Cattle-off practice match 43F (push equation)

Points

• Topic Link
• 1e18 data not \ (O (n) \) tolerance repellent, then push equation, the official solution, wherein little error formula
• And no pretreatment mu, directly determined according to the number of prime numbers to positive and negative

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

int T;
ll k, q, n, m;
int mark[25];

void pre() {
    mark[2] = mark[3] = mark[5] = mark[7] = mark[11] = mark[13] = mark[17] = mark[19] = 1;
}

int main() {
    pre();
    for (scanf("%d", &T); T--;) {
        scanf("%lld %lld %lld %lld", &k, &q, &n, &m);
        if (!k) {//特判
            puts("QAQ"); continue;
        }
        ll h = n, P = 1;
        int v[10], t = 0;
        for (int i = 2; i <= m; i++)
            if (mark[i])//m内质数
                v[t++] = i, P *= i;
        for (int i = 1; i < (1 << t); i++) {//枚举组合
            int tmp = 1, cnt = 0;
            for (int j = 0; j < t; j++) {
                if ((i >> j) & 1)
                    cnt++, tmp *= v[j];
            }
            ll s = (cnt & 1) ? -1 : 1;//奇数个时mu为负
            h += s * n / tmp;
        }
        bool flag = (min(h + k - 1, n) >= q);
        puts(flag ? "Yes" : "QAQ");
    }
}