message stringlengths 2 44.5k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 42 109k | cluster float64 5 5 | __index_level_0__ int64 84 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
We have N+1 integers: 10^{100}, 10^{100}+1, ..., 10^{100}+N.
We will choose K or more of these integers. Find the number of possible values of the sum of the chosen numbers, modulo (10^9+7).
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq K \leq N+1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
Output
Print the number of possible values of the sum, modulo (10^9+7).
Examples
Input
3 2
Output
10
Input
200000 200001
Output
1
Input
141421 35623
Output
220280457 | instruction | 0 | 39,388 | 5 | 78,776 |
"Correct Solution:
```
N, K = map(int, input().split())
sm = 0
for i in range(K, N+2):
sm += (N+N-i+1)*i//2 -(i-1)*i//2 + 1
print(sm%(10**9+7))
``` | output | 1 | 39,388 | 5 | 78,777 |
Provide a correct Python 3 solution for this coding contest problem.
We have N+1 integers: 10^{100}, 10^{100}+1, ..., 10^{100}+N.
We will choose K or more of these integers. Find the number of possible values of the sum of the chosen numbers, modulo (10^9+7).
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq K \leq N+1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
Output
Print the number of possible values of the sum, modulo (10^9+7).
Examples
Input
3 2
Output
10
Input
200000 200001
Output
1
Input
141421 35623
Output
220280457 | instruction | 0 | 39,389 | 5 | 78,778 |
"Correct Solution:
```
n,k = map(int,input().split())
mod = 10**9+7
print(((n**3+3*n**2+8*n+2*(k-1)**3-8*(k-1)-3*(k-1)**2*n-3*(k-1)*n)//6+1) % mod)
``` | output | 1 | 39,389 | 5 | 78,779 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N+1 integers: 10^{100}, 10^{100}+1, ..., 10^{100}+N.
We will choose K or more of these integers. Find the number of possible values of the sum of the chosen numbers, modulo (10^9+7).
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq K \leq N+1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
Output
Print the number of possible values of the sum, modulo (10^9+7).
Examples
Input
3 2
Output
10
Input
200000 200001
Output
1
Input
141421 35623
Output
220280457
Submitted Solution:
```
N, K=map(int,input().split())
print(int(((N+1)*(N+K+1)*(N-K+2)/2+N-K+2-(N+1)*(N+2)*(2*N+3)/6+(K-1)*K*(2*K-1)/6)%(10**9+7)))
``` | instruction | 0 | 39,390 | 5 | 78,780 |
Yes | output | 1 | 39,390 | 5 | 78,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N+1 integers: 10^{100}, 10^{100}+1, ..., 10^{100}+N.
We will choose K or more of these integers. Find the number of possible values of the sum of the chosen numbers, modulo (10^9+7).
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq K \leq N+1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
Output
Print the number of possible values of the sum, modulo (10^9+7).
Examples
Input
3 2
Output
10
Input
200000 200001
Output
1
Input
141421 35623
Output
220280457
Submitted Solution:
```
N, K = map(int,input().split())
ans = 0
for i in range(K,N+2):
ans += (2*N+1-i)*i//2-(i-1)*i//2+1
print(ans%(10**9+7))
``` | instruction | 0 | 39,391 | 5 | 78,782 |
Yes | output | 1 | 39,391 | 5 | 78,783 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N+1 integers: 10^{100}, 10^{100}+1, ..., 10^{100}+N.
We will choose K or more of these integers. Find the number of possible values of the sum of the chosen numbers, modulo (10^9+7).
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq K \leq N+1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
Output
Print the number of possible values of the sum, modulo (10^9+7).
Examples
Input
3 2
Output
10
Input
200000 200001
Output
1
Input
141421 35623
Output
220280457
Submitted Solution:
```
N,K = map(int, input().split())
div = 10**9 + 7
ans = 0
for i in range(K, N+2):
ans += i + N*i - i**2 + 1
print(ans%div)
``` | instruction | 0 | 39,392 | 5 | 78,784 |
Yes | output | 1 | 39,392 | 5 | 78,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N+1 integers: 10^{100}, 10^{100}+1, ..., 10^{100}+N.
We will choose K or more of these integers. Find the number of possible values of the sum of the chosen numbers, modulo (10^9+7).
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq K \leq N+1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
Output
Print the number of possible values of the sum, modulo (10^9+7).
Examples
Input
3 2
Output
10
Input
200000 200001
Output
1
Input
141421 35623
Output
220280457
Submitted Solution:
```
n,k=map(int,input().split())
print((n-k+2)*((-k-~n)*(n+2*k)+6)//6%(10**9+7))
``` | instruction | 0 | 39,393 | 5 | 78,786 |
Yes | output | 1 | 39,393 | 5 | 78,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N+1 integers: 10^{100}, 10^{100}+1, ..., 10^{100}+N.
We will choose K or more of these integers. Find the number of possible values of the sum of the chosen numbers, modulo (10^9+7).
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq K \leq N+1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
Output
Print the number of possible values of the sum, modulo (10^9+7).
Examples
Input
3 2
Output
10
Input
200000 200001
Output
1
Input
141421 35623
Output
220280457
Submitted Solution:
```
from scipy.special import comb
N, K = map(int, input().split())
result = 0
if (N + 1) % 2 != 0:
for i in range(K, N + 2):
sycle = comb(N + 1, i, exact=True) - comb((N + 1) // 2, i // 2, exact=True) + 1
print(sycle)
result += sycle % (10 ** 9 + 7)
else:
for i in range(K, N + 2):
if i % 2 == 0:
sycle = (
comb(N + 1, i, exact=True) - comb((N + 1) // 2, i // 2, exact=True) + 1
)
else:
sycle = comb(N + 1, i, exact=True)
result += sycle % (10 ** 9 + 7)
print(sycle)
print(result % (10 ** 9 + 7))
``` | instruction | 0 | 39,394 | 5 | 78,788 |
No | output | 1 | 39,394 | 5 | 78,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N+1 integers: 10^{100}, 10^{100}+1, ..., 10^{100}+N.
We will choose K or more of these integers. Find the number of possible values of the sum of the chosen numbers, modulo (10^9+7).
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq K \leq N+1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
Output
Print the number of possible values of the sum, modulo (10^9+7).
Examples
Input
3 2
Output
10
Input
200000 200001
Output
1
Input
141421 35623
Output
220280457
Submitted Solution:
```
N, K = map(int, input().split())
c = 0
for i in range(K, N + 1 + 1):
ma = ((N + N - (i - 1)) * i) / 2
mi = ((0 + i - 1) * i) / 2
c += int(ma - mi + 1)
c %= (10 ** 9 + 7)
print(c)
``` | instruction | 0 | 39,395 | 5 | 78,790 |
No | output | 1 | 39,395 | 5 | 78,791 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N+1 integers: 10^{100}, 10^{100}+1, ..., 10^{100}+N.
We will choose K or more of these integers. Find the number of possible values of the sum of the chosen numbers, modulo (10^9+7).
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq K \leq N+1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
Output
Print the number of possible values of the sum, modulo (10^9+7).
Examples
Input
3 2
Output
10
Input
200000 200001
Output
1
Input
141421 35623
Output
220280457
Submitted Solution:
```
n, k = map(int, input().split())
MOD = 10**9 + 7
# for k = n
result = 1 % MOD
cand = [i for i in range(0, n+1)]
for ki in range(k, n+1):
if ki == 1:
result += (n+1) % MOD
elif ki == 2:
result += (2*n-1) % MOD
else:
minans = (ki-1) * ((ki-1)//2)
maxans = (2*n-ki+1) * (ki//2)
if ki % 2 == 1:
minans += int((ki-1)/2)
maxans += int((2*n-ki+1)/2)
result += (maxans - minans + 1) % MOD
print(result % MOD)
``` | instruction | 0 | 39,396 | 5 | 78,792 |
No | output | 1 | 39,396 | 5 | 78,793 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N+1 integers: 10^{100}, 10^{100}+1, ..., 10^{100}+N.
We will choose K or more of these integers. Find the number of possible values of the sum of the chosen numbers, modulo (10^9+7).
Constraints
* 1 \leq N \leq 2\times 10^5
* 1 \leq K \leq N+1
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N K
Output
Print the number of possible values of the sum, modulo (10^9+7).
Examples
Input
3 2
Output
10
Input
200000 200001
Output
1
Input
141421 35623
Output
220280457
Submitted Solution:
```
n,k=map(int,input().split())
ans=0
x=list(range(n+1))
for i in range(k,n+1):
A=sum(x[n-i+1:n+1])
B=sum(x[0:i])
ans+=A-B+1
print(ans)
``` | instruction | 0 | 39,397 | 5 | 78,794 |
No | output | 1 | 39,397 | 5 | 78,795 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer N.
Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.
* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.
Constraints
* 2 \leq N \leq 500
Input
Input is given from Standard Input in the following format:
N
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Example
Input
2
Output
4 7
23 10 | instruction | 0 | 39,446 | 5 | 78,892 |
"Correct Solution:
```
import fractions
import sys
def lcm(a,b):
return a * b // fractions.gcd(a,b)
N = int(input())
if N == 2:
print ("4 7")
print ("23 10")
sys.exit()
lis = [2]
now = 3
while len(lis) <= 2 * N:
flag = True
for i in lis:
if now % i == 0:
flag = False
break
if flag:
lis.append(now)
now += 2
#print (lis)
ans = [ [1] * N for i in range(N) ]
for i in range(N):
for j in range(N):
if (i + j) % 2 == 0:
ans[i][j] *= lis[(i+j) // 2]
if N % 2 == 1:
for i in range(N):
for j in range(N):
if (i + abs(N-1 - j)) % 2 == 0:
ans[i][j] *= lis[ -1 - (i + abs(N-1 - j)) // 2 ]
else:
for i in range(N):
for j in range(N):
if (i + abs(N-1 - j)) % 2 == 1:
ans[i][j] *= lis[ -1 - ((i + abs(N-1 - j) -1 ) // 2)]
#for i in range(N):
# print (" ".join(map(str,ans[i])))
for i in range(N):
for j in range(N):
if (i + j) % 2 == 1:
t = 1
if i > 0:
t = lcm(t,ans[i-1][j])
if j > 0:
t = lcm(t,ans[i][j-1])
if i < N-1:
t = lcm(t,ans[i+1][j])
if j < N-1:
t = lcm(t,ans[i][j+1])
ans[i][j] = t+1
for i in range(N):
print (" ".join(map(str,ans[i])))
``` | output | 1 | 39,446 | 5 | 78,893 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer N.
Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.
* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.
