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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

/*	author:  Kite_kuma
	created: 2020.09.11 13:05:07 */

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

#pragma region aliases

#define rep(i, n) for(long long i = 0; i < (n); i++)
#define rrep(i, n) for(long long i = (n)-1; i > -1; i--)
#define Rep(i, m, n) for(long long i = (m); i < (n); i++)
#define rRep(i, m, n) for(long long i = (n)-1; i >= (m); i--)
#define REP(i, m, n, p) for(long long i = m; i < n; i += p)
#define foa(s, v) for(auto &s : v)
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define bcnt(n) __builtin_popcountll(n)
#define endk endl
#define ednl endl
#define enld endl

using ll = long long;
using ld = long double;
using ull = unsigned long long;
using vb = vector<bool>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using vvvi = vector<vector<vector<int>>>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using mll = map<long long, long long>;
using pll = pair<long long, long long>;
using qll = queue<long long>;
using sll = set<long long>;
using vpll = vector<pair<long long, long long>>;
template <class T = ll>
using V = vector<T>;
template <class T = ll>
using VV = V<V<T>>;
template <class T = ll>
using VVV = V<V<V<T>>>;
//昇順pq(小さい方から取り出す)
template <class T = ll>
using pqup = priority_queue<T, vector<T>, greater<T>>;
//降順pq(大きい方から取り出す)
template <class T = ll>
using pqdn = priority_queue<T>;

#pragma endregion

#pragma region constants

long long const limLL = 9223372036854775807;  // POW(2,63)-1 ~ 9.22e18
long long const dekai = 3e16;
const long double pi = acos(-1);
const char el = '\n';
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int ddx[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
int ddy[8] = {-1, 0, 1, -1, 1, -1, 0, 1};

const int mod = 1000000007;
// const int mod = 998244353;

#pragma endregion

#pragma region basic_procedure

template <class T>
inline bool isin(T x, T lef, T rig) {
	return ((lef <= x) && (x < rig));
}

template <class T>
inline bool chmin(T &a, T b) {
	if(a > b) {
		a = b;
		return true;
	}
	return false;
}
template <class T>
inline bool chmax(T &a, T b) {
	if(a < b) {
		a = b;
		return true;
	}
	return false;
}

void Yes(bool f = 1) { cout << (f ? "Yes" : "No") << "\n"; }
void No() { cout << "No\n"; }
void YES(bool f = 1) { cout << (f ? "YES" : "NO") << "\n"; }
void NO() { cout << "NO\n"; }
template <class T>
void drop(T answer) {
	cout << answer << "\n";
	exit(0);
}
void err() {
	cout << -1 << "\n";
	exit(0);
}

vector<long long> vin(long long n) {  //整数n個の入力を受け取ってベクトルに突っ込んで返す
	vector<long long> v(n);
	for(long long i = 0; i < n; i++) {
		cin >> v[i];
	}
	return v;
}

//ベクトルの出力(検証済)
// vectorの中身を出力する 答えの出力に利用可能
template <class T>
void vout(vector<T> &v, bool tate = 0) {
	if(v.size() > 0) {
		for(auto it = v.begin(); it < v.end(); it++) {
			cout << *it;
			if(it != v.end() - 1) {
				if(tate)
					cout << endl;
				else
					cout << " ";
			}
		}
	}
	cout << endl;
}

template <class T>
void add(vector<T> &v, T val) {	 //ベクトルの各要素に加算
	for(auto &a : v) a += val;
	return;
}

// vectorの中身を数える map<要素,個数>を返す
template <class T>
map<T, long long> cntv(vector<T> v) {
	map<T, long long> m;
	for(auto &g : v) {
		if(m.count(g))
			m[g]++;
		else
			m[g] = 1;
	}
	return m;
}

//配列圧縮(検証済)
//{1,36,1,3,8,-2,-92}を
//{2, 5,2,3,4, 1,  0}にする
template <class T>
vector<long long> press(vector<T> &v) {
	long long n = v.size();
	vector<long long> w(n);
	map<T, long long> m;
	for(T &p : v) m[p] = 0;
	long long i = 0;
	for(auto &p : m) {
		p.second = i;
		i++;
	}
	for(long long i = 0; i < n; i++) w.at(i) = m[v.at(i)];
	return w;
}

template <class T>
T divup(T a, T b) {
	//端数繰りあがり割り算
	assert(b != 0);
	T x = abs(a);
	T y = abs(b);
	T z = (x + y - 1) / y;
	if((a < 0 && b > 0) || (a > 0 && b < 0))
		return -z;
	else if(a == 0)
		return 0;
	else
		return z;
}

