HDU 2841 Visible Trees

Visible Trees

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 2841
64-bit integer IO format: %I64d      Java class name: Main

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 $n(1leq m, nleq 100000)$

 

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

 

解题：容斥原理

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 200010;
vector<int>g[maxn];
void init(){
    for(int i = 2; i < maxn; ++i){
        for(int j = 1; i*j < maxn; ++j)
            g[i*j].push_back(i);
    }
}
int main(){
    int kase,n,m;
    init();
    scanf("%d",&kase);
    while(kase--){
        scanf("%d%d",&n,&m);
        LL ret = 0;
        for(int i = 2; i <= n; ++i){
            int cnt = 0,ans = 0;
            for(int j = 1,t = g[i].size();j < (1<<t); ++j){
                 LL tmp = 1;
                 for(int k = cnt = 0; k < t; ++k){
                    if((j>>k)&1){
                        cnt++;
                        tmp *= g[i][k];
                    }
                }
                if(cnt&1) ans += m/tmp;
                else ans -= m/tmp;
            }
            ret += m - ans;
        }
        printf("%I64d
",ret + m);
    }
    return 0;
}