杭电2841 Visible Trees（容斥原理第二题）

Visible Trees

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1610 Accepted Submission(s): 664



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

/*
题目大意：从1---n与1---m的矩阵，问从(0,0)点看，如果是一排只能看到第一个点，能看到多少个点，相当于所有点与(0,0)点连线，能得到多少条线。可知，lcm相当于斜率，当(x/lcm,y/lcm)时，不在一条线上
挺经典的，容斥原理，相当于求1---n与1---m的互质的点的个数  加油！！！
Time:2014-12-17 17:43
*/
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long LL;
const int MAX=100000+10;
vector<int>fac[MAX];
bool vis[MAX];
void Del(){
    memset(vis,0,sizeof(vis));
    for(int i=0;i<MAX;i++)fac[i].clear();

    for(int i=2;i<MAX;i++){
        if(!vis[i]){
                fac[i].push_back(i);
            for(int j=i+i;j<MAX;j+=i){
                    vis[j]=true;
                fac[j].push_back(i);
            }
        }
    }
}
LL Get_ans(int n,int m){
    LL cnt=0;
    int k=fac
.size();
    int maxn=1<<k;
    for(int x=1;x<maxn;x++){
        int num=0,val=1;
        for(int i=0;i<k;i++){
            if(x&(1<<i)){
                num++;
                val*=fac
[i];
            }
        }
        if(num&1)
            cnt+=m/val;
        else
            cnt-=m/val;
    }
    return m-cnt;
}
int main(){
    int T,n,m;
    Del();
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        if(n>m)swap(n,m);//后边是循环1<<m次，如果差的多话，可以减少很多循环次数
        LL ans=m;//注意：n=1时，有m个，不是n个
       for(int i=2;i<=n;i++)
            ans+=Get_ans(i,m);

        printf("%lld\n",ans);
    }
return 0;
}