題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=6287
首先預處理序列，把所有序列值因數分解，存入vector
對於每組查詢，就只需要二分查找因子數是否滿足了
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <cstring>
using namespace std;
typedef long long ll;
static const int MAX_N = 1e5 + 5;
static const ll Mod = 233;
static const int N = 105;
static const int INF = 0x3f3f3f3f;
vector<int>vec[MAX_N];
void pre_deal(int x, int ind) { //預處理存儲因子對應序列編號
for (int i = 2; i * i <= x; ++i) {
while (x % i == 0) {
vec[i].push_back(ind);
x /= i;
}
}
if (x > 1) vec[x].push_back(ind);
}
int get_div(int l, int r, int i) { //r到l中因子i的數量
return upper_bound(vec[i].begin(), vec[i].end(), r) - upper_bound(vec[i].begin(), vec[i].end(), l - 1);
}
bool solve(int l, int r, int d) {
for (int i = 2; i * i <= d; ++i) {
int cnt = 0;
while (d % i == 0) {
cnt++;
d /= i;
}
if (cnt > get_div(l, r, i)) return false;
}
//存在不可分解的因子，特判
if (d > 1 && get_div(l, r, d) <= 0) return false;
return true;
}
int main() {
/*freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);*/
int T;
scanf("%d", &T);
while (T--) {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 0; i < MAX_N; ++i) vec[i].clear();
for (int i = 1; i <= n; ++i) {
int v;
scanf("%d", &v);
pre_deal(v, i);
}
while (m--) {
int l, r, d;
scanf("%d%d%d", &l, &r, &d);
if (solve(l, r, d)) puts("Yes");
else puts("No");
}
}
return 0;
}