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After inventing Turing Tree, 3xian always felt boring when solving problems about intervals, because Turing Tree could easily have the solution. As well, wily 3xian made lots of new problems about intervals. So, today, this sick thing happens again…

Now given a sequence of N numbers A1, A2, …, AN and a number of Queries(i, j) (1≤i≤j≤N). For each Query(i, j), you are to caculate the sum of distinct values in the subsequence Ai, Ai+1, …, Aj.

• Line 1: N (1 ≤ N ≤ 30,000).
• Line 2: N integers A1, A2, …, AN (0 ≤ Ai ≤ 1,000,000,000).
• Line 3: Q (1 ≤ Q ≤ 100,000), the number of Queries.
• Next Q lines: each line contains 2 integers i, j representing a Query (1 ≤ i ≤ j ≤ N).
  Output
  For each Query, print the sum of distinct values of the specified subsequence in one line.
  Sample Input
  2
  3
  1 1 4
  2
  1 2
  2 3
  5
  1 1 2 1 3
  3
  1 5
  2 4
  3 5
  Sample Output
  1
  5
  6
  3
  6
  求区间不同数的和。本质上和求区间不同数的种类数是一样的。
  代码如下：

#include
   
    #define ll long longusing namespace std;const int maxx=3e5+100;struct node{
   	int l;	int r;	ll sum;}p[maxx<<2];struct Node{
   	int l;	int r;	int id;	bool operator <(const Node &a)const{
   		return a.r>r;	}}b[maxx];ll a[maxx],ans[maxx];int n,m;inline void pushup(int cur){
   	p[cur].sum=p[cur<<1].sum+p[cur<<1|1].sum;}inline void build(int l,int r,int cur){
   	p[cur].l=l;	p[cur].r=r;	p[cur].sum=0;	if(l==r) return ;	int mid=l+r>>1;	build(l,mid,cur<<1);	build(mid+1,r,cur<<1|1);}inline void update(int pos,ll v,int cur){
   	int L=p[cur].l;	int R=p[cur].r;	if(L==R)	{
   		p[cur].sum+=v;		return ;	}	int mid=L+R>>1;	if(pos<=mid) update(pos,v,cur<<1);	else update(pos,v,cur<<1|1);	pushup(cur);}inline ll query(int l,int r,int cur){
   	int L=p[cur].l;	int R=p[cur].r;	if(l<=L&&R<=r) return p[cur].sum;	int mid=L+R>>1;	if(r<=mid) return query(l,r,cur<<1);	else if(l>mid) return query(l,r,cur<<1|1);	else return query(l,mid,cur<<1)+query(mid+1,r,cur<<1|1);}int main(){
   	int t;	scanf("%d",&t);	while(t--)	{
   		scanf("%d",&n);		for(int i=1;i<=n;i++) scanf("%lld",&a[i]);		scanf("%d",&m);		for(int i=1;i<=m;i++) scanf("%d%d",&b[i].l,&b[i].r),b[i].id=i;		sort(b+1,b+1+m);map
    
      mp;		build(1,n,1);		for(int i=1,j=1;i<=n&&j<=m;i++)		{
   			if(mp[a[i]]) update(mp[a[i]],-a[i],1);			update(i,a[i],1);			mp[a[i]]=i;			while(j<=m&&b[j].r==i)			{
   				ans[b[j].id]=query(b[j].l,b[j].r,1);				j++;			}		}		for(int i=1;i<=m;i++) printf("%lld\n",ans[i]);	}}
    
   

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