codeforces 545D Queue_codeforce545d-优快云博客

本文链接：https://blog.youkuaiyun.com/zhang__liuchen/article/details/78367992

博客介绍了Codeforces 545D问题，内容涉及如何通过调整队列中人员顺序来最大化不失望的顾客数量。通过对每个人所需服务时间的比较，以时间排序并进行分析，可以找到使大多数人等待时间不超过服务时间的策略。

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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.

题意：判断最多能有多少人接受服务。

分析：第i个人需要t[i]的服务，那就和它之前的所有时间和比较，如果小于等于当前的服务时间，就可以等，当然了，需要比较，以时间排序。

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5;
int main()
{
    int n;
    int t[N + 5];
    scanf("%d", &n);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d", &t[i]);
    }
    sort(t + 1, t + n + 1);
    int a[N + 5]={0};
    a[1] = t[1];
    int k = 1;
    for(int i = 2; i <= n; i++)
    {
        if(a[k] <= t[i])
        {
            a[k + 1] = a[k] + t[i];
            k++;
        }
    }
    cout << k << endl;
    return 0;
}