Hdu2841容斥原理

人在(0,0)点，问n*m的矩阵上的点有多少可以与人直接可见，其实就是矩阵上点与(0,0)点的形成的斜率种类数。

#pragma comment(linker,"/STACK:102400000,102400000")
#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<iostream>
#include<string>
#include<queue>
#include<stack>
#include<list>
#include<stdlib.h>
#include<algorithm>
#include<vector>
#include<map>
#include<cstring>
#include<set>
using namespace std;
typedef long long LL;

LL cnt, ans, m;
LL q[20];
void gao(LL x, LL pre, LL flag, LL key)
{
    if (x == cnt) {
        if (flag == 0) ans -= key / pre;
        else ans += key / pre;
        return;
    }
    gao(x + 1, pre, flag, key);
    gao(x + 1, pre*q[x], flag ^ 1, key);
}

LL ask(LL n) {
    cnt = 0;
    LL t = n;
    ans = 0;
    for (LL i = 2; i*i <= t; i++) {
        if (t%i)  continue;
        while (t%i == 0) t /= i;
        q[cnt++] = i;
    }
    if (t > 1) q[cnt++] = t;
    gao(0, 1, 1, m);
    return ans;

}

int main()
{
    LL T, n;
    cin >> T;
    while (T--) {
        cin >> n >> m;
        if (m < n) swap(n, m);
        LL sum = 0;
        for (LL i = 1; i <= n; i++) {
            sum += ask(i);
        }
        cout << sum << endl;
    }
    return 0;
}