codeforces#303div2_D 贪心

codeforces#303div2_D 贪心

D. Queue

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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).

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.

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.

题意：如题。。。

思路：贪心，先排序，逐个判断a[i]，不符条件者换到最后。这个贪心策略基于一个事实，但a[i]>a[i+1]时，若a[i]<s[i-1]，有a[i+1]<a[i]<s[i-1]<s[i+1]，即第i个不符，第i+1个若比第i个大，必然不符。碰到贪心题脑子又短路了。。。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<map>
#include<string>
#include<stack>
#include<queue>
#include<vector>

using namespace std;

const int maxn=1000100;
const int INF=(1<<29);
typedef long long ll;

int n;
ll a[maxn];

int main()
{
    while(cin>>n){
        for(int i=1;i<=n;i++){
            scanf("%I64d",&a[i]);
        }
        sort(a+1,a+n+1);
        a[0]=0;
        ll s=0;
        int cnt=0;
        for(int i=1;i<=n;i++){
            if(s<=a[i]){
                cnt++;
                s+=a[i];
            }
        }
        cout<<cnt<<endl;
    }
    return 0;
}