HDU2841 （队列容斥）

Visible TreesTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2213    Accepted Submission(s): 908

Problem Description

There are many trees forming a m * n grid, the grid starts from (1,1). Farmer Sherlock is standing at (0,0) point. He wonders how many trees he can see.
If two trees and Sherlock are in one line, Farmer Sherlock can only see the tree nearest to him.

 

Input

The first line contains one integer t, represents the number of test cases. Then there are multiple test cases. For each test case there is one line containing two integers m and n(1 ≤ m, n ≤ 100000)

 

Output

For each test case output one line represents the number of trees Farmer Sherlock can see.

 

Sample Input

2 1 1 2 3

 

Sample Output

1 5

 

Source

2009 Multi-University Training Contest 3 - Host by WHU 

 

其实感觉这个代码写搓了  但是确实是利用队列的一种方式  重点理解dfs容斥方式

传送门 http://www.cnblogs.com/hsd-/p/5008912.html

 

#include<bits/stdc++.h>
using namespace std;
__int64 n;
__int64 a,b;
__int64  slove( __int64 m,__int64 gg)
{
    __int64  que[1000];
    __int64  a[1000];
    memset(que,0,sizeof(que));
    memset(a,0,sizeof(a));
    __int64 sum=0;
    __int64 t=0;
    __int64 ss=0;
    for(__int64 i=2;i*i<=m;i++)
    {
        if(m%i==0)
        {
            que[ss++]=i;
            while(m%i==0)
                m=m/i;
        }
    }
    if(m>1)
    que[ss++]=m;
    a[t++]=-1;
    for(__int64 i=0;i<ss;i++)
    {
         int k=t;
         for(__int64 j=0;j<k;j++)
         {
             a[t++]=que[i]*a[j]*(-1);
         }
    }
    for(__int64 i=1;i<t;i++)
       sum=sum+gg/a[i];
    return sum;
}
int main()
{
    while(scanf("%I64d",&n)!=EOF)
    {
        for(__int64 i=1;i<=n;i++)
        {
            __int64 re=0;
            scanf("%I64d %I64d",&a,&b);
            for(__int64 j=1;j<=a;j++)
                re+=slove(j,b);
            printf("%I64d
",a*b-re);
        }
    }
    return 0;
}