ural 1356. Something Easier 哥德巴赫猜想

1356. Something Easier

Time limit: 1.0 second
Memory limit: 64 MB

 

“How do physicists define prime numbers? Very easily: prime numbers are the number 2 and all the odd numbers greater than 2. They may show that this definition corresponds to the mathematical one: 3 is prime, 5 is prime, 7 is prime… 9? 9 is certainly not prime. Then: 11 is prime, 13 is prime. So 9 is the experiment mistake.”

From mathematical analysis course

Once physicist and mathematician argued how many prime numbers one needed for the purpose that their sum was equal to N. One said that it wasn’t known and the other that 3 was always enough. The question is how many.

Input

The first line contains T, an amount of tests. Then T lines with integer N follow (0 ≤ T ≤ 20; 2 ≤ N ≤ 10⁹).

Output

For each test in a separate line you should output prime numbers so that their sum equals to N. An amount of such prime numbers is to be minimal possible.

Sample

input	output
7 2 27 85 192 14983 3 7	2 23 2 2 2 83 11 181 14983 3 7

Problem Author: Aleksandr Bikbaev
Problem Source: USU Junior Championship March'2005

思路：（1） 当n为偶数的时候， n > 4  n可以分解成两个素数的和

　　　（2）当n为奇数的时候， 如果n是素数就直接输出；

　　　如果n-2也是素数， 可以直接输出 2 n-2；

　　　那么把奇数n分解为三个奇素数；

很好奇这么高的复杂度，竟然跑的这么快啊

#include <iostream>
#include <cstdio>
using namespace std;

bool fun(int x){
    if(x <= 1) return false;
    for(int i = 2; i*i <= x; i++){
        if(x%i == 0){
            return false;
        }
    }
    return true;
}

int main() {
    //freopen("in.txt", "r", stdin);
    int t, n;
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        if(n%2 == 0){
            if(n == 2){
                puts("2");
            }else if(n == 4){
                puts("2 2");
            }else {
                for(int i = 3; i < n; i += 2){
                    if(fun(i) && fun(n-i)){
                        printf("%d %d
", i, n-i);
                        break;
                    }
                }
            }
        }else {
            if(fun(n)) printf("%d
", n);
            else {
                if(fun(n-2)){
                    printf("2 %d
", n-2);
                }else {
                    int flag = 1;
                    for(int i = 3; i < n; i += 2){
                        for(int j = 3; j < n - i; j += 2){
                            if(fun(i) && fun(j) && fun(n-i-j)){
                                if(flag) {
                                printf("%d %d %d
", i, j, n-i-j);
                                flag = 0;
                                break;
                                }
                            }
                        }
                        if(!flag) break;
                    }
                }
            }
        }
    }
    return 0;
}