CodeForces 303D 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).（10的五次方）

The next line contains n integers ti (1 ≤ ti ≤ 109)（10的9次方）, 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.

讲的是很多人排队，第 i 个人需要的时间是 a[i] ，需要等待的时间是排在他前面人的时间总和，如果总和超过 a[i]，那么那个人会不高兴，问怎样安排才能让最多人高兴，起初看完题目以为要调换那么多个人，因此在代码里多了一项换人的代码，然后果断超时，再后来发现，原来只要从前往后扫，比总和大的，就加上去，计数加一，比总和小的直接跳过，就行了，总的来说，不用想太多。

#include<iostream>#include<algorithm>using namespace std;int a[100010];int main(){    int n;    while(cin>>n)    {        for(int i=0;i<n;i++)  cin>>a[i];        sort(a,a+n);        long long flag=a[0];        int count=1;        for(int i=1;i<n;i++)        {            if(a[i]>=flag)            {                flag+=a[i];                count++;            }        }          cout<<count<<endl;    }    return 0;}