[HDOJ1695]GCD

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1695

　　两个区间内各取一个数，使这两个数的gcd为确定值，统计符合条件的数的对数。（认为(a,b)和(b,a)是同一对）

　　

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <cmath>

using namespace std;

typedef long long LL;
const int maxn = 100010;

LL a, b, c, d, k;
vector<LL> fac[maxn];
int vis[maxn];

inline LL min(LL &a, LL &b) {
    return a > b ? b : a;
}

inline void swap(LL &a, LL &b) {
    a = a ^ b;
    b = a ^ b;
    a = a ^ b;
}

void init() {
    memset(vis, 0, sizeof(vis));
    for(int i = 0; i < maxn; i++) {
        fac[i].clear();
    }
    for(int i = 2; i < maxn; i+=2) {
        fac[i].push_back(2);
    }
    for(int i = 3; i < maxn; i+=2) {
        if(!vis[i]) {
            for(int j = i; j < maxn; j+=i) {
                vis[j] = 1;
                fac[j].push_back(i);
            }
        }
    }
}

LL inc(int j, int w) {
    int cnt = 0;
    LL m = 1;
    for(int i = 0; i < fac[w]. size(); i++) {
        if(j & (1 << i)) {
            cnt++;
            m *= fac[w][i];
        }
    }
    int sum = k / m;
    if(cnt % 2 == 1) {
        return sum;
    }
    else {
        return -sum;
    }
}

LL solve() {
    LL ans = 0;
    b /= k; 
    d /= k;
    if(b > d) {
        swap(b, d);
    }
    for(int i = 1; i <= d; i++) {
        k = min((LL)i, (LL)b);
        ans += k;
        int nn = 1 << fac[i].size();
        for(int j = 1; j < nn; j++) {
            ans -= inc(j, i);
        }
    }
    return ans;
}

int main() {
    init();
    int T;
    scanf("%d", &T);
    for(int i = 1; i <= T; i++) {
        scanf("%d %d %d %d %d", &a, &b, &c, &d, &k);
        if(k == 0) {
            printf("Case %d: 0\n", i);
            continue;
        }
        LL ans = solve();
        printf("Case %d: %I64d\n", i, ans);
    }
    return 0;
}