ZOJ Monthly, August 2012 - A Alice's present MAP函数

Alice's present

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Time Limit: 5 Seconds      Memory Limit: 65536 KB

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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 ton. Each time Alice chooses an interval fromi to j in the sequence ( includei 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 intergerm ( 1 ≤m ≤ 50,000 ),indicate how many intervals Alice will query. Then followed by m lines, each line contains two integeru,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

51 2 3 1 231 41 53 561 2 3 3 2 141 42 53 64 6

Sample Output

12OK333OK

Hint

Alice will check each interval from right to left, don't make mistakes.

题意：

输入n个数  在输入m个询问区间 问区间内从右向左看有没有重复的数字 如果有 输出第一个重复的 如果没有 输出OK

思路：对每个区间进行访问  利用map函数的一对一 的性质  利用其find函数 看是否有的已经存在 如果不存在 则把 该数 和1 添加进map（）

如果找到了就说明重复了 直接输出该数即可

#include<stdio.h>#include<map>#include<cstring>using namespace std;int a[500500];map<int,int>mp;int main(){int n,i,m,left,right;while(scanf("%d",&n)!=EOF){for(i=1;i<=n;i++)scanf("%d",&a[i]);scanf("%d",&m);while(m--){scanf("%d%d",&left,&right);mp.clear();for(i=right;i>=left;i--){if(mp.find(a[i])==mp.end())mp[a[i]]=1;else{break;}}if(i==left-1)printf("OK\n");elseprintf("%d\n",a[i]);}printf("\n");}return 0;}