Constraints
* 2 \leq N \leq 500
Input
Input is given from Standard Input in the following format:
N
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Example
Input
2
Output
4 7
23 10 | instruction | 0 | 39,448 | 5 | 78,896 |
"Correct Solution:
```
N = int(input())
if N == 2:
print(2, 3)
print(5, 4)
exit()
a = [[1] * N for _ in range(N)]
PN = [2]
L = 3
while len(PN) < N*2:
c = True
for p in PN:
if L % p == 0: c = False
if c == True: PN.append(L)
L += 2
def P1(k):
return PN[(k//2)]
def P2(k):
if k >= 0:
return PN[(k//2) + N]
else:
return PN[(k//2)]
for i in range(N):
for j in range(N):
if (i + j) % 2 == 0:
a[i][j] = P1(i + j) * P2(i - j)
else:
a[i][j] += P1(i + j + 1) * P1(i + j - 1) * P2(i - j + 1) * P2(i - j - 1)
for a_i in a:
print(*a_i)
``` | output | 1 | 39,448 | 5 | 78,897 |
Provide a correct Python 3 solution for this coding contest problem.
You are given an integer N.
Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.
* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.
Constraints
* 2 \leq N \leq 500
Input
Input is given from Standard Input in the following format:
N
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Example
Input
2
Output
4 7
23 10 | instruction | 0 | 39,453 | 5 | 78,906 |
"Correct Solution:
```
N = int(input())
if N == 2:
print(4,7)
print(23,10)
exit()
MAXN = 8000
sieve = [0,0] + [1]*MAXN
primes = []
p = 2
while p*p <= MAXN:
if sieve[p]:
primes.append(p)
for q in range(2*p,MAXN+1,p):
sieve[q] = 0
p += 1
for q in range(p,MAXN):
if sieve[q]:
primes.append(q)
mem = [[[] for j in range(N)] for i in range(N)]
pi = 0
for i in range(0,2*N-1,4):
for x in range(N):
y = i-x
if y >= N: continue
if y < 0: break
mem[y][x].append(primes[pi])
pi += 1
for i in reversed(range(2,2*N-1,4)):
for x in range(N):
y = i-x
if y >= N: continue
if y < 0: break
mem[y][x].append(primes[pi])
pi += 1
ini = -((N-1)//2*2)
for i in range(ini,N,4):
for x in range(N):
y = x-i
if y < 0: continue
if y >= N: break
mem[y][x].append(primes[pi])
pi += 1
for i in range(ini+2,N,4):
for x in range(N):
y = x-i
if y < 0: continue
if y >= N: break
mem[y][x].append(primes[pi])
pi += 1
ans = [[None]*N for i in range(N)]
dxy = [(1,0),(0,1),(-1,0),(0,-1)]
for y in range(N):
for x in range(N):
if (x+y)%2:
pset = set()
for dx,dy in dxy:
nx,ny = x+dx,y+dy
if not 0 <= nx < N: continue
if not 0 <= ny < N: continue
pset.add(mem[ny][nx][0])
pset.add(mem[ny][nx][1])
a = 1
for p in pset:
a *= p
ans[y][x] = a+1
else:
ans[y][x] = mem[y][x][0] * mem[y][x][1]
for row in ans:
print(*row)
``` | output | 1 | 39,453 | 5 | 78,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer N.
Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.
* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.
Constraints
* 2 \leq N \leq 500
Input
Input is given from Standard Input in the following format:
N
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Example
Input
2
Output
4 7
23 10
Submitted Solution:
```
N = int(input())
xzero = 3
yzero = 5
y = N+1
biggest = 8*(int(N/4)+1)*(int(N/4)+1)+4
a = [[0 for i in range(N+4)] for j in range(N+4)]
fours = [4*i for i in range((int(N/4)+1))]
odds = [2*i for i in range(int((N+1)/2))]
evens = [2*i+1 for i in range(int(N/2))]
if N==2:
print("4 7")
print("23 10")
else:
x = xzero
for i in fours:
for j in fours:
a[i][j] = x
a[i+1][j+1] = 2*x
a[i][j+2]=biggest-x
a[i+1][j+3] = 2*(biggest-x)
a[i+2][j] = biggest-x-(int(N/4)+1)*2
a[i+3][j+1] = 2*(biggest-x-(int(N/4)+1)*2)
a[i+2][j+2] = x+(int(N/4)+1)*2
a[i+3][j+3] = 2*(x+(int(N/4)+1)*2)
x += 2
x+=(int(N/4)+1)*2
for i in odds:
for j in evens:
ans = 1
ans *= a[i][j - 1]
ans *= a[i][j + 1]
if i > 0:
ans *= a[i - 1][j]
else:
ans *= 2
ans += 1
a[i][j] = ans
for i in evens:
for j in odds:
ans = 1
ans *= a[i - 1][j]
ans *= a[i + 1][j]
if j > 0:
ans *= a[i][j - 1]
else:
ans *= 2
ans += 1
a[i][j] = ans
a = [i[0:N] for i in a[0:N]]
for i in range(N):
print(" ".join(str(j) for j in a[i]))
``` | instruction | 0 | 39,454 | 5 | 78,908 |
Yes | output | 1 | 39,454 | 5 | 78,909 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer N.
Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.
* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.
Constraints
* 2 \leq N \leq 500
Input
Input is given from Standard Input in the following format:
N
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Example
Input
2
Output
4 7
23 10
Submitted Solution:
```
import numpy as np
N = int(input())
def search(num1,num2, m):
i = 1
while(True):
ans = i * num1 + m
if ans % num2 == m:
return ans
i += 1
m = 3
ans = np.zeros((N,N), dtype = np.uint64)
ans[0,0] = 4
ans[0,1] = 7
a,b = 0,1
for i in range(N-1):
if i % 2 == 0:
for j in range(i+1):
a += 1
if ans[a, b-1]== 0:
ans[a,b] = ans[a-1, b] + m
else:
ans[a,b] = search(ans[a-1, b], ans[a, b-1], m)
for j in range(i+1):
b -= 1
ans[a,b] = search(ans[a, b+1],ans[a-1, b], m)
if i+2 < N:
a += 1
ans[a,b] = ans[a-1,b] + m
else:
for j in range(i+1):
b += 1
if ans[a-1, b] == 0:
ans[a,b] = ans[a,b-1]+m
else:
ans[a,b] = search(ans[a, b-1], ans[a-1, b], m)
for j in range(i+1):
a -= 1
ans[a,b] = search(ans[a+1, b], ans[a, b-1], m)
if i + 2 < N:
b += 1
ans[a,b] = ans[a, b-1] + m
for i in range(N):
a = list(ans[i,:])
print(' '.join(map(str, a)))
``` | instruction | 0 | 39,455 | 5 | 78,910 |
No | output | 1 | 39,455 | 5 | 78,911 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer N.
Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.
* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.
Constraints
* 2 \leq N \leq 500
Input
Input is given from Standard Input in the following format:
N
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Example
Input
2
Output
4 7
23 10
Submitted Solution:
```
from math import gcd
Primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999, 2003, 2011, 2017, 2027, 2029, 2039, 2053, 2063, 2069, 2081, 2083, 2087, 2089, 2099, 2111, 2113, 2129, 2131, 2137, 2141, 2143, 2153, 2161, 2179, 2203, 2207, 2213, 2221, 2237, 2239, 2243, 2251, 2267, 2269, 2273, 2281, 2287, 2293, 2297, 2309, 2311, 2333, 2339, 2341, 2347, 2351, 2357, 2371, 2377, 2381, 2383, 2389, 2393, 2399, 2411, 2417, 2423, 2437, 2441, 2447, 2459, 2467, 2473, 2477, 2503, 2521, 2531, 2539, 2543, 2549, 2551, 2557, 2579, 2591, 2593, 2609, 2617, 2621, 2633, 2647, 2657, 2659, 2663, 2671, 2677, 2683, 2687, 2689, 2693, 2699, 2707, 2711, 2713, 2719, 2729, 2731, 2741, 2749, 2753, 2767, 2777, 2789, 2791, 2797, 2801, 2803, 2819, 2833, 2837, 2843, 2851, 2857, 2861, 2879, 2887, 2897, 2903, 2909, 2917, 2927, 2939, 2953, 2957, 2963, 2969, 2971, 2999, 3001, 3011, 3019, 3023, 3037, 3041, 3049, 3061, 3067, 3079, 3083, 3089, 3109, 3119, 3121, 3137, 3163, 3167, 3169, 3181, 3187, 3191, 3203, 3209, 3217, 3221, 3229, 3251, 3253, 3257, 3259, 3271, 3299, 3301, 3307, 3313, 3319, 3323, 3329, 3331, 3343, 3347, 3359, 3361, 3371, 3373, 3389, 3391, 3407, 3413, 3433, 3449, 3457, 3461, 3463, 3467, 3469, 3491, 3499, 3511, 3517, 3527, 3529, 3533, 3539, 3541, 3547, 3557, 3559, 3571, 3581, 3583, 3593, 3607, 3613, 3617, 3623, 3631, 3637, 3643, 3659, 3671, 3673, 3677, 3691, 3697, 3701, 3709, 3719, 3727, 3733, 3739, 3761, 3767, 3769, 3779, 3793, 3797, 3803, 3821, 3823, 3833, 3847, 3851, 3853, 3863, 3877, 3881, 3889, 3907, 3911, 3917, 3919, 3923, 3929, 3931, 3943, 3947, 3967, 3989, 4001, 4003, 4007, 4013, 4019, 4021, 4027, 4049, 4051, 4057, 4073, 4079, 4091, 4093, 4099, 4111, 4127, 4129, 4133, 4139, 4153, 4157, 4159, 4177, 4201, 4211, 4217, 4219, 4229, 4231, 4241, 4243, 4253, 4259, 4261, 4271, 4273, 4283, 4289, 4297, 4327, 4337, 4339, 4349, 4357, 4363, 4373, 4391, 4397, 4409, 4421, 4423, 4441, 4447, 4451, 4457, 4463, 4481, 4483, 4493, 4507, 4513, 4517, 4519, 4523, 4547, 4549, 4561, 4567, 4583, 4591, 4597, 4603, 4621, 4637, 4639, 4643, 4649, 4651, 4657, 4663, 4673, 4679, 4691, 4703, 4721, 4723, 4729, 4733, 4751, 4759, 4783, 4787, 4789, 4793, 4799, 4801, 4813, 4817, 4831, 4861, 4871, 4877, 4889, 4903, 4909, 4919, 4931, 4933, 4937, 4943, 4951, 4957, 4967, 4969, 4973, 4987, 4993, 4999, 5003, 5009, 5011, 5021, 5023, 5039, 5051, 5059, 5077, 5081, 5087, 5099, 5101, 5107, 5113, 5119, 5147, 5153, 5167, 5171, 5179, 5189, 5197, 5209, 5227, 5231, 5233, 5237, 5261, 5273, 5279, 5281, 5297, 5303, 5309, 5323, 5333, 5347, 5351, 5381, 5387, 5393, 5399, 5407, 5413, 5417, 5419, 5431, 5437, 5441, 5443, 5449, 5471, 5477, 5479, 5483, 5501, 5503, 5507, 5519, 5521, 5527, 5531, 5557, 5563, 5569, 5573, 5581, 5591, 5623, 5639, 5641, 5647, 5651, 5653, 5657, 5659, 5669, 5683, 5689, 5693, 5701, 5711, 5717, 5737, 5741, 5743, 5749, 5779, 5783, 5791, 