)
print(traversal(root_id))
solve()
``` | output | 1 | 72,438 | 7 | 144,877 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Mobiles are widely known as moving works of art. The IOI Japan Committee has decided to create mobiles to publicize JOI. JOI public relations mobiles are sticks, strings, and weights. It is constructed as follows using the three types of elements of.
* One end of the bar is painted blue and the other end is painted red.
* The rod is hung with a string with one point other than both ends as a fulcrum.
* Both the length from the fulcrum to the red end and the length from the fulcrum to the blue end are positive integers.
* At both ends of the rod, hang a weight or another rod with a string.
* The weight is hung on one end of one of the rods using a string.
* Nothing hangs on the weight.
* The weight of the weight is a positive integer.
* Only one of the strings is tied at one end to the fulcrum of that bar to hang one bar, the other end is not tied to any other component. Meet one or the other.
* Connect the end of a bar to the fulcrum of a bar.
* Connect the end of a rod to a weight.
However, all rods need to be balanced. The weights of the rods and strings are negligibly light, so consider that the weights of the rods and strings are all 0. About the stick,
(Total weight of weights suspended below the red end of the rod) × (Length from the fulcrum of the rod to the red end) = (Hung below the blue end of the rod) Total weight of the weight) × (length from the fulcrum of the rod to the end of the blue)
If, then the bar is balanced.
<image>
The length and how to tie the rods to make up the mobile has already been decided, but the weight of the weight has not been decided yet. The lighter the mobile, the easier it is to hang, so I want to make the mobile as light as possible. As mentioned above, while balancing all the rods, find a way to attach a weight that minimizes the total weight of the mobile, and create a program that outputs the total weight of the mobile at that time. Information about the configuration is given.
* Number of bars n
* Information for each bar (bar numbers 1 to n)
* Ratio of the length from the fulcrum to the red end to the length from the fulcrum to the blue end
* Number of rods to hang on the red end (0 for hanging weights)
* Number of rods to hang on the blue edge (0 for hanging weights)
<image>
input
The input consists of multiple datasets. Each dataset is given in the following format. The input ends on a line containing one zero.
The first line contains the number of bars n used in the mobile. The following n lines (1 ≤ n ≤ 100) contain the data for each bar. I + 1st line In (1 ≤ i ≤ n), four integers p, q, r, b are written with a blank as a delimiter, and on the bar i, the length from the fulcrum to the red end and the length from the fulcrum to the blue end. The ratio is p: q, the number of the bar hanging at the red end is r, and the number of the bar hanging at the blue end is b, where bar number 0 is the weight. In any input, if the minimum value of the weight of the mobile is w and the maximum value of the positive integer used to express the ratio in the input is L, then wL <231 Meet.
The number of datasets does not exceed 15.
output
The weight of the mobile is output to one line for each data set.
Examples
Input
1
6 9 0 0
4
3 2 0 4
1 3 0 0
4 4 2 1
2 2 0 0
0
Output
5
40
Input
None
Output
None | instruction | 0 | 72,439 | 7 | 144,878 |
"Correct Solution:
```
from math import gcd
def lcm(i, j):
return i * j // gcd(i, j)
def main():
while True:
n = int(input())
if not n: break
lst = [[]]
weight = [-1 for i in range(n + 1)]
def setWeight(n):
p, q, r, b = lst[n]
if weight[n] != -1: pass
elif r == 0 and b == 0:
weight[n] = (p + q) // gcd(p, q)
elif r == 0:
setWeight(b)
l = lcm(weight[b] * q, p)
weight[n] = l // q + l // p
elif b == 0:
setWeight(r)
l = lcm(weight[r] * p, q)
weight[n] = l // p + l // q
else:
if weight[r] == -1: setWeight(r)
if weight[b] == -1: setWeight(b)
l = lcm(weight[r] * p, weight[b] * q)
weight[n] = l // p + l // q
for i in range(n):
lst.append(list(map(int,input().split())))
for i in range(0, n):
if weight[i + 1] == -1:
setWeight(i + 1)
print(max(weight))
if __name__ == '__main__':
main()
``` | output | 1 | 72,439 | 7 | 144,879 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Flatland is inhabited by pixels of three colors: red, green and blue. We know that if two pixels of different colors meet in a violent fight, only one of them survives the fight (that is, the total number of pixels decreases by one). Besides, if pixels of colors x and y (x ≠ y) meet in a violent fight, then the pixel that survives the fight immediately changes its color to z (z ≠ x; z ≠ y). Pixels of the same color are friends, so they don't fight.
The King of Flatland knows that his land will be peaceful and prosperous when the pixels are of the same color. For each of the three colors you know the number of pixels of this color that inhabit Flatland. Help the king and determine whether fights can bring peace and prosperity to the country and if it is possible, find the minimum number of fights needed to make the land peaceful and prosperous.
Input
The first line contains three space-separated integers a, b and c (0 ≤ a, b, c ≤ 231; a + b + c > 0) — the number of red, green and blue pixels, correspondingly.
Output
Print a single number — the minimum number of pixel fights before the country becomes peaceful and prosperous. If making the country peaceful and prosperous is impossible, print -1.
Examples
Input
1 1 1
Output
1
Input
3 1 0
Output
3
Note
In the first test sample the country needs only one fight to achieve peace and prosperity. Besides, it can be any fight whatsoever. For example, let's assume that the green and the blue pixels fight, then the surviving pixel will be red. As a result, after the fight there are two red pixels. There won't be other pixels.
