补题链接：Here
算法相关文章：主席树
显然是可以离线之后fenwick维护。
因为不喜欢离线，所以直接主席树了。
每次找到对应区间，然后相当于就是区间sum的问题了。
#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10, M = N * 40;
int n, m, rt[N], lc[M], rc[M], sum[M], cnt;
#define mid (l+r>>1)
void change (int l, int r, int &x, int y, int k) {
sum[x = ++cnt] = sum[y] + 1, lc[x] = lc[y], rc[x] = rc[y];
if (l == r) return ;
if (k <= mid) change (l, mid, lc[x], lc[y], k);
else change (mid + 1, r, rc[x], rc[y], k);
}
int ask (int l, int r, int x, int y, int ql, int qr) {
if (l == ql && r == qr) return sum[y] - sum[x];
if (qr <= mid) return ask (l, mid, lc[x], lc[y], ql, qr);
else if (ql > mid) return ask (mid + 1, r, rc[x], rc[y], ql, qr);
else return ask (l, mid, lc[x], lc[y], ql, mid) + ask (mid + 1, r, rc[x], rc[y], mid + 1, qr);
}
signed main() {
cin >> n >> m;
for (int i = 1, x; i <= n; i++) scanf ("%d", &x), change (1, 1e5, rt[i], rt[i - 1], x);
for (int i = 1, l, r, k; i <= m; i++) scanf ("%d %d %d", &l, &r, &k), printf ("%d
", ask (1, 1e5, rt[l - 1], rt[r], 1, k) );
return 0;
}