[HAOI2011]PROBLEM B

[HAOI2011]PROBLEM B

推式子

代码

/*
  Author : lifehappy
*/
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int N = 1e6 + 10;

int mu[N], prime[N], cnt, n, m;

bool st[N];

void init() {
    mu[1] = 1;
    for(int i = 2; i < N; i++) {
        if(!st[i]) {
            prime[++cnt] = i;
            mu[i] = -1;
        }
        for(int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {
            st[i * prime[j]] = 1;
            if(i % prime[j] == 0) {
                break;
            }
            mu[i * prime[j]] = -mu[i];
        }
    }
    for(int i = 1; i < N; i++) {
        mu[i] += mu[i - 1];
    }
}

ll f(int n, int m) {
    if(n > m) swap(n, m);
    ll ans = 0;
    for(int l = 1, r; l <= n; l = r + 1) {
        r = min(n / (n / l), m / (m / l));
        ans += 1ll * (mu[r] - mu[l - 1]) * (n / l) * (m / l);
    }
    return ans;
}

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    init();
    int T;
    scanf("%d", &T);
    while(T--) {
        int a, b, c, d, k;
        scanf("%d %d %d %d %d", &a, &b, &c, &d, &k);
        printf("%lld\n", f(b / k, d / k) - f((a - 1) / k, d / k) - f(b / k, (c - 1) / k) + f((a - 1) / k, (c - 1) / k));
    }
    return 0;
}