Hdu4135容斥原理

求A到B之间有多少个数能与N互质。

学了下容斥原理的写法， 想将N分解质因数，然后容斥原理，N - 单个质因数倍数个数+2个质因数倍数的个数-3个质因数的个数......

#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;
vector<int> q;
LL ans;
void dfs(LL x, LL now, LL flag,LL key)
{
    if (x == q.size()) {
        if (flag == 0) ans -= key / now;
        else ans += key / now;
        return;
    }
    dfs(x + 1, now, flag, key);
    dfs(x + 1, now*q[x], flag ^ 1, key);
}

LL solve(LL x)
{
    ans = 0;
    dfs(0, 1, 1, x);
    return ans;
}


int main()
{
    int Icase = 0;
    int T,  n;
    LL a, b;
    cin >> T;
    while (T--){
        ans = 0;
        cin >> a >> b >> n;
        q.clear();
        int t = n;
        for (int i = 2; i*i <= t; i++) {
            if (t%i) continue;
            while (t%i == 0) t /= i;
            q.push_back(i);
        }
        if (t > 1) q.push_back(t);
        printf("Case #%d: ", ++Icase);
        cout << solve(b) - solve(a - 1) << endl;
    }
    return 0;
}