求一个序列中全部数字的和，当中数值同样的仅仅能计算一次。

先储存全部的请求，然后依照它们的右边界排序，在查询的同一时候更新区间。这里事实上有一点点DP的味道，在它进行某个查询之前，保证全部的反复数字(除去最后一个)都被删除光了，而且有不能影响其它查询，所以呢，仅仅能从近期的那个操作进行计算。1次query就可以

#include<iostream>
#include<stdio.h>
#include<string>
#include<string.h>
#include<algorithm>
#include<vector>

using namespace std;

const int maxn = 50050;
const int maxm = 200050;
vector<int> vec;
__int64 res[maxm];
__int64 ans;
int pre[1000050];

void init()
{
	vec.clear();
	vec.push_back(0);	
	memset(pre, 0, sizeof(pre)); 
}

struct line{
	int left, right, id;
	bool operator<(const line &y)const
	{
		return right < y.right;
	}
}q[maxm];

struct node{
	int left, right;
	__int64 sum;
	int mid(){
		return (left+right)/2;
	}
}tree[maxn*4];

void buildTree(int left, int right, int rt)
{
	tree[rt].left = left;
	tree[rt].right = right;
	if(left == right){
		int a;
		scanf("%d", &a);
		vec.push_back(a);
		return ;
	}
	int mid = tree[rt].mid();
	buildTree(left, mid, rt<<1);
	buildTree(mid+1, right, rt<<1|1);
	tree[rt].sum = tree[rt<<1].sum + tree[rt<<1|1].sum;	
}

void update(int left, int right, int rt, int pos, int add)
{
	if(left == right){
		tree[rt].sum = add;
		return ;
	}
	int mid = tree[rt].mid();
	if(pos <= mid)
		update(left, mid, rt<<1, pos, add);
	if(pos > mid)
		update(mid+1, right, rt<<1|1, pos, add);
	tree[rt].sum = tree[rt<<1].sum + tree[rt<<1|1].sum;
}

void query(int left, int right, int rt, int L, int R)
{
	if(L <= left && R >= right){
		ans += tree[rt].sum;
		return ;
	}
	int mid = tree[rt].mid();
	if(L <= mid)
		query(left, mid, rt<<1, L, R);
	if(R > mid)
		query(mid+1, right, rt<<1|1, L, R);
}

int main()
{
	int T, M, N;
	scanf("%d", &T);
	while(T --)
	{
		scanf("%d", &N);
		
		init();
		buildTree(1, N, 1);
		
		scanf("%d", &M);
		for(int i = 1; i <= M; i ++)
		{
			scanf("%d%d", &q[i].left, &q[i].right);
			q[i].id = i;
		}
		sort(q+1, q+M+1);//排序是从index=1開始的M个数 
		
		int qn = 0;
		for(int i = 1; i <= N && qn <=M; i ++)
		{
			if(pre[vec[i]] != 0){
				update(1, N, 1, pre[vec[i]], 0);
			}
			update(1, N, 1, i, vec[i]);
			while(qn <= M && q[qn].right <= i){
				ans = 0;
				if(q[qn].right == i){
					query(1, N, 1, q[qn].left, q[qn].right);
					res[q[qn].id] = ans;
				}
				qn ++;
			}
			pre[vec[i]] = i; 
		}
		for(int i = 1; i <= M; i ++) printf("%I64d\n", res[i]);
	}
	
	return 0;
}