message stringlengths 2 57.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 61 108k | cluster float64 22 22 | __index_level_0__ int64 122 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.
Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.
Constraints
* 1 \leq K \leq 200
* K is an integer.
Input
Input is given from Standard Input in the following format:
K
Output
Print the value of \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.
Examples
Input
2
Output
9
Input
200
Output
10813692
Submitted Solution:
```
import math
x=0;y=0
K=int(input())
for i in range(1,K+1):
for j in range(1,K+1):
y=math.gcd(i,j)
for h in range(1,K+1):
x=x+math.gcd(y,h)
print(x)
``` | instruction | 0 | 4,833 | 22 | 9,666 |
Yes | output | 1 | 4,833 | 22 | 9,667 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.
Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.
Constraints
* 1 \leq K \leq 200
* K is an integer.
Input
Input is given from Standard Input in the following format:
K
Output
Print the value of \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.
Examples
Input
2
Output
9
Input
200
Output
10813692
Submitted Solution:
```
from math import gcd
K=int(input())
S=0
for a in range(1,K+1):
for b in range(1,K+1):
for c in range(1,K+1):
S+=gcd(a,gcd(b,c))
print(S)
``` | instruction | 0 | 4,834 | 22 | 9,668 |
Yes | output | 1 | 4,834 | 22 | 9,669 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.
Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.
Constraints
* 1 \leq K \leq 200
* K is an integer.
Input
Input is given from Standard Input in the following format:
K
Output
Print the value of \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.
Examples
Input
2
Output
9
Input
200
Output
10813692
Submitted Solution:
```
import math
K=int(input())
ans=0
for i in range(1,K+1):
for j in range(1,K+1):
for k in range(1,K+1):
ans+=math.gcd(i,math.gcd(j,k))
print(ans)
``` | instruction | 0 | 4,835 | 22 | 9,670 |
Yes | output | 1 | 4,835 | 22 | 9,671 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.
Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.
Constraints
* 1 \leq K \leq 200
* K is an integer.
Input
Input is given from Standard Input in the following format:
K
Output
Print the value of \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.
Examples
Input
2
Output
9
Input
200
Output
10813692
Submitted Solution:
```
import math as mt
import itertools as iters
K = int(input())
A = 0
for a,b,c in iters.product(range(1,K+1),repeat=3):
A += mt.gcd(a,mt.gcd(b,c))
print(A)
``` | instruction | 0 | 4,836 | 22 | 9,672 |
Yes | output | 1 | 4,836 | 22 | 9,673 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.
Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.
Constraints
* 1 \leq K \leq 200
* K is an integer.
Input
Input is given from Standard Input in the following format:
K
Output
Print the value of \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.
Examples
Input
2
Output
9
Input
200
Output
10813692
Submitted Solution:
```
import math
from functools import reduce
#入力:N(int:整数)
def input1():
return int(input())
#入力:N,M(int:整数)
def input2():
return map(int,input().split())
#入力:[n1,n2,...nk](int:整数配列)
def input_array():
return list(map(int,input().split()))
def gcd(*numbers):
return reduce(math.gcd, numbers)
def gcd_list(numbers):
return reduce(math.gcd, numbers)
k=input1()
result=0
for i in range(1,k+1):
for j in range(1,k+1):
for k in range(1,k+1):
result+=gcd(i,j,k)
print(result)
``` | instruction | 0 | 4,837 | 22 | 9,674 |
No | output | 1 | 4,837 | 22 | 9,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.
Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.
Constraints
* 1 \leq K \leq 200
* K is an integer.
Input
Input is given from Standard Input in the following format:
K
Output
Print the value of \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.
Examples
Input
2
Output
9
Input
200
Output
10813692
Submitted Solution:
```
import math
from functools import reduce
K = int(input())
summ = 0
# 組み合わせの数
# a, b, cが同じは1pattern
# a, b, cのうち、２つが同じパターンは３
# a, b, cのうち、全て異なるパターンは6
# つまり、３つの数の組み合わせの数を求め、各パターン倍すればよい
# タプルで集合を作成する
st = set()
for a in range(1, K+1):
for b in range(1, K+1):
for c in range(1, K+1):
sort_abc = sorted([a, b, c])
st.add(tuple(sort_abc))
# print(st)
for s in st:
if s[0] == s[1] == s[2]:
summ += reduce(math.gcd, s)
# print(0, s)
elif s[0] == s[1] or s[0] == s[2] or s[1] == s[2]:
summ += 3*reduce(math.gcd, s)
# print(1, s)
else:
summ += 6*reduce(math.gcd, s)
# print(2, s)
print(summ)
``` | instruction | 0 | 4,838 | 22 | 9,676 |
No | output | 1 | 4,838 | 22 | 9,677 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.
Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.
Constraints
* 1 \leq K \leq 200
* K is an integer.
Input
Input is given from Standard Input in the following format:
K
Output
Print the value of \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.
Examples
Input
2
Output
9
Input
200
Output
10813692
Submitted Solution:
```
import math
def gcd(*numbers):
return reduce(math.gcd, numbers)
K = int(input())
ans = 0
for a in range(1, K+1):
for b in range(1, K+1):
for c in range(1, K+1):
ans += math.gcd(math.gcd(a, b), c)
print(ans)
``` | instruction | 0 | 4,839 | 22 | 9,678 |
No | output | 1 | 4,839 | 22 | 9,679 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.
Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.
Constraints
* 1 \leq K \leq 200
* K is an integer.
Input
Input is given from Standard Input in the following format:
K
Output
Print the value of \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.
Examples
Input
2
Output
9
Input
200
Output
10813692
Submitted Solution:
```
import math
from functools import reduce
def gcd(*numbers):
return reduce(math.gcd,numbers)
K=int(input())
s=0
for a in range(1,K+1):
for b in range(1,K+1):
for c in range(1,K+1):
s+=gcd(a,b,c)
print(s)
``` | instruction | 0 | 4,840 | 22 | 9,680 |
No | output | 1 | 4,840 | 22 | 9,681 |
Provide a correct Python 3 solution for this coding contest problem.
Since I got tired to write long problem statements, I decided to make this problem statement short. For given positive integer L, how many pairs of positive integers a, b (a ≤ b) such that LCM(a, b) = L are there? Here, LCM(a, b) stands for the least common multiple of a and b.
Constraints
* 1 ≤ L ≤ 1012
Input
For each dataset, an integer L is given in a line. Input terminates when L = 0.
Output
For each dataset, output the number of pairs of a and b.
