Codeforces Round #303 (Div. 2) D 贪心

链接：戳这里

D. 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).

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：

n个人在排队买单，每个人都知道自己结账所花费的时间，但是这里有一个失望值。当等待的时间超过自己结账的时间，这个人会感到很失望。现在给你一个操作，任意交换队列里面两个人的位置，可以操作多次。

问使得失望人的个数最小的操作。输出不失望的人数

思路：

先按时间排个序，发现一个人失望的时候，则直接把他丢最后去，因为反正已经失望了，不如去最后等吧

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
int n;
ll a[100100];
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%I64d",&a[i]);
    sort(a+1,a+n+1);
    ll sum=0;
    int ans=0;
    for(int i=1;i<=n;i++){
        if(sum>a[i]){
            ans++;
        } else
        sum+=a[i];
    }
    cout<<n-ans<<endl;
    return 0;
}