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：

#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;
}