Example
Input
12
9
2
0
Output
8
3
2 | instruction | 0 | 4,980 | 22 | 9,960 |
"Correct Solution:
```
from collections import defaultdict
def soinnsuubunnkai(n):
dic = defaultdict(int)
i = 2
while i * i <= n:
while n % i == 0:
dic[i] += 1
n //= i
i += 1
if n != 1:
dic[n] += 1
return list(dic.values())
def saiki(values, score, ind, end):
if ind == end:
return score
return saiki(values, score * values[ind], ind + 1, end) * 2+ \
saiki(values, score, ind + 1, end)
while True:
n = int(input())
if n == 0:
break
values = soinnsuubunnkai(n)
print((saiki(values, 1, 0, len(values)) + 1) // 2)
``` | output | 1 | 4,980 | 22 | 9,961 |
Provide a correct Python 3 solution for this coding contest problem.
Since I got tired to write long problem statements, I decided to make this problem statement short. For given positive integer L, how many pairs of positive integers a, b (a ≤ b) such that LCM(a, b) = L are there? Here, LCM(a, b) stands for the least common multiple of a and b.
Constraints
* 1 ≤ L ≤ 1012
Input
For each dataset, an integer L is given in a line. Input terminates when L = 0.
Output
For each dataset, output the number of pairs of a and b.
Example
Input
12
9
2
0
Output
8
3
2 | instruction | 0 | 4,981 | 22 | 9,962 |
"Correct Solution:
```
# AOJ 1060: No Story
# Python3 2018.6.8 bal4u
MAX = 1000004
ptbl = [ 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 ]
def sieve():
for p in ptbl:
for i in range(p*p, MAX, p): tbl[i] = 1
for i in range(997, MAX, 2):
if tbl[i] == 0: ptbl.append(i)
def prime_factor(n):
power = []
if (n & 1) == 0:
c = 0
while True:
n >>= 1
c += 1
if n & 1: break
power.append(c)
if n <= 1: return power
if n <= MAX and tbl[n] == 0:
power.append(1)
return power
k = int(n**0.5)
for p in ptbl:
if n <= 1: break
if p > k or (n <= MAX and tbl[n] == 0):
power.append(1)
break
if n % p: continue
c = 0
while True:
n //= p
c += 1
if n % p: break
power.append(c)
return power
tbl = [0]*MAX
sieve()
while True:
n = int(input())
if n == 0: break
if n == 1:
print(1)
continue
if n <= MAX and (n & 1) and tbl[n] == 0:
print(2)
continue
power = prime_factor(n)
ans = 1
for p in power: ans = ans*(1+(p<<1))
print((ans+1)>>1)
``` | output | 1 | 4,981 | 22 | 9,963 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | instruction | 0 | 5,161 | 22 | 10,322 |
Tags: greedy, math, number theory
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
ok = False
c = 2
while not ok and c*c*c <= n:
if n % c != 0:
c += 1
continue
# a * b = n / c
# a > b > c
b = c+1
while not ok and b*b <= (n // c):
if (n // c) % b != 0:
b += 1
continue
a = n // (c * b)
if a > b:
print('YES')
print(a, b, c)
ok = True
b += 1
c += 1
if not ok:
print('NO')
``` | output | 1 | 5,161 | 22 | 10,323 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | instruction | 0 | 5,162 | 22 | 10,324 |
Tags: greedy, math, number theory
Correct Solution:
```
from functools import reduce
def fac(n):
return sorted(set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
t=int(input())
for i in range(t):
n=int(input())
x=fac(n)
x=x[1:-1]
flag=1
if len(x)<3:
flag=0
else:
a,b=x[0],0
for j in range(1,len(x)):
if n%(a*x[j])==0:
b=x[j]
break
if b:
te=a*b
if n%te==0:
c=n//te
if a!=b and a!=c and b!=c and a>=2 and b>=2 and c>=2:
pass
else:
flag=0
else:
flag=0
else:
flag=0
if flag:
pass
print("YES")
print(a,b,c)
else:
print("NO")
``` | output | 1 | 5,162 | 22 | 10,325 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | instruction | 0 | 5,163 | 22 | 10,326 |
Tags: greedy, math, number theory
Correct Solution:
```
import math
if __name__ == "__main__":
t = int(input())
for _ in range(t):
n, a, b = int(input()), 0, 0
i = 1
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0 and i != n // i:
n, a = n // i, i
break
for j in range(i, int(math.sqrt(n)) + 1):
if n % j == 0:
if a != j and a != n // j and j != n // j:
n, b = n // j, j
break
if a != 0 and b != 0:
print("YES")
print(a, b, n)
else:
print("NO")
``` | output | 1 | 5,163 | 22 | 10,327 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | instruction | 0 | 5,164 | 22 | 10,328 |
Tags: greedy, math, number theory
Correct Solution:
```
import math
for i in range(int(input())):
n=int(input())
l=[]
count=0
for i in range(2,int(math.sqrt(n))+1):
if n%i==0:
l.append(i)
n=n//i
break
for i in range(2,int(math.sqrt(n))+1):
if n%i==0 and len(l)==1 and l[0]!=i:
l.append(i)
n=n//i
count=1
break
flag=0
if count==1:
if n!=l[0] and n!=l[1]:
l.append(n)
else:
flag=1
if len(l)==3 and flag==0:
print("YES")
print(*l)
else:
print("NO")
``` | output | 1 | 5,164 | 22 | 10,329 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | instruction | 0 | 5,165 | 22 | 10,330 |
Tags: greedy, math, number theory
Correct Solution:
```
# cook your dish here
import math
for t in range(int(input())):
n=int(input())
flag=False
for i in range(2,int(math.sqrt(n))+1):
if n%i==0:
x=i
yy=n//i
for j in range(i+1,int(math.sqrt(yy))+1):
if yy%j==0:
y=j
z=yy//j
if z>=2 and z!=y and z!=x:
flag=True
l=[x,y,z]
print("YES")
print(*l)
break
if flag:
break
if flag==False:
print("NO")
``` | output | 1 | 5,165 | 22 | 10,331 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | instruction | 0 | 5,166 | 22 | 10,332 |
Tags: greedy, math, number theory
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sat Jan 25 10:20:51 2020
@author: Anthony
"""
def abc(num):
threshold=num**0.5
myDivisors=[]
potentialDiv=2
current=num
while(potentialDiv<=threshold and current!=1):
