POJ 2371 Questions and answers

The database of the Pentagon contains a top-secret information. We don‘t know what the information is — you know, it‘s top-secret, — but we know the format of its representation. It is extremely simple. We don‘t know why, but all the data is coded by the natural numbers from 1 up to 5000. The size of the main base (we‘ll denote it be N) is rather big — it may contain up to 100 000 those numbers. The database is to process quickly every query. The most often query is: "Which element is i-th by its value?"— with i being a natural number in a range from 1 to N.

Your program is to play a role of a controller of the database. In the other words, it should be able to process quickly queries like this.

The standard input of the problem consists of two parts. At first, a database is written, and then there‘s a sequence of queries. The format of database is very simple: in the first line there‘s a number N, in the next N lines there are numbers of the database one in each line in an arbitrary order. A sequence of queries is written simply as well: in the first line of the sequence a number of queries K (1 <= K <= 100) is written, and in the next K lines there are queries one in each line. The query "Which element is i-th by its value?" is coded by the number i. A database is separated from a sequence of queries by the string of three symbols "#".

The output should consist of K lines. In each line there should be an answer to the corresponding query. The answer to the query "i" is an element from the database, which is i-th by its value (in the order from the least up to the greatest element).

5
7
121
123
7
121
###
4
3
3
2
5

121
121
7
123

排序，输出第K项
CODE:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define REP(i, s, n) for(int i = s; i <= n; i ++)
#define REP_(i, s, n) for(int i = n; i >= s; i --)
#define MAX_N 100000 + 10

using namespace std;

int main(){
    int n, a[MAX_N];
    scanf("%d", &n);
    REP(i, 1, n) scanf("%d", &a[i]);
    sort(a + 1, a + 1 + n);
    int m, x;
    
    char s[4]; cin >> s;
    scanf("%d", &m);
    while(m --){
        scanf("%d", &x);
        printf("%d\n", a[x]);
    }
    return 0;
}