传送门

Time Limit: 3 Sec  Memory Limit: 128 MB

Description

Alice likes playing games. So she will take part in the movements of M within N days, and each game is represented in an integer between 1 and M. Roommates have Q magic questions: How many different kinds of games does Alice participate between Lth day and Rth day(including Lth day and Rth day)?

Input

You will be given a number of cases; each case contains blocks of several lines. The first line contains 2 numbers of N and M. The second line contains N numbers implying the game numbers that Alice take part in within N days. The third line contains a number of Q. Then Q lines is entered. Each line contain two numbers of L and R.

1≤N,M,Q≤100000

Output

There should be Q output lines per test case containing Q answers required.

Sample Input

5 3 1 2 3 2 2 3 1 4 2 4 1 5

Sample Output

3 2 3

HINT


这是今年校赛的K题,一道经典题目,但现场没A。

在线可以用主席树,目前还不会。有一个巧妙的利用数状数组的离线解法,比较好写。

要点是:

1.将查询按右端点从小到大排序。

2.将每个数上一次出现的位置记录下来。当这个数再次出现时,将它上次出现位置上的计数消除。

Implementation:

主体是个双指针。

#include <bits/stdc++.h>
using namespace std; const int N(1e5+);
int n, m, q, a[N], pos[N], bit[N], ans[N]; void add(int x, int v){
for(; x<=n; bit[x]+=v, x+=x&-x);
} int sum(int x){
int res=;
for(; x; res+=bit[x], x-=x&-x);
return res;
} struct P{
int l, r, id;
P(int l, int r, int id):l(l),r(r),id(id){}
P(){};
bool operator<(const P&b)const{return r<b.r;}
}p[N]; int main(){
// ios::sync_with_stdio(false);
for(; ~scanf("%d%d", &n, &m); ){
for(int i=; i<=n; i++) scanf("%d", a+i);
scanf("%d", &q);
for(int l, r, i=; i<q; i++) scanf("%d%d", &l, &r), p[i]={l, r, i}; sort(p, p+q); //error-prone
memset(bit, , sizeof(bit));
memset(pos, , sizeof(pos));
for(int i=, j=, k; j<q&&i<=n; ){ //error-prone
for(; i<=p[j].r; i++){
if(pos[a[i]]) add(pos[a[i]], -);
pos[a[i]]=i;
add(i, );
}
for(k=j; k<q&&p[k].r==p[j].r; k++) //error-prone
ans[p[k].id]=sum(p[k].r)-sum(p[k].l-);
j=k;
}
for(int i=; i<q; i++) printf("%d\n", ans[i]); //error-prone
}
return ;
}

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