5801, 5807, 5813, 5821, 5827, 5839, 5843, 5849, 5851, 5857, 5861, 5867, 5869, 5879, 5881, 5897, 5903, 5923, 5927, 5939, 5953, 5981, 5987, 6007, 6011, 6029, 6037, 6043, 6047, 6053, 6067, 6073, 6079, 6089, 6091, 6101, 6113, 6121, 6131, 6133, 6143, 6151, 6163, 6173, 6197, 6199, 6203, 6211, 6217, 6221, 6229, 6247, 6257, 6263, 6269, 6271, 6277, 6287, 6299, 6301, 6311, 6317, 6323, 6329, 6337, 6343, 6353, 6359, 6361, 6367, 6373, 6379, 6389, 6397, 6421, 6427, 6449, 6451, 6469, 6473, 6481, 6491, 6521, 6529, 6547, 6551, 6553, 6563, 6569, 6571, 6577, 6581, 6599, 6607, 6619, 6637, 6653, 6659, 6661, 6673, 6679, 6689, 6691, 6701, 6703, 6709, 6719, 6733, 6737, 6761, 6763, 6779, 6781, 6791, 6793, 6803, 6823, 6827, 6829, 6833, 6841, 6857, 6863, 6869, 6871, 6883, 6899, 6907, 6911, 6917, 6947, 6949, 6959, 6961, 6967, 6971, 6977, 6983, 6991, 6997, 7001, 7013, 7019, 7027, 7039, 7043, 7057, 7069, 7079, 7103, 7109, 7121, 7127, 7129, 7151, 7159, 7177, 7187, 7193, 7207, 7211, 7213, 7219, 7229, 7237, 7243, 7247, 7253, 7283, 7297, 7307, 7309, 7321, 7331, 7333, 7349, 7351, 7369, 7393, 7411, 7417, 7433, 7451, 7457, 7459, 7477, 7481, 7487, 7489, 7499, 7507, 7517, 7523, 7529, 7537, 7541, 7547, 7549, 7559, 7561, 7573, 7577, 7583, 7589, 7591, 7603, 7607, 7621, 7639, 7643, 7649, 7669, 7673, 7681, 7687, 7691, 7699, 7703, 7717, 7723, 7727, 7741, 7753, 7757, 7759, 7789, 7793, 7817, 7823, 7829, 7841, 7853, 7867, 7873, 7877, 7879, 7883, 7901, 7907, 7919, 7927, 7933, 7937, 7949, 7951, 7963, 7993, 8009, 8011, 8017, 8039, 8053, 8059, 8069, 8081, 8087, 8089, 8093, 8101, 8111, 8117, 8123, 8147, 8161, 8167, 8171, 8179, 8191, 8209, 8219, 8221, 8231, 8233, 8237, 8243, 8263, 8269, 8273, 8287, 8291, 8293, 8297, 8311, 8317, 8329, 8353, 8363, 8369, 8377, 8387, 8389, 8419, 8423, 8429, 8431, 8443, 8447, 8461, 8467, 8501, 8513, 8521, 8527, 8537, 8539, 8543, 8563, 8573, 8581, 8597, 8599, 8609, 8623, 8627, 8629, 8641, 8647, 8663, 8669, 8677, 8681, 8689, 8693, 8699, 8707, 8713, 8719, 8731, 8737, 8741, 8747, 8753, 8761, 8779, 8783, 8803, 8807, 8819, 8821, 8831, 8837, 8839, 8849, 8861, 8863, 8867, 8887, 8893, 8923, 8929, 8933, 8941, 8951, 8963, 8969, 8971, 8999, 9001, 9007, 9011, 9013, 9029, 9041, 9043, 9049, 9059, 9067, 9091, 9103, 9109, 9127, 9133, 9137, 9151, 9157, 9161, 9173, 9181, 9187, 9199, 9203, 9209, 9221, 9227, 9239, 9241, 9257, 9277, 9281, 9283, 9293, 9311, 9319, 9323, 9337, 9341, 9343, 9349, 9371, 9377, 9391, 9397, 9403, 9413, 9419, 9421, 9431, 9433, 9437, 9439, 9461, 9463, 9467, 9473, 9479, 9491, 9497, 9511, 9521, 9533, 9539, 9547, 9551, 9587, 9601, 9613, 9619, 9623, 9629, 9631, 9643, 9649, 9661, 9677, 9679, 9689, 9697, 9719, 9721, 9733, 9739, 9743, 9749, 9767, 9769, 9781, 9787, 9791, 9803, 9811, 9817, 9829, 9833, 9839, 9851, 9857, 9859, 9871, 9883, 9887, 9901, 9907, 9923, 9929, 9931, 9941, 9949, 9967, 9973]
N = int(input())
a = []
for i in range(N):
a.append([])
for j in range(N):
if (i + j) % 2 == 0:
a[i].append(Primes[250 + (i - j) // 2] * Primes[500 + (i + j) // 2])
else:
a[i].append(0)
for i in range(N):
for j in range(N):
if (i + j) % 2 != 0:
mul = 1
if i > 0:
mul = mul * a[i-1][j] // gcd(mul,a[i-1][j])
if j > 0:
mul = mul * a[i][j-1] // gcd(mul,a[i][j-1])
if i < N-1:
mul = mul * a[i+1][j] // gcd(mul,a[i+1][j])
if j < N - 1:
mul = mul * a[i][j+1] // gcd(mul,a[i][j+1])
mul += 1
a[i][j] = mul
for row in a:
print(" ".join(map(str,row)))
``` | instruction | 0 | 39,456 | 5 | 78,912 |
No | output | 1 | 39,456 | 5 | 78,913 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer N.
Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.
* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.
Constraints
* 2 \leq N \leq 500
Input
Input is given from Standard Input in the following format:
N
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Example
Input
2
Output
4 7
23 10
Submitted Solution:
```
import math, itertools
def genprime():
m = 10000
primes = []
vis = [True] * m
for i in range(2, m):
if vis[i]:
primes.append(i)
yield i
for p in primes:
foo = i * p
if foo >= m:
break
vis[foo] = False
if i % p == 0:
break
n = int(input())
res = [[1 for i in range(n)] for j in range(n)]
coord = [(i, j) for i in range(n) for j in range(n) if (i + j) % 2 == 0]
gen = genprime()
``` | instruction | 0 | 39,457 | 5 | 78,914 |
No | output | 1 | 39,457 | 5 | 78,915 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer N.
Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.
* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.
Constraints
* 2 \leq N \leq 500
Input
Input is given from Standard Input in the following format:
N
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Output
Print your solution in the following format:
a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}
Example
Input
2
Output
4 7
23 10
Submitted Solution:
```
import itertools
``` | instruction | 0 | 39,458 | 5 | 78,916 |
No | output | 1 | 39,458 | 5 | 78,917 |
Provide a correct Python 3 solution for this coding contest problem.
You have an integer sequence of length N: a_1, a_2, ..., a_N.
You repeatedly perform the following operation until the length of the sequence becomes 1:
* First, choose an element of the sequence.
* If that element is at either end of the sequence, delete the element.
* If that element is not at either end of the sequence, replace the element with the sum of the two elements that are adjacent to it. Then, delete those two elements.
You would like to maximize the final element that remains in the sequence.
Find the maximum possible value of the final element, and the way to achieve it.
Constraints
* All input values are integers.
* 2 \leq N \leq 1000
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
* In the first line, print the maximum possible value of the final element in the sequence.
* In the second line, print the number of operations that you perform.
* In the (2+i)-th line, if the element chosen in the i-th operation is the x-th element from the left in the sequence at that moment, print x.
* If there are multiple ways to achieve the maximum value of the final element, any of them may be printed.
Examples
Input
5
1 4 3 7 5
Output
11
3
1
4
2
Input
4
100 100 -1 100
Output
200
2
3
1
Input
6
-1 -2 -3 1 2 3
Output
4
3
2
1
2
Input
9
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
4
2
2
2
2 | instruction | 0 | 39,459 | 5 | 78,918 |
"Correct Solution:
```
#!/usr/bin/env python3
def solve(n, aa):
opa = []
a_max = max(aa)
if a_max <= 0:
print(a_max)
x = aa.index(a_max) + 1
for i in range(n, x, -1):
opa.append(i)
for i in range(1, x):
opa.append(1)
else:
sum0 = 0
sum1 = 0
for i in range(n):
a = aa[i]
if 0 < a:
if i % 2 == 0:
sum0 += a
else:
sum1 += a
print(max(sum0, sum1))
used = [False] * n
i0 = 1 if sum0 < sum1 else 0
for i in range(i0, n, 2):
if 0 < aa[i]:
used[i] = True
k = n
for i in range(n - 1, -1, -1):
if not used[i]:
opa.append(i + 1)
k -= 1
else:
break
j = 0
for i in range(k):
if not used[i]:
opa.append(1)
j += 1
else:
break
c = 0
for i in range(k - 1, j - 1, -1):
if used[i]:
c = 0
else:
c += 1
if c == 3:
opa.append(i - j + 2)
k -= 2
c = 1
for i in range((k - j) // 2):
opa.append(2)
print(len(opa))
for v in opa:
print(v)
def main():
n = input()
n = int(n)
aa = list(map(int, input().split()))
solve(n, aa)
if __name__ == '__main__':
main()
``` | output | 1 | 39,459 | 5 | 78,919 |
Provide a correct Python 3 solution for this coding contest problem.
You have an integer sequence of length N: a_1, a_2, ..., a_N.
You repeatedly perform the following operation until the length of the sequence becomes 1:
* First, choose an element of the sequence.
* If that element is at either end of the sequence, delete the element.
* If that element is not at either end of the sequence, replace the element with the sum of the two elements that are adjacent to it. Then, delete those two elements.
You would like to maximize the final element that remains in the sequence.
Find the maximum possible value of the final element, and the way to achieve it.
Constraints
* All input values are integers.
* 2 \leq N \leq 1000
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
* In the first line, print the maximum possible value of the final element in the sequence.
* In the second line, print the number of operations that you perform.
* In the (2+i)-th line, if the element chosen in the i-th operation is the x-th element from the left in the sequence at that moment, print x.
* If there are multiple ways to achieve the maximum value of the final element, any of them may be printed.