In the second sample the following sequence of fights is possible: red and blue, green and red, red and blue. As a result, after all fights there is one green pixel left. | instruction | 0 | 72,779 | 7 | 145,558 |
Tags: constructive algorithms, math
Correct Solution:
```
a = list(map(int,input().split()))
a.sort()
if int(a[1] % 2 == 0)+int(a[0] % 2 == 0)==1:
print(a[2])
else:
print(int((((a[1]-a[0])+(a[1]+a[0]))/2)))
``` | output | 1 | 72,779 | 7 | 145,559 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Flatland is inhabited by pixels of three colors: red, green and blue. We know that if two pixels of different colors meet in a violent fight, only one of them survives the fight (that is, the total number of pixels decreases by one). Besides, if pixels of colors x and y (x ≠ y) meet in a violent fight, then the pixel that survives the fight immediately changes its color to z (z ≠ x; z ≠ y). Pixels of the same color are friends, so they don't fight.
The King of Flatland knows that his land will be peaceful and prosperous when the pixels are of the same color. For each of the three colors you know the number of pixels of this color that inhabit Flatland. Help the king and determine whether fights can bring peace and prosperity to the country and if it is possible, find the minimum number of fights needed to make the land peaceful and prosperous.
Input
The first line contains three space-separated integers a, b and c (0 ≤ a, b, c ≤ 231; a + b + c > 0) — the number of red, green and blue pixels, correspondingly.
Output
Print a single number — the minimum number of pixel fights before the country becomes peaceful and prosperous. If making the country peaceful and prosperous is impossible, print -1.
Examples
Input
1 1 1
Output
1
Input
3 1 0
Output
3
Note
In the first test sample the country needs only one fight to achieve peace and prosperity. Besides, it can be any fight whatsoever. For example, let's assume that the green and the blue pixels fight, then the surviving pixel will be red. As a result, after the fight there are two red pixels. There won't be other pixels.
In the second sample the following sequence of fights is possible: red and blue, green and red, red and blue. As a result, after all fights there is one green pixel left. | instruction | 0 | 72,780 | 7 | 145,560 |
Tags: constructive algorithms, math
Correct Solution:
```
import sys
input=sys.stdin.readline
l=input().split()
li=[int(i) for i in l]
maxa=10**18
li.sort()
if((li[2]-li[1])%2==0):
maxa=min(maxa,li[2])
if((li[2]-li[0])%2==0):
maxa=min(maxa,li[2])
if((li[1]-li[0])%2==0):
maxa=min(maxa,li[1])
if(maxa==10**18):
print(-1)
else:
print(maxa)
``` | output | 1 | 72,780 | 7 | 145,561 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Flatland is inhabited by pixels of three colors: red, green and blue. We know that if two pixels of different colors meet in a violent fight, only one of them survives the fight (that is, the total number of pixels decreases by one). Besides, if pixels of colors x and y (x ≠ y) meet in a violent fight, then the pixel that survives the fight immediately changes its color to z (z ≠ x; z ≠ y). Pixels of the same color are friends, so they don't fight.
The King of Flatland knows that his land will be peaceful and prosperous when the pixels are of the same color. For each of the three colors you know the number of pixels of this color that inhabit Flatland. Help the king and determine whether fights can bring peace and prosperity to the country and if it is possible, find the minimum number of fights needed to make the land peaceful and prosperous.
Input
The first line contains three space-separated integers a, b and c (0 ≤ a, b, c ≤ 231; a + b + c > 0) — the number of red, green and blue pixels, correspondingly.
Output
Print a single number — the minimum number of pixel fights before the country becomes peaceful and prosperous. If making the country peaceful and prosperous is impossible, print -1.
Examples
Input
1 1 1
Output
1
Input
3 1 0
Output
3
Note
In the first test sample the country needs only one fight to achieve peace and prosperity. Besides, it can be any fight whatsoever. For example, let's assume that the green and the blue pixels fight, then the surviving pixel will be red. As a result, after the fight there are two red pixels. There won't be other pixels.
In the second sample the following sequence of fights is possible: red and blue, green and red, red and blue. As a result, after all fights there is one green pixel left. | instruction | 0 | 72,781 | 7 | 145,562 |
Tags: constructive algorithms, math
Correct Solution:
```
def pixel(input):
red, green, blue = input.split()
red = int(red)
green = int(green)
blue = int(blue)
plist = [red,green,blue]
plist.sort()
lowestChet = is_chet(plist[0])
midChet = is_chet(plist[1])
if ((midChet + lowestChet) == 2) or ((midChet + lowestChet) == 0):
return plist[1]
else:
return plist[2]
def is_chet(n):
return 1 if n % 2 == 0 else 0
print(pixel(input()))
``` | output | 1 | 72,781 | 7 | 145,563 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Flatland is inhabited by pixels of three colors: red, green and blue. We know that if two pixels of different colors meet in a violent fight, only one of them survives the fight (that is, the total number of pixels decreases by one). Besides, if pixels of colors x and y (x ≠ y) meet in a violent fight, then the pixel that survives the fight immediately changes its color to z (z ≠ x; z ≠ y). Pixels of the same color are friends, so they don't fight.
The King of Flatland knows that his land will be peaceful and prosperous when the pixels are of the same color. For each of the three colors you know the number of pixels of this color that inhabit Flatland. Help the king and determine whether fights can bring peace and prosperity to the country and if it is possible, find the minimum number of fights needed to make the land peaceful and prosperous.
Input
The first line contains three space-separated integers a, b and c (0 ≤ a, b, c ≤ 231; a + b + c > 0) — the number of red, green and blue pixels, correspondingly.
Output
Print a single number — the minimum number of pixel fights before the country becomes peaceful and prosperous. If making the country peaceful and prosperous is impossible, print -1.
