Codeforces-545D. Queue

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本文介绍了一个通过调整服务队列中顾客顺序来最大化顾客满意度的问题。针对每个顾客的服务时间和等待时间，通过算法实现顾客顺序的最佳排列，使得不满顾客的数量降到最低。

                          D. Queue
                 time limit per test1 second
               memory limit per test256 megabytes
                     inputstandard input
                    outputstandard output

Little girl Susie went shopping with her mom and she wondered how to improve service quality.

There are n people in the queue. For each person we know time ti needed to serve him. A person will be disappointed if the time he waits is more than the time needed to serve him. The time a person waits is the total time when all the people who stand in the queue in front of him are served. Susie thought that if we swap some people in the queue, then we can decrease the number of people who are disappointed.

Help Susie find out what is the maximum number of not disappointed people can be achieved by swapping people in the queue.

Input
The first line contains integer n (1 ≤ n ≤ 105).

The next line contains n integers ti (1 ≤ ti ≤ 109), separated by spaces.

Output
Print a single number — the maximum number of not disappointed people in the queue.

Examples
input
5
15 2 1 5 3
output
4
Note
Value 4 is achieved at such an arrangement, for example: 1, 2, 3, 5, 15. Thus, you can make everything feel not disappointed except for the person with time 5.

#include<iostream>
#include<algorithm>
using namespace std;
int a[100005];
int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        scanf("%d", &a[i]);
    sort(a, a + n);
    long long sum = 0;
    int maximum = 0;
    for (int i = 0; i < n; i++)
    {
        if (a[i] >= sum)
        {
            maximum++;
            sum += a[i];
        }
    }
    printf("%d\n", maximum);
    return 0;
}