Examples
Input
5
1 4 3 7 5
Output
11
3
1
4
2
Input
4
100 100 -1 100
Output
200
2
3
1
Input
6
-1 -2 -3 1 2 3
Output
4
3
2
1
2
Input
9
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
4
2
2
2
2 | instruction | 0 | 39,460 | 5 | 78,920 |
"Correct Solution:
```
def examC():
ans = 0
print(ans)
return
def examD():
ans = 0
print(ans)
return
def examE():
N = I()
A = LI()
ans = -inf
S = [0]*(N+1)
fr = [-1]*N
best = -1
for i,a in enumerate(A):
S[i] = a
for j in range(i):
if ((i-j)%2==0):
if (S[j]+a>S[i]):
fr[i] = j
S[i] = S[j] + a
if (S[i]>ans):
ans = S[i]
best = i
#print(best)
V = []
for i in range(best+1,N)[::-1]:
V.append(i+1)
i = best
while(fr[i]>=0):
f = fr[i]
#print(i,f)
while(f<i):
V.append(1+(i+f)//2)
i -= 2
for _ in range(i):
V.append(1)
print(ans)
print(len(V))
for v in V:
print(v)
return
def examF():
ans = 0
print(ans)
return
from decimal import getcontext,Decimal as dec
import sys,bisect,itertools,heapq,math,random
from copy import deepcopy
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
def I(): return int(input())
def LI(): return list(map(int,sys.stdin.readline().split()))
def DI(): return dec(input())
def LDI(): return list(map(dec,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,_ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = dec("0.000000000001")
alphabet = [chr(ord('a') + i) for i in range(26)]
alphabet_convert = {chr(ord('a') + i): i for i in range(26)}
getcontext().prec = 28
sys.setrecursionlimit(10**7)
if __name__ == '__main__':
examE()
"""
142
12 9 1445 0 1
asd dfg hj o o
aidn
"""
``` | output | 1 | 39,460 | 5 | 78,921 |
Provide a correct Python 3 solution for this coding contest problem.
You have an integer sequence of length N: a_1, a_2, ..., a_N.
You repeatedly perform the following operation until the length of the sequence becomes 1:
* First, choose an element of the sequence.
* If that element is at either end of the sequence, delete the element.
* If that element is not at either end of the sequence, replace the element with the sum of the two elements that are adjacent to it. Then, delete those two elements.
You would like to maximize the final element that remains in the sequence.
Find the maximum possible value of the final element, and the way to achieve it.
Constraints
* All input values are integers.
* 2 \leq N \leq 1000
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
* In the first line, print the maximum possible value of the final element in the sequence.
* In the second line, print the number of operations that you perform.
* In the (2+i)-th line, if the element chosen in the i-th operation is the x-th element from the left in the sequence at that moment, print x.
* If there are multiple ways to achieve the maximum value of the final element, any of them may be printed.
Examples
Input
5
1 4 3 7 5
Output
11
3
1
4
2
Input
4
100 100 -1 100
Output
200
2
3
1
Input
6
-1 -2 -3 1 2 3
Output
4
3
2
1
2
Input
9
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
4
2
2
2
2 | instruction | 0 | 39,461 | 5 | 78,922 |
"Correct Solution:
```
from collections import defaultdict, Counter
from itertools import product, groupby, count, permutations, combinations
from math import pi, sqrt
from collections import deque
from bisect import bisect, bisect_left, bisect_right
from string import ascii_lowercase
from functools import lru_cache
import sys
sys.setrecursionlimit(10000)
INF = float("inf")
YES, Yes, yes, NO, No, no = "YES", "Yes", "yes", "NO", "No", "no"
dy4, dx4 = [0, 1, 0, -1], [1, 0, -1, 0]
dy8, dx8 = [0, -1, 0, 1, 1, -1, -1, 1], [1, 0, -1, 0, 1, 1, -1, -1]
def inside(y, x, H, W):
return 0 <= y < H and 0 <= x < W
def ceil(a, b):
return (a + b - 1) // b
def check(A, operations):
c = A[:]
for o in operations:
if o == 1:
A = A[1:]
elif o == len(A):
A = A[:len(A) - 1]
else:
o -= 1
A = A[:o - 1] + [A[o - 1] + A[o + 1]] + A[o + 2:]
if len(A) != 1:
print(c)
print(operations)
assert(len(A) == 1)
return A[0]
def solve(N, A):
if max(A) < 0:
ans = max(A)
idx = A.index(ans)
operations = []
for i in range(N, idx + 1, -1):
operations.append(i)
for i in range(1, idx + 1):
operations.append(1)
return ans, operations
ans = 0
middle = []
start, end = None, None
for i in range(0, N, 2):
if A[i] >= 0:
ans += A[i]
if start is None:
start = i
end = i
else:
if start is not None:
middle.append(i)
operations = []
num = 0
for i in range(end + 1, N):
operations.append(N - num)
num += 1
for i in middle[::-1]:
if start < i < end:
operations.append(i + 1)
num += 2
for i in range(start):
operations.append(1)
num += 1
for i in range((N - num) // 2):
operations.append(2)
return ans, operations
def main():
N = int(input())
A = list(map(int, input().split()))
ans1, ope1 = solve(N, A[:])
ans2, ope2 = solve(N - 1, A[1:])
ope2 = [1] + ope2
if ans1 > ans2:
print(ans1)
print(len(ope1))
print(*ope1, sep="\n")
else:
print(ans2)
print(len(ope2))
print(*ope2, sep="\n")
if __name__ == '__main__':
main()
``` | output | 1 | 39,461 | 5 | 78,923 |
Provide a correct Python 3 solution for this coding contest problem.
You have an integer sequence of length N: a_1, a_2, ..., a_N.
You repeatedly perform the following operation until the length of the sequence becomes 1:
* First, choose an element of the sequence.
* If that element is at either end of the sequence, delete the element.
* If that element is not at either end of the sequence, replace the element with the sum of the two elements that are adjacent to it. Then, delete those two elements.
You would like to maximize the final element that remains in the sequence.
Find the maximum possible value of the final element, and the way to achieve it.
Constraints
* All input values are integers.
* 2 \leq N \leq 1000
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
* In the first line, print the maximum possible value of the final element in the sequence.
* In the second line, print the number of operations that you perform.
* In the (2+i)-th line, if the element chosen in the i-th operation is the x-th element from the left in the sequence at that moment, print x.
* If there are multiple ways to achieve the maximum value of the final element, any of them may be printed.
Examples
Input
5
1 4 3 7 5
Output
11
3
1
4
2
Input
4
100 100 -1 100
Output
200
2
3
1
Input
6
-1 -2 -3 1 2 3
Output
4
3
2
1
2
Input
9
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
4
2
2
2
2 | instruction | 0 | 39,462 | 5 | 78,924 |
"Correct Solution:
```
def solve(n, aaa):
odd_idx, even_idx = [], []
odd_sum, even_sum = 0, 0
for i in range(0, n, 2):
if aaa[i] > 0:
even_idx.append(i)
even_sum += aaa[i]
for i in range(1, n, 2):
if aaa[i] > 0:
odd_idx.append(i)
odd_sum += aaa[i]
if odd_sum < even_sum:
ans = even_sum
idx = even_idx
else:
ans = odd_sum
idx = odd_idx
if ans == 0:
import numpy as np
i = np.argmax(aaa)
ans = aaa[i]
buf = list(range(n, i + 1, -1)) + [1] * i
return ans, buf
j = idx[-1]
buf = list(range(n, j + 1, -1))
for i in idx[-2::-1]:
buf.extend(range((i + j) // 2 + 1, i + 1, -1))
j = i
buf += [1] * idx[0]
return ans, buf
n = int(input())
aaa = list(map(int, input().split()))
ans, buf = solve(n, aaa)
print(ans)
print(len(buf))
print('\n'.join(map(str, buf)))
``` | output | 1 | 39,462 | 5 | 78,925 |
Provide a correct Python 3 solution for this coding contest problem.
You have an integer sequence of length N: a_1, a_2, ..., a_N.
You repeatedly perform the following operation until the length of the sequence becomes 1:
* First, choose an element of the sequence.
* If that element is at either end of the sequence, delete the element.
* If that element is not at either end of the sequence, replace the element with the sum of the two elements that are adjacent to it. Then, delete those two elements.
You would like to maximize the final element that remains in the sequence.
Find the maximum possible value of the final element, and the way to achieve it.
Constraints
* All input values are integers.
* 2 \leq N \leq 1000
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
* In the first line, print the maximum possible value of the final element in the sequence.
* In the second line, print the number of operations that you perform.
* In the (2+i)-th line, if the element chosen in the i-th operation is the x-th element from the left in the sequence at that moment, print x.
* If there are multiple ways to achieve the maximum value of the final element, any of them may be printed.
Examples
Input
5
1 4 3 7 5
Output
11
3
1
4
2
Input
4
100 100 -1 100
Output
200
2
3
1
Input
6
-1 -2 -3 1 2 3
Output
4
3
2
1
2
Input
9
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
4
2
2
2
2 | instruction | 0 | 39,463 | 5 | 78,926 |
"Correct Solution:
```
from collections import deque
N=int(input())
As=[int(i) for i in input().split()]
neg_odd,neg_even=deque([]),deque([])
S_odd,S_even=0,0
for i,A in enumerate(As):
if i%2==0:
if A>=0:
S_even+=A
else:
neg_even.append(i)
if i%2==1:
if A>=0:
S_odd+=A
else:
neg_odd.append(i)
ans=[]
A_size=N
if S_odd==S_even==0:
a=max(As)
print(a)
i=0
while As[i]!=a:
ans.append(1)
i+=1
A_size-=1
while A_size>1:
ans.append(A_size)
A_size-=1
elif S_odd<=S_even:
print(S_even)
if N%2==0:
ans.append(N)
A_size-=1
if neg_even:
t=0
while A_size-1==neg_even[-1]:
neg_even.pop()
ans.append(A_size)
ans.append(A_size-1)
A_size-=2
if not neg_even:
break
if neg_even:
while t==neg_even[0]:
neg_even.popleft()
ans.append(1)
ans.append(1)
t+=2
A_size-=2
if not neg_even:
break
for a in neg_even:
ans.append(a+1-t)
A_size-=2
t+=2
while A_size>1:
ans.append(2)
A_size-=2
else:
print(S_odd)
if N%2==1:
ans.append(N)
A_size-=1
ans.append(1)
A_size-=1
if neg_odd:
t=1
while A_size==neg_odd[-1]:
neg_odd.pop()
ans.append(A_size)
ans.append(A_size-1)
A_size-=2
if not neg_odd:
break
if neg_odd:
while t==neg_odd[0]:
neg_odd.popleft()
ans.append(1)
ans.append(1)
t+=2
A_size-=2
if not neg_odd:
break
for a in neg_odd:
ans.append(a-t+1)
A_size-=2
t+=2
while A_size>1:
ans.append(2)
A_size-=2
print(len(ans))
for a in ans:
print(a)
``` | output | 1 | 39,463 | 5 | 78,927 |
Provide a correct Python 3 solution for this coding contest problem.
You have an integer sequence of length N: a_1, a_2, ..., a_N.
You repeatedly perform the following operation until the length of the sequence becomes 1:
* First, choose an element of the sequence.
* If that element is at either end of the sequence, delete the element.
* If that element is not at either end of the sequence, replace the element with the sum of the two elements that are adjacent to it. Then, delete those two elements.
You would like to maximize the final element that remains in the sequence.
Find the maximum possible value of the final element, and the way to achieve it.
Constraints
* All input values are integers.
