Codeforce 545D. Queue

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.

从小到大排序后贪心，如果到当前人是不满意的话就跳过(也就是相当于把他换到后面)，否则就去服务他，sum+=ti

#include <cstdio>
#include <cctype>
#include <stdlib.h>
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
using namespace std;
typedef long long LL;

int n;
LL sum = 0;
LL t[100008];

int main() {
  // freopen("test.in","r",stdin);
  cin >> n;
  for (int i=0;i<n;i++){
    cin >> t[i];
  }
  sort(t,t+n);
  int total = 0;
  for (int i=0;i<n;i++){
    // cout << sum << " " << t[i] << endl;
    if (sum > t[i]){
      continue;
    }
    else {
      total ++;
    }
    sum += t[i];
  }
  cout << total;
  return 0;
}