Problem:
Score : 600 points

Problem Statement

We have a sequence of k numbers: d_0,d_1,...,d_{k - 1}.

Process the following q queries in order:

The i-th query contains three integers n_i, x_i, and m_i.
Let a_0,a_1,...,a_{n_i - 1} be the following sequence of n_i numbers: \begin{eqnarray} a_j = \begin{cases} x_i & ( j = 0 ) \\ a_{j - 1} + d_{(j - 1)~\textrm{mod}~k} & ( 0 < j \leq n_i - 1 ) \end{cases}\end{eqnarray}
Print the number of j~(0 \leq j < n_i - 1) such that (a_j~\textrm{mod}~m_i) < (a_{j + 1}~\textrm{mod}~m_i).

Here (y~\textrm{mod}~z) denotes the remainder of y divided by z, for two integers y and z~(z > 0).

Constraints

All values in input are integers.

1 \leq k, q \leq 5000

0 \leq d_i \leq 10^9

2 \leq n_i \leq 10^9

0 \leq x_i \leq 10^9

2 \leq m_i \leq 10^9

Input

Input is given from Standard Input in the following format:

k q
d_0 d_1 ... d_{k - 1}
n_1 x_1 m_1
n_2 x_2 m_2
:
n_q x_q m_q

Output

Print q lines.

The i-th line should contain the response to the i-th query.

Sample Input 1

3 1
3 1 4
5 3 2

Sample Output 1

1

For the first query, the sequence {a_j} will be 3,6,7,11,14.

(a_0~\textrm{mod}~2) > (a_1~\textrm{mod}~2)

(a_1~\textrm{mod}~2) < (a_2~\textrm{mod}~2)

(a_2~\textrm{mod}~2) = (a_3~\textrm{mod}~2)

(a_3~\textrm{mod}~2) > (a_4~\textrm{mod}~2)

Thus, the response to this query should be 1.

Sample Input 2

7 3
27 18 28 18 28 46 1000000000
1000000000 1 7
1000000000 2 10
1000000000 3 12

Sample Output 2

224489796
214285714
559523809

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<functional>
#include <random>
#include<iostream>
#include<iomanip>
using namespace std;

long long k, q;
long long d[5000];
long long n[5000];
long long x[5000];
long long m[5000];
long long sum;
long long sum_all;
long long diff;
int main() {
	cin >> k >> q;
	for (int i = 0; i < k; i++) {
		cin >> d[i];
	}

	for (int i = 0; i < q; i++) {
		cin >> n[i] >> x[i] >> m[i];
	}
	for (int i = 0; i < q; i++) {
		sum = (x[i] % m[i]);
		sum_all = 0;
		diff = 0;
		for (int j = 0; j < k; j++) {
			sum_all += d[j] % m[i];
			if (d[j] % m[i] == 0) {
				diff += 1;
			}
		}
		sum += sum_all * ((n[i] - 1) / k);
		diff = diff * ((n[i] - 1) / k);
		for (int j = 0; j < (n[i] - 1) % k; j++) {
			sum += d[j] % m[i];
			if (d[j] % m[i] == 0) {
				diff += 1;
			}
		}
//		cout << sum << " ";
//		if (sum < 0) { n[i] += 1; }
		cout << n[i] - 1 - sum / m[i] - diff << "\n";
	}
}