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

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

 and r

输出描述:

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

示例1

输入

复制

输出

复制

2
1
3

备注:

* 1 ≤ n, q ≤ 10

⁵

* 1 ≤ a

 ≤ n* 1 ≤ l

, r

 ≤ 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\n",ans[i]);}
}
}

转载于:https://www.cnblogs.com/zyf3855923/p/9357571.html

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Different Integers

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输出描述:

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输出

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