Examples
Input
1 1 1
Output
1
Input
3 1 0
Output
3
Note
In the first test sample the country needs only one fight to achieve peace and prosperity. Besides, it can be any fight whatsoever. For example, let's assume that the green and the blue pixels fight, then the surviving pixel will be red. As a result, after the fight there are two red pixels. There won't be other pixels.
In the second sample the following sequence of fights is possible: red and blue, green and red, red and blue. As a result, after all fights there is one green pixel left. | instruction | 0 | 72,782 | 7 | 145,564 |
Tags: constructive algorithms, math
Correct Solution:
```
a = list(map(int,input().split()))
def calc(a):
return int((((a[1]-a[0])+(a[1]+a[0]))/2))
a.sort()
if a[1] % 2 == 0 and a[0] % 2 == 0:
print(calc(a))
elif a[1] % 2 == 0 or a[0] % 2 == 0:
print(a[2])
else:
print(calc(a))
``` | output | 1 | 72,782 | 7 | 145,565 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Flatland is inhabited by pixels of three colors: red, green and blue. We know that if two pixels of different colors meet in a violent fight, only one of them survives the fight (that is, the total number of pixels decreases by one). Besides, if pixels of colors x and y (x ≠ y) meet in a violent fight, then the pixel that survives the fight immediately changes its color to z (z ≠ x; z ≠ y). Pixels of the same color are friends, so they don't fight.
The King of Flatland knows that his land will be peaceful and prosperous when the pixels are of the same color. For each of the three colors you know the number of pixels of this color that inhabit Flatland. Help the king and determine whether fights can bring peace and prosperity to the country and if it is possible, find the minimum number of fights needed to make the land peaceful and prosperous.
Input
The first line contains three space-separated integers a, b and c (0 ≤ a, b, c ≤ 231; a + b + c > 0) — the number of red, green and blue pixels, correspondingly.
Output
Print a single number — the minimum number of pixel fights before the country becomes peaceful and prosperous. If making the country peaceful and prosperous is impossible, print -1.
Examples
Input
1 1 1
Output
1
Input
3 1 0
Output
3
Note
In the first test sample the country needs only one fight to achieve peace and prosperity. Besides, it can be any fight whatsoever. For example, let's assume that the green and the blue pixels fight, then the surviving pixel will be red. As a result, after the fight there are two red pixels. There won't be other pixels.
In the second sample the following sequence of fights is possible: red and blue, green and red, red and blue. As a result, after all fights there is one green pixel left.
Submitted Solution:
```
a = list(map(int,input().split()))
a.sort()
if a[1] % 2 == 0 and not a[0] % 2 == 0:
print(a[2])
else:
print(int((((a[1]-a[0])+(a[1]+a[0]))/2)))
``` | instruction | 0 | 72,783 | 7 | 145,566 |
No | output | 1 | 72,783 | 7 | 145,567 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Flatland is inhabited by pixels of three colors: red, green and blue. We know that if two pixels of different colors meet in a violent fight, only one of them survives the fight (that is, the total number of pixels decreases by one). Besides, if pixels of colors x and y (x ≠ y) meet in a violent fight, then the pixel that survives the fight immediately changes its color to z (z ≠ x; z ≠ y). Pixels of the same color are friends, so they don't fight.
The King of Flatland knows that his land will be peaceful and prosperous when the pixels are of the same color. For each of the three colors you know the number of pixels of this color that inhabit Flatland. Help the king and determine whether fights can bring peace and prosperity to the country and if it is possible, find the minimum number of fights needed to make the land peaceful and prosperous.
Input
The first line contains three space-separated integers a, b and c (0 ≤ a, b, c ≤ 231; a + b + c > 0) — the number of red, green and blue pixels, correspondingly.
Output
Print a single number — the minimum number of pixel fights before the country becomes peaceful and prosperous. If making the country peaceful and prosperous is impossible, print -1.
Examples
Input
1 1 1
Output
1
Input
3 1 0
Output
3
Note
In the first test sample the country needs only one fight to achieve peace and prosperity. Besides, it can be any fight whatsoever. For example, let's assume that the green and the blue pixels fight, then the surviving pixel will be red. As a result, after the fight there are two red pixels. There won't be other pixels.
In the second sample the following sequence of fights is possible: red and blue, green and red, red and blue. As a result, after all fights there is one green pixel left.
Submitted Solution:
```
def pixel(input):
red, green, blue = input.split()
red = int(red)
green = int(green)
blue = int(blue)
srednee = (red + green + blue) // 3
ost = (red + green + blue) % 3
if srednee in [red,green,blue] and ost == 0:
return srednee
closest = min([red,green,blue], key=lambda x: abs(x - srednee))
#just for lol
if closest in [red,green,blue] and srednee > 10000:
return closest
for item in [red, green, blue]:
if item > (srednee+ost):
return item
print(pixel(input()))
``` | instruction | 0 | 72,784 | 7 | 145,568 |
No | output | 1 | 72,784 | 7 | 145,569 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Flatland is inhabited by pixels of three colors: red, green and blue. We know that if two pixels of different colors meet in a violent fight, only one of them survives the fight (that is, the total number of pixels decreases by one). Besides, if pixels of colors x and y (x ≠ y) meet in a violent fight, then the pixel that survives the fight immediately changes its color to z (z ≠ x; z ≠ y). Pixels of the same color are friends, so they don't fight.
The King of Flatland knows that his land will be peaceful and prosperous when the pixels are of the same color. For each of the three colors you know the number of pixels of this color that inhabit Flatland. Help the king and determine whether fights can bring peace and prosperity to the country and if it is possible, find the minimum number of fights needed to make the land peaceful and prosperous.
Input
The first line contains three space-separated integers a, b and c (0 ≤ a, b, c ≤ 231; a + b + c > 0) — the number of red, green and blue pixels, correspondingly.
