Given an array of N integers indexed from 1 to N, and q queries, each in the form i j, you have to find the number of distinct integers from index i to j (inclusive).
Input
Input starts with an integer T (≤ 5), denoting the number of test cases.
The first line of a case is a blank line. The next line contains two integers N (1 ≤ N ≤ 105), q (1 ≤ q ≤ 50000). The next line contains N space separated integers forming the array. There integers range in [0, 105].
Each of the next q lines will contain a query which is in the form i j (1 ≤ i ≤ j ≤ N).
Output
For each test case, print the case number in a single line. Then for each query you have to print a line containing number of distinct integers from index i to j.
Sample Input
1
8 5
1 1 1 2 3 5 1 2
1 8
2 3
3 6
4 5
4 8
Sample Output
Case 1:
4
1
4
2
4
Note
Dataset is huge. Use faster I/O methods.
题解：N个整数，q组查询，查询从i到j中不同整数的数目。
莫队就是需要排序，分块，四个while循环，节省很多时间
模板如下：
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
const int N=100005;
int a[N],ans,c[N],b[N];
struct node
{
int l,r,id,pos;
} q[N];
bool cmp(node a,node b)
{
if(a.pos==b.pos)
return a.r<b.r;
return a.pos<b.pos;
}
void update(int x,int p)
{
if(x==0)
return;
if(c[a[x]])
ans-=1;
c[a[x]]+=p;
if(c[a[x]])
ans+=1;
}
int main()
{
int t,T=1;
scanf("%d",&t);
while(t--)
{
memset(c,0,sizeof(c));
int n,m;
scanf("%d%d",&n,&m);
int p=sqrt(n);
for(int i=1; i<=n; i++)
scanf("%d",&a[i]);
for(int i=0; i<m; i++)
{
scanf("%d%d",&q[i].l,&q[i].r);
q[i].id=i;
q[i].pos=q[i].l/p;
}
sort(q,q+m,cmp);
int l=0,r=0;
ans=0;
for(int i=0; i<m; i++)
{
while(l<q[i].l)
update(l++,-1);
while(l>q[i].l)
update(--l,1);
while(r>q[i].r)
update(r--,-1);
while(r<q[i].r)
update(++r,1);
b[q[i].id]=ans;
}
printf("Case %d:\n",T++);
for(int i=0; i<m; i++)
printf("%d\n",b[i]);
}
return 0;
}