【HDU

B - Necklace

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,yx,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,RL,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

261 2 3 4 3 531 23 52 661 1 1 2 3 531 12 43 5

Sample Output

3714136

题意：有n颗珠子，每颗珠子都有其价值，现在给你m次询问，计算区间内的珠子价值为多少，且相同价值的珠子只能被计算一次。

分析：很明显的一道线段树，不同的是每个相同的值只能计算一次。我们可以先记录问题，并按照区间右端点从小到大排序，这样在后面计算问题的答案时便于删除区间中相同的值，而不对其他问题产生影响。用map记录相同价值的值和其所在的位置。

代码如下：

#include <set>#include <map>#include <queue>#include <cmath>#include <vector>#include <cctype>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define lson l, m, rt << 1#define rson m+1, r, rt << 1|1#define mod 835672545#define INF 0x3f3f3f3f#define LL long longusing namespace std;const int N = 50005;const int MX = 200005;int n, m;LL a[N], ans[MX];struct node{    int index, l, r;}q[MX];struct tree{    int l, r;    LL w;}T[N<<2];void build(int l, int r, int rt){    T[rt].l = l;    T[rt].r = r;    T[rt].w = 0;    if(l == r)  return ;    int m = (l + r) >> 1;    build(l, m, rt << 1);    build(m+1, r, rt << 1|1);}void update(int p, LL v, int rt){    T[rt].w += v;    if(T[rt].l == p && T[rt].r == p)    return ;    int m = (T[rt].l + T[rt].r) >> 1;    if(p <= m)    update(p, v, rt << 1);    else    update(p, v, rt << 1|1);}LL query(int l, int r, int rt){    if(l <= T[rt].l && T[rt].r <= r){        return T[rt].w;    }    LL ret = 0;    int m = (T[rt].l + T[rt].r) >> 1;    if(l <= m)  ret += query(l, r, rt << 1);    if(r > m)  ret += query(l, r, rt << 1|1);    return ret;}bool cmp(node a, node b){    return a.r < b.r;}map<LL, int> mp;int main(){    int t;    scanf("%d", &t);    while(t--){        mp.clear();        scanf("%d", &n);        for(int i = 1; i <= n; i++){            scanf("%lld", &a[i]);        }        build(1, n, 1);        scanf("%d", &m);        for(int i = 1; i <= m; i++){            scanf("%d%d", &q[i].l, &q[i].r);            q[i].index = i;        }        sort(q+1, q+m+1, cmp);        int id = 1;        for(int i = 1; i <= n; i++){            update(i, a[i], 1);            if(mp[a[i]])    update(mp[a[i]], -a[i], 1);            mp[a[i]] = i;            while(id <= m && q[id].r == i){                ans[q[id].index] = query(q[id].l, q[id].r, 1);                id++;            }        }        for(int i = 1; i <= m; i++){            printf("%lld\n", ans[i]);        }    }    return 0;}