if current//potentialDiv == current/potentialDiv:
myDivisors.append(potentialDiv)
current/=potentialDiv
potentialDiv+=1
if len(myDivisors)>=2 and (current not in myDivisors):
myDivisors.append(int(current))
return myDivisors
if len(myDivisors)>=3 and current==1:
return myDivisors
else:
return False
repeat=int(input())
for i in range(0,repeat):
temp=abc(int(input()))
if temp:
for i in range(0,len(temp)):
for j in range(i+1,len(temp)):
if (temp[i]*temp[j] not in temp) and len(temp)>3:
temp[i]*=temp[j]
temp[j]=1
safiye=[]
for x in temp:
if x!=1:
safiye.append(x)
if len(safiye)==3:
print("YES")
for x in safiye:
print(x,end=" ")
else:
print("NO")
else:
print("NO")
``` | output | 1 | 5,166 | 22 | 10,333 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | instruction | 0 | 5,167 | 22 | 10,334 |
Tags: greedy, math, number theory
Correct Solution:
```
for i in range(int(input())):
n = int(input())
f = []
for j in range(2, int(n**(1/2))):
while n % j == 0:
f.append(j)
n = int(n/j)
if n > 1:
f.append(n)
if len(f) >= 3:
if len(f) == 3:
if f[0] != f[1] and f[1] != f[2] and f[2] != f[0]:
print('YES')
print(f[0],f[1],f[2])
else:
print('NO')
else:
f0 = f[0]
f1 = 1
f2 = 1
if f[1] == f[0]:
f1 = f[1] * f[2]
for j in range(3, len(f)):
f2 *= f[j]
else:
f1 = f[1]
for j in range(2, len(f)):
f2 *= f[j]
if f0 != f1 and f1 != f2 and f2 != f0:
print('YES')
print(f0, int(f1), int(f2))
else:
print('NO')
else:
print('NO')
``` | output | 1 | 5,167 | 22 | 10,335 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823 | instruction | 0 | 5,168 | 22 | 10,336 |
Tags: greedy, math, number theory
Correct Solution:
```
"""T=int(input())
for _ in range(0,T):
n=int(input())
a,b=map(int,input().split())
s=input()
s=[int(x) for x in input().split()]
for i in range(0,len(s)):
a,b=map(int,input().split())"""
import math
T=int(input())
for _ in range(0,T):
N=int(input())
n=N
L=[]
while (n % 2 == 0):
L.append(2)
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while (n % i== 0):
L.append(i)
n = n // i
if (n > 2):
L.append(n)
if(len(L)<3):
print('NO')
else:
t1=L[0]
t2=L[1]
t3=1
if(L[1]==L[0]):
t2=L[1]*L[2]
for j in range(3,len(L)):
t3*=L[j]
else:
for j in range(2,len(L)):
t3*=L[j]
if(t1!=t2 and t2!=t3 and t1!=t3 and t1>1 and t2>1 and t3>1):
print('YES')
print(t1,t2,t3)
else:
print('NO')
``` | output | 1 | 5,168 | 22 | 10,337 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823
Submitted Solution:
```
import sys
import math
input=sys.stdin.readline
# A function to print all prime factors of
# a given number n
def primeFactors(n):
# Print the number of two's that divide n
while n % 2 == 0:
n = n / 2
return(2,n)
# n must be odd at this point
# so a skip of 2 ( i = i + 2) can be used
for i in range(3,int(math.sqrt(n))+1,2):
# while i divides n , print i ad divide n
while n % i== 0:
n = n / i
return(i,n)
# Condition if n is a prime
# number greater than 2
return(-1,-1)
t=int(input())
for _ in range(t):
n=int(input())
a,n=primeFactors(n)
b=-1
c=-1
flag=0
if(a==-1):
print("NO")
else:
i=2
while(i<=math.sqrt(n)):
if(n%i==0):
if(n//i>=2):
if(i!=a and(n//i!=a and i!=n//i)):
b=i
c=n//i
flag=1
i+=1
if(flag==1):
print("YES")
print(a,int(b),int(c))
else:
print("NO")
``` | instruction | 0 | 5,170 | 22 | 10,340 |
Yes | output | 1 | 5,170 | 22 | 10,341 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given one integer number n. Find three distinct integers a, b, c such that 2 ≤ a, b, c and a ⋅ b ⋅ c = n or say that it is impossible to do it.
If there are several answers, you can print any.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
The next n lines describe test cases. The i-th test case is given on a new line as one integer n (2 ≤ n ≤ 10^9).
Output
For each test case, print the answer on it. Print "NO" if it is impossible to represent n as a ⋅ b ⋅ c for some distinct integers a, b, c such that 2 ≤ a, b, c.
Otherwise, print "YES" and any possible such representation.
Example
Input
5
64
32
97
2
12345
Output
YES
2 4 8
NO
NO
NO
YES
3 5 823
Submitted Solution:
```
import math
from collections import defaultdict
def primeFactors(n):
d=defaultdict(int)
while n % 2 == 0:
d[2]+=1
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
d[i]+=1
n = n / i
if n > 2:
d[n]+=1
return d
t=int(input())
for i in range(t):
n=int(input())
d= primeFactors(n)
# print(d)
if len( d.keys() )>=3:
print("YES")
s=[]
ww=1
for j in list(d.keys())[:2]:
s.append( int(j**d[j]) )
ww*=j**(d[j]-1)
for j in list(d.keys())[2:]:
ww*= int(j**d[j])
s.append(ww)
print(*s)
elif len(list(d.keys()))==1:
w,w1 = int(list(d.keys())[0]), int(d[list(d.keys())[0]])
if w1>=6:
print("YES")
ans = "{} {} {}".format(w,w**2,w**(w1-3))
print(ans)
else:
print("NO")
elif len(list(d.keys()))==2:
keys= list(map(int,list(d.keys())))
value = list(map(int,list(d.values())))
value1 = sorted(list(d.values()))
if sum(value)>=4:
ans = "{} {} {}".format( keys[0], keys[1], keys[0]**(d[value1[0]]-1) * keys[1]**(d[value1[1]]-1) )
print(ans)
else:
print("NO")
``` | instruction | 0 | 5,173 | 22 | 10,346 |
No | output | 1 | 5,173 | 22 | 10,347 |
Provide a correct Python 3 solution for this coding contest problem.
Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 100.
Input
Given multiple datasets. Each dataset is given the integer n (n ≤ 10000). When n is 0, it is the last input. The number of datasets does not exceed 50.
Output
For n in each dataset, print s on one line.
