D. Dima and Lisa
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Dima loves representing an odd number as the sum of multiple primes, and Lisa loves it when there are at most three primes. Help them to represent the given number as the sum of at most than three primes.
More formally, you are given an odd numer n. Find a set of numbers pi (1 ≤ i ≤ k), such that
1 ≤ k ≤ 3
pi is a prime
The numbers pi do not necessarily have to be distinct. It is guaranteed that at least one possible solution exists.
Input
The single line contains an odd number n (3 ≤ n < 109).
Output
In the first line print k (1 ≤ k ≤ 3), showing how many numbers are in the representation you found.
In the second line print numbers pi in any order. If there are multiple possible solutions, you can print any of them.
Examples
input
27
output
3
5 11 11
Note
A prime is an integer strictly larger than one that is divisible only by one and by itself.
題意:
給出一個奇數,將其分解成若干個素數,素數個數不能超過3個.
解題思路:
根據哥德巴赫猜想,一個大於2的偶數可以分解成兩個素數.
那麼把n減去3,然後暴力查找即可.
AC代碼:
#include<bits/stdc++.h>
using namespace std;
bool isPrime(int n)
{
if(n == 1) return 0;
int sqrt_n = sqrt(n);
for(int i = 2;i <= sqrt_n;i++)
if(n % i == 0) return 0;
return 1;
}
int main()
{
int n;
scanf("%d",&n);
if(isPrime(n))
{
printf("1\n%d",n);
}
else
{
printf("3\n3 ");
n -= 3;
for(int i = 2;i <= n;i++)
if(isPrime(i) && isPrime(n-i)) {printf("%d %d",i,n-i);break;}
}
return 0;
}
