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最新推荐文章于 2020-03-06 15:47:54 发布

SH_Tomorrow

最新推荐文章于 2020-03-06 15:47:54 发布

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分类专栏： 2015年暑假集训周赛题目文章标签：数据结构 C语言算法循环

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小苏西与妈妈购物时思考如何通过调整排队顺序来提高服务效率和顾客满意度。文章探讨了如何通过合理安排顾客排队顺序，减少等待时间过长导致的不满情况，以实现最大化的顾客满意。

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Description

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 Input

Input

5
15 2 1 5 3

Output

Hint

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.

///这个题因为数字过大，所以非常有技巧，如果用错了方法，虽然结果对，但提交就TLE，需要用到sort函数。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 100100
int a[N];
int main()
{
    int n,i,j;
     scanf("%d",&n);
    for(i=0; i<=n-1; i++)
        {
            scanf("%d",&a[i]);
        }
    sort(a,a+n);
    int k=1;
   int sum=a[0];
    for(i=1;i<=n-1;i++)
    {
      if(sum<=a[i])//判断是否符合条件
      {
        sum=sum+a[i];
        k++;
      }
    }

    printf("%d\n",k);
    return 0;
}