long long POW(long long a, long long n) {
	long long res = 1;
	while(n > 0) {
		if(n & 1) res = res * a;
		a = a * a;
		n >>= 1;
	}
	return res;
}

template <class T>
int sgn(T x) {	//符号関数
	if(x < 0) return -1;
	if(x == 0) return 0;
	return 1;
}

long long modpow(long long a, long long n, long long mod) {	 // a^n mod
	if(mod == 1) return 0LL;
	long long res = 1;
	while(n > 0) {
		if(n & 1) res = res * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}

// a * x % mod == __gcd(a,mod)なるxを返す
// a が modの倍数でないことが条件
long long modinv(long long a, long long mod) {
	long long b = mod, u = 1, v = 0;
	while(b) {
		long long t = a / b;
		a -= t * b;
		swap(a, b);
		u -= t * v;
		swap(u, v);
	}
	u %= mod;
	if(u < 0) u += mod;
	return u;
}

vvll comb(100, vll(100, -1));
long long com(long long n, long long k) {  //普通の二項計数(overflowに注意)
	assert(n < 100 && k < 100);
	if(n < k || k < 0 || n < 0) return 0;
	if(comb[n][k] != -1) return comb[n][k];
	ll res;
	if(n - k < k)
		res = com(n, n - k);
	else if(k == 0)
		res = 1;
	else
		res = com(n - 1, k - 1) + com(n - 1, k);
	comb[n][k] = res;
	return res;
}

// nCk modを求める
const ll MAX = 5100000;
// この値は求める二項計数の値に応じて変える
// MAX=3*10^7のとき1900msほど、ほぼ比例
// MAX=5*10^6程度ならそれほど気にしなくてよい(300ms程)
long long fac[MAX], finv[MAX], inv[MAX];

void cominit() {
	// テーブルを作る前処理
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for(ll i = 2; i < MAX; i++) {
		fac[i] = fac[i - 1] * i % mod;
		inv[i] = mod - inv[mod % i] * (mod / i) % mod;
		finv[i] = finv[i - 1] * inv[i] % mod;
	}
}
long long commod(ll n, ll k) {	// 二項係数計算
	if(n < k) return 0;
	if(n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % mod) % mod;
}
long long pmod(ll n, ll k) {  //順列計算
	if(n < k) return 0;
	if(n < 0 || k < 0) return 0;
	return fac[n] * finv[n - k] % mod;
}
long long hmod(ll n, ll k) {  // nHk計算
	// n個の区別しないoを区別するk個の箱に入れる方法の総数
	//(n+k-1)C(k-1)と等しい
	return commod(n + k - 1, n);
}
#pragma endregion

#pragma region input
#define VEC(type, name, size) \
	vector<type> name(size);  \
	INPUT(name)
#define VVEC(type, name, h, w)                     \
	vector<vector<type>> name(h, vector<type>(w)); \
	INPUT(name)
#define INT(...)     \
	int __VA_ARGS__; \
	INPUT(__VA_ARGS__)
#define LL(...)            \
	long long __VA_ARGS__; \
	INPUT(__VA_ARGS__)
#define STR(...)        \
	string __VA_ARGS__; \
	INPUT(__VA_ARGS__)
#define CHAR(...)     \
	char __VA_ARGS__; \
	INPUT(__VA_ARGS__)
#define DOUBLE(...)     \
	double __VA_ARGS__; \
	INPUT(__VA_ARGS__)
#define LD(...)              \
	long double __VA_ARGS__; \
	INPUT(__VA_ARGS__)

template <class T>
void scan(T &a) {
	cin >> a;
}
template <class T>
void scan(vector<T> &a) {
	for(auto &i : a) scan(i);
}
template <class T, class L>
void scan(pair<T, L> &p) {
	scan(p.first);
	scan(p.second);
}
void INPUT() {}
template <class Head, class... Tail>
void INPUT(Head &head, Tail &... tail) {
	scan(head);
	INPUT(tail...);
}
template <class T>
inline void print(T x) {
	cout << x << '\n';
}

template <typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) {
	is >> p.first >> p.second;
	return is;
}
#pragma endregion

#pragma region debug

#pragma region output
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
	os << p.first << " " << p.second;
	return os;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
	for(int i = 0; i < (int)v.size(); i++) {
		if(i) os << " ";
		os << v[i];
	}
	return os;
}
#pragma endregion