Example
Input
2
9
0
Output
5
100 | instruction | 0 | 5,713 | 22 | 11,426 |
"Correct Solution:
```
from math import sqrt, ceil
from itertools import accumulate
N = 120000
temp = [True]*(N+1)
temp[0] = temp[1] = False
for i in range(2, ceil(sqrt(N+1))):
if temp[i]:
temp[i+i::i] = [False]*(len(temp[i+i::i]))
cumsum = [i for i in range(N) if temp[i]]
cumsum = list(accumulate(cumsum))
while True:
n = int(input())
if not n: break
print(cumsum[n-1])
``` | output | 1 | 5,713 | 22 | 11,427 |
Provide a correct Python 3 solution for this coding contest problem.
Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 100.
Input
Given multiple datasets. Each dataset is given the integer n (n ≤ 10000). When n is 0, it is the last input. The number of datasets does not exceed 50.
Output
For n in each dataset, print s on one line.
Example
Input
2
9
0
Output
5
100 | instruction | 0 | 5,714 | 22 | 11,428 |
"Correct Solution:
```
# Aizu Problem 0053: Sum of Prime Numbers
#
import sys, math, os, bisect
# read input:
PYDEV = os.environ.get('PYDEV')
if PYDEV=="True":
sys.stdin = open("sample-input.txt", "rt")
def primes2(n):
""" Input n>=6, Returns a list of primes, 2 <= p < n """
n, correction = n-n%6+6, 2-(n%6>1)
sieve = [True] * (n//3)
for i in range(1,int(n**0.5)//3+1):
if sieve[i]:
k=3*i+1|1
sieve[ k*k//3 ::2*k] = [False] * ((n//6-k*k//6-1)//k+1)
sieve[k*(k-2*(i&1)+4)//3::2*k] = [False] * ((n//6-k*(k-2*(i&1)+4)//6-1)//k+1)
return [2,3] + [3*i+1|1 for i in range(1,n//3-correction) if sieve[i]]
primes = primes2(110000)
S = []
s = 0
for p in primes:
s += p
S.append(s)
while True:
n = int(input())
if n == 0:
break
print(S[n-1])
``` | output | 1 | 5,714 | 22 | 11,429 |
Provide a correct Python 3 solution for this coding contest problem.
Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 100.
Input
Given multiple datasets. Each dataset is given the integer n (n ≤ 10000). When n is 0, it is the last input. The number of datasets does not exceed 50.
Output
For n in each dataset, print s on one line.
Example
Input
2
9
0
Output
5
100 | instruction | 0 | 5,715 | 22 | 11,430 |
"Correct Solution:
```
def isPrime(x):
if x == 2:
return True
if x < 2 or x % 2 == 0:
return False
i, root_x = 3, int(pow(x, 0.5))
while i <= root_x:
if x % i == 0:
return False
i += 2
return True
primes = [2]
for i in range(3, 104730):
if isPrime(i):
primes.append(primes[-1]+i)
while True:
n = int(input())
if n == 0:
break
print(primes[n-1])
``` | output | 1 | 5,715 | 22 | 11,431 |
Provide a correct Python 3 solution for this coding contest problem.
Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 100.
Input
Given multiple datasets. Each dataset is given the integer n (n ≤ 10000). When n is 0, it is the last input. The number of datasets does not exceed 50.
Output
For n in each dataset, print s on one line.
Example
Input
2
9
0
Output
5
100 | instruction | 0 | 5,716 | 22 | 11,432 |
"Correct Solution:
```
num = 200000
L = [True] * (num+1)
L[0] = False
L[1] = False
for i in range( 2, int(num**0.5)+ 2 ):
if not L[i]:
continue
for j in range(i*2, num+1, i):
L[j] = False
p = [ x for x in range(num+1) if L[x] ]
while True:
n = int(input())
if n == 0:
break
print(sum(p[0:n]))
``` | output | 1 | 5,716 | 22 | 11,433 |
Provide a correct Python 3 solution for this coding contest problem.
Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 100.
Input
Given multiple datasets. Each dataset is given the integer n (n ≤ 10000). When n is 0, it is the last input. The number of datasets does not exceed 50.
Output
For n in each dataset, print s on one line.
Example
Input
2
9
0
Output
5
100 | instruction | 0 | 5,717 | 22 | 11,434 |
"Correct Solution:
```
# AOJ 0053: Sum of Prime Numbers
# Python3 2018.6.15 bal4u
MAX = 104729 # 10000th prime
SQRT = 323 # sqrt(MAX)
prime = [True for i in range(MAX+2)]
def sieve():
for i in range(2, MAX, 2):
prime[i] = False
for i in range(3, SQRT, 2):
if prime[i] is True:
for j in range(i*i, MAX, i):
prime[j] = False
sieve()
sum = [0 for i in range(MAX+2)]
j = 2
sum[1] = 2;
for i in range(3, MAX+1, 2):
if prime[i] is True:
sum[j] = sum[j-1] + i
j += 1
while True:
n = int(input())
if n == 0: break
print(sum[n])
``` | output | 1 | 5,717 | 22 | 11,435 |
Provide a correct Python 3 solution for this coding contest problem.
Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 100.
Input
Given multiple datasets. Each dataset is given the integer n (n ≤ 10000). When n is 0, it is the last input. The number of datasets does not exceed 50.
Output
For n in each dataset, print s on one line.
Example
Input
2
9
0
Output
5
100 | instruction | 0 | 5,718 | 22 | 11,436 |
"Correct Solution:
```
def make_ps(n):
nums = [True] * n
nums[0] = nums[1] = False
p = 2
sqrt = n ** 0.5
while p < sqrt:
for i in range(2 * p, n, p):
nums[i] = False
while True:
p += 1
if nums[p]:
break
return [i for i in range(n) if nums[i]]
ps = make_ps(105000)
while True:
n = int(input())
if not n:
break
print(sum(ps[:n]))
``` | output | 1 | 5,718 | 22 | 11,437 |
Provide a correct Python 3 solution for this coding contest problem.
Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 100.
Input
Given multiple datasets. Each dataset is given the integer n (n ≤ 10000). When n is 0, it is the last input. The number of datasets does not exceed 50.
Output
For n in each dataset, print s on one line.
Example
Input
2
9
0
Output
5
100 | instruction | 0 | 5,719 | 22 | 11,438 |
"Correct Solution:
```
from itertools import *
n=104730;a=list(range(n));a[:2]=0,0
for i in range(2,323):a[i*2::i]=[0]*len(a[i*2::i])
p=list(compress(range(n),a))
print('\n'.join(str(sum(p[:int(e)]))for e in iter(input,'0')))
``` | output | 1 | 5,719 | 22 | 11,439 |
Provide a correct Python 3 solution for this coding contest problem.
Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 100.
Input
Given multiple datasets. Each dataset is given the integer n (n ≤ 10000). When n is 0, it is the last input. The number of datasets does not exceed 50.
