ZOJ 3940 Modulo Query

Modulo Query

Time Limit: 2 Seconds Memory Limit: 65536 KB

One day, Peter came across a function which looks like:

F(1, X) = X mod A₁.
F(i, X) = F(i - 1, X) mod A_i, 2 ≤ i ≤ N.

Where A is an integer array of length N , X is a non-negative integer no greater than M .

Peter wants to know the number of solutions for equation F(N, X) = Y, where Y is a given number.

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains two integers N and M (2 ≤ N ≤ 10⁵, 0 ≤ M ≤ 10⁹).

The second line contains N integers: A₁, A₂, ..., A_N (1 ≤ A_i ≤ 10⁹).

The third line contains an integer Q (1 ≤ Q ≤ 10⁵) - the number of queries. Each of the following Q lines contains an integer Y_i (0 ≤ Y_i ≤ 10⁹), which means Peter wants to know the number of solutions for equation F(N, X) = Y_i.

Output

For each test cases, output an integer S = (1 ⋅ Z₁ + 2 ⋅ Z₂ + ... + Q ⋅ Z_Q) mod (10⁹ + 7), where Z_i is the answer for the i-th query.

Sample Input

Sample Output

Hint

The answer for each query is: 4, 2, 0, 0, 0.

Author: LIN, Xi

Source: The 13th Zhejiang Provincial Collegiate Programming Contest

没想到是这么暴力的题，比赛的时候没敢做。。。

#include<queue>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
const int mod = 1e9 + 7;
const int maxn = 4e5 + 10;
int T, q, n, m, x, sum[maxn], b[maxn], ans, sz;struct point
{int x, y;point(int x = 0, int y = 0) :x(x), y(y) {}bool operator<(const point&a)const{return x < a.x;}
}a[maxn];int main()
{scanf("%d", &T);while (T--){scanf("%d%d", &n, &m);priority_queue<point> p;p.push(point(m, 1));while (n--){scanf("%d", &x);while (p.top().x >= x){point q = p.top();   p.pop();while (!p.empty() && p.top().x == q.x) q.y += p.top().y, p.pop();p.push(point(x - 1, (q.x + 1) / x*q.y));if ((q.x + 1) % x)p.push(point(q.x%x, q.y));}}sz = 1;   ans = 0;while (!p.empty()){point q = p.top(); p.pop();while (!p.empty() && p.top().x == q.x) q.y += p.top().y, p.pop();a[sz++] = q;}sort(a + 1, a + sz);for (int i = 1; i < sz; i++){sum[i] = sum[i - 1] + a[i].y;b[i] = a[i].x;}scanf("%d", &q);for (int i = 1; i <= q; i++){scanf("%d", &x);int k = lower_bound(b + 1, b + sz, x) - b;(ans += (LL)(sum[sz - 1] - sum[k - 1])*i%mod) %= mod;}printf("%d\n", ans);}return 0;
}