* 2 \leq N \leq 1000
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
* In the first line, print the maximum possible value of the final element in the sequence.
* In the second line, print the number of operations that you perform.
* In the (2+i)-th line, if the element chosen in the i-th operation is the x-th element from the left in the sequence at that moment, print x.
* If there are multiple ways to achieve the maximum value of the final element, any of them may be printed.
Examples
Input
5
1 4 3 7 5
Output
11
3
1
4
2
Input
4
100 100 -1 100
Output
200
2
3
1
Input
6
-1 -2 -3 1 2 3
Output
4
3
2
1
2
Input
9
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
4
2
2
2
2 | instruction | 0 | 39,464 | 5 | 78,928 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(10 ** 6)
input = sys.stdin.readline
def main():
n = int(input())
aa = list(map(int, input().split()))
s0 = 0
s = 0
i0 = []
ii = []
for i, a in enumerate(aa):
if a < 0: continue
if i % 2:
s += a
ii += [i]
else:
s0 += a
i0 += [i]
# 正の数がなかった場合
if not ii + i0:
s = max(aa)
print(s)
print(n - 1)
for i in range(n - 1, -1, -1):
if aa[i] == s:
break
print(i + 1)
for _ in range(i):
print(1)
exit()
if s0 > s:
s = s0
ii = i0
# print(s, ii)
# 右削除
ans = list(range(n - 1, ii[-1], -1))
# 中心部連結
for j in range(len(ii) - 2, -1, -1):
l, r = ii[j], ii[j + 1]
ans += list(range((l + r) // 2, l, -1))
# 左削除
ans += [0] * ii[0]
# print(ans)
print(s)
print(len(ans))
for a in ans:
print(a + 1)
main()
``` | output | 1 | 39,464 | 5 | 78,929 |
Provide a correct Python 3 solution for this coding contest problem.
You have an integer sequence of length N: a_1, a_2, ..., a_N.
You repeatedly perform the following operation until the length of the sequence becomes 1:
* First, choose an element of the sequence.
* If that element is at either end of the sequence, delete the element.
* If that element is not at either end of the sequence, replace the element with the sum of the two elements that are adjacent to it. Then, delete those two elements.
You would like to maximize the final element that remains in the sequence.
Find the maximum possible value of the final element, and the way to achieve it.
Constraints
* All input values are integers.
* 2 \leq N \leq 1000
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
* In the first line, print the maximum possible value of the final element in the sequence.
* In the second line, print the number of operations that you perform.
* In the (2+i)-th line, if the element chosen in the i-th operation is the x-th element from the left in the sequence at that moment, print x.
* If there are multiple ways to achieve the maximum value of the final element, any of them may be printed.
Examples
Input
5
1 4 3 7 5
Output
11
3
1
4
2
Input
4
100 100 -1 100
Output
200
2
3
1
Input
6
-1 -2 -3 1 2 3
Output
4
3
2
1
2
Input
9
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
4
2
2
2
2 | instruction | 0 | 39,465 | 5 | 78,930 |
"Correct Solution:
```
N=int(input())
A=list(map(int,input().split()))
even=[0 for i in range(N)]
odd=[0 for i in range(N)]
even[0]=max(0,A[0])
for i in range(1,N):
if i%2==1:
even[i]=even[i-1]
odd[i]=odd[i-1]+max(0,A[i])
else:
even[i]=even[i-1]+max(0,A[i])
odd[i]=odd[i-1]
eval=-10**20
L=-1
R=-1
for i in range(N):
for j in range(i,N):
if j==i:
test=A[i]
if eval<test:
eval=test
L,R=i,i
elif (j-i)%2==0:
if i%2==0:
test=even[j]-even[i-1]*(i!=0)+A[j]*(A[j]<0)+A[i]*(A[i]<0)
if eval<test:
eval=test
L,R=i,j
else:
test=odd[j]-odd[i-1]*(i!=0)+A[j]*(A[j]<0)+A[i]*(A[i]<0)
if eval<test:
eval=test
L,R=i,j
print(eval)
print(N-1-R+L+(R-L)//2)
#print(L,R)
for i in range(N-1,R,-1):
print(i+1)
for i in range(L):
print(1)
for i in range(L+2,R-1,2):
if A[i]>0:
print(2)
else:
print(3)
if L!=R:
print(2)
``` | output | 1 | 39,465 | 5 | 78,931 |
Provide a correct Python 3 solution for this coding contest problem.
You have an integer sequence of length N: a_1, a_2, ..., a_N.
You repeatedly perform the following operation until the length of the sequence becomes 1:
* First, choose an element of the sequence.
* If that element is at either end of the sequence, delete the element.
* If that element is not at either end of the sequence, replace the element with the sum of the two elements that are adjacent to it. Then, delete those two elements.
You would like to maximize the final element that remains in the sequence.
Find the maximum possible value of the final element, and the way to achieve it.
Constraints
* All input values are integers.
* 2 \leq N \leq 1000
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
* In the first line, print the maximum possible value of the final element in the sequence.
* In the second line, print the number of operations that you perform.
* In the (2+i)-th line, if the element chosen in the i-th operation is the x-th element from the left in the sequence at that moment, print x.
* If there are multiple ways to achieve the maximum value of the final element, any of them may be printed.
Examples
Input
5
1 4 3 7 5
Output
11
3
1
4
2
Input
4
100 100 -1 100
Output
200
2
3
1
Input
6
-1 -2 -3 1 2 3
Output
4
3
2
1
2
Input
9
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
4
2
2
2
2 | instruction | 0 | 39,466 | 5 | 78,932 |
"Correct Solution:
```
N = int(input())
A_ = list(map(int, input().split()))
ma = -float("inf")
ama = -1
for i, a in enumerate(A_):
if a>0:
break
if ma < a:
ma = a
ama = i
else:
print(ma)
print(N-1)
Ans = []
for i in range(N, ama+1, -1):
Ans.append(i)
for i in range(ama):
Ans.append(1)
print("\n".join(map(str, Ans)))
exit()
A = A_[:]
R1 = []
an1 = 0
for i in range(0, N, 2):
a = A[i]
if a>=0:
an1 += a
else:
R1.append(i)
idx_r1 = i
R2 = []
an2 = 0
for i in range(1, N, 2):
a = A[i]
if a>=0:
an2 += a
else:
R2.append(i)
idx_r2 = i
ans, R, idx_r = (an1, R1, idx_r1) if an1 > an2 else (an2, R2, idx_r2)
idx_r += 1
print(ans)
Ans = []
for idx in R[::-1]:
del A[idx]
if 1<=idx<len(A):
A[idx-1] += A.pop(idx)
elif idx==0:
idx_r -= 1
Ans.append(idx+1)
for idx in range(idx_r, -1, -2):
if 0<=idx<len(A):
del A[idx]
if 1<=idx<len(A):
A[idx-1] += A.pop(idx)
Ans.append(idx+1)
for idx in Ans:
del A_[idx-1]
if 1<idx<=len(A):
A[idx-2] += A.pop(idx-1)
# print(Ans)
assert len(A)==1 and A[0]==ans
print(len(Ans))
print("\n".join(map(str, Ans)))
``` | output | 1 | 39,466 | 5 | 78,933 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have an integer sequence of length N: a_1, a_2, ..., a_N.
You repeatedly perform the following operation until the length of the sequence becomes 1:
* First, choose an element of the sequence.
* If that element is at either end of the sequence, delete the element.
* If that element is not at either end of the sequence, replace the element with the sum of the two elements that are adjacent to it. Then, delete those two elements.
You would like to maximize the final element that remains in the sequence.
Find the maximum possible value of the final element, and the way to achieve it.
Constraints
* All input values are integers.
* 2 \leq N \leq 1000
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
* In the first line, print the maximum possible value of the final element in the sequence.
* In the second line, print the number of operations that you perform.
* In the (2+i)-th line, if the element chosen in the i-th operation is the x-th element from the left in the sequence at that moment, print x.
* If there are multiple ways to achieve the maximum value of the final element, any of them may be printed.
Examples
Input
5
1 4 3 7 5
Output
11
3
1
4
2
Input
4
100 100 -1 100
Output
200
2
3
1
Input
6
-1 -2 -3 1 2 3
Output
4
3
2
1
2
Input
9
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
4
2
2
2
2
Submitted Solution:
```
from collections import defaultdict, Counter
from itertools import product, groupby, count, permutations, combinations
from math import pi, sqrt
from collections import deque
from bisect import bisect, bisect_left, bisect_right
from string import ascii_lowercase
from functools import lru_cache
import sys
sys.setrecursionlimit(10000)
INF = float("inf")
YES, Yes, yes, NO, No, no = "YES", "Yes", "yes", "NO", "No", "no"
dy4, dx4 = [0, 1, 0, -1], [1, 0, -1, 0]
dy8, dx8 = [0, -1, 0, 1, 1, -1, -1, 1], [1, 0, -1, 0, 1, 1, -1, -1]
def inside(y, x, H, W):
return 0 <= y < H and 0 <= x < W
def ceil(a, b):
return (a + b - 1) // b
def check(A, operations):
c = A[:]
for o in operations:
if o == 1:
A = A[1:]
elif o == len(A):
A = A[:len(A) - 1]
else:
o -= 1
A = A[:o - 1] + [A[o - 1] + A[o + 1]] + A[o + 2:]
if len(A) != 1:
print(c)
print(operations)
assert(len(A) == 1)
return A[0]
def solve(N, A):
ans1 = 0
del1 = []
ok1 = False
start1, end1 = None, None
for i in range(0, N, 2):
if A[i] >= 0:
ans1 += A[i]
ok1 = True
if start1 is None:
start1 = i
end1 = i
else:
if ok1:
del1.append(i)
ans2 = 0
del2 = [1]
ok2 = False
start2 = None
end2 = None
for i in range(1, N, 2):
if A[i] >= 0:
ans2 += A[i]
ok2 = True
if start2 is None:
start2 = i
end2 = i
else:
if ok2:
del2.append(i)
ans, dele, start, end = None, None, None, None
if ok1 and ok2:
if ans1 > ans2:
ans, dele, start, end = ans1, del1, start1, end1
else:
ans, dele, start, end = ans2, del2, start2, end2
elif ok1:
ans, dele, start, end = ans1, del1, start1, end1
elif ok2:
ans, dele, start, end = ans2, del2, start2, end2
if not ok1 and not ok2:
ans = max(A)
idx = A.index(ans)
a = []
for i in range(N, idx + 1, -1):
a.append(i)
for i in range(1, idx + 1):
a.append(1)
return ans, a
else:
a = []
num = 0
for i in range(end + 1, N):
a.append(N - num)
num += 1
for i in dele[::-1]:
if start < i < end:
a.append(i + 1)
num += 2
for i in range(start):
a.append(1)
num += 1
for i in range((N - num) // 2) :
a.append(2)
return ans, a
def main():
N = int(input())
A = list(map(int, input().split()))
ans, ope = solve(N, A[:])
print(ans)
print(len(ope))
print(*ope, sep="\n")
if __name__ == '__main__':
main()
``` | instruction | 0 | 39,467 | 5 | 78,934 |
Yes | output | 1 | 39,467 | 5 | 78,935 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have an integer sequence of length N: a_1, a_2, ..., a_N.
