Little Elephant and Array - CodeForces 220 B 树状数组

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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：在一个序列中，从l到r的x数中出现的次数正好是x次的情况有多少。

思路：离线做，先处理要求的区间，按r递增排序，然后每次处理序列中的一个数的时候，找出+1所在的区间，并将之前的区间-1恢复，用差分的方式处理树状数组。

AC代码如下：

#include<cstdio>#include<cstring>#include<map>#include<vector>#include<algorithm>using namespace std;map<int,int> match;int tree[100010],n,m,num,ans[100010],val[100010],f[100010],p[100010];vector<int> vc[100010];struct node{    int l,r,pos;}op[100010];bool cmp(node a,node b){    return a.r<b.r;}int lowbit(int x){    return x&(-x);}void update(int x,int val){    for(;x<=n;x+=lowbit(x))       tree[x]+=val;}int query(int x){    int ret=0;    for(;x>0;x-=lowbit(x))       ret+=tree[x];    return ret;}int main(){    int i,j,k,pos;    scanf("%d%d",&n,&m);    for(i=1;i<=n;i++)    {        scanf("%d",&val[i]);        f[i]=val[i];    }    sort(f+1,f+1+n);    for(i=1;i<=n;i++)       if(f[i]!=f[i-1])       {            num++;            match[f[i]]=num;       }    for(i=1;i<=m;i++)    {        scanf("%d%d",&op[i].l,&op[i].r);        op[i].pos=i;    }    for(i=1;i<=num;i++)       vc[i].push_back(0);    sort(op+1,op+1+m,cmp);    op[m+1].r=0;    pos=1;    for(i=1;i<=n;i++)    {        k=match[val[i]];        p[k]++;        vc[k].push_back(i);        if(p[k]>=val[i])        {            update(vc[k][p[k]-val[i]]+1,1);            update(vc[k][p[k]-val[i]+1]+1,-1);            if(p[k]>val[i])            {                update(vc[k][p[k]-val[i]-1]+1,-1);                update(vc[k][p[k]-val[i]]+1,1);            }        }        while(op[pos].r==i)        {            ans[op[pos].pos]=query(op[pos].l);            pos++;        }    }    for(i=1;i<=m;i++)       printf("%d\n",ans[i]);}

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