Output
For n in each dataset, print s on one line.
Example
Input
2
9
0
Output
5
100 | instruction | 0 | 5,720 | 22 | 11,440 |
"Correct Solution:
```
a=[True]*104743
p=[]
for i in range(2,int(104743**0.5)+1):
if a[i]:
for j in range(i*i,104743,i):a[j]=False
for i in range(2,104730):
if a[i]:p.append(i)
while 1:
n=int(input())
if n==0:break
print(sum(p[:n]))
``` | output | 1 | 5,720 | 22 | 11,441 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 100.
Input
Given multiple datasets. Each dataset is given the integer n (n ≤ 10000). When n is 0, it is the last input. The number of datasets does not exceed 50.
Output
For n in each dataset, print s on one line.
Example
Input
2
9
0
Output
5
100
Submitted Solution:
```
import math
M = 104729
lst = list(range(M+1))
for i in range(0, M+1, 2):
lst[i] = 0
lst[1] = 0
lst[2] = 2
for i in range(3, math.ceil(math.sqrt(M+1)), 2):
if lst[i]:
for j in range(i**2, M+1, i):
lst[j] = 0
anslst = [i for i in lst if i != 0]
while True:
n = int(input())
if n == 0 :
break
print(sum(anslst[:n]))
``` | instruction | 0 | 5,721 | 22 | 11,442 |
Yes | output | 1 | 5,721 | 22 | 11,443 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 100.
Input
Given multiple datasets. Each dataset is given the integer n (n ≤ 10000). When n is 0, it is the last input. The number of datasets does not exceed 50.
Output
For n in each dataset, print s on one line.
Example
Input
2
9
0
Output
5
100
Submitted Solution:
```
prim=[True]*1000000
prim[0]=prim[1]=False
for i in range(2,350):
if prim[i]:
for j in range(i*2,110000,i):
prim[j]=False
prime=[i for i,j in enumerate(prim) if j==True]
while True:
n=int(input())
if n==0:
break
print(sum(prime[:n]))
``` | instruction | 0 | 5,722 | 22 | 11,444 |
Yes | output | 1 | 5,722 | 22 | 11,445 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 100.
Input
Given multiple datasets. Each dataset is given the integer n (n ≤ 10000). When n is 0, it is the last input. The number of datasets does not exceed 50.
Output
For n in each dataset, print s on one line.
Example
Input
2
9
0
Output
5
100
Submitted Solution:
```
import sys
f = sys.stdin
def create_prime(n):
prime = [1] * (n + 1)
prime[:2] = [0, 0]
for i in range(len(prime)):
if prime[i]:
for j in range(2 * i, len(prime), i):
prime[j] = 0
return prime
prime = create_prime(200000)
while True:
n = int(f.readline())
if n == 0:
break
s = 0
for i in range(len(prime)):
if prime[i]:
s += i
n -= 1
if n == 0:
print(s)
break
``` | instruction | 0 | 5,723 | 22 | 11,446 |
Yes | output | 1 | 5,723 | 22 | 11,447 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 100.
Input
Given multiple datasets. Each dataset is given the integer n (n ≤ 10000). When n is 0, it is the last input. The number of datasets does not exceed 50.
Output
For n in each dataset, print s on one line.
Example
Input
2
9
0
Output
5
100
Submitted Solution:
```
def sieve(n):
p=[True]*n
p[0]=p[1]=False
for i in range(2,int(n**0.5)+1):
if p[i]:
for j in range(i*i,n,i):
p[j]=False
prime =[i for i in range(2,n) if p[i]]
return prime
def function(n):
return sum(A[:n])
A=sieve(110000)
while True:
n=int(input())
if n==0:
break
print(function(n))
``` | instruction | 0 | 5,724 | 22 | 11,448 |
Yes | output | 1 | 5,724 | 22 | 11,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 100.
Input
Given multiple datasets. Each dataset is given the integer n (n ≤ 10000). When n is 0, it is the last input. The number of datasets does not exceed 50.
Output
For n in each dataset, print s on one line.
Example
Input
2
9
0
Output
5
100
Submitted Solution:
```
def create_prime_list(limit):
""" ??¨??????????????????????????§limit?????§????´???°?????????????±???????
https://ja.wikipedia.org/wiki/%E3%82%A8%E3%83%A9%E3%83%88%E3%82%B9%E3%83%86%E3%83%8D%E3%82%B9%E3%81%AE%E7%AF%A9
"""
x = limit**0.5
primes = []
#print('x={0}'.format(x))
nums = [x for x in range(2, limit+1)]
while nums[0]<=x:
primes.append(nums[0])
current_prime = nums[0]
nums = [x for x in nums if x%current_prime != 0]
primes.extend(nums)
#print(primes)
return primes
def solve(n):
primes = create_prime_list(110000)
primes = primes[:n]
return primes
def is_prime(q, k=50):
""" ??????????????????????´???°?????????????????¨??????????????????????????°????´???°???????????????????????????
http://d.hatena.ne.jp/pashango_p/20090704/1246692091
"""
import random
q = abs(q)
# ?¨???????????????§?????????????????§????????????????????????
if q == 2: return True
if q < 2 or q & 1 == 0: return False
# n-1=2^s*d??¨????????????a?????´??°???d????\???°)???d????±???????
d = (q - 1) >> 1
while d & 1 == 0:
d >>= 1
# ?????????k?????°?????????
for i in range(k):
a = random.randint(1, q - 1)
t = d
y = pow(a, t, q)
# [0,s-1]??????????????????????????§??????
while t != q - 1 and y != 1 and y != q - 1:
y = pow(y, 2, q)
t <<= 1
if y != q - 1 and t & 1 == 0:
return False
return True
def find_nth_prime(n):
""" n???????????§????´???°????±??????? """
count = 0
num = 2
result = []
while True:
if is_prime(num):
count += 1
result.append(num)
if count >= n:
break
num += 1
#print(count, num)
print(result)
return result
if __name__ == '__main__':
# ??????????????\???
while True:
data = int(input())
if data == 0:
break
# ????????¨???????????¨???
# result = find_nth_prime(data)
result = solve(data)
print(sum(result))
``` | instruction | 0 | 5,725 | 22 | 11,450 |
No | output | 1 | 5,725 | 22 | 11,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let p (i) be the i-th prime number from the smallest. For example, 7 is the fourth prime number from the smallest, 2, 3, 5, 7, so p (4) = 7.
Given n, the sum of p (i) from i = 1 to n s
s = p (1) + p (2) + .... + p (n)
Create a program that outputs. For example, when n = 9, s = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 100.
