Codeforces Round #136 (Div. 2) 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 numbers alj, 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.

Sample test(s)

input

7 23 1 2 2 3 3 71 73 4

output

31

题意：给你n个数和m个区间，求区间上是否存在t个数值为t的数，每个区间输出一列。

代码：

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespace std;int n,m,i,j,k,a[100005],b[100005],c[100005];int d[505][100005],s[100005],L,R;int main(){    scanf("%d%d",&n,&m);  {    for(i=1;i<=n;i++)      { scanf("%d",&a[i]);        if(a[i]<=n&&(++b[a[i]]==a[i]))           c[a[i]]=1;          }//筛选出整个区间上数值为a的数的个数大于等于a的数    int t=0;     for(i=1;i<=n;i++)      {          if(c[i]==0) continue;          for(j=1;j<=n;j++)           if(a[j]==i)            d[t][j]=d[t][j-1]+1;           else             d[t][j]=d[t][j-1];          s[t++]=i;      }//求出t在0~j区间上个个数，用d[t][j]表示，另外对数据进行离散化。    while(m--)     {        int ans=0;         scanf("%d%d",&L,&R);         for(i=0;i<t;i++)           {             if(d[i][R]-d[i][L-1]==s[i])                   ans++;           }         cout<<ans<<endl;     }  }return 0;}