#pragma region view

template <typename T>
void view(const T e) {
	std::cerr << e;
}

template <typename T, typename U>
void view(const std::pair<T, U> &p) {
	std::cerr << "(";
	view(p.first);
	cerr << ", ";
	view(p.second);
	cerr << ")";
}

template <typename T>
void view(std::set<T> &s) {
	if(s.empty()) {
		cerr << "{ }";
		return;
	}
	std::cerr << "{ ";
	for(auto &t : s) {
		view(t);
		std::cerr << ", ";
	}
	std::cerr << "\b\b }";
}

template <typename T>
void view(const std::vector<T> &v) {
	if(v.empty()) {
		cerr << "{ }";
		return;
	}
	std::cerr << "{ ";
	for(const auto &e : v) {
		view(e);
		std::cerr << ", ";
	}
	std::cerr << "\b\b }";
}

template <typename T>
void view(const std::vector<std::vector<T>> &vv) {
	std::cerr << "{\n";
	for(const auto &v : vv) {
		std::cerr << "\t";
		view(v);
		cerr << "\n";
	}
	std::cerr << "}";
}

template <typename T, typename U>
void view(const std::vector<std::pair<T, U>> &v) {
	std::cerr << "{\n";
	for(const auto &c : v) {
		std::cerr << "\t(";
		view(c.first);
		cerr << ", ";
		view(c.second);
		cerr << ")\n";
	}
	std::cerr << "}";
}

template <typename T, typename U>
void view(const std::map<T, U> &m) {
	std::cerr << "{\n";
	for(auto &t : m) {
		std::cerr << "\t[";
		view(t.first);
		cerr << "] : ";
		view(t.second);
		cerr << "\n";
	}
	std::cerr << "}";
}

#pragma endregion

// when debugging : g++ foo.cpp -DLOCAL
#ifdef LOCAL
void debug_out() {}
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
	view(H);
	cerr << ", ";
	debug_out(T...);
}
#define debug(...)                                           \
	do {                                                     \
		cerr << __LINE__ << " [" << #__VA_ARGS__ << "] : ["; \
		debug_out(__VA_ARGS__);                              \
		cerr << "\b\b]\n";                                   \
	} while(0)
#define dump(x)                                 \
	do {                                        \
		cerr << __LINE__ << " " << #x << " : "; \
		view(x);                                \
		cerr << "\n";                           \
	} while(0)
#else
#define debug(...) (void(0))
#define dump(x) (void(0))
#endif

#pragma endregion

/**
 * @brief Sum-Of-Floor-Of-Linear(一次関数の床関数の和)
 * ∑[i = 0,1,...,n-1]((a*i+b)//m)
 */
template <typename T>
T sum_of_floor_of_linear(const T &n, const T &m, T a, T b) {
	T ret = 0;
	if(a >= m) ret += (n - 1) * n * (a / m) / 2, a %= m;
	if(b >= m) ret += n * (b / m), b %= m;
	T y = (a * n + b) / m;
	if(y == 0) return ret;
	T x = y * m - b;
	ret += (n - (x + a - 1) / a) * y;
	ret += sum_of_floor_of_linear(y, a, m, (a - x % a) % a);
	return ret;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout << fixed << setprecision(15);
	srand((unsigned)time(NULL));

	INT(k, q);

	VEC(ll, d, k);
	ll sum = 0;
	rep(i, k) sum += d[i];
	vector<ll> ans(q);

	rep(i, q) {
		ll res = 0;
		LL(n, x, m);
		ll add = sum % m;
		vll a(k+1);
		a[0] = x % m;
		rep(i, k) { a[i + 1] = (a[i] + d[i]) % m; }
		dump(a);
		rep(i, k) {
			int nex = i + 1;
			ll now = a[i];
			ll aft = a[nex];
			ll loop = min((n - i - 1) / k + 1, (n - nex - 1) / k + 1);
			ll res1, res2;
			res1 = sum_of_floor_of_linear(loop, m, add, now);
			res2 = sum_of_floor_of_linear(loop, m, add, aft);
			debug(i, nex, now, aft);
			debug(loop, res1, res2);
			ll add = 0;
			if(now == aft) {
				add = 0;
			} else if(now < aft) {
				add = loop - (res2 - res1);
			} else {
				add = res1 - res2;
			}
			dump(add);
			res += add;
		}

		ans[i] = res;
	}

	vout(ans, 1);

	return 0;
}