UVa 10168 - Summation of Four Primes

题目：判断给定的数字能否写成4个素数的和。

分析：数论。1.如果N < 8 不成立；2.如果 N >= 8 时，如果N为奇数则令N = 2 + 3 + K，

            N为偶数则令N = 2 + 2 + K这样K为偶数且K>=4，这时满足哥德巴赫猜想：

            一个偶数可以拆成2个素数和的形式。通过循环拆解K即可。

            素数判定可以利用筛法打表，或者直接按照随机判定算法判定。

注意：数组不要越界。

打表法：

#include <iostream>
#include <cstdlib>
#include <cstring>

using namespace std;

int visit[10000001];
int prime[700001];

int main()
{
	memset( visit, 0, sizeof(visit) );
	int count = 0;
	visit[0] = visit[1] = 1;
	for ( int i = 2 ; i < 10000000 ; ++ i )
		if ( !visit[i] ) {
			for ( int j = i<<1 ; j < 10000000 ; j += i )
				visit[j] = 1;
			prime[count ++] = i;
		}
	
	int n;
	while ( cin >> n ) {
		if ( n < 8  ) cout << "Impossible." << endl;
		else if ( n%2 ) {
			for ( int i = 0 ; i < count ; ++ i )
				if ( !visit[n-5-prime[i]] ) {
					cout << "2 3 " << prime[i] << " " << n-5-prime[i] << endl;
					break;
				}
		}else {
			for ( int i = 0 ; i < count ; ++ i )
				if ( !visit[n-4-prime[i]] ) {
					cout << "2 2 " << prime[i] << " " << n-4-prime[i] << endl;
					break;
				}
		}
	}
	
	return 0;
}

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cmath>

using namespace std;

typedef long long LL;

LL  powmod( int a, int x, int n )
{
	if ( x == 1 ) return a+0LL;
	LL t = powmod( a, x/2, n );
	if ( x%2 ) return ((t*t)%n*a)%n;
	else return (t*t)%n;
}

int witness( int a, int n )  
{  
	LL  u = n-1LL,d,x;
	int t = 0;
	for ( ; !(u&0x1) ; u >>= 1 ) ++ t;
	d = powmod( a, u, n );
	while ( t -- > 0 ) {
		x = d; d = (d*d)%n;
		if ( d == 1 && x != 1 && x != n-1 ) return 1;
	}
    return (d != 1);
}  
  
int miller( int n, int s = 10 )  
{  
	if ( n == 2 ) return 1;
	if ( n == 1 || !(n&1) ) return 0;
    for ( int i = 0 ; i < s ; ++ i )  
        if ( witness( rand()%(n-1)+1, n ) )  
            return 0;  
    return 1;  
}

int main()
{
	int n;
	while ( cin >> n ) {
		if ( n < 8  ) cout << "Impossible." << endl;
		else if ( n <= 9 ) cout << "2 2 2 " << n-6 << endl;
		else {
			if ( n%2 ) cout << "2 3 ";
			else cout << "2 2 ";
			
			n = n-n%2-4;
			for ( int i = 3 ; i <= n ; i += 2 )
				if ( miller(i) && miller(n-i) ) {
					cout << i << " " << n-i << endl;
					break;
				}
		}
	}
	
	return 0;
}

UVa 10168 - Summation of Four Primes,布布扣,bubuko.com

UVa 10168 - Summation of Four Primes

原文：http://blog.csdn.net/mobius_strip/article/details/21879957