牛客网暑期ACM多校训练营（第一场）

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

时间限制：C/C++ 2秒，其他语言4秒
空间限制：C/C++ 524288K，其他语言1048576K
64bit IO Format: %lld

题目描述

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.


骚操作:直接把数组*2 然后求1-L,R-N 就变成 求R-L+N之间的不同数的个数了；注意，主席树会TLE；要用线段数组+离线

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <map>
 
using namespace std;
const int maxn=300000+5;
map<int,int> mp;
int data[maxn];
int a[maxn];
int ans[200000+5];
struct node{
    int l,r,id;
    bool operator<(node t)const{
        return r<t.r;
    }
}q[200000+5];
int sum(int i){
    int ans=0;
    while(i>0){
        ans+=data[i];
        i-=i&-i;
    }
    return ans;
}
void add(int i,int x){
    while(i<maxn){
        data[i]+=x;
        i+=i&-i;
    }
}
 
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        fill(data,data+n*2+2,0);
        mp.clear();
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            a[i+n]=a[i];
        }
        n=n*2;
 
 
        for(int i=0;i<m;i++){
            int x,y;
            scanf("%d%d",&x,&y);
            q[i].l=y;
            q[i].r=x+n/2;
            q[i].id=i;
        }
        sort(q,q+m);
        int pre=1;
        for(int i=0;i<m;i++){
            for(int j=pre;j<=q[i].r;j++){
                if(mp[a[j]]!=0){
                    add(mp[a[j]],-1);
                }
                add(j,1);
                mp[a[j]]=j;
            }
            pre=q[i].r+1;
            ans[q[i].id]=sum(q[i].r)-sum(q[i].l-1);
        }
        for(int i=0;i<m;i++){
            printf("%d
",ans[i]);
        }
    }
    return 0;
}

也可以莫队加上读入挂

#include<bits/stdc++.h>
 
using namespace std;
int n,q,a[100005];
int L,R,ans;
struct node{
    int l,r,id;
};
node temp[100005];
int sum[1000005];
int anw[1000005];
int block[100005];
int read()
{
    char ch=' ';
    int ans=0;
    while(ch<'0' || ch>'9')
        ch=getchar();
    while(ch<='9' && ch>='0')
    {
        ans=ans*10+ch-'0';
        ch=getchar();
    }
    return ans;
}
int cmp(node a,node b){
    if(block[a.l]==block[b.l])return block[a.r]<block[b.r];
    return block[a.l]<block[b.l];
}
void add(int x){
    if(sum[x]==0)ans++;sum[x]++;
}
void del(int x){
    sum[x]--;if(sum[x]==0)ans--;
}
int main()
{
 
    ios::sync_with_stdio(false);
    while(~scanf("%d%d",&n,&q)){
        memset(sum,0,sizeof(sum));
        for(int i=1;i<=n;i++){
            a[i]=read();
            block[i]=i/sqrt(n);
        }
        for(int i=1;i<=q;i++){
            temp[i].l=read();
            temp[i].r=read();
            temp[i].id=i;
        }
        sort(temp+1,temp+1+q,cmp);
        L=0;R=n+1;ans=0;
        for(int i=1;i<=q;i++){
            while(L<temp[i].l)L++,add(a[L]);
            while(R>temp[i].r)R--,add(a[R]);
            while(L>temp[i].l)del(a[L]),L--;
            while(R<temp[i].r)del(a[R]),R++;
            anw[temp[i].id]=ans;
        }
        for(int i=1;i<=q;i++)printf("%d
",anw[i]);
 
    }
    return 0;
}