Necklace
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2467 Accepted Submission(s): 873
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.
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
/*分析：对于所有的询问先离线储存好然后排序，
然后对输入的数组进行扫描，对于任何数都只保存最后出现的位置，把原来的位置赋值为0
这样在扫描的过程中进行求和并且保存最后的位置使得后面包含该数的区间要不只包含一次该数或者不包含
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <iomanip>
#define INF 99999999
typedef __int64 LL;
using namespace std;
const int MAX=50000+10;
int n,q,size;
LL c[MAX],a[MAX],sum[MAX*4];
int hash[MAX],mp[MAX];
struct Node{
int l,r,id;
bool operator<(const Node &temp)const{
return r<temp.r;
}
}s[MAX*4];
int lowbit(int x){
return x&(-x);
}
void Update(int x,int d){
while(x<=n){
c[x]+=d;
x+=lowbit(x);
}
}
LL Query(int x){
LL sum=0;
while(x>0){
sum+=c[x];
x-=lowbit(x);
}
return sum;
}
int search(int x){
int left=1,right=size,mid;
while(left<=right){//>=x的第一个数
mid=left+right>>1;
if(hash[mid]<x)left=mid+1;
else right=mid-1;
}
return right+1;
}
int main(){
int t,x;
scanf("%d",&t);
while(t--){
memset(mp,0,sizeof mp);
memset(c,0,sizeof c);
scanf("%d",&n);
for(int i=1;i<=n;++i)scanf("%d",&a[i]),hash[i]=a[i];
sort(hash+1,hash+1+n);
size=1;
for(int i=2;i<=n;++i)if(hash[i] != hash[size])hash[++size]=hash[i];
scanf("%d",&q);
for(int i=1;i<=q;++i){
scanf("%d%d",&s[i].l,&s[i].r);
s[i].id=i;
}
sort(s+1,s+1+q);
int i=1,j=1;
while(i<=n && j<=q){
int k=search(a[i]);
if(mp[k])Update(mp[k],-a[i]);
Update(i,a[i]);
mp[k]=i;
if(s[j].r == i){
LL ans=Query(i);
while(s[j].r == i){
LL temp=Query(s[j].l-1);
sum[s[j].id]=ans-temp;
++j;
}
}
++i;
}
for(int i=1;i<=q;++i)printf("%I64d\n",sum[i]);
}
return 0;
}