这是我要解决的问题，我正在使用The Fact That Prefix Sum[i] - Prefix Sum[i-1]导致频率大于零的线索来识别不同的数字，然后我正在消除频率，但即使使用 BIT，我也得到了 TLE
给定一个包含 n 个数字 a1、a2、...、an 和一些 d 查询的序列。
d-query 是一对 (i, j) (1 ≤ i ≤ j ≤ n)。
对于每个 d-query (i, j)，您必须返回子序列 ai、ai+1、...、aj 中不同元素的数量。
Input
Line 1: n (1 ≤ n ≤ 30000).
Line 2: n numbers a1, a2, ..., an (1 ≤ ai ≤ 106).
Line 3: q (1 ≤ q ≤ 200000), the number of d-queries.
In the next q lines, each line contains 2 numbers i, j
representing a d-query (1 ≤ i ≤ j ≤ n).
Output
For each d-query (i, j), print the number of distinct elements in the
subsequence ai, ai+1, ..., aj in a single line.
Example
Input
5
1 1 2 1 3
3
1 5
2 4
3 5
Output
3
2
3
代码是：
#include <iostream>
#include <algorithm>
#include <vector>
#include <stdlib.h>
#include <stdio.h>
typedef long long int ll;
using namespace std;
void update(ll n, ll val, vector<ll> &b);
ll read(ll n,vector<ll> &b);
ll readsingle(ll n,vector<ll> &b);
void map(vector<ll> &a,vector<ll> &b,ll n) /**** RElative Mapping ***/
{
ll temp;
a.clear();
b.clear();
for(ll i=0; i<n; i++)
{
cin>>temp;
a.push_back(temp);
b.push_back(temp);
}
sort(b.begin(),b.end());
for(ll i=0; i<n; i++)
*(a.begin()+i) = (lower_bound(b.begin(),b.end(),a[i])-b.begin())+1;
b.assign(n+1,0);
}
int main()
{
ll n;
cin>>n;
vector<ll> a,b;
map(a,b,n);
ll t;
cin>>t;
while(t--)
{
ll l ,u;
b.assign(n+1,0);
cin>>l>>u;
l--;/*** Reduce For Zero Based INdex ****/
u--;
for(ll i=l;i<=u;i++)
update(a[i],1,b);
ll cont=0;
for(ll i=l;i<=u;i++)
if(readsingle(a[i],b)>0)
{
cont++;
update(a[i],-readsingle(a[i],b),b); /***Eliminate The Frequency */
}
cout<<cont<<endl;
}
return 0;
}
ll readsingle(ll n,vector<ll> &b)
{
return read(n,b)-read(n-1,b);
}
ll read(ll n,vector<ll> &b)
{
ll sum=0;
for(; n; sum+=b[n],n-=n&-n);
return sum;
}
void update(ll n, ll val, vector<ll> &b)
{
for(; n<=b.size(); b[n]+=val,n+=n&-n);
}