//HDOJ 3333 Turing Tree 线段树 单点更新 成段查询/*题意：求某区间没所有值不同的数的总和思路：先对所有的询问按照区间末尾排序 然后从序列前面开始遍历，当遇到相同的元素的时候 将前面的元素删除，这样保证有重复的元素一定出现在距离区间末尾最近的 转化后就变成简单的单点更新 区间求和了 数据较大 要离散化处理*/#include<stdio.h>#include<string.h>#include<stdlib.h>#define N 30005#define M 100005#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,rint T,n,m,cnt;__int64 sum[N<<2],ans[M];int num[N],s[N],vis[N];struct node{int from;int to;int id;}q[M];int cmp(const void *a,const void *b){return *(int *)a - *(int *)b;}int cmp1(const void *a,const void *b){return (*(node *)a).to - (*(node *)b).to;}void Pushup(int rt){sum[rt] = sum[rt<<1] + sum[rt<<1|1];}void Build(int rt,int l,int r){if(l == r){scanf("%d",&num[cnt]);s[cnt] = num[cnt];sum[rt] = num[cnt++];return;}int mid = (l + r) >> 1;Build(lson);Build(rson);Pushup(rt);}void Update(int rt,int l,int r,int x,int c){if(l == r){sum[rt] += c;return ;}int mid = (l + r) >> 1;if(x <= mid)Update(lson,x,c);elseUpdate(rson,x,c);Pushup(rt);}__int64 Query(int rt,int l,int r,int L,int R){if(L <= l && R >= r)return sum[rt];int mid = (r + l) >> 1;__int64 res = 0;if(L <= mid) res += Query(lson,L,R);if(R > mid ) res += Query(rson,L,R);return res;}void Debug(int n){int i;for(i = 1; i <= n; ++i)printf("sum[%d] %I64d\n",i,sum[i]);}int Bin(int key){int l=0,r = cnt,mid;while(r >= l){mid = (l + r) >> 1;if(s[mid] == key) return mid;if(s[mid] < key) l = mid + 1;else r = mid - 1;}return -1;}void Init(){int i;cnt = 0;scanf("%d",&n);Build(1,1,n);//Debug(10);qsort(s,n,sizeof(s[0]),cmp);for(i = cnt = 1; i < n; ++i)if(s[i]!= s[i-1])s[cnt++] = s[i];--cnt;scanf("%d",&m);for(i = 0; i < m; ++i){scanf("%d %d",&q[i].from,&q[i].to);q[i].id = i;}qsort(q,m,sizeof(q[0]),cmp1);}void Solve(){int i,j,cur;j = 0;memset(vis,-1,sizeof(vis));for(i = 0; i < n; ++i){cur = Bin(num[i]);if(vis[cur] == -1)vis[cur] = i+1;else{Update(1,1,n,vis[cur],-num[i]);vis[cur] = i+1;}for(;j < m; ++j){if(q[j].to == i+1){ans[q[j].id] = Query(1,1,n,q[j].from,q[j].to);}elsebreak;}}}void Print(){int i;for(i = 0; i < m; ++i)printf("%I64d\n",ans[i]);}int main(){scanf("%d",&T);while(T--){Init();Solve();Print();}return 0;}
