Description
On an alien planet, every extraterrestrial is born with a number. If the sum of two numbers is a prime number, then two extraterrestrials can be friends. But every extraterrestrial can only has at most one friend. You are given all number of the extraterrestrials, please determining the maximum number of friend pair.
Input
There are several test cases.
Each test start with positive integers N(1 ≤ N ≤ 100), which means there are N extraterrestrials on the alien planet.
The following N lines, each line contains a positive integer pi ( 2 ≤ pi ≤10^18),indicate the i-th extraterrestrial is born with pi number.
The input will finish with the end of file.
Output
For each the case, your program will output maximum number of friend pair.
Sample Input
3 2 2 3 4 2 5 3 8
Sample Output
1 2
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<time.h>
using namespace std;
typedef unsigned long long LL;
LL sum[101];
int a[101][101];
int n,vis[101],cn[101];
LL multi(LL a,LL b,LL n) //加法代替乘法，防止溢出 long long
{
LL s=0;
while(b)
{
if(b%2==1) s=(s+a)%n;
b/=2;
a=(a+a)%n;
}
return s;
}
LL mod(LL a,LL b,LL n)
{
LL s=1;
while(b)
{
if(b%2==1) s=multi(s,a,n);//<span style="color:#ff0000;">s=(s*a)%n;</span>
b/=2;
a=multi(a,a,n);
}
return s;
}
//拉宾米勒数素数
bool test(LL n)
{
LL a,x;
int k=0,i,j;
if(n<2) return 0;
if(n==2) return 1;//2是唯一的偶素数，直接返回
if(n%2==0) return 0;
srand((LL)time(0));
for(i=0;i<10;i++) //测试最大次数
{
a=rand()%(n-2)+2; //生成一个2到n-2的数字
x=mod(a,n-1,n); //快速幂（a^n-1）%n==1 费马小定理（a为整数，a,n互质)
if(x!=1)//结果不是1，不是素数
return 0;
}
return 1;//是素数
}
int funxx(int u)
{
int i;
for(i=1;i<=n;i++)
{
if(a[u][i]&&!vis[i])
{
vis[i]=1;
if(cn[i]==-1||funxx(cn[i]))
{
cn[i]=u;
return 1;
}
}
}
return 0;
}
int hun()
{
int i,count=0;
memset(cn,-1,sizeof(cn));
for(i=1;i<=n;i++)
{
memset(vis,0,sizeof(vis));
if(funxx(i))
count++;
}
return count;
}
int main()
{
int i,j;
while(scanf("%d",&n)==1)
{
memset(a,0,sizeof(a));
for(i=1;i<=n;i++)
scanf("%lld",&sum[i]);
for(i=1;i<=n;i++)
{
for(j=i+1;j<=n;j++)
{
if(test(sum[i]+sum[j])) {a[i][j]=1;a[j][i]=1;}
}
}
printf("%d\n",hun()/2);//二分图
}
return 0;
}
原文地址：http://blog.csdn.net/u012346225/article/details/45912559
