原题：
Problem Description
Mery has a beautiful necklace. The necklace is made up of N magic balls. Each ball has a beautiful value. The balls with the same beautiful value look the same, so if two or more balls have the same beautiful value, we just count it once. We define the beautiful value of some interval [x,y] as F(x,y). F(x,y) is calculated as the sum of the beautiful value from the xth ball to the yth ball and the same value is ONLY COUNTED ONCE. For example, if the necklace is 1 1 1 2 3 1, we have F(1,3)=1, F(2,4)=3, F(2,6)=6.
Now Mery thinks the necklace is too long. She plans to take some continuous part of the necklace to build a new one. She wants to know each of the beautiful value of M continuous parts of the necklace. She will give you M intervals [L,R] (1<=L<=R<=N) and you must tell her F(L,R) of them.
Now Mery thinks the necklace is too long. She plans to take some continuous part of the necklace to build a new one. She wants to know each of the beautiful value of M continuous parts of the necklace. She will give you M intervals [L,R] (1<=L<=R<=N) and you must tell her F(L,R) of them.
Input
The first line is T(T<=10), representing the number of test cases. For each case, the first line is a number N,1 <=N <=50000, indicating the number of the magic balls. The second line contains N non-negative integer numbers not greater 1000000, representing the beautiful value of the N balls. The third line has a number M, 1 <=M <=200000, meaning the nunber of the queries. Each of the next M lines contains L and R, the query.
Output
For each query, output a line contains an integer number, representing the result of the query.
Sample Input
2 6 1 2 3 4 3 5 3 1 2 3 5 2 6 6 1 1 1 2 3 5 3 1 1 2 4 3 5
Sample Output
3 7 14 1 3 6
题目链接：http://acm.hdu.edu.cn/webcontest/contest_showproblem.php?cid=11842&pid=1006&ojid=0&hide=1&problem=Problem%20F
题意：
给出n个数，给出m个区间，让求这一区间的元素和。
如果有一样的数字，只算一遍。
例如Sample 2，1 2 区间段和是1.
思路：求区间段和，首先想到的就是用树状数组。那么怎么考虑一样的数字的问题呢？就要用到离线处理了。首先将所求区间和进行储存并记录其出现的初始位置。然后对所有区间按照右端点进行排序。然后用一个变量进行记录。如果该元素之前出现过，那么在之前位置往后减去该元素，然后再在当前位置后加上该元素。直到区间结束，更新过后求结果。对结果进行保存。
源代码：
/*1006离线处理。先将区间保存，然后按照右端点升序排列。离线处理要记录初始的区间位置然后从头开始查找。若该点的元素出现过，则从原来位置开始减去该元素。若没有出现过，则加上该元素。最后求和。*/
#include <iostream>
#include <cstring>
#include <algorithm>
#include <stdio.h>
#include <vector>
#define MAX 1000000
using namespace std;
long long C[MAX];
long long a[MAX];
long long flag[MAX];
long long jieguo[MAX];
struct Qujian
{int i;int j;int id;
}qujian[MAX];
bool cmp(Qujian a,Qujian b)
{return a.j < b.j;
}
long long lowbit(long long x)
{return x & (-x);
}
long long getsum(long long x)
{long long sum = 0;while (x > 0){sum += C[x];x -= lowbit(x);}return sum;
}
void add(long long x,long long v)
{while (x < MAX){C[x] += v;x += lowbit(x);}
}
int main()
{int n;scanf("%d",&n);while (n--){int m;memset(flag,0,sizeof(flag));memset(C,0,sizeof(C));memset(a,0,sizeof(a));scanf("%d",&m);for (int i = 1; i <= m; i++)scanf("%lld",&a[i]);int k;scanf("%d",&k);for (int p = 1; p <= k; p++){scanf("%d%d",&qujian[p].i,&qujian[p].j);qujian[p].id = p;}sort(qujian+1,qujian+k+1,cmp);//for (int i = 1; i <= k; i++)//{// cout << qujian[i].i << endl << qujian[i].j << endl;// cout << qujian[i].id << endl;//}long long flag1 = 1;for (int i = 1; i <= k; i++){while (flag1 <= qujian[i].j){long long b;b = a[flag1];if (flag[b])add(flag[b],-b);add(flag1,b);flag[b] = flag1;flag1++;}//cout << getsum(qujian[i].j) << endl << getsum(qujian[i]. i - 1)<< endl;jieguo[qujian[i].id] = getsum(qujian[i].j) - getsum(qujian[i].i-1);}for (int i = 1; i <= k; i++)printf("%lld\n",jieguo[i]);}return 0;
}