Different Integers

链接：https://www.nowcoder.com/acm/contest/139/J
来源：牛客网

Given a sequence of integers a₁, a₂, ..., a_n and q pairs of integers (l₁, r₁), (l₂, r₂), ..., (l_q, r_q), find count(l₁, r₁), count(l₂, r₂), ..., count(l_q, r_q) where count(i, j) is the number of different integers among a₁, a₂, ..., a_i, a_j, a_{j + 1}, ..., a_n.

输入描述:

The input consists of several test cases and is terminated by end-of-file.
The first line of each test cases contains two integers n and q.
The second line contains n integers a

₁

, a

₂

, ..., a

_n

.
The i-th of the following q lines contains two integers l

_i

 and r

_i

.

输出描述:

For each test case, print q integers which denote the result.

示例1

输入

复制

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

输出

复制

2
1
3

备注:

* 1 ≤ n, q ≤ 10

⁵

* 1 ≤ a

_i

 ≤ n
* 1 ≤ l

_i

, r

_i

 ≤ n
* The number of test cases does not exceed 10.

不行了。n忘乘了2，卡了一小时。
就是延长数组一倍，树状数组维护。

#include <bits/stdc++.h>
#define maxn 100005
using namespace std;
int n,m;
int vis[maxn*2],a[maxn*2],tree[maxn*2];
struct node
{
    int l,r,id;
    bool operator <(const node &b)
    {
        return r<b.r;
    }
}query[maxn];
int lowbit(int x)
{
    return x&(-x);
}

int sum(int x)
//int query(int x)
{
    int res=0;
    while(x)
    {
        res+=tree[x];
        x-=lowbit(x);
    }
    return res;
}
void add(int x,int val)
{
    while(x<=n)
    {
        tree[x]+=val;
        x+=lowbit(x);
    }
}
int ans[maxn];
int main()
{
    while(~scanf("%d%d",&n,&m))
    {   int i;
        memset(tree,0,sizeof(tree));
        memset(vis,0,sizeof(vis));
        for(i=1;i<=n;i++)
        {
          scanf("%d",&a[i]);
          a[i+n]=a[i];
        }
        for(i=1;i<=m;i++)
        {   int u,v;
            scanf("%d%d",&u,&v);
            query[i].l=v;
            query[i].r=u+n;
            query[i].id=i;
        }
        sort(query+1,query+1+m);
        int cur=1;
        n*=2;
        for(i=1;i<=m;i++)
        {
            for(int j=cur;j<=query[i].r;j++)
            {
                if(vis[a[j]]!=0)
                {
                    add(vis[a[j]],-1);
                }
                add(j,1);
                vis[a[j]]=j;
            }
            cur=query[i].r+1;
            //cout<<query[i].r<<" "<<sum(query[i].r)<<endl;
            //cout<<query[i].l<<" "<<sum(query[i].l-1)<<endl;
            ans[query[i].id]=sum(query[i].r)-sum(query[i].l-1);
        }
        for(i=1;i<=m;i++)
       {
            printf("%d
",ans[i]);
        }
}
}