Hdu1695容斥原理

就是计算  区间[1,b/k] [1,d/k] 互质的对数，(3,4) 和(4,3)算一种。每次只加比他大的就行了。

#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;


int ans;
vector<int> q;
int m;
void gao(int x, int pre, int flag, int key)
{
    if (x == q.size()) {
        if (flag) ans += key / pre;
        else ans -= key / pre;
        return;
    }
    gao(x + 1, pre, flag, key);
    gao(x + 1, pre*q[x], flag ^ 1, key);
}

int ask(int n)
{
    int t = n;
    q.clear();
    for (int i = 2; i*i <= t; i++) {
        if (t%i) continue;
        while (t%i == 0) t /= i; q.push_back(i);
    }
    ans = 0;
    if (t>1) q.push_back(t);
    gao(0, 1, 1, m);
    int cc = ans; ans = 0;
    gao(0, 1, 1, n-1);   // 1 和本身互质
    return cc - ans;
}

int main()
{
    int T, a, b, c, d, k;
    cin >> T;
    int Icase = 0;
    while (T--) {
        scanf("%d%d%d%d%d", &a, &b, &c, &d, &k);
        if (k == 0) {
            printf("Case %d: 0
", ++Icase);
            continue;
        }
        b /= k; d /= k;
        if (d < b) swap(b, d);
        m = d;
        LL sum = 0;
        for (int i = 1; i <= b; i++) {
            sum += ask(i);
        }
        printf("Case %d: ", ++Icase);
        cout << sum << endl;//开始强行加一，会出现等于0的情况。
    }
    return 0;
}