Input
Given multiple datasets. Each dataset is given the integer n (n ≤ 10000). When n is 0, it is the last input. The number of datasets does not exceed 50.
Output
For n in each dataset, print s on one line.
Example
Input
2
9
0
Output
5
100
Submitted Solution:
```
import sys
def gen1():
yield 2
l = [2]
count = 2
while True:
count += 1
flag = 1
for i in l:
if i > count ** (1/2):
break
else:
if count % i == 0:
flag = 0
break
if flag:
l.append(count)
yield count
for line in sys.stdin:
g = gen1()
n = int(line)
if n == 0:
break
s = 0
for i in range(n):
a = g.__next__()
s += a
print(a)
print(s)
``` | instruction | 0 | 5,726 | 22 | 11,452 |
No | output | 1 | 5,726 | 22 | 11,453 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer. | instruction | 0 | 5,942 | 22 | 11,884 |
Tags: brute force, dp, games, greedy
Correct Solution:
```
from bisect import bisect_left, bisect_right
from sys import stdin, stdout
R = lambda : stdin.readline().strip()
RL = lambda f=None: list(map(f, R().split(' '))) if f else list(R().split(' '))
output = lambda x: stdout.write(str(x) + '\n')
output_list = lambda x: output(' '.join(map(str, x)))
n = int(R())
a = RL(int)
ass = []
for i in range(0, n, 2):
ass.append(a[i])
for i in range(1, n, 2):
ass.append(a[i])
ass += ass
k = (n+1)//2
sm = sum(ass[:k])
ans = sm
for i in range(k, len(ass)):
sm += ass[i]
sm -= ass[i-k]
ans = max(sm, ans)
output(ans)
``` | output | 1 | 5,942 | 22 | 11,885 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer. | instruction | 0 | 5,943 | 22 | 11,886 |
Tags: brute force, dp, games, greedy
Correct Solution:
```
n=int(input())
a=[int(x) for x in input().split()]
sumo=0
for i in range(0,n,2):
sumo+=a[i]
temo=sumo
for i in range(1,n,2):
temo+=a[i]-a[i-1]
if temo>sumo: sumo=temo
for i in range(0,n,2):
temo+=a[i]-a[i-1]
if temo>sumo: sumo=temo
print(sumo)
``` | output | 1 | 5,943 | 22 | 11,887 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer. | instruction | 0 | 5,944 | 22 | 11,888 |
Tags: brute force, dp, games, greedy
Correct Solution:
```
n = int(input())
l = list(map(int, input().split()))
count = 0
for i in range(0, n, 2):
count += l[i]
m = count
for i in range(0, 2*n-2, 2):
count = count - l[i%n] + l[(i+1)%n]
if count > m: m = count
print(m)
``` | output | 1 | 5,944 | 22 | 11,889 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer. | instruction | 0 | 5,945 | 22 | 11,890 |
Tags: brute force, dp, games, greedy
Correct Solution:
```
n=int(input())
a=list(map(lambda x:int(x),input().split()))
new_a=[]
for i in range(0,len(a),2):
new_a.append(a[i])
for j in range(1,len(a),2):
new_a.append(a[j])
length=(n+1)//2
new_a=new_a+new_a[0:length-1]
window=new_a[0:length]
prefix_sum=[new_a[0]]
for i in range(1,len(new_a)):
prefix_sum.append(prefix_sum[i-1]+new_a[i])
s=prefix_sum[length-1]
answer=s
for i in range(length,len(new_a)):
s=s-new_a[i-length]+new_a[i]
answer=max(answer,s)
print(answer)
``` | output | 1 | 5,945 | 22 | 11,891 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer. | instruction | 0 | 5,946 | 22 | 11,892 |
Tags: brute force, dp, games, greedy
Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
if n != 1:
maxCircle = 0
maxCircleCand = a[0]
for i in range(1,n,2):
maxCircleCand += a[i]
maxCircle = maxCircleCand
for i in range(1,n,2):
maxCircleCand -= a[i]
maxCircleCand += a[i+1]
if maxCircle < maxCircleCand:
maxCircle = maxCircleCand
maxCircleCand = 0
maxCircleCand += a[1]
for i in range(2,n,2):
maxCircleCand += a[i]
if maxCircle < maxCircleCand:
maxCircle = maxCircleCand
for i in range(2,n-1,2):
maxCircleCand -= a[i]
maxCircleCand += a[i+1]
if maxCircle < maxCircleCand:
maxCircle = maxCircleCand
print(maxCircle)
else:
print(a[0])
``` | output | 1 | 5,946 | 22 | 11,893 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer. | instruction | 0 | 5,947 | 22 | 11,894 |
Tags: brute force, dp, games, greedy
Correct Solution:
```
#!/usr/bin/pypy3
n = int(input())
a = list(map(int, input().split()))
c = 0
for i in range(0, n, 2):
c += a[i]
ans = c
for i in range(0, 2 * (n - 1), 2):
c = c + a[(i + 1) % n] - a[i % n]
ans = max(ans, c)
print(ans)
``` | output | 1 | 5,947 | 22 | 11,895 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer. | instruction | 0 | 5,948 | 22 | 11,896 |
Tags: brute force, dp, games, greedy
Correct Solution:
```
from sys import stdin, stdout
int_in = lambda: int(stdin.readline())
arr_in = lambda: [int(x) for x in stdin.readline().split()]
mat_in = lambda rows: [arr_in() for _ in range(rows)]
str_in = lambda: stdin.readline().strip()
out = lambda o: stdout.write("{}\n".format(o))
arr_out = lambda o: out(" ".join(map(str, o)))
bool_out = lambda o: out("YES" if o else "NO")
tests = lambda: range(1, int_in() + 1)
case_out = lambda i, o: out("Case #{}: {}".format(i, o))
# https://codeforces.com/contest/1372/problem/D
def solve(n, a):
if n == 1:
return a[0]
arr = []
for i in range(0, len(a), 2):
arr.append(a[i])
for i in range(1, len(a), 2):
arr.append(a[i])
arr += arr
need = (n + 1) // 2
curr = 0
for i in range(need):
curr += arr[i]
best = curr
for i in range(1, len(arr) - need):
curr -= arr[i - 1]
curr += arr[need + i - 1]
best = max(best, curr)
return best
if __name__ == "__main__":
n = int_in()
a = arr_in()
out(solve(n, a))
``` | output | 1 | 5,948 | 22 | 11,897 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer. | instruction | 0 | 5,949 | 22 | 11,898 |
Tags: brute force, dp, games, greedy
Correct Solution:
```
import sys
import math as mt
input=sys.stdin.readline
I=lambda:list(map(int,input().split()))
n,=I()
l=I()
ar=[]
for i in range(0,n,2):
ar.append(l[i])
for i in range(1,n,2):
ar.append(l[i])
k=(n+1)//2
s=sum(ar[:k])
ans=s
ar+=ar
for i in range(k,2*n):
s-=ar[(i-k)]
s+=ar[i]
ans=max(ans,s)
print(ans)
