CSU 1552-Friends（大数判断素数+二分匹配）

1552：Friends

Time Limit: 3 Sec  Memory Limit: 256 MBSubmit: 707  Solved: 191[Submit][Status][WebBoard]

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

思路：

这题用普通的素数判断是会超时的，所以要用大数的素数判断（拉宾米勒测试 ），之后就是一个二分匹配的模板就行了。

AC代码：

#include<iostream>#include<algorithm>#include<cstring>#include<string>#include<cstdio>#include<vector>#include<cmath>#include<time.h>using namespace std;#define T 1100#define inf 0x3f3f3f3f#define mod 1000000007#define lson (rt<<1)#define rson (rt<<1|1)typedef unsigned long long ll;int V;int g[T][T];int cx[T],cy[T];int mk[T];//*******************//拉宾米勒测试ll MIN;ll mult_mod(ll a,ll b,ll n){ll s=0;while(b){if(b&1) s=(s+a)%n;a=(a+a)%n;b>>=1;}return s;}ll pow_mod(ll a,ll b,ll n){ll s=1;while(b){if(b&1) s=mult_mod(s,a,n);a=mult_mod(a,a,n);b>>=1;}return s;}bool Prime(ll n){ll u=n-1,pre,x;int i,j,k=0;if(n==2||n==3||n==5||n==7||n==11)  return 1;if(n==1||(!(n%2))||(!(n%3))||(!(n%5))||(!(n%7))||(!(n%11)))   return 0;for(;!(u&1);k++,u>>=1);srand((ll)time(0));for(i=0;i<5;i++){x=rand()%(n-2)+2;x=pow_mod(x,u,n);pre=x;for(j=0;j<k;j++){x=mult_mod(x,x,n);if(x==1&&pre!=1&&pre!=(n-1))return 0;pre=x;}if(x!=1)  return false;}return true;}//************************void add_edge(int u,int v){g[u][v]=1;g[v][u]=1;}int dfs(int v){for(int i=0;i<V;++i){if(g[v][i]&&!mk[i]){mk[i]=1;if(cy[i]==-1||dfs(cy[i])){cx[v]=i;cy[i]=v;return 1;}}}return 0;}int bipartite_matching(){int res=0;memset(cx,-1,sizeof(cx));memset(cy,-1,sizeof(cy));for(int i=0;i<V;++i){if(cx[i]==-1){memset(mk,0,sizeof(mk));res+=dfs(i);}}return res;}int main(){#ifdef zscfreopen("input.txt","r",stdin);#endifint i,j;ll a[T];while(~scanf("%d",&V)){memset(g,0,sizeof(g));for(i=0;i<V;++i)scanf("%lld",&a[i]);for(i=0;i<V;++i){for(j=i+1;j<V;++j){ll tmp = a[i]+a[j];if(Prime(tmp)){add_edge(i,j);}}}printf("%d\n",bipartite_matching()/2);}return 0;}