CSU1552: Friends（快速判断大数是不是素数+二分匹配）

发布时间：2020-12-14 02:40:11 所属栏目：大数据来源：网络整理

导读：1552: Friends Time Limit:? 3 Sec?? Memory Limit:? 256 MB Submit:? 187?? Solved:? 43 [ Submit][ Status][ Web Board] Description On an alien planet,every extraterrestrial is born with a number. If the sum of two numbers is a prime number,the

1552: Friends

Time Limit:?3 Sec?? Memory Limit:?256 MB
Submit:?187?? Solved:?43
[ Submit][ Status][ Web Board]

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

Sample Output

1
2

HINT

当遇到匹配问题时，而且只能用一次，这时应想一下二分最大匹配。

#include<stdio.h>
#include<math.h>
#include <algorithm>
#include<string.h>
#include <time.h>
using namespace std;
#define ll long long
//-----大素数的判断-------
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 n)
{
    ll s=1;
    while(b)
    {
        if(b&1) s=mult_mod(s,a,n);
        a=mult_mod(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,ok[105][105],mark[105],vist[105];
int DFS(int x)
{
    for(int i=1;i<=n;i++)
    if(ok[x][i]&&vist[i]==0)
    {
        vist[i]=1;
        if(mark[i]==0||DFS(mark[i]))
        {
            mark[i]=x; return 1;
        }
    }
    return 0;
}
int main()
{

    ll a[105];
    while(scanf("%d",&n)>0)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
        }
        memset(ok,sizeof(ok));
        for(int i=1;i<n;i++)
        for(int j=i+1;j<=n;j++)
        if(Prime(a[i]+a[j]))
            ok[i][j]=ok[j][i]=1;

        int ans=0;
        memset(mark,sizeof(mark));
        for(int i=1;i<=n;i++)
        {
            memset(vist,sizeof(vist));
            ans+=DFS(i);
        }

        printf("%dn",ans/2);
    }
}

（编辑：李大同）

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