CodeForces 584D - Dima and Lisa(数论)

最新推荐文章于 2024-05-23 08:50:51 发布

原创最新推荐文章于 2024-05-23 08:50:51 发布 · 646 阅读

0 ·

CC 4.0 BY-SA版权

CodeForces 专栏收录该内容

65 篇文章

订阅专栏

本文介绍了一种将奇数分解为最多三个素数之和的算法实现，并提供了完整的AC代码示例。根据哥德巴赫猜想，通过检查是否为素数及暴力查找的方式实现了这一目标。

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.

题意:
给出一个奇数,将其分解成若干个素数,素数个数不能超过3个.

解题思路:
根据哥德巴赫猜想,一个大于2的偶数可以分解成两个素数.
那么把n减去3,然后暴力查找即可.

AC代码:

#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;
}