ZOJ 3633 Alice's present【线段树】

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.

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

1

2

OK



3

3

3

OK

题目大意：给出n个数，在区间[x, y]内找出从右向左出现不止一次的数，输出最右端的数字；没有输出OK;
思路：区间查询，就该往线段树方向上想；那么当百每个数字的左端离他最近的数字就录下来，在区间【x, y】内寻找最大值，那么就是右端起最先重复出现的数字，但是必须要　　　　满足最大的数maxnum>=x, 否则是属于没有重复数字的，输出ok;

代码如下：

View Code

#include<stdio.h>

#include<string.h>

#include<iostream>

#include<map>

#include<math.h> 

#include<algorithm>

using namespace std;

#define N  500005

int dpmax[N][19], a[N]; 

int main()

{

    int i, j, n, m, x, y;

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

    {

        map<int, int>M; 

        memset(a, 0, sizeof(a)); 

        for(i=1; i<=n; i++)

        {

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

            dpmax[i][0]=M[a[i]];

            M[a[i]]=i;

        } 

        int mm=(int)floor(log(1.0*n)/log(2.0)); //n==500000时, mm=18 

        for(j=1; j<=mm; j++)

            for(i=n; i>=1; i--)

                if((i+(1<<(j-1)))<=n)

                    dpmax[i][j]=max(dpmax[i][j-1], dpmax[i+(1<<(j-1))][j-1]);

         scanf("%d", &m);

        for(i=1; i<=m; i++)

        {

            scanf("%d%d", &x, &y);

            int  mid=(int)floor(log(y*1.0-x+1)/log(2.0));

            int maxnum=max(dpmax[x][mid], dpmax[y-(1<<mid)+1][mid]);

            if(maxnum<x)

                printf("OK\n");

            else

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

        }

        printf("\n"); 

    }

    return 0;

}