csu 1552: Friends（拉宾米勒测试+二分匹配）

1552: Friends

Time Limit: 3 Sec   Memory Limit: 256 MB

Submit: 129   Solved: 24

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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      

题意很简单。。。

主要的是大数判断是否是质数  这里用到的是拉宾米勒测试 不懂的请百度

另外 在二分图匹配时  要建无向图。。。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <string.h>
#include <cstdlib>
#include <time.h>
#define ll long long
 
using namespace std;
 
ll MIN;
//ll gcd(ll a,ll b)
//{
//    if(b==0) return a;
//    else return (b,a%b);
//}
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;
}
 
int 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 0;
    }
    return 1;
}
 
int N;
int mp[110][110];
int linker[110];
int vis[110];
int ans;
 
bool dfs(int u)
{
    for(int i=0;i<N;i++)
    {
        if(mp[u][i]&&!vis[i])
        {
            vis[i]=1;
            if(linker[i]==-1||dfs(linker[i]))
            {
                linker[i]=u;
                return 1;
            }
        }
    }
    return 0;
}
 
void hungary()
{
    memset(linker,-1,sizeof(linker));
    for(int i=0;i<N;i++)
    {
        memset(vis,0,sizeof(vis));
        if(dfs(i))
        {
            ans++;
//            cout<<"ans  "<<ans<<endl;
        }
    }
}
 
ll A[110];
 
int main()
{
//    freopen("F.txt","r",stdin);
    while(scanf("%d",&N)!=EOF)
    {
        for(int i=0;i<N;i++)
        {
            scanf("%lld",&A[i]);
        }
        memset(mp,0,sizeof(mp));
        for(int i=0;i<N;i++)
            for(int j=i+1;j<N;j++)
            {
//                if(i==j)  continue;
                if(Prime(A[i]+A[j]))
                    mp[i][j]= mp[j][i] =1;
            }
//            for(int j=i+1;j<N;j++)
//                if(Miller_Rabin(a[i]+a[j]))
//                {
//                    mp[i][j]=mp[j][i]=1;
//                    cout<<i<<"  "<<j<<"  "<<mp[i][j]<<endl;
//                }
        ans=0;
        hungary();
        printf("%d\n",ans/2);
    }
    return 0;
}