Output
Print a single number — the minimum number of pixel fights before the country becomes peaceful and prosperous. If making the country peaceful and prosperous is impossible, print -1.
Examples
Input
1 1 1
Output
1
Input
3 1 0
Output
3
Note
In the first test sample the country needs only one fight to achieve peace and prosperity. Besides, it can be any fight whatsoever. For example, let's assume that the green and the blue pixels fight, then the surviving pixel will be red. As a result, after the fight there are two red pixels. There won't be other pixels.
In the second sample the following sequence of fights is possible: red and blue, green and red, red and blue. As a result, after all fights there is one green pixel left.
Submitted Solution:
```
import operator
def pixel(input):
red, green, blue = input.split()
pixels = {
'red': int(red),
'green': int(green),
'blue': int(blue)
}
battles = 0
while True:
is_mir = len([x for x in [pixels['red'], pixels['green'], pixels['blue']] if x == 0]) > 1
if is_mir:
break
min_pixel = get_min_pixel(pixels)
if min_pixel == 'red':
pixels['red'] += 1
pixels['blue'] -= 1
pixels['green'] -= 1
pass
elif min_pixel == 'green':
pixels['red'] -= 1
pixels['blue'] -= 1
pixels['green'] += 1
elif min_pixel == 'blue':
pixels['red'] -= 1
pixels['blue'] += 1
pixels['green'] -= 1
battles += 1
return battles
def get_min_pixel(pixels):
return min(pixels.items(), key=operator.itemgetter(1))[0]
print(pixel(input()))
``` | instruction | 0 | 72,785 | 7 | 145,570 |
No | output | 1 | 72,785 | 7 | 145,571 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Flatland is inhabited by pixels of three colors: red, green and blue. We know that if two pixels of different colors meet in a violent fight, only one of them survives the fight (that is, the total number of pixels decreases by one). Besides, if pixels of colors x and y (x ≠ y) meet in a violent fight, then the pixel that survives the fight immediately changes its color to z (z ≠ x; z ≠ y). Pixels of the same color are friends, so they don't fight.
The King of Flatland knows that his land will be peaceful and prosperous when the pixels are of the same color. For each of the three colors you know the number of pixels of this color that inhabit Flatland. Help the king and determine whether fights can bring peace and prosperity to the country and if it is possible, find the minimum number of fights needed to make the land peaceful and prosperous.
Input
The first line contains three space-separated integers a, b and c (0 ≤ a, b, c ≤ 231; a + b + c > 0) — the number of red, green and blue pixels, correspondingly.
Output
Print a single number — the minimum number of pixel fights before the country becomes peaceful and prosperous. If making the country peaceful and prosperous is impossible, print -1.
Examples
Input
1 1 1
Output
1
Input
3 1 0
Output
3
Note
In the first test sample the country needs only one fight to achieve peace and prosperity. Besides, it can be any fight whatsoever. For example, let's assume that the green and the blue pixels fight, then the surviving pixel will be red. As a result, after the fight there are two red pixels. There won't be other pixels.
In the second sample the following sequence of fights is possible: red and blue, green and red, red and blue. As a result, after all fights there is one green pixel left.
Submitted Solution:
```
pop = input().split(' ')
pops = []
for p in pop:
pops.append(int(p))
#[4,1,1]
def ss(lst): #second smallest
if len(lst)==2 and lst[0]<=lst[1]:return lst[1]
return ss(lst[:-1]) if lst[0]<=lst[-1]>=lst[1] else ss([lst[-1]]+lst[:-1])
def gi(num): #get index
return pops.index(num)
def idf(): #ideal fight
v = min([i for i in pops if i > 0])
w = ss([i for i in pops if i > 0])
g = [i for i in pops if i is not v and i is not w][0]
pops[gi(v)] -= 1
pops[gi(w)] -= 1
pops[gi(g)] += 1
def vf():
md = [i for i in pops if i is not max(pops) and i is not min(pops)][0]
mx = max([i for i in pops if i > 0])
g = [i for i in pops if i is not md and i is not mx][0]
pops[gi(md)] -= 1
pops[gi(mx)] -= 1
pops[gi(g)] += 1
def unstable():
for i in pops:
if i == 0:
return False
return True
atk = 0
try:
while (len([i for i in pops if i > 0])>1):
while unstable():
idf()
atk+=1
if len([i for i in pops if i > 0]) == 2:
vf()
atk+=1
except:
atk = -1
print(atk)
``` | instruction | 0 | 72,786 | 7 | 145,572 |
No | output | 1 | 72,786 | 7 | 145,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland crossword is a puzzle that is solved on a square grid with n rows and n columns. Initially all the cells are white.
To solve the puzzle one has to color some cells on the border of the grid black in such a way that:
* exactly U cells in the top row are black;
* exactly R cells in the rightmost column are black;
* exactly D cells in the bottom row are black;
* exactly L cells in the leftmost column are black.
Note that you can color zero cells black and leave every cell white.
Your task is to check if there exists a solution to the given puzzle.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains 5 integers n, U, R, D, L (2 ≤ n ≤ 100; 0 ≤ U, R, D, L ≤ n).
