G-Queue(CF-545D

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 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.

题意：多组数据。先输入n，表示有n个人在排队，接下来输入n个数，表示第i个人要被服务的时间。如果一个人等待

的时间多于自己被服务的时间他就会不满意，等待的时间即排在他之前的所有人被服务的时间。求怎么排队才能使满

意的人更多，满意的人数为多少。

思路：按升序排一次序即可，然后再判断自身服务的时间是否大于等于自己要等待的时间，要是是就++，否则就让自

己被服务的时间为0。

Sample Input

5

15 2 1 5 3

Sample Output

4

<pre name="code" class="cpp">#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#define MAX 2000000
using namespace std;
struct node
{
    long long int time,flag;
    bool operator < (const struct node aa)const       //按服务时间升序排序
    {
        return time<aa.time;
    }
} sum[MAX];
int main()
{
    //freopen("lalala.text","r",stdin);
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0; i<n; i++)
            scanf("%I64d",&sum[i].time);
        sort(sum,sum+n);
        sum[0].flag=0;                     //第一个人需要等待的时间是0
        long long int ans=1;
        for(int i=1; i<n; i++)
        {
            sum[i].flag=sum[i-1].time+sum[i-1].flag;
            if(sum[i].time>=sum[i].flag)             //如果这个人的被服务时间大于等于等待时间，满意的人数就++
                ans++;
            else
                sum[i].time=0;                          //否则就让该人的被服务时间等于0
        }
        printf("%I64d\n",ans);
    }
    return 0;
}