ZOJ 3940 Modulo Query
One day, Peter came across a function which looks like:
- F(1, X) = X mod A1.
- F(i, X) = F(i - 1, X) mod Ai, 2 ≤ i ≤ N.
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, ..., AN (1 ≤ Ai ≤ 109).
The third line contains an integer Q (1 ≤ Q ≤ 105) - the number of queries. Each of the following Q lines contains an integer Yi (0 ≤ Yi ≤ 109), which means Peter wants to know the number of solutions for equation F(N, X) = Yi.
Output
For each test cases, output an integer S = (1 ⋅ Z1 + 2 ⋅ Z2 + ... + Q ⋅ ZQ) mod (109 + 7), where Zi is the answer for the i-th query.
Sample Input
1 3 5 3 2 4 5 0 1 2 3 4
Sample Output
8
Hint
The answer for each query is: 4, 2, 0, 0, 0.
Author: LIN, XiSource: 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;
}