``` | output | 1 | 5,949 | 22 | 11,899 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Submitted Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
eacc = [0]
oacc = [0]
for i in range(n):
eacc.append(eacc[-1]+(a[i] if i%2==0 else 0))
oacc.append(oacc[-1]+(a[i] if i%2==1 else 0))
ans = 0
for i in range(n):
if i%2==0:
ans = max(ans, oacc[i]+eacc[n]-eacc[i])
else:
ans = max(ans, eacc[i]+oacc[n]-oacc[i])
print(ans)
``` | instruction | 0 | 5,950 | 22 | 11,900 |
Yes | output | 1 | 5,950 | 22 | 11,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
a0=[a[i] for i in range(n) if i%2==0]
a1=[a[i] for i in range(n) if i%2==1]
a=a0+a1
l=n//2+1
ans=sum(a[:l])
ans_sub=sum(a[:l])
for i in range(n):
ans_sub-=a[i]
ans_sub+=a[(i+l)%n]
ans=max(ans,ans_sub)
print(ans)
``` | instruction | 0 | 5,952 | 22 | 11,904 |
Yes | output | 1 | 5,952 | 22 | 11,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Submitted Solution:
```
from sys import stdin
input = stdin.buffer.readline
n = int(input())
*a, = map(int, input().split())
s1 = sum(a[i] for i in range(0, n, 2))
s2 = sum(a[i] for i in range(1, n, 2)) + a[0]
ans = max(s1, s2)
for i in range(0, n, 2):
s1 -= a[i] - a[(i + 1) % n]
ans = max(ans, s1)
for i in range(1, n, 2):
s2 -= a[i] - a[(i + 1) % n]
ans = max(ans, s2)
print(ans)
``` | instruction | 0 | 5,953 | 22 | 11,906 |
Yes | output | 1 | 5,953 | 22 | 11,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Submitted Solution:
```
from sys import stdin
input = stdin.buffer.readline
n = int(input())
*a, = map(int, input().split())
s1 = sum(a[i] for i in range(0, n, 2))
s2 = sum(a[i] for i in range(1, n, 2)) + a[0]
ans = max(s1, s2)
for i in range(1, n, 2):
s1 += a[i] - a[(i + 1) % n]
ans = max(ans, s1)
for i in range(0, n, 2):
s1 += a[i] - a[(i + 1) % n]
ans = max(ans, s1)
print(ans)
``` | instruction | 0 | 5,955 | 22 | 11,910 |
No | output | 1 | 5,955 | 22 | 11,911 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Danny, the local Math Maniac, is fascinated by circles, Omkar's most recent creation. Help him solve this circle problem!
You are given n nonnegative integers a_1, a_2, ..., a_n arranged in a circle, where n must be odd (ie. n-1 is divisible by 2). Formally, for all i such that 2 ≤ i ≤ n, the elements a_{i - 1} and a_i are considered to be adjacent, and a_n and a_1 are also considered to be adjacent. In one operation, you pick a number on the circle, replace it with the sum of the two elements adjacent to it, and then delete the two adjacent elements from the circle. This is repeated until only one number remains in the circle, which we call the circular value.
Help Danny find the maximum possible circular value after some sequences of operations.
Input
The first line contains one odd integer n (1 ≤ n < 2 ⋅ 10^5, n is odd) — the initial size of the circle.
The second line contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the initial numbers in the circle.
Output
Output the maximum possible circular value after applying some sequence of operations to the given circle.
Examples
Input
3
7 10 2
Output
17
Input
1
4
Output
4
Note
For the first test case, here's how a circular value of 17 is obtained:
Pick the number at index 3. The sum of adjacent elements equals 17. Delete 7 and 10 from the circle and replace 2 with 17.
Note that the answer may not fit in a 32-bit integer.
Submitted Solution:
```
n = int(input())
l1 = [int(x) for x in input().split()]
temp=0
for i in range(len(l1)):
temp = max(temp,sum(l1[i::2])+max(l1[i-1::-2]))
print(temp)
``` | instruction | 0 | 5,956 | 22 | 11,912 |
No | output | 1 | 5,956 | 22 | 11,913 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.
In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.
A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.
Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.
In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.
It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.
If there are different possible orders with a correct answer, print any of them.
Example
Input
3
6
4
30
Output
2 3 6
1
2 4
0
2 30 6 3 15 5 10
0
Note
In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.
In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.
In the third test case all divisors of 30 greater than 1 can be placed in some order so that there are no two adjacent numbers that are coprime. | instruction | 0 | 5,974 | 22 | 11,948 |
Tags: constructive algorithms, implementation, math, number theory
Correct Solution:
```
import math
import random
def cnt(x):
fac = {}
y = x
while (x > 1):
i = 2
flag = False
while (i*i <= x):
if (x % i == 0):
fac[i] = 0
while (x % i == 0):
x //= i
fac[i] += 1
flag = True
break
i += 1
if (not flag):
fac[x] = 1
break
f = set()
i = 2
while (i*i <= y):
if (y % i == 0):
f.add(i)
if (y // i != i):
f.add(y//i)
i += 1
f.add(y)
return fac,f
t = int(input())
while (t):
n = int(input())
mp,f = cnt(n)
primes = list(mp.keys())
if (len(primes) == 1):
for x in f:
print(x,end=' ')
print('\n0')
elif (len(primes) == 2):
if (mp[primes[0]] == 1 and mp[primes[1]] == 1):
print(primes[0], primes[1], n)
print(1)
else:
a,b = primes
print(a,a*b,b,end=' ')
f.discard(a)
f.discard(a*b)
f.discard(b)
for i in range(2,mp[b]+1):
print(b**i,end=' ')
for i in range(1,mp[b]+1):
for j in range(1,mp[a]+1):
if (i != 1 or j != 1):
print(b**i * a**j, end=' ')
for i in range(2,mp[a]+1):
print(a**i,end=' ')
print('\n0')
else:
for i in range(len(primes)):
a,b = primes[i],primes[(i+1)%len(primes)]
f.discard(a)
f.discard(a*b)
ff = {}
for p in primes:
ff[p] = set()
for x in list(f):
if (x % p == 0):
ff[p].add(x)
f.discard(x)
for i in range(len(primes)):
a,b = primes[i],primes[(i+1)%len(primes)]
print(a,end=' ')
for v in ff[a]:
print(v,end=' ')
print(a*b,end=' ')
print('\n0')
t -= 1
``` | output | 1 | 5,974 | 22 | 11,949 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.
