Problem Description

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.

 

Input

The first line is an integer T (1 ≤ T ≤ 10), indecating the number of testcases below.
For each case, the input format will be like this:
* 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

 

Sample Output

 

Author

3xian@GDUT

 

Source

 

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#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <vector>

#include <algorithm>

#include <map>

#define maxn 55555*4

#define LL long long

#define inf 0x3f3f3f3f3f3f3f3f3f3f

#define lowbit(x)   x & (-x)

using namespace std;

//int visit[maxn];

//int VISIT[maxn];

map <LL,LL >  visit;

map <LL ,LL > VISIT;

LL  A[maxn];

int n,cnt;

struct node

{

    int L,R,id;

};

node Q[maxn];

LL c[maxn];

LL ans[maxn];

void init()

{

    memset(c,,sizeof(c));

    //memset(visit,0,sizeof(visit));

     //memset(VISIT,0,sizeof(VISIT));

    visit.clear();

    VISIT.clear();

    cnt=;

}

LL sum(int x)

{

    LL ret=;

    while(x>)

    {

      ret+=c[x];

      x-=(x&(-x) );

    }

    return ret;

}

void add(int x,LL d)

{

   while(x<=n)

   {

      c[x]+=d;

      x+=(x& (-x) );

   }

}

void solve()

{

    for(int i=;i<=n;i++)

    {

        if(visit[A[i]]!=)

        {

            add(visit[A[i]],-A[i]);

        }

        visit[A[i]]=i;

        add(i,A[i]);

        if(VISIT[i]!=)

        {

            for(int j=;j<=VISIT[i];j++)

            {

                cnt++;

                ans[Q[cnt].id]=sum(Q[cnt].R)-sum(Q[cnt].L-);

                //printf("%lld\n",ans[Q[cnt].id]);

            }

        }

    }

}

bool cmp(node a,node b)

{

  if(a.R==b.R)

     return a.L<b.L;

   else

    return a.R<b.R;

}

int main()

{

//freopen("test.txt","r",stdin);

    int t,q;

    scanf("%d",&t);

    while(t--)

    {

         init();

         scanf("%d",&n);

         for(int i=;i<=n;i++)

            scanf("%lld",&A[i]);

         scanf("%d",&q);

         for(int i=;i<=q;i++)

         {

             scanf("%d%d",&Q[i].L,&Q[i].R);

             Q[i].id=i;

             // printf("%d\n",Q[i].id);

             ++VISIT[Q[i].R];

         }

         sort(Q+,Q+q+,cmp);

        /* for(int i=1;i<=q;i++)

         {

            printf("%d\n",Q[i].id);

         }*/

         solve();

         for(int i=;i<=q;i++)

         {

            printf("%lld\n",ans[i]);

         }

    }

    return ;

}