Output
For each testcase print "YES" if the solution exists and "NO" otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
5 2 5 3 1
3 0 0 0 0
4 4 1 4 0
2 1 1 1 1
Output
YES
YES
NO
YES
Note
Here are possible solutions to testcases 1, 2 and 4:
<image> | instruction | 0 | 73,560 | 7 | 147,120 |
Tags: bitmasks, brute force, greedy, implementation
Correct Solution:
```
import sys
t = int(input())
for i in range(t):
n, u, r, d, l = map(int, sys.stdin.readline().rstrip().split())
l_must = 0
r_must = 0
u_must = 0
d_must = 0
must = 0
if n-u < 2:
if n-u == 0:
must += 2
l_must += 1
r_must += 1
else:
must += 1
if n-d < 2:
if n-d == 0:
l_must += 1
r_must += 1
must += 2
else:
must += 1
if r+l < must:
print("NO")
continue
elif l < l_must or r < r_must:
print("NO")
continue
must = 0
if n-r < 2:
if n-r == 0:
must += 2
u_must += 1
d_must += 1
else:
must += 1
if n-l < 2:
if n-l == 0:
must += 2
u_must += 1
d_must += 1
else:
must += 1
if u+d < must:
print("NO")
continue
elif u < u_must or d < d_must:
print("NO")
continue
print("YES")
``` | output | 1 | 73,560 | 7 | 147,121 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland crossword is a puzzle that is solved on a square grid with n rows and n columns. Initially all the cells are white.
To solve the puzzle one has to color some cells on the border of the grid black in such a way that:
* exactly U cells in the top row are black;
* exactly R cells in the rightmost column are black;
* exactly D cells in the bottom row are black;
* exactly L cells in the leftmost column are black.
Note that you can color zero cells black and leave every cell white.
Your task is to check if there exists a solution to the given puzzle.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains 5 integers n, U, R, D, L (2 ≤ n ≤ 100; 0 ≤ U, R, D, L ≤ n).
Output
For each testcase print "YES" if the solution exists and "NO" otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
5 2 5 3 1
3 0 0 0 0
4 4 1 4 0
2 1 1 1 1
Output
YES
YES
NO
YES
Note
Here are possible solutions to testcases 1, 2 and 4:
<image> | instruction | 0 | 73,561 | 7 | 147,122 |
Tags: bitmasks, brute force, greedy, implementation
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
def mul_input():
return map(int,input().split())
def arr_input():
return list(map(int,input().split()))
def solve():
n,u,r,d,l=mul_input()
u1=u
r1=r
d1=d
l1=l
if u==n:
l1-=1
r1-=1
elif u==n-1:
if r1>l1:
r1-=1
else:
l1-=1
if d==n:
l1-=1
r1-=1
elif d==n-1:
if r1>l1:
r1-=1
else:
l1-=1
if r==n:
u1-=1
d1-=1
elif r==n-1:
if u1>d1:
u1-=1
else:
d1-=1
if l==n:
u1-=1
d1-=1
elif l==n-1:
if u1>d1:
u1-=1
else:
d1-=1
if u1>=0 and d1>=0 and l1>=0 and r1>=0:
print("YES")
else:
print("NO")
def main():
test=1
test=int(input())
for _ in range(test):
solve()
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | output | 1 | 73,561 | 7 | 147,123 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland crossword is a puzzle that is solved on a square grid with n rows and n columns. Initially all the cells are white.
To solve the puzzle one has to color some cells on the border of the grid black in such a way that:
* exactly U cells in the top row are black;
* exactly R cells in the rightmost column are black;
* exactly D cells in the bottom row are black;
* exactly L cells in the leftmost column are black.
Note that you can color zero cells black and leave every cell white.
Your task is to check if there exists a solution to the given puzzle.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains 5 integers n, U, R, D, L (2 ≤ n ≤ 100; 0 ≤ U, R, D, L ≤ n).
Output
For each testcase print "YES" if the solution exists and "NO" otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
5 2 5 3 1
3 0 0 0 0
4 4 1 4 0
2 1 1 1 1
Output
YES
YES
NO
YES
Note
Here are possible solutions to testcases 1, 2 and 4:
<image> | instruction | 0 | 73,562 | 7 | 147,124 |
Tags: bitmasks, brute force, greedy, implementation
Correct Solution:
```
def solve(n,u,r,d,l):
u1=u
#Squares to be painted on each side
r1,d1,l1=r,d,l
if(u==n):
r1-=1
l1-=1
if(u==n-1):
if(l1>r1):
l1-=1
else:
r1-=1
if(r==n):
u1-=1
d1-=1
if(r==n-1):
if(u1>d1):
u1-=1
else:
d1-=1
if(d==n):
l1-=1
r1-=1
if(d==n-1):
if(l1>r1):
l1-=1
else:
r1-=1
if(l==n):
u1-=1
d1-=1
if(l==n-1):
if(u1>d1):
u1-=1
else:
d1-=1
if(u1<0 or r1<0 or d1<0 or l1<0 ):
return "NO"
else:
return "YES"
t= int(input())
for _ in range(t):
n,u,r,d,l = map(int,input().split())
print(solve(n,u,r,d,l))
``` | output | 1 | 73,562 | 7 | 147,125 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland crossword is a puzzle that is solved on a square grid with n rows and n columns. Initially all the cells are white.
To solve the puzzle one has to color some cells on the border of the grid black in such a way that:
* exactly U cells in the top row are black;
* exactly R cells in the rightmost column are black;
* exactly D cells in the bottom row are black;
* exactly L cells in the leftmost column are black.
Note that you can color zero cells black and leave every cell white.
Your task is to check if there exists a solution to the given puzzle.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains 5 integers n, U, R, D, L (2 ≤ n ≤ 100; 0 ≤ U, R, D, L ≤ n).