In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.
A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.
Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.
In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.
It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.
If there are different possible orders with a correct answer, print any of them.
Example
Input
3
6
4
30
Output
2 3 6
1
2 4
0
2 30 6 3 15 5 10
0
Note
In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.
In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.
In the third test case all divisors of 30 greater than 1 can be placed in some order so that there are no two adjacent numbers that are coprime. | instruction | 0 | 5,975 | 22 | 11,950 |
Tags: constructive algorithms, implementation, math, number theory
Correct Solution:
```
import sys
input=sys.stdin.readline
t=int(input())
for you in range(t):
n=int(input())
z=n
primes=[]
i=2
while(i*i<=z):
if(z%i==0):
primes.append(i)
while(z%i==0):
z=z//i
i+=1
if(z!=1):
primes.append(z)
hashi=dict()
for i in primes:
hashi[i]=[]
hashinew=dict()
new=[]
k=len(primes)
hasho=dict()
if(k>2):
for i in range(k):
new.append(primes[i]*primes[(i+1)%k])
hasho[primes[i]*primes[(i+1)%k]]=1
if(k==2):
hasho[primes[0]*primes[1]]=1
i=2
while(i*i<=n):
if(n%i==0):
num1=i
num2=n//i
if(num1 not in hasho):
for j in primes:
if(num1%j==0):
break
hashi[j].append(num1)
if(num2!=num1 and num2 not in hasho):
for j in primes:
if(num2%j==0):
break
hashi[j].append(num2)
i+=1
for j in primes:
if(n%j==0):
break
hashi[j].append(n)
done=dict()
if(len(primes)==1):
for i in hashi[primes[0]]:
print(i,end=" ")
print()
print(0)
continue
if(len(primes)==2):
if(primes[0]*primes[1]==n):
print(primes[0],primes[1],n)
print(1)
else:
for i in hashi[primes[0]]:
print(i,end=" ")
for i in hashi[primes[1]]:
print(i,end=" ")
print(primes[0]*primes[1],end=" ")
print()
print(0)
continue
for i in range(k):
for j in hashi[primes[i]]:
print(j,end=" ")
ko=primes[i]*primes[(i+1)%k]
print(ko,end=" ")
print()
print(0)
``` | output | 1 | 5,975 | 22 | 11,951 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An agent called Cypher is decrypting a message, that contains a [composite number](https://en.wikipedia.org/wiki/Composite_number) n. All divisors of n, which are greater than 1, are placed in a circle. Cypher can choose the initial order of numbers in the circle.
In one move Cypher can choose two adjacent numbers in a circle and insert their [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) between them. He can do that move as many times as needed.
A message is decrypted, if every two adjacent numbers are not coprime. Note that for such constraints it's always possible to decrypt the message.
Find the minimal number of moves that Cypher should do to decrypt the message, and show the initial order of numbers in the circle for that.
Input
The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines describe each test case.
In a single line of each test case description, there is a single composite number n (4 ≤ n ≤ 10^9) — the number from the message.
It's guaranteed that the total number of divisors of n for all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case in the first line output the initial order of divisors, which are greater than 1, in the circle. In the second line output, the minimal number of moves needed to decrypt the message.
If there are different possible orders with a correct answer, print any of them.
Example
Input
3
6
4
30
Output
2 3 6
1
2 4
0
2 30 6 3 15 5 10
0
Note
In the first test case 6 has three divisors, which are greater than 1: 2, 3, 6. Regardless of the initial order, numbers 2 and 3 are adjacent, so it's needed to place their least common multiple between them. After that the circle becomes 2, 6, 3, 6, and every two adjacent numbers are not coprime.
In the second test case 4 has two divisors greater than 1: 2, 4, and they are not coprime, so any initial order is correct, and it's not needed to place any least common multiples.
In the third test case all divisors of 30 greater than 1 can be placed in some order so that there are no two adjacent numbers that are coprime. | instruction | 0 | 5,976 | 22 | 11,952 |
Tags: constructive algorithms, implementation, math, number theory
Correct Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop,heapify
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
from itertools import accumulate
from functools import lru_cache
M = mod = 998244353
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n')]
def li3():return [int(i) for i in input().rstrip('\n')]
def isprime(n):
for j in range(2, int(n ** 0.5) + 1):
if n % j == 0:return 0
return 1
for _ in range(val()):
n = val()
l1 = factors(n)[1:]
l = []
for j in l1:
if isprime(j):l.append(j)
l1 = set(l1)
l1 -= set(l)
# print(l, l1)
d = defaultdict(set)
for j in range(len(l)):
for i in sorted(list(l1)):
if i % l[j] == 0 and i % l[j - 1] == 0:
d[tuple(sorted([l[j], l[j - 1]]))].add(i)
l1.remove(i)
break
# print(l, l1)
for j in range(len(l)):
for i in sorted(list(l1)):
if i % l[j] == 0 and i % l[j - 1] == 0:
d[tuple(sorted([l[j], l[j - 1]]))].add(i)
l1.remove(i)
# print(l, l1, d)
only = defaultdict(list)
for j in range(len(l)):
for i in sorted(list(l1)):
if i % l[j] == 0:
only[l[j]].append(i)
l1.remove(i)
fin = []
if len(l) == 2:
fin.append(l[0])
for j in only[l[0]]:fin.append(j)
for i in range(len(l)):
for j in list(d[tuple(sorted([l[i], l[(i + 1) % len(l)]]))]):
fin.append(j)
d[tuple(sorted([l[i], l[(i + 1) % len(l)]]))].remove(j)
if i != len(l) - 1:break
if i != len(l) - 1:
fin.append(l[i + 1])
for j in only[l[i + 1]]:
fin.append(j)
else:
fin.append(l[0])
for j in only[l[0]]:fin.append(j)
for i in range(len(l)):
for j in d[tuple(sorted([l[i], l[(i + 1) % len(l)]]))]:
fin.append(j)
if i != len(l) - 1:
fin.append(l[i + 1])
for j in only[l[i + 1]]:
fin.append(j)
ans = 0
for i in range(len(fin)):
if math.gcd(fin[i], fin[i - 1]) == 1:ans += 1
print(*fin)
print(ans)
``` | output | 1 | 5,976 | 22 | 11,953 |
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