You repeatedly perform the following operation until the length of the sequence becomes 1:
* First, choose an element of the sequence.
* If that element is at either end of the sequence, delete the element.
* If that element is not at either end of the sequence, replace the element with the sum of the two elements that are adjacent to it. Then, delete those two elements.
You would like to maximize the final element that remains in the sequence.
Find the maximum possible value of the final element, and the way to achieve it.
Constraints
* All input values are integers.
* 2 \leq N \leq 1000
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
* In the first line, print the maximum possible value of the final element in the sequence.
* In the second line, print the number of operations that you perform.
* In the (2+i)-th line, if the element chosen in the i-th operation is the x-th element from the left in the sequence at that moment, print x.
* If there are multiple ways to achieve the maximum value of the final element, any of them may be printed.
Examples
Input
5
1 4 3 7 5
Output
11
3
1
4
2
Input
4
100 100 -1 100
Output
200
2
3
1
Input
6
-1 -2 -3 1 2 3
Output
4
3
2
1
2
Input
9
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
4
2
2
2
2
Submitted Solution:
```
N=int(input())
A=list(map(int,input().split()))
data=[[],[]]
flag=[[],[]]
for i in range(N):
data[i%2].append(A[i])
flag[0].append(i%2)
flag[1].append((i+1)%2)
num0=0
for i in range(len(data[0])):
u=data[0][i]
if u>0:
num0+=u
else:
flag[0][2*i]=1
if num0==0:
num0=max(data[0])
flag[0]=[1 for i in range(N)]
for i in range(len(data[0])):
u=data[0][i]
if u==num0:
flag[0][2*i]=0
break
num1=0
for i in range(len(data[1])):
u=data[1][i]
if u>0:
num1+=u
else:
flag[1][2*i+1]=1
if num1==0:
num1=max(data[1])
flag[1]=[1 for i in range(N)]
for i in range(len(data[1])):
u=data[1][i]
if u==num1:
flag[1][2*i+1]=0
break
if num0>num1:
print(num0)
H=flag[0]
else:
print(num1)
H=flag[1]
ans=[]
ddd=0
for i in range(N):
if H[i]==1:
ans.append(1)
ddd+=1
else:
H=H[i:]
break
H=[0]+H
kkk=N-ddd
while True:
if H[kkk]==1:
ans.append(kkk)
kkk-=1
else:
break
while kkk>0:
if H[kkk]==0:
kkk-=1
else:
cnt=0
while H[kkk]==1:
kkk-=1
cnt+=1
for j in range((cnt+1)//2,0,-1):
ans.append(kkk+j)
print(len(ans))
for u in ans:
print(u)
``` | instruction | 0 | 39,468 | 5 | 78,936 |
Yes | output | 1 | 39,468 | 5 | 78,937 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have an integer sequence of length N: a_1, a_2, ..., a_N.
You repeatedly perform the following operation until the length of the sequence becomes 1:
* First, choose an element of the sequence.
* If that element is at either end of the sequence, delete the element.
* If that element is not at either end of the sequence, replace the element with the sum of the two elements that are adjacent to it. Then, delete those two elements.
You would like to maximize the final element that remains in the sequence.
Find the maximum possible value of the final element, and the way to achieve it.
Constraints
* All input values are integers.
* 2 \leq N \leq 1000
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
* In the first line, print the maximum possible value of the final element in the sequence.
* In the second line, print the number of operations that you perform.
* In the (2+i)-th line, if the element chosen in the i-th operation is the x-th element from the left in the sequence at that moment, print x.
* If there are multiple ways to achieve the maximum value of the final element, any of them may be printed.
Examples
Input
5
1 4 3 7 5
Output
11
3
1
4
2
Input
4
100 100 -1 100
Output
200
2
3
1
Input
6
-1 -2 -3 1 2 3
Output
4
3
2
1
2
Input
9
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
4
2
2
2
2
Submitted Solution:
```
from itertools import accumulate
n = int(input())
a = list(map(int,input().split()))
x = a[::2]
y = a[1::2]
lx = len(x)
ly = len(y)
sumx = sum([max(x[i],0) for i in range(lx)])
sumy = sum([max(y[i],0) for i in range(ly)])
ansls = []
if sumx >= sumy:
ans = sumx
flg = 0
else:
ans = sumy
flg = 1
if sumx == sumy == 0:
ans = max(a)
t = a.index(max(a))
print(ans)
print(n-1)
for i in range(n,t+1,-1):
print(i)
for i in range(t):
print(1)
exit()
if flg:
x = y
lx = ly
ansls.append(1)
n -= 1
ls = [0]
fp = -1
fptmp = 0
for i in range(lx):
if x[i] >= 0:
ls.append(1)
if fp == -1:
fp = fptmp
else:
ls.append(-1)
fptmp += 1
pnt = (n+1)//2*2-1
if pnt < n:
ansls.append(n)
flgb = 0
while pnt > 0:
px = (pnt+1)//2
if not flgb and ls[px] < 0:
ansls.append(pnt)
ansls.append(pnt-1)
ls.pop()
pnt -= 2
elif px <= fp:
ansls.append(1)
ansls.append(1)
pnt -= 2
elif ls[px] > 0:
flgb = 1
if px-1 <= fp:
pnt -= 2
elif ls[px-1] < 0:
ansls.append(pnt-2)
ls.pop()
ls.pop()
ls.append(1)
pnt -= 2
else:
ansls.append(pnt-1)
ls.pop()
pnt -= 2
print(ans)
print(len(ansls))
print(*ansls,sep="\n")
``` | instruction | 0 | 39,469 | 5 | 78,938 |
Yes | output | 1 | 39,469 | 5 | 78,939 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have an integer sequence of length N: a_1, a_2, ..., a_N.
You repeatedly perform the following operation until the length of the sequence becomes 1:
* First, choose an element of the sequence.
* If that element is at either end of the sequence, delete the element.
* If that element is not at either end of the sequence, replace the element with the sum of the two elements that are adjacent to it. Then, delete those two elements.
You would like to maximize the final element that remains in the sequence.
Find the maximum possible value of the final element, and the way to achieve it.
Constraints
* All input values are integers.
* 2 \leq N \leq 1000
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
* In the first line, print the maximum possible value of the final element in the sequence.
* In the second line, print the number of operations that you perform.
* In the (2+i)-th line, if the element chosen in the i-th operation is the x-th element from the left in the sequence at that moment, print x.
* If there are multiple ways to achieve the maximum value of the final element, any of them may be printed.
Examples
Input
5
1 4 3 7 5
Output
11
3
1
4
2
Input
4
100 100 -1 100
Output
200
2
3
1
Input
6
-1 -2 -3 1 2 3
Output
4
3
2
1
2
Input
9
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
4
2
2
2
2
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
ans1 = sum([a[i] for i in range(0, n, 2) if a[i] > 0])
ans2 = sum([a[i] for i in range(1, n, 2) if a[i] > 0])
if max(ans1, ans2) == 0:
print(max(a))
print(n-1)
i_max = a.index(max(a))
print(*list(range(n, i_max+1, -1)), sep="\n")
if i_max:
print(*[1 for _ in range(i_max)], sep="\n")
elif ans1 >= ans2:
print(ans1)
op = []
left = 0
while a[left] <= 0:
left += 2
right = True
in_seg = False
for i in range(n-1, -1, -1):
if right:
if i%2 == 1 or a[i] <= 0:
op.append(i+1)
else:
right = False
elif i < left:
op += [1] * (i+1)
break
else:
if in_seg == False and i%2 == 1:
in_seg = True
seg_right = i
elif i%2 == 0 and a[i] > 0:
in_seg = False
op += list(range((i+seg_right+3)//2, i+1, -1))
print(len(op))
print(*op, sep="\n")
else:
print(ans2)
op = []
left = 1
while a[left] <= 0:
left += 2
right = True
in_seg = False
for i in range(n-1, -1, -1):
if right:
if i%2 == 0 or a[i] <= 0:
op.append(i+1)
else:
right = False
elif i < left:
op += [1] * (i+1)
break
else:
if in_seg == False and i%2 == 0:
in_seg = True
seg_right = i
elif i%2 == 1 and a[i] > 0:
in_seg = False
op += list(range((i+seg_right+3)//2, i+1, -1))
print(len(op))
print(*op, sep="\n")
``` | instruction | 0 | 39,470 | 5 | 78,940 |
Yes | output | 1 | 39,470 | 5 | 78,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have an integer sequence of length N: a_1, a_2, ..., a_N.
You repeatedly perform the following operation until the length of the sequence becomes 1:
* First, choose an element of the sequence.
* If that element is at either end of the sequence, delete the element.
* If that element is not at either end of the sequence, replace the element with the sum of the two elements that are adjacent to it. Then, delete those two elements.
You would like to maximize the final element that remains in the sequence.
Find the maximum possible value of the final element, and the way to achieve it.
Constraints
* All input values are integers.
* 2 \leq N \leq 1000
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
* In the first line, print the maximum possible value of the final element in the sequence.
* In the second line, print the number of operations that you perform.
* In the (2+i)-th line, if the element chosen in the i-th operation is the x-th element from the left in the sequence at that moment, print x.
* If there are multiple ways to achieve the maximum value of the final element, any of them may be printed.
Examples
Input
5
1 4 3 7 5
Output
11
3
1
4
2
Input
4
100 100 -1 100
Output
200
2
3
1
Input
6
-1 -2 -3 1 2 3
Output
4
3
2
1
2
Input
9
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
4
2
2
2
2
Submitted Solution:
```
n = int(input())
a = []
for i in range(n):
a.append(int(input()))
# a = [3,5,-12,-2,-2,8,9,-3,7,5]
# n = len(a)
# print(a)
odd_max = 0
even_max = 0
for i in range(n):
if a[i] > 0:
if i % 2 == 0:
odd_max += a[i]
else:
even_max +=a[i]
# print(odd_max,even_max)
out_max = max(odd_max,even_max)
neg = []
flag = 1 #奇1偶0
if odd_max < even_max: flag = 0
# print(flag)
for i in range(n-1,-1,-1):
if a[i] < 0 and i % 2 != flag:
neg.append(i)
# print(neg)
# def p(index):
# a.pop(index)
# a[index - 1] += a[index]
# a.pop(index)
answer = []
for i in range(len(neg)):
index = neg[i]
# p(index)
answer.append(index+1)
# print(a)
if n % 2 == 0:
k = n - len(neg) * 2 - 1
# print(k)
for i in range(k,0,-2):
# print(i)
answer.append(i+flag)
# print(a)
else:
k = n - len(neg) * 2 - flag
# print(k)
for i in range(k,0,-2):
# print(i)
answer.append(i)
# print(a)
if n % 2 == 1 and flag == 0:
step = n//2+1
else:
step = n//2
print(step)
print(out_max)
for i in range(step):
print(answer[i])
``` | instruction | 0 | 39,471 | 5 | 78,942 |
No | output | 1 | 39,471 | 5 | 78,943 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have an integer sequence of length N: a_1, a_2, ..., a_N.
