C++ :

/* std：候选值（出现次数 >= 自身代号的值，数量 O(sqrt n)）+ 二分统计区间出现次数 */
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
int n, m, K = 0, ncan = 0;
int raw[100005], a[100005];
int srt[100005], cand[100005];
vector<int> occ[100005];
long long val[100005];
int main(){
    if(scanf("%d", &n) != 1) return 0;
    for(int i = 1; i <= n; i++){ scanf("%d", &raw[i]); srt[i] = raw[i]; }
    sort(srt + 1, srt + n + 1);
    K = (int)(unique(srt + 1, srt + n + 1) - (srt + 1));
    for(int i = 1; i <= n; i++){
        int id = (int)(lower_bound(srt + 1, srt + K + 1, raw[i]) - srt);
        a[i] = id;
        val[id] = raw[i];
        occ[id].push_back(i);
    }
    for(int id = 1; id <= K; id++)
        if((long long)occ[id].size() >= val[id]) cand[ncan++] = id;
    scanf("%d", &m);
    while(m--){
        int L, R, k;
        scanf("%d %d %d", &L, &R, &k);
        int good = 0;
        for(int t = 0; t < ncan; t++){
            vector<int>& v = occ[cand[t]];
            int c = (int)(upper_bound(v.begin(), v.end(), R) - lower_bound(v.begin(), v.end(), L));
            if((long long)c >= val[cand[t]]) good++;
        }
        long long ans = 1;
        for(int j = 0; j < k; j++) ans = ans * (good - j) / (j + 1);
        printf("%lld\n", ans);
    }
    return 0;
}