HDU 2841 Visible Trees

这道题，是求n*m的矩形内，有多少对互质的数（x,y）即Gcd( x ,y ) ==1;

这个题首先用到欧拉的质因子分解，再利用容斥定理；

View Code

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<algorithm>

#include<cmath>

#include<queue>

#include<set>

#include<map>

#include<cstring>

#include<vector>

using namespace std;

class Node

{

public:

      int cnt,num[20];    

}node[100024];

void Prime( )

{

    int prime[10000],cnt=0;

    bool hash[50024]={0};

    int t = (int)sqrt( 100000.0 ) + 1;

    for( int i = 3 ; i <=t ; i+=2 )

    {

         if( !hash[i>>1] )

         {

            int x = i<<1;

            for( int j = i * i ; j <=100001; j += x )

                 hash[j>>1] = true;        

         }    

    }    

    prime[cnt++] = 2;

    for( int i = 1 ;i <= 50000 ; i++ )

    {

        if( !hash[i] )

        {

            prime[cnt++] = ( i << 1 ) + 1;

        }    

    }

    for( int i = 2; i <= 100000 ; i++ )

    {

        t = i;

        node[i].cnt = 0;

        for( int j = 0 ; j < cnt; j ++ )

        {

             if( prime[j] > t ) break;

             if( t % prime[j] == 0 )

             {

                 node[i].num[node[i].cnt++] = prime[j];

                 while( t % prime[j] == 0 )

                        t /= prime[j];

             }    

        }

    }

}

long long DFS( int t , int m , int k )

{

    long long ans = 0;

    for( int i = t ; i < node[k].cnt ;i ++ )

    {

         ans += m/node[k].num[i] - DFS( i + 1 , m/node[k].num[i] , k ); 

    }

    return ans;

}

int main(  )

{

    int n ,m,Case;

    Prime();

    while( scanf( "%d\n",&Case ) ==1 )

    {

         while( Case-- )

         {

             long long ans = 0;

             scanf( "%d %d",&n,&m );

             for( int i = 1; i <= n ; i++ )

                  ans += m - DFS( 0 , m , i );    

            printf( "%I64d\n",ans );    

         }    

    }

    //system( "pause" );

    return 0;

}