D. Queue

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#D. Queue

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小苏西购物排队体验，探索如何通过调整队伍顺序减少顾客等待时间，从而提升服务满意度。通过排序优化，实现顾客等待时间与服务时间匹配，最大化不感到失望的顾客数量。

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 t_i 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 t_i (1 ≤ t_i ≤ 10⁹), separated by spaces.

Output

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

Sample test(s)

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.

////这个题用快排测试数据都不过，TLE，所以只能用sort排序（我也是醉了），排完序之后，定义s=0,让s与a[i]比较，如果s<=a[i],就让sum++，并且让s的值更新为s+a[i]，同时利用循环，这是一个比较省时的很好的想法，比用两个for循环好，但是由于测试数据太大，两个for只会TLE（我用这个方法做了一遍，结果不是TLE，而是WA，妈呀！感觉做的挺对的，为什么WA呢，到底哪里错了，求抱大腿！

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;

int main()
{
    int n, a[100010];
    scanf("%d", &n);
    for(int i=0; i<n; i++)
    {
        scanf("%d", &a[i]);
    }
    sort(a,a+n);
    int sum=0, s=0;
    for(int i=0; i<n; i++)
    {
        if(s<=a[i])
        {
            s=s+a[i];
            sum++;
        }
    }
    printf("%d\n", sum);
    return 0;
}