Output
For each testcase print "YES" if the solution exists and "NO" otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
5 2 5 3 1
3 0 0 0 0
4 4 1 4 0
2 1 1 1 1
Output
YES
YES
NO
YES
Note
Here are possible solutions to testcases 1, 2 and 4:
<image> | instruction | 0 | 73,563 | 7 | 147,126 |
Tags: bitmasks, brute force, greedy, implementation
Correct Solution:
```
# from __future__ import print_function
# import sys
# def eprint(*args, **kwargs): print(*args, file=sys.stderr, **kwargs)
def solution(n,u,r,d,l):
maxfilled=u+r+d+l
minfilled=u+r+d+l-4
found = False
# for i in range(2**(4*(n-1))):
for i in range(16):
ur,rd,dl,lu=(int(x) for x in bin(i)[2:].zfill(4))
# eprint(ur,rd,dl,lu)
n2=n-2
if (u-ur-lu<=n2) and (r-ur-rd<=n2) and (d-rd-dl<=n2) and (l-dl-lu<=n2)\
and (u-ur-lu>=0) and (r-ur-rd>=0) and (d-rd-dl>=0) and (l-dl-lu>=0):
# eprint(n2,u,r,d,l,ur,rd,dl,lu)
found=True
break
if found:
print("YES")
else:
print("NO")
def main():
for _ in range(int(input())):
solution(*map(int,input().split()))
if __name__ == "__main__":
main()
``` | output | 1 | 73,563 | 7 | 147,127 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland crossword is a puzzle that is solved on a square grid with n rows and n columns. Initially all the cells are white.
To solve the puzzle one has to color some cells on the border of the grid black in such a way that:
* exactly U cells in the top row are black;
* exactly R cells in the rightmost column are black;
* exactly D cells in the bottom row are black;
* exactly L cells in the leftmost column are black.
Note that you can color zero cells black and leave every cell white.
Your task is to check if there exists a solution to the given puzzle.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains 5 integers n, U, R, D, L (2 ≤ n ≤ 100; 0 ≤ U, R, D, L ≤ n).
Output
For each testcase print "YES" if the solution exists and "NO" otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
5 2 5 3 1
3 0 0 0 0
4 4 1 4 0
2 1 1 1 1
Output
YES
YES
NO
YES
Note
Here are possible solutions to testcases 1, 2 and 4:
<image> | instruction | 0 | 73,564 | 7 | 147,128 |
Tags: bitmasks, brute force, greedy, implementation
Correct Solution:
```
tt = int(input())
while tt:
n, U, R, D, L = map(int, input().split())
for mask in range(16):
rU, rR, rD, rL = U, R, D, L
if mask & 1:
rU -= 1
rL -= 1
if mask & 2:
rL -= 1
rD -= 1
if mask & 4:
rD -= 1
rR -= 1
if mask & 8:
rR -= 1
rU -= 1
if min(rU, rR, rD, rL) >= 0 and max(rU, rR, rD, rL) <= n - 2:
print("YES")
break
else:
print("NO")
tt -= 1
``` | output | 1 | 73,564 | 7 | 147,129 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland crossword is a puzzle that is solved on a square grid with n rows and n columns. Initially all the cells are white.
To solve the puzzle one has to color some cells on the border of the grid black in such a way that:
* exactly U cells in the top row are black;
* exactly R cells in the rightmost column are black;
* exactly D cells in the bottom row are black;
* exactly L cells in the leftmost column are black.
Note that you can color zero cells black and leave every cell white.
Your task is to check if there exists a solution to the given puzzle.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains 5 integers n, U, R, D, L (2 ≤ n ≤ 100; 0 ≤ U, R, D, L ≤ n).
Output
For each testcase print "YES" if the solution exists and "NO" otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
5 2 5 3 1
3 0 0 0 0
4 4 1 4 0
2 1 1 1 1
Output
YES
YES
NO
YES
Note
Here are possible solutions to testcases 1, 2 and 4:
<image> | instruction | 0 | 73,565 | 7 | 147,130 |
Tags: bitmasks, brute force, greedy, implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
for f in range(int(input())):
n,u,r,d,l=map(int,input().split())
mnv=mnh=0
if u>=n-1:
mnh+=u-n+2
if r>=n-1:
mnv+=r-n+2
if d>=n-1:
mnh+=d-n+2
if l>=n-1:
mnv+=l-n+2
ans="YES"
if u+d<mnv:
ans="NO"
if r+l<mnh:
ans="NO"
if (u==n or d==n) and (r==0 or l==0):
ans="NO"
if (r==n or l==n) and (u==0 or d==0):
ans="NO"
if (u==n and d==n) and (r==1 or l==1):
ans="NO"
if (r==n and l==n) and (u==1 or d==1):
ans="NO"
print(ans)
``` | output | 1 | 73,565 | 7 | 147,131 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland crossword is a puzzle that is solved on a square grid with n rows and n columns. Initially all the cells are white.
To solve the puzzle one has to color some cells on the border of the grid black in such a way that:
* exactly U cells in the top row are black;
* exactly R cells in the rightmost column are black;
* exactly D cells in the bottom row are black;
* exactly L cells in the leftmost column are black.
Note that you can color zero cells black and leave every cell white.
Your task is to check if there exists a solution to the given puzzle.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains 5 integers n, U, R, D, L (2 ≤ n ≤ 100; 0 ≤ U, R, D, L ≤ n).