You repeatedly perform the following operation until the length of the sequence becomes 1:
* First, choose an element of the sequence.
* If that element is at either end of the sequence, delete the element.
* If that element is not at either end of the sequence, replace the element with the sum of the two elements that are adjacent to it. Then, delete those two elements.
You would like to maximize the final element that remains in the sequence.
Find the maximum possible value of the final element, and the way to achieve it.
Constraints
* All input values are integers.
* 2 \leq N \leq 1000
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
* In the first line, print the maximum possible value of the final element in the sequence.
* In the second line, print the number of operations that you perform.
* In the (2+i)-th line, if the element chosen in the i-th operation is the x-th element from the left in the sequence at that moment, print x.
* If there are multiple ways to achieve the maximum value of the final element, any of them may be printed.
Examples
Input
5
1 4 3 7 5
Output
11
3
1
4
2
Input
4
100 100 -1 100
Output
200
2
3
1
Input
6
-1 -2 -3 1 2 3
Output
4
3
2
1
2
Input
9
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
4
2
2
2
2
Submitted Solution:
```
N = int(input())
A = list(map(int, input().split()))
sum_even = sum(map(lambda i: A[i] if A[i] > 0 else 0, range(1, N, 2)))
sum_odd = sum(map(lambda i: A[i] if A[i] > 0 else 0, range(0, N, 2)))
if sum_even == 0 and sum_odd == 0:
max_ = max(A)
print(max_)
index = A.index(max_)
print(len(A)-1)
for _ in range(index):
print(1)
for j in range(N-index-1):
print(N - index - j)
exit()
print(max(sum_even, sum_odd))
EVEN = sum_even >= sum_odd
ans = []
if EVEN:
A = A[1:]
ans.append(1)
if len(A)%2 == 0:
ans.append(len(A))
A = A[:-1]
while len(A) > 1:
if A[0] <= 0:
ans.append(1)
ans.append(1)
A = A[2:]
else:
if A[2] <= 0:
ans.append(3)
A.pop(2)
if len(A) == 2:
ans.append(2)
A.pop(1)
else:
ans.append(3)
A.pop(2)
else:
ans.append(2)
A.pop(1)
A[0] += A.pop(1)
print(len(ans))
for a in ans:
print(a)
``` | instruction | 0 | 39,472 | 5 | 78,944 |
No | output | 1 | 39,472 | 5 | 78,945 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have an integer sequence of length N: a_1, a_2, ..., a_N.
You repeatedly perform the following operation until the length of the sequence becomes 1:
* First, choose an element of the sequence.
* If that element is at either end of the sequence, delete the element.
* If that element is not at either end of the sequence, replace the element with the sum of the two elements that are adjacent to it. Then, delete those two elements.
You would like to maximize the final element that remains in the sequence.
Find the maximum possible value of the final element, and the way to achieve it.
Constraints
* All input values are integers.
* 2 \leq N \leq 1000
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
* In the first line, print the maximum possible value of the final element in the sequence.
* In the second line, print the number of operations that you perform.
* In the (2+i)-th line, if the element chosen in the i-th operation is the x-th element from the left in the sequence at that moment, print x.
* If there are multiple ways to achieve the maximum value of the final element, any of them may be printed.
Examples
Input
5
1 4 3 7 5
Output
11
3
1
4
2
Input
4
100 100 -1 100
Output
200
2
3
1
Input
6
-1 -2 -3 1 2 3
Output
4
3
2
1
2
Input
9
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
4
2
2
2
2
Submitted Solution:
```
N = int(input())
A = [int(a) for a in input().split()]
if max(A) <= 0:
ma = - 10 ** 9
for i, a in enumerate(A):
if a > ma:
ma = a
mai = i
print(ma)
print(N - 1)
for i in range(mai+1, N)[::-1]:
print(i+1)
for i in range(mai):
print(1)
else:
E = sum([a for a in A[::2] if a > 0])
O = sum([a for a in A[1::2] if a > 0])
ANS = []
if O > E:
ANS.append(1)
A = A[1:]
if len(A) % 2 == 0:
ANS.append(len(A))
A.pop()
while len(A) > 1:
if A[0] < 0:
ANS.append(1)
ANS.append(1)
A = A[2:]
elif A[-1] < 0:
ANS.append(len(A))
A.pop()
ANS.append(len(A))
A.pop()
elif A[-3] < 0:
ANS.append(len(A) - 2)
A[-3] = A.pop()
A.pop()
else:
ANS.append(len(A) - 1)
A[-3] += A[-1]
A.pop()
A.pop()
print(A[0])
print(len(ANS))
for a in ANS:
print(a)
``` | instruction | 0 | 39,473 | 5 | 78,946 |
No | output | 1 | 39,473 | 5 | 78,947 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have an integer sequence of length N: a_1, a_2, ..., a_N.
You repeatedly perform the following operation until the length of the sequence becomes 1:
* First, choose an element of the sequence.
* If that element is at either end of the sequence, delete the element.
* If that element is not at either end of the sequence, replace the element with the sum of the two elements that are adjacent to it. Then, delete those two elements.
You would like to maximize the final element that remains in the sequence.
Find the maximum possible value of the final element, and the way to achieve it.
Constraints
* All input values are integers.
* 2 \leq N \leq 1000
* |a_i| \leq 10^9
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
* In the first line, print the maximum possible value of the final element in the sequence.
* In the second line, print the number of operations that you perform.
* In the (2+i)-th line, if the element chosen in the i-th operation is the x-th element from the left in the sequence at that moment, print x.
* If there are multiple ways to achieve the maximum value of the final element, any of them may be printed.
Examples
Input
5
1 4 3 7 5
Output
11
3
1
4
2
Input
4
100 100 -1 100
Output
200
2
3
1
Input
6
-1 -2 -3 1 2 3
Output
4
3
2
1
2
Input
9
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
Output
5000000000
4
2
2
2
2
Submitted Solution:
```
N = int(input())
A_ = list(map(int, input().split()))
ma = -float("inf")
ama = -1
for i, a in enumerate(A_):
if a>0:
break
if ma < a:
ma = a
ama = i
else:
print(ma)
print(N-1)
Ans = []
for i in range(N, ama+1, -1):
Ans.append(i)
for i in range(ama):
Ans.append(1)
print("\n".join(map(str, Ans)))
exit()
A = A_[:]
R1 = []
an1 = 0
for i in range(0, N, 2):
a = A[i]
if a>=0:
an1 += a
else:
R1.append(i)
idx_r1 = i
R2 = []
an2 = 0
for i in range(1, N, 2):
a = A[i]
if a>=0:
an2 += a
else:
R2.append(i)
idx_r2 = i
ans, R, idx_r = (an1, R1, idx_r1) if an1 > an2 else (an2, R2, idx_r2)
idx_r += 1
print(ans)
Ans = []
for idx in R[::-1]:
del A[idx]
if 1<=idx<len(A):
A[idx-1] += A.pop(idx)
Ans.append(idx+1)
for idx in range(idx_r, -1, -2):
if 0<=idx<len(A):
del A[idx]
if 1<=idx<len(A):
A[idx-1] += A.pop(idx)
Ans.append(idx+1)
for idx in Ans:
del A_[idx-1]
if 1<idx<=len(A):
A[idx-2] += A.pop(idx-1)
assert len(A)==1 and A[0]==ans
print(len(Ans))
print("\n".join(map(str, Ans)))
``` | instruction | 0 | 39,474 | 5 | 78,948 |
No | output | 1 | 39,474 | 5 | 78,949 |
Provide a correct Python 3 solution for this coding contest problem.
You are given three integers, A, B and C.
Among them, two are the same, but the remaining one is different from the rest.
For example, when A=5,B=7,C=5, A and C are the same, but B is different.
Find the one that is different from the rest among the given three integers.
Constraints
* -100 \leq A,B,C \leq 100
* A, B and C are integers.
* The input satisfies the condition in the statement.
Input
Input is given from Standard Input in the following format:
A B C
Output
Among A, B and C, print the integer that is different from the rest.
Examples
Input
5 7 5
Output
7
Input
1 1 7
Output
7
Input
-100 100 100
Output
-100 | instruction | 0 | 39,475 | 5 | 78,950 |
"Correct Solution:
```
s = list(map(int, input().split()))
for i in s:
if s.count(i)==1:
print(i)
``` | output | 1 | 39,475 | 5 | 78,951 |
Provide a correct Python 3 solution for this coding contest problem.
You are given three integers, A, B and C.
Among them, two are the same, but the remaining one is different from the rest.
For example, when A=5,B=7,C=5, A and C are the same, but B is different.
Find the one that is different from the rest among the given three integers.
Constraints
* -100 \leq A,B,C \leq 100
* A, B and C are integers.
* The input satisfies the condition in the statement.
Input
Input is given from Standard Input in the following format:
A B C
Output
Among A, B and C, print the integer that is different from the rest.
Examples
Input
5 7 5
Output
7
Input
1 1 7
Output
7
Input
-100 100 100
Output
-100 | instruction | 0 | 39,476 | 5 | 78,952 |
"Correct Solution:
```
A=list(map(int,input().split()))
A=sorted(A)
print(A[0] if A[1]==A[2] else A[2])
``` | output | 1 | 39,476 | 5 | 78,953 |
Provide a correct Python 3 solution for this coding contest problem.
You are given three integers, A, B and C.
Among them, two are the same, but the remaining one is different from the rest.
For example, when A=5,B=7,C=5, A and C are the same, but B is different.
Find the one that is different from the rest among the given three integers.
Constraints
* -100 \leq A,B,C \leq 100
* A, B and C are integers.
* The input satisfies the condition in the statement.
Input
Input is given from Standard Input in the following format:
A B C
Output
Among A, B and C, print the integer that is different from the rest.
Examples
Input
5 7 5
Output
7
Input
1 1 7
Output
7
Input
-100 100 100
Output
-100 | instruction | 0 | 39,477 | 5 | 78,954 |
"Correct Solution:
```
L=sorted(list(map(int, input().split())))
if L[0]==L[1]:
print(L[2])
else:
print(L[0])
``` | output | 1 | 39,477 | 5 | 78,955 |
Provide a correct Python 3 solution for this coding contest problem.
You are given three integers, A, B and C.
Among them, two are the same, but the remaining one is different from the rest.
For example, when A=5,B=7,C=5, A and C are the same, but B is different.
Find the one that is different from the rest among the given three integers.
Constraints
* -100 \leq A,B,C \leq 100
* A, B and C are integers.
* The input satisfies the condition in the statement.
Input
Input is given from Standard Input in the following format:
A B C
Output
Among A, B and C, print the integer that is different from the rest.
