利用id来进行树状数组，而不是离散化以后的val HDU 4417 离线+树状数组

题目大意：给你一个长度为n的数组，问[L,R]之间<=val的个数

思路：就像标题说的那样就行了。树状数组不一定是离散化以后的区间，而可以是id

//看看会不会爆int!数组会不会少了一维！
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
const int maxn = 1e5 + ;
int tree[maxn], ans[maxn];
int n, m;
pair<int, int> a[maxn];///val pos
pair<pair<int, int>, pair<int, int> > b[maxn];///val pos [l,r]
int lowbit(int x){return x & -x;}

void update(int x){
    for (int i = x; i <= n; i += lowbit(i))
        tree[i] += ;
}

int sum(int x){
    int ans = ;
    for (int i = x; i > ; i -= lowbit(i)){
        ans += tree[i];
    }
    return ans;
}

int main(){
    int kase = ;
    int t; cin >> t;
    while (t--){
        scanf("%d%d", &n, &m);
        for (int i = ; i <= n; i++){
            scanf("%d", &a[i].first);
            a[i].second = i;
        }
        sort(a + , a +  + n);
        for (int i = ; i <= m; i++){
            int l, r, val;
            scanf("%d%d%d", &l, &r, &val);
            l++, r++;
            b[i] = mk(mk(val, i), mk(l, r));
        }
        sort(b + , b +  + m);
        memset(tree, , sizeof(tree));
        printf("Case %d:\n", ++kase);
        int j = ;
        for (int i = ; i <= n && j <= m; ){
            while (i <= n && b[j].first.first >= a[i].first){
                update(a[i].second);
                i++;
            }
            while (j <= m && b[j].first.first < a[i].first){
                ans[b[j].first.second] = sum(b[j].second.second) - sum(b[j].second.first - );
                j++;
            }
        }
        while (j <= m){
            ans[b[j].first.second] = sum(b[j].second.second) - sum(b[j].second.first - );
            j++;
        }
        for (int i = ; i <= m; i++) {
            printf("%d\n", ans[i]);
        }
    }
    return ;
}