POJ_2371 Questions and answers

[align=center]Questions and answers[/align]

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9465		Accepted: 5027

Description
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.
Input
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 "#".
Output
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).
Sample Input

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

Sample Output

121
121
7
123

题意：给出一些数，再从这些书中分别取出第K大的数。比如 {4,4,5,6,6} 取 K ={1, 2} 第一次后{4,5,6,6} 第二次后 {4,6,6}

思路：计数排序加查找，因为数据范围为1-5000，可以用基数排序，具体代码如下：

#include<iostream>
#include<string>
using namespace std;
int main()
{
int n,k,tmp,i,z,sum;
int a[5005];
string t;
for(i = 0; i < 5005; i++)
a[i] = 0;
cin >> n;
for(i = 1; i <= n; i++)
{
cin >> tmp;
a[tmp]++;
}
cin >> t;
cin >> k;
for(i = 1; i <= k; i++)
{
cin >> tmp;
sum = a[1];
z = 1;
while(sum < tmp)
{
z++;
sum += a[z];
}
cout << z << endl;
}

return 0;
}