Examples
Input
5 7 5
Output
7
Input
1 1 7
Output
7
Input
-100 100 100
Output
-100 | instruction | 0 | 39,478 | 5 | 78,956 |
"Correct Solution:
```
a,b,c=map(int,input().split())
print(a) if b==c else print(b) if a==c else print(c)
``` | output | 1 | 39,478 | 5 | 78,957 |
Provide a correct Python 3 solution for this coding contest problem.
You are given three integers, A, B and C.
Among them, two are the same, but the remaining one is different from the rest.
For example, when A=5,B=7,C=5, A and C are the same, but B is different.
Find the one that is different from the rest among the given three integers.
Constraints
* -100 \leq A,B,C \leq 100
* A, B and C are integers.
* The input satisfies the condition in the statement.
Input
Input is given from Standard Input in the following format:
A B C
Output
Among A, B and C, print the integer that is different from the rest.
Examples
Input
5 7 5
Output
7
Input
1 1 7
Output
7
Input
-100 100 100
Output
-100 | instruction | 0 | 39,479 | 5 | 78,958 |
"Correct Solution:
```
a = input().split()
a.sort()
if a[0] == a[1]:
print(a[2])
else:
print(a[0])
``` | output | 1 | 39,479 | 5 | 78,959 |
Provide a correct Python 3 solution for this coding contest problem.
You are given three integers, A, B and C.
Among them, two are the same, but the remaining one is different from the rest.
For example, when A=5,B=7,C=5, A and C are the same, but B is different.
Find the one that is different from the rest among the given three integers.
Constraints
* -100 \leq A,B,C \leq 100
* A, B and C are integers.
* The input satisfies the condition in the statement.
Input
Input is given from Standard Input in the following format:
A B C
Output
Among A, B and C, print the integer that is different from the rest.
Examples
Input
5 7 5
Output
7
Input
1 1 7
Output
7
Input
-100 100 100
Output
-100 | instruction | 0 | 39,480 | 5 | 78,960 |
"Correct Solution:
```
a,b,c = sorted(map(int, input().split()))
print(a if b==c else c)
``` | output | 1 | 39,480 | 5 | 78,961 |
Provide a correct Python 3 solution for this coding contest problem.
You are given three integers, A, B and C.
Among them, two are the same, but the remaining one is different from the rest.
For example, when A=5,B=7,C=5, A and C are the same, but B is different.
Find the one that is different from the rest among the given three integers.
Constraints
* -100 \leq A,B,C \leq 100
* A, B and C are integers.
* The input satisfies the condition in the statement.
Input
Input is given from Standard Input in the following format:
A B C
Output
Among A, B and C, print the integer that is different from the rest.
Examples
Input
5 7 5
Output
7
Input
1 1 7
Output
7
Input
-100 100 100
Output
-100 | instruction | 0 | 39,481 | 5 | 78,962 |
"Correct Solution:
```
a=list(input().split())
a.sort()
if a[0]==a[1]:
print(a[2])
else:
print(a[0])
``` | output | 1 | 39,481 | 5 | 78,963 |
Provide a correct Python 3 solution for this coding contest problem.
You are given three integers, A, B and C.
Among them, two are the same, but the remaining one is different from the rest.
For example, when A=5,B=7,C=5, A and C are the same, but B is different.
Find the one that is different from the rest among the given three integers.
Constraints
* -100 \leq A,B,C \leq 100
* A, B and C are integers.
* The input satisfies the condition in the statement.
Input
Input is given from Standard Input in the following format:
A B C
Output
Among A, B and C, print the integer that is different from the rest.
Examples
Input
5 7 5
Output
7
Input
1 1 7
Output
7
Input
-100 100 100
Output
-100 | instruction | 0 | 39,482 | 5 | 78,964 |
"Correct Solution:
```
a,b,c = sorted(map(int,input().split()))
print( c if a == b else a)
``` | output | 1 | 39,482 | 5 | 78,965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three integers, A, B and C.
Among them, two are the same, but the remaining one is different from the rest.
For example, when A=5,B=7,C=5, A and C are the same, but B is different.
Find the one that is different from the rest among the given three integers.
Constraints
* -100 \leq A,B,C \leq 100
* A, B and C are integers.
* The input satisfies the condition in the statement.
Input
Input is given from Standard Input in the following format:
A B C
Output
Among A, B and C, print the integer that is different from the rest.
Examples
Input
5 7 5
Output
7
Input
1 1 7
Output
7
Input
-100 100 100
Output
-100
Submitted Solution:
```
a,b,c = map(int,input().split())
print(c if a==b else a if b==c else b)
``` | instruction | 0 | 39,483 | 5 | 78,966 |
Yes | output | 1 | 39,483 | 5 | 78,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three integers, A, B and C.
Among them, two are the same, but the remaining one is different from the rest.
For example, when A=5,B=7,C=5, A and C are the same, but B is different.
Find the one that is different from the rest among the given three integers.
Constraints
* -100 \leq A,B,C \leq 100
* A, B and C are integers.
* The input satisfies the condition in the statement.
Input
Input is given from Standard Input in the following format:
A B C
Output
Among A, B and C, print the integer that is different from the rest.
Examples
Input
5 7 5
Output
7
Input
1 1 7
Output
7
Input
-100 100 100
Output
-100
Submitted Solution:
```
a = list(map(int, input().split()))
a.sort()
print(a[2] if a[0] == a[1] else a[0])
``` | instruction | 0 | 39,484 | 5 | 78,968 |
Yes | output | 1 | 39,484 | 5 | 78,969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three integers, A, B and C.
Among them, two are the same, but the remaining one is different from the rest.
For example, when A=5,B=7,C=5, A and C are the same, but B is different.
Find the one that is different from the rest among the given three integers.
Constraints
* -100 \leq A,B,C \leq 100
* A, B and C are integers.
* The input satisfies the condition in the statement.
Input
Input is given from Standard Input in the following format:
A B C
Output
Among A, B and C, print the integer that is different from the rest.
Examples
Input
5 7 5
Output
7
Input
1 1 7
Output
7
Input
-100 100 100
Output
-100
Submitted Solution:
```
a = [int(_) for _ in input().split()]
for i in a:
if a.count(i) == 1:
print(i)
``` | instruction | 0 | 39,485 | 5 | 78,970 |
Yes | output | 1 | 39,485 | 5 | 78,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three integers, A, B and C.
Among them, two are the same, but the remaining one is different from the rest.
For example, when A=5,B=7,C=5, A and C are the same, but B is different.
Find the one that is different from the rest among the given three integers.
Constraints
* -100 \leq A,B,C \leq 100
* A, B and C are integers.
* The input satisfies the condition in the statement.
Input
Input is given from Standard Input in the following format:
A B C
Output
Among A, B and C, print the integer that is different from the rest.
Examples
Input
5 7 5
Output
7
Input
1 1 7
Output
7
Input
-100 100 100
Output
-100
Submitted Solution:
```
A, B, C = sorted(map(int, input().split()))
print(A if A != B else C)
``` | instruction | 0 | 39,486 | 5 | 78,972 |
Yes | output | 1 | 39,486 | 5 | 78,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three integers, A, B and C.
Among them, two are the same, but the remaining one is different from the rest.
For example, when A=5,B=7,C=5, A and C are the same, but B is different.
Find the one that is different from the rest among the given three integers.
Constraints
* -100 \leq A,B,C \leq 100
* A, B and C are integers.
* The input satisfies the condition in the statement.
Input
Input is given from Standard Input in the following format:
A B C
Output
Among A, B and C, print the integer that is different from the rest.
Examples
Input
5 7 5
Output
7
Input
1 1 7
Output
7
Input
-100 100 100
Output
-100
Submitted Solution:
```
a,b,c = map(int,input().split())
print('a' if b==c else 'b' if a==c else 'c' if a==b)
``` | instruction | 0 | 39,487 | 5 | 78,974 |
No | output | 1 | 39,487 | 5 | 78,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three integers, A, B and C.
Among them, two are the same, but the remaining one is different from the rest.
For example, when A=5,B=7,C=5, A and C are the same, but B is different.
Find the one that is different from the rest among the given three integers.
Constraints
* -100 \leq A,B,C \leq 100
* A, B and C are integers.
* The input satisfies the condition in the statement.
Input
Input is given from Standard Input in the following format:
A B C
Output
Among A, B and C, print the integer that is different from the rest.
Examples
Input
5 7 5
Output
7
Input
1 1 7
Output
7
Input
-100 100 100
Output
-100
Submitted Solution:
```
from itertools import* #permutations("AB",repeat=2) combinations("AB",2) product("AB",2) groupby accumulate
import sys
#input=sys.stdin.readline #危険！基本オフにしろ！
sys.setrecursionlimit(10**6)
def main():
n,k=map(int,input().split())
grid=[tuple(map(int,input().split()))for i in range(n)]
x=set(s for s,t in grid)
y=set(t for s,t in grid)
ans=[]
for i in combinations(x,2):
for j in combinations(y,2):
xx=max(i)
xxx=min(i)
yy=max(j)
yyy=min(j)
if sum(xxx<=s<=xx and yyy<=t<=yy for s,t in grid)>=k:
ans.append(abs((yy-yyy)*(xx-xxx)))
print(min(ans))
if __name__ == '__main__':
main()
``` | instruction | 0 | 39,488 | 5 | 78,976 |
No | output | 1 | 39,488 | 5 | 78,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three integers, A, B and C.
Among them, two are the same, but the remaining one is different from the rest.
For example, when A=5,B=7,C=5, A and C are the same, but B is different.
Find the one that is different from the rest among the given three integers.
Constraints
* -100 \leq A,B,C \leq 100
* A, B and C are integers.
* The input satisfies the condition in the statement.
Input
Input is given from Standard Input in the following format:
A B C
Output
Among A, B and C, print the integer that is different from the rest.
Examples
Input
5 7 5
Output
7
Input
1 1 7
Output
7
Input
-100 100 100
Output
-100
Submitted Solution:
```
A, B, C = map(int, input().split())
if A == B:
print('C')
elif B == C:
print('A')
else:
print('B)
``` | instruction | 0 | 39,489 | 5 | 78,978 |
No | output | 1 | 39,489 | 5 | 78,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three integers, A, B and C.
Among them, two are the same, but the remaining one is different from the rest.
For example, when A=5,B=7,C=5, A and C are the same, but B is different.
Find the one that is different from the rest among the given three integers.
Constraints
* -100 \leq A,B,C \leq 100
* A, B and C are integers.
* The input satisfies the condition in the statement.
Input
Input is given from Standard Input in the following format:
A B C
Output
Among A, B and C, print the integer that is different from the rest.
Examples
Input
5 7 5
Output
7
Input
1 1 7
Output
7
Input
-100 100 100
Output
-100
Submitted Solution:
```
a=list(map(int,input().split()))
for i in range(2):
if a.count(a[i])!=1:
del a[i]
print(int(a[0]))
``` | instruction | 0 | 39,490 | 5 | 78,980 |
No | output | 1 | 39,490 | 5 | 78,981 |
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