CodeForces 220B Little Elephant and Array

B. 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 2
3 1 2 2 3 3 7
1 7
3 4
Output
3
1

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

#if ( _win32 || __win32__ )
    #define lld "%i64d"
#else
    #define lld "%lld"
#endif

#define MP make_pair
#define PB push_back
#define INT_INF 0x3fffffff
#define LL_INF 0x3fffffffffffffff
#define EPS 1e-12
#define MOD 1000000007
#define PI 3.14159265358979323846
#define N 100010
#define E 100010

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef unsigned int Uint;
typedef double DB;

int a[N] , cnt[N];
bool is[N];

int s[500][N];
int val[500];

int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(cnt,0,sizeof(cnt));
        memset(is,0,sizeof(is));
        for(int i=1; i<=n; i++)
        {
            scanf("%d",a+i);
            if(a[i]<=n && ++cnt[a[i]]==a[i]) is[a[i]]=1;
        }
        int tot=0;
        memset(s,0,sizeof(s));
        memset(val,0,sizeof(val));
        for(int i=1; i<=n; i++)
        {
            if(!is[i]) continue;
            for(int j=1; j<=n; j++)
            {
                s[tot][j]=s[tot][j-1];
                if(a[j]==i) s[tot][j]++;
            }
            val[tot++]=i;
        }
        for(int ca=1,L,R; ca<=m; ca++)
        {
            scanf("%d%d",&L,&R);
            int ans=0;
            for(int i=0; i


你可能感兴趣的:(ACM_CodeForces)