Zoj 3633 Alice's present

Alice'spresent

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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 ofthem has a number tip on it's back, the tip can be treated as apositive integer. (the number can be repeated). One day, Alicehears that her best friend Marisa's birthday is coming , so shedecides to sent Marisa some dolls for present. Alice puts her dollsin a row and marks them from 1 to n. Each time Alicechooses an interval from i to j in thesequence ( include i and j ) , and thenchecks the number tips on dolls in the interval from right to left.If any number appears more than once , Alice will treat thisinterval as unsuitable. Otherwise, this interval will be treated assuitable.

This work is so boring and it will waste Alice a lot of time. SoAlice 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 Aliceowns.

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. Thenfollowed by m lines, each line contains two integer u,v ( 1≤ u< v≤n ),indicate the left endpoint and right endpoint of theinterval. 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 morethan 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

时间复杂度 nlogn 

  空间复杂度n
这个题数据不强，直接用个数组来保存而不用map也能水过
若数据强，它输入的数据 less than 2^31-1，开不了这么大的数组

#include<stdio.h>
#include<map>
using namespace std;

#define MAXN 511111

map<int,int>num;
int d[MAXN],c[MAXN];       //d数组存储位置，c数组存储数

int main(void)
{
    int n,i,tmp,m;
//    freopen("d:\\in.txt","r",stdin);
    map<int,int>::iterator a;
    while(scanf("%d",&n)==1)
    {
        d[0]=-1;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&tmp);
            a=num.find(tmp);
            if(a!=num.end())
            {
                if(a->second>d[i-1])
                {
                    d[i]=a->second;
                    c[i]=a->first;
                }
                else
                {
                    d[i]=d[i-1];
                    c[i]=c[i-1];
                }
                a->second=i;
            }
            else
            {
                d[i]=d[i-1];
                c[i]=c[i-1];
                num.insert(pair<int,int>(tmp,i));
            }
        }
        scanf("%d",&m);
        int b1,b2;
        for(i=1;i<=m;i++)
        {
            scanf("%d%d",&b1,&b2);
            if(d[b2]>=b1)
                printf("%d\n",c[b2]);
            else
                printf("OK\n");
        }
        printf("\n");
        num.clear();
    }
    return 0;
}