Output
For each testcase print "YES" if the solution exists and "NO" otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
5 2 5 3 1
3 0 0 0 0
4 4 1 4 0
2 1 1 1 1
Output
YES
YES
NO
YES
Note
Here are possible solutions to testcases 1, 2 and 4:
<image> | instruction | 0 | 73,566 | 7 | 147,132 |
Tags: bitmasks, brute force, greedy, implementation
Correct Solution:
```
t = int(input())
for k in range(t):
n, U, R, D, L = map(int, input().split())
pr = 0
for i in range(16):
m = []
for j in range(4):
dd = (i >> j) & 1
m.append(dd)
ul, ur, dl, dr = m[0], m[1], m[2], m[3]
p1 = 0 <= U - ul - ur <= n - 2
p2 = 0 <= R - dr - ur <= n - 2
p3 = 0 <= D - dl - dr <= n - 2
p4 = 0 <= L - ul - dl <= n - 2
if p1 and p2 and p3 and p4 and p1 == True:
print('YES')
pr = 1
break
if pr == 0:
print('NO')
``` | output | 1 | 73,566 | 7 | 147,133 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland crossword is a puzzle that is solved on a square grid with n rows and n columns. Initially all the cells are white.
To solve the puzzle one has to color some cells on the border of the grid black in such a way that:
* exactly U cells in the top row are black;
* exactly R cells in the rightmost column are black;
* exactly D cells in the bottom row are black;
* exactly L cells in the leftmost column are black.
Note that you can color zero cells black and leave every cell white.
Your task is to check if there exists a solution to the given puzzle.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains 5 integers n, U, R, D, L (2 ≤ n ≤ 100; 0 ≤ U, R, D, L ≤ n).
Output
For each testcase print "YES" if the solution exists and "NO" otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
5 2 5 3 1
3 0 0 0 0
4 4 1 4 0
2 1 1 1 1
Output
YES
YES
NO
YES
Note
Here are possible solutions to testcases 1, 2 and 4:
<image> | instruction | 0 | 73,567 | 7 | 147,134 |
Tags: bitmasks, brute force, greedy, implementation
Correct Solution:
```
def solution():
n, u, r, d, l = map(int, input().split())
if max(u, r, d, l) <= n - 2:
print('YES')
elif max(u, r, d, l) == n - 1:
if u == n - 1 and d == n - 1 and (l + r < 2):
print('NO')
elif r == n - 1 and l == n - 1 and (u + d < 2):
print('NO')
elif (u == n - 1 or d == n - 1) and (l + r < 1):
print('NO')
elif (l == n - 1 or r == n - 1) and (u + d < 1):
print('NO')
else:
print('YES')
else:
if u == n and d == n and (l < 2 or r < 2):
print('NO')
elif l == n and r == n and (u < 2 or d < 2):
print('NO')
elif (u == n and d == n - 1 or d == n and u == n - 1) and (l + r < 3):
print('NO')
elif (r == n and l == n - 1 or l == n and r == n - 1) and (u + d < 3):
print('NO')
elif (u == n or d == n) and (l == 0 or r == 0):
print('NO')
elif (l == n or r == n) and (u == 0 or d == 0):
print('NO')
else:
print('YES')
for t in range(int(input())):
solution()
``` | output | 1 | 73,567 | 7 | 147,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland crossword is a puzzle that is solved on a square grid with n rows and n columns. Initially all the cells are white.
To solve the puzzle one has to color some cells on the border of the grid black in such a way that:
* exactly U cells in the top row are black;
* exactly R cells in the rightmost column are black;
* exactly D cells in the bottom row are black;
* exactly L cells in the leftmost column are black.
Note that you can color zero cells black and leave every cell white.
Your task is to check if there exists a solution to the given puzzle.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains 5 integers n, U, R, D, L (2 ≤ n ≤ 100; 0 ≤ U, R, D, L ≤ n).
Output
For each testcase print "YES" if the solution exists and "NO" otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
5 2 5 3 1
3 0 0 0 0
4 4 1 4 0
2 1 1 1 1
Output
YES
YES
NO
YES
Note
Here are possible solutions to testcases 1, 2 and 4:
<image>
Submitted Solution:
```
"""
- math -
from math import factorial
from math import gcd
from math import pi
large_p = 10**9 + 7
from itertools import accumulate # 累積和
from operator import mul
- other -
from collections import Counter
from itertools import combinations
from itertools import combinations_with_replacement
from itertools import permutations
from operator import itemgetter
from functools import reduce
- list -
from copy import deepcopy
from collections import deque
from heapq import heapify, heappop, heappush
- run -
sys.setrecursionlimit(10**6)
from time import sleep
from numba import njit
import numpy as np
@njit('(i4[::1],)', cache=True)
- output -
p2d = lambda x: print(*x, sep="\n")
p2line = lambda x: print(*x, sep=" ")
print(*list1, sep="\n")
- input -
import sys
read = sys.stdin.read
readline = sys.stdin.readline
readlines = sys.stdin.readlines
= [tuple(map(int, input().split())) for _ in range(n)]
= [tuple(map(int, readline().split())) for _ in range(n)]
m = map(int, read().split()) /ab = zip(m, m) /for a, b in ab:
ps = [line.split() for line in readlines()] # inputの最後まで読んでしまうので注意。
n, *a = map(int, read().split())
a = tuple(map(int, read().split())) # 1行１文字などの入力に。
a = list(map(int, readlines()))
a = np.array(read().split(), np.int32) # int。全入力をread。
a = np.array(list(read().rstrip())) #文字
a = np.array(readline().split(), np.int64) #int。１行read。
a, b = map(int, input().split())
n, k, *x = map(int, read().split())
"""
# ATC03
import sys
read = sys.stdin.read
from itertools import product
def main():
t = int(input())
pros = tuple(product((1, 0), repeat=4))
cases = tuple(map(int, read().split()))
for i1 in range(t):
n, u, r, d, l = cases[i1*5:i1*5+5]
for pro in pros:
j = [False] * 4
j[0] = 0 <= u - pro[0] - pro[1] <= n - 2
j[1] = 0 <= r - pro[1] - pro[2] <= n - 2
j[2] = 0 <= d - pro[2] - pro[3] <= n - 2
j[3] = 0 <= l - pro[3] - pro[0] <= n - 2
if all(j):
print('YES')
break
else:
print('NO')
if __name__ == '__main__':
main()
``` | instruction | 0 | 73,568 | 7 | 147,136 |
Yes | output | 1 | 73,568 | 7 | 147,137 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland crossword is a puzzle that is solved on a square grid with n rows and n columns. Initially all the cells are white.
