UVA10168 Summation of Four Primes【筛选法】

Euler proved in one of his classic theorems that prime numbers are infinite in number. But can everynumber be expressed as a summation of four positive primes? I don’t know the answer. May be youcan help!!! I want your solution to be very efficient as I have a 386 machine at home. But the time limitspecified above is for a Pentium III 800 machine. The definition of prime number for this problem is“A prime number is a positive number which has exactly two distinct integer factors”. As for example37 is prime as it has exactly two distinct integer factors 37 and 1.

Input

The input contains one integer number N (N ≤ 10000000) in every line. This is the number you willhave to express as a summation of four primes. Input is terminated by end of file.

Output

For each line of input there is one line of output, which contains four prime numbers according tothe given condition. If the number cannot be expressed as a summation of four prime numbers printthe line ‘Impossible.’ in a single line. There can be multiple solutions. Any good solution will beaccepted.

Sample Input

Sample Output

3 11 3 7

3 7 13 13

11 11 17 7

问题链接：UVA10168 Summation of Four Primes

问题简述：（略）

问题分析：

　　这个问题是哥德巴赫猜想的一个扩展。

　　筛选法是必要的，先找出素数。

　　对于n < 8，则n不能表示为４个素数之和；

　　若n为偶数，则两个素数为2和2，剩下部分的n（n-4）再分解；

　　若n为奇数，则两个素数为2和3，剩下部分的n（n-5）再分解。

程序说明：（略）

题记：（略）

参考链接：（略）

AC的C＋＋语言程序如下：

/* UVA10168 Summation of Four Primes */#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>using namespace std;const int N = 1e7;
const int SQRTN = ceil(sqrt((double) N));
bool isPrime[N + 1];// Eratosthenes筛选法
void esieve(void)
{memset(isPrime, true, sizeof(isPrime));isPrime[0] = isPrime[1] = false;for(int i=2; i<=SQRTN; i++) {if(isPrime[i]) {for(int j=i*i; j<=N; j+=i)  //筛选isPrime[j] = false;}}
}int main()
{esieve();int n;while(~scanf("%d", &n)) {if(n < 8)printf("Impossible.\n");else if(n % 2 == 1) {/* 奇数 */for(int i=2; i<n-5; i++)if(isPrime[i] && isPrime[n - 5 - i]) {printf("2 3 %d %d\n", i, n - 5 - i);break;}} else {/* 偶数 */for(int i=2; i<n-4; i++)if(isPrime[i] && isPrime[n - 4 - i]) {printf("2 2 %d %d\n", i, n - 4 - i);break;}}}return 0;
}

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