题目链接：hdu3333

题目大意：给一段n长度的数字序列，以及q长度的区间询问，问区间不同数字大小之和。

解题思路：跟区间数种类这题类似只不过种类数改成不同种类数字之和。树状数组改成维护不同种类数字之和。注意数字之和会爆int，C数组设置成long long

AC代码：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <string>
#include <deque>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <vector>
#include <utility>

using namespace std;
typedef long long ll;
const int Vmaxn = (int)3e4+5;
const int Qmaxn = (int)1e5+5;

struct Node {
	int l,r;
	int id;
	bool operator < (const Node& A) const {
		return this->r < A.r;
	}
};

ll c[Vmaxn],ans[Qmaxn];
int n,q,val[Vmaxn];
Node qry[Qmaxn];
map<int, int > vis;

inline int lowbit (int x) {
	return x & (-x);
}

void add (int idx, ll num) {
	while (idx <= Vmaxn) {
		c[idx] += num;
		idx += lowbit(idx);
	}
}

ll get_sum (int idx) {
	ll ans = 0;
	while (idx > 0) {
		ans += c[idx];
		idx -= lowbit(idx);
	}
	return ans;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	int T;
	cin >> T;
	for (int k = 1; k <= T; k++) {
		cin >> n;
		for (int i = 1; i <= n; i++) {
			cin >> val[i];
		}
		cin >> q;
		for (int i = 1; i <= q; i++) {
			cin >> qry[i].l >> qry[i].r;
			qry[i].id = i;
		}
		sort(qry + 1, qry + 1 + q);
		memset(c, 0, sizeof(c));
		vis.clear();
		
		int point = 1;
		for (int i = 1; i <= q; i++) {
			while (point <= qry[i].r) { // 更新
				if(vis.find(val[point]) != vis.end()) {
					add(vis[val[point]], -val[point]);
				}
				vis[val[point]] = point;
				add(vis[val[point]], val[point]);
				point ++; //注意别遗漏
			}
			ans[qry[i].id] = get_sum(qry[i].r) - get_sum(qry[i].l- 1);
		}
		
		for (int i = 1; i <= q; i++) {
			cout << ans[i] << '\n';
		}
	}
	return 0;
}