This problem can be solve in simpler O(NsqrtN) solution, but I will describe O(NlogN) one.
We will solve this problem in offline. For each x (0 ≤ x < n) we should keep all the queries that end in x. Iterate that x from 0 to n - 1. Also we need to keep some array D such that for current x Dl + Dl + 1 + ... + Dx will be the answer for query [l;x]. To keep D correct, before the processing all queries that end in x, we need to update D. Let t be the current integer in A, i. e. Ax, and vector P be the list of indices of previous occurences of t (0-based numeration of vector). Then, if |P| ≥ t, you need to add 1 to DP[|P| - t], because this position is now the first (from right) that contains exactly t occurences of t. After that, if |P| > t, you need to subtract 2 from DP[|P| - t - 1], in order to close current interval and cancel previous. Finally, if |P| > t + 1, then you need additionally add 1 to DP[|P| - t - 2] to cancel previous close of the interval.
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;
#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)
const int MAXN = 1e5+100;
int n,m;
int a[MAXN],c[MAXN],ans[MAXN];
struct Query
{
int l,r,id;
bool operator < (const Query &t) const {return r<t.r;}
}q[MAXN];
inline int lowbit(int x){return x&(-x);}
void add(int i, int v)
{
while(i<=n)
{
c[i]+=v;
i+=lowbit(i);
}
}
int sum(int x)
{
int ret=0;
while(x>0)
{
ret+=c[x];
x-=lowbit(x);
}
return ret;
}
int main()
{
int sz;
while(~scanf("%d%d",&n,&m))
{
vector<int>data[MAXN];
CL(c,0);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int i=1;i<=m;i++)
{
scanf("%d%d",&q[i].l,&q[i].r);
q[i].id=i;
}
sort(q+1,q+1+m);
for(int i=1,k=1;i<=n;i++)
{
if(a[i]<=n)
{
data[a[i]].push_back(i);
sz=data[a[i]].size();
if(sz>=a[i])
{
add(data[a[i]][sz-a[i]],1);
if(sz>a[i])add(data[a[i]][sz-a[i]-1],-2);
if(sz>a[i]+1)add(data[a[i]][sz-a[i]-2],1);
}
}
while(q[k].r==i && k<=m)
{
ans[q[k].id]=sum(q[k].r)-sum(q[k].l-1);
k++;
}
}
for(int i=1;i<=m;i++)
printf("%d
",ans[i]);
}
return 0;
}
用于调试理解的及及加了凝视的代码
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;
#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)
const int MAXN = 1e5+100;
int n,m;
int a[MAXN],c[MAXN],ans[MAXN];
struct Query
{
int l,r,id;
bool operator < (const Query &t) const {return r<t.r;}
}q[MAXN];
inline int lowbit(int x){return x&(-x);}
void add(int i, int v)
{
while(i<=n)
{
c[i]+=v;
i+=lowbit(i);
}
}
int sum(int x)
{
int ret=0;
while(x>0)
{
ret+=c[x];
x-=lowbit(x);
}
return ret;
}
int main()
{
int sz;
while(~scanf("%d%d",&n,&m))
{
vector<int>data[MAXN];
CL(c,0);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int i=1;i<=m;i++)
{
scanf("%d%d",&q[i].l,&q[i].r);
q[i].id=i;
}
sort(q+1,q+1+m);
for(int i=1,k=1;i<=n;i++)
{
if(a[i]<=n)
{
data[a[i]].push_back(i);
sz=data[a[i]].size();
if(sz>=a[i])
{
add(data[a[i]][sz-a[i]],1);//从右往左第a[i]次出现a[i]的位置+1
if(sz>a[i])add(data[a[i]][sz-a[i]-1],-2);
//从右往左第a[i]+1次出现a[i]的位置 -2,
//由于当Sz==a[i]的时候，这个位置已经被加过1。此次读到i的时候。
//从右往左第a[i]次出现a[i]的位置也被+1。
//那么查询第a[i]+1次出现a[i]的位置到i。答案就是-2+1+1=0，
//查询第a[i]次出现a[i]的位置到i,答案就是1
if(sz>a[i]+1)add(data[a[i]][sz-a[i]-2],1);
//从右往左第a[i]+2次出现a[i]的位置 +1,之前被+1-2，所以变成0
//这三行代码维护出来，从当前的i往左数，第a[i]次出现a[i]的位置总是1
//第a[i]+1次出现a[i]的位置总是-1，第a[i]+2及很多其它次的位置总是0,这样以i为右端点的区间的查询结果就都对了
}
}
while(q[k].r==i && k<=m)
{
/////////////
printf("#i=%d#
",i);
for(int j=0;j<=n;j++)
printf("c[%d]=%d
",j,c[j]);
//////////////
ans[q[k].id]=sum(q[k].r)-sum(q[k].l-1);
k++;
}
}
for(int i=1;i<=m;i++)
printf("%d
",ans[i]);
}
return 0;
}