To solve the puzzle one has to color some cells on the border of the grid black in such a way that:
* exactly U cells in the top row are black;
* exactly R cells in the rightmost column are black;
* exactly D cells in the bottom row are black;
* exactly L cells in the leftmost column are black.
Note that you can color zero cells black and leave every cell white.
Your task is to check if there exists a solution to the given puzzle.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains 5 integers n, U, R, D, L (2 ≤ n ≤ 100; 0 ≤ U, R, D, L ≤ n).
Output
For each testcase print "YES" if the solution exists and "NO" otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
5 2 5 3 1
3 0 0 0 0
4 4 1 4 0
2 1 1 1 1
Output
YES
YES
NO
YES
Note
Here are possible solutions to testcases 1, 2 and 4:
<image>
Submitted Solution:
```
t=int(input())
def check(com,seq,n):
ft=0
l=[]
for i in range(4):
app=seq[i]-com[i]
if(app<0):
ft+=1
l.append(app)
if(ft==0 and n not in l and n-1 not in l):
return 1
return 0
for lp in range(t):
n,U,R,D,L=[int(y) for y in input().split()]
nc=(U,R,D,L)
#1 2
#3 4
l=[]
for k1 in range(2):
for k2 in range(2):
for k3 in range(2):
for k4 in range(2):
l.append((k1+k2,k2+k4,k3+k4,k1+k3))
for com in l:
if(check(com,nc,n)==1):
print("YES")
break
else:
print("NO")
``` | instruction | 0 | 73,569 | 7 | 147,138 |
Yes | output | 1 | 73,569 | 7 | 147,139 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland crossword is a puzzle that is solved on a square grid with n rows and n columns. Initially all the cells are white.
To solve the puzzle one has to color some cells on the border of the grid black in such a way that:
* exactly U cells in the top row are black;
* exactly R cells in the rightmost column are black;
* exactly D cells in the bottom row are black;
* exactly L cells in the leftmost column are black.
Note that you can color zero cells black and leave every cell white.
Your task is to check if there exists a solution to the given puzzle.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains 5 integers n, U, R, D, L (2 ≤ n ≤ 100; 0 ≤ U, R, D, L ≤ n).
Output
For each testcase print "YES" if the solution exists and "NO" otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
5 2 5 3 1
3 0 0 0 0
4 4 1 4 0
2 1 1 1 1
Output
YES
YES
NO
YES
Note
Here are possible solutions to testcases 1, 2 and 4:
<image>
Submitted Solution:
```
for _ in range(int(input())):
n, u, r, d, l = map(int, input().split())
arr = [u,r,d,l]
req = [0,0,0,0]
for i in range(4):
if arr[i]==n:
try:
req[i-1]+=1
except:
req[3]+=1
try:
req[i+1]+=1
except:
req[0]+=1
elif arr[i] == n-1:
if i-1<0:
left = 3
else:
left = i-1
if i+1>3:
right = 0
else:
right = i+1
if arr[left]>=arr[right]:
req[left]+=1
else:
req[right]+=1
flag = 0
for i in range(4):
if arr[i]<req[i]:
flag = 1
if flag == 0:
print("YES")
else:
print("NO")
``` | instruction | 0 | 73,570 | 7 | 147,140 |
Yes | output | 1 | 73,570 | 7 | 147,141 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland crossword is a puzzle that is solved on a square grid with n rows and n columns. Initially all the cells are white.
To solve the puzzle one has to color some cells on the border of the grid black in such a way that:
* exactly U cells in the top row are black;
* exactly R cells in the rightmost column are black;
* exactly D cells in the bottom row are black;
* exactly L cells in the leftmost column are black.
Note that you can color zero cells black and leave every cell white.
Your task is to check if there exists a solution to the given puzzle.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains 5 integers n, U, R, D, L (2 ≤ n ≤ 100; 0 ≤ U, R, D, L ≤ n).
Output
For each testcase print "YES" if the solution exists and "NO" otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
5 2 5 3 1
3 0 0 0 0
4 4 1 4 0
2 1 1 1 1
Output
YES
YES
NO
YES
Note
Here are possible solutions to testcases 1, 2 and 4:
<image>
Submitted Solution:
```
def gen_case():
l = [0, 0, 0, 0, 0]
while l[4] == 0:
l[0] += 1
for i in range(len(l) - 1):
l[i + 1] += l[i] // 2
l[i] %= 2
yield {key: elem for key, elem in zip(['UL', 'UR', 'DR', 'DL'], l[0:4])}
return
def check(case, n, u, r, d, l):
if case['UL'] == 1:
u -= 1
l -= 1
if case['UR'] == 1:
u -= 1
r -= 1
if case['DR'] == 1:
d -= 1
r -= 1
if case['DL'] == 1:
d -= 1
l -= 1
n -= 2
for elem in (u, r, d, l):
if elem < 0 or elem > n:
return False
return True
def solve():
for _ in range(int(input())):
n, u, r, d, l = map(int, input().split())
print(
{True: 'YES', False: 'NO'}
[True in [check(case, n, u, r, d, l) for case in gen_case()]]
)
solve()
``` | instruction | 0 | 73,571 | 7 | 147,142 |
Yes | output | 1 | 73,571 | 7 | 147,143 |
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