Codeforces 221 D. Little Elephant and Array

D. Little Elephant and Array

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The Little Elephant loves playing with arrays. He has array a, consisting of n positive integers, indexed from 1 to n. Let's denote the number with index i as ai.

Additionally the Little Elephant has m queries to the array, each query is characterised by a pair of integers lj and rj (1 ≤ lj ≤ rj ≤ n). For each query lj, rj the Little Elephant has to count, how many numbers x exist, such that number x occurs exactly x times among numbersalj, alj + 1, ..., arj.

Help the Little Elephant to count the answers to all queries.

Input

The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the size of array a and the number of queries to it. The next line contains n space-separated positive integers a1, a2, ..., an (1 ≤ ai ≤ 109). Next m lines contain descriptions of queries, one per line. The j-th of these lines contains the description of the j-th query as two space-separated integers lj and rj (1 ≤ lj ≤ rj ≤ n).

Output

In m lines print m integers — the answers to the queries. The j-th line should contain the answer to the j-th query.

Examples

input

7 2
3 1 2 2 3 3 7
1 7
3 4

output

3
1

题意：询问区间[l,r]内出现次数等于它本身的数的个数

莫队算法

#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,m,siz,ans;
int a[100001],hash[100001];
int sum[100001];
struct node
{
    int bl,id;
    int l,r,k;
}e[100001];
bool cmp(node p,node q)
{
    if(p.bl!=q.bl) return p.bl<q.bl;
    return p.r<q.r;
}
bool cmp2(node p,node q)
{
    return p.id<q.id;
}
void update(int pos,int w)
{
    if(sum[a[pos]]!=hash[a[pos]]&&sum[a[pos]]+w==hash[a[pos]])    ans++;
    else if(sum[a[pos]]==hash[a[pos]]&&sum[a[pos]]+w!=hash[a[pos]]) ans--;
    sum[a[pos]]+=w;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]),hash[i]=a[i];
    siz=sqrt(n);
    sort(hash+1,hash+n+1);
    int cnt=unique(hash+1,hash+n+1)-(hash+1);
    for(int i=1;i<=n;i++) a[i]=lower_bound(hash+1,hash+cnt+1,a[i])-hash;
    int ll,rr;
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&ll,&rr);
        e[i].id=i;
        e[i].bl=(ll-1)/siz+1;
        e[i].l=ll;
        e[i].r=rr;
    }
    sort(e+1,e+m+1,cmp);
    int l=1,r=0,opl,opr;
    for(int i=1;i<=m;i++)
    {
        opl=e[i].l; opr=e[i].r;
        while(l>opl)  update(--l,1);
        while(l<opl)  update(l++,-1);
        while(r<opr)  update(++r,1);
        while(r>opr)  update(r--,-1); 
        e[i].k=ans;
    }    
    sort(e+1,e+m+1,cmp2);
    for(int i=1;i<=m;i++) printf("%d\n",e[i].k);
}