ZOJ 3633 Alice's present 倍增区间查询最大值

Alice's present

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=95030#problem/A

Description

As a doll master, Alice owns a wide range of dolls, and each of them has a number tip on it's back, the tip can be treated as a positive integer. (the number can be repeated). One day, Alice hears that her best friend Marisa's birthday is coming , so she decides to sent Marisa some dolls for present. Alice puts her dolls in a row and marks them from 1 to n. Each time Alice chooses an interval from i to j in the sequence ( include i and j ) , and then checks the number tips on dolls in the interval from right to left. If any number appears more than once , Alice will treat this interval as unsuitable. Otherwise, this interval will be treated as suitable.

This work is so boring and it will waste Alice a lot of time. So Alice asks you for help .

Input

There are multiple test cases. For each test case:

The first line contains an integer n ( 3≤ n ≤ 500,000) ,indicate the number of dolls which Alice owns.

The second line contains n positive integers , decribe the number tips on dolls. All of them are less than 2^31-1. The third line contains an interger m ( 1 ≤ m ≤ 50,000 ),indicate how many intervals Alice will query. Then followed by m lines, each line contains two integer u, v ( 1≤ u< v≤ n ),indicate the left endpoint and right endpoint of the interval. Process to the end of input.

,Ci，即此题的初始分值、每分钟减少的分值、dxy做这道题需要花费的时间。

Output

For each test case:

For each query, If this interval is suitable , print one line "OK". Otherwise, print one line ,the integer which appears more than once first.

Print an blank line after each case.

Sample Input

5

1 2 3 1 2

3

1 4

1 5

3 5

6

1 2 3 3 2 1

4

1 4

2 5

3 6

4 6

Sample Output

HINT

题意

查询一个区间是否有重复，以及从右边开始，第一个重复的数是什么

题解：

直接记录pre就好了

然后st区间查询最大值就好了

代码：

#include<stdio.h>

#include<iostream>

#include<cstring>

#include<map>

using namespace std;

#define maxn 500500

int a[];

int dp[maxn][];

int mm[maxn];

int val[maxn];

void initrmp(int n)

{

    mm[]=-;

    for(int i=;i<=n;i++)

    {

        mm[i]=((i&(i-))==)?mm[i-]+:mm[i-];

        dp[i][]=val[i];

    }

    for(int j = ;j<=mm[n];j++)

        for(int i=;i+(<<j)-<=n;i++)

            dp[i][j]=max(dp[i][j-],dp[i+(<<(j-))][j-]);

}

int query(int l,int r)

{

    int k = mm[r-l+];

    return max(dp[l][k],dp[r-(<<k)+][k]);

}

map<int,int> H;

int main()

{

    int n;

    while(scanf("%d",&n)!=EOF)

    {

        memset(mm,,sizeof(mm));

        memset(val,,sizeof(val));

        memset(dp,,sizeof(dp));

        memset(a,,sizeof(a));

        H.clear();

        //int n;scanf("%d",&n);

        for(int i=;i<=n;i++)

            scanf("%d",&a[i]);

        for(int i=;i<=n;i++)

        {

            val[i]=H[a[i]];

            H[a[i]]=i;

        }

        initrmp(n);

        int m;scanf("%d",&m);

        while(m--)

        {

            int l ,r;

            scanf("%d%d",&l,&r);

            int p = query(l,r);

            if(p<l)printf("OK\n");

            else printf("%d\n",a[p]);

        }

        printf("\n");

    }

}