Codeforces Round #303 (Div. 2) D. Queue 傻逼题

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转载最新推荐文章于 2025-08-21 10:38:03 发布 · 124 阅读

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#数据结构与算法

本博客探讨了一个关于C语言中木刻工队问题的解决方案，详细介绍了如何通过优化队列顺序来减少不满意的人数。通过实例分析，解释了如何通过简单的排序算法来解决实际问题，并提供了代码实现。

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C. Woodcutters

Time Limit: 20 Sec Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/545/problem/D

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 t_i 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 t_i (1 ≤ t_i ≤ 10⁹), separated by spaces.

Output

Print a single number — the maximum number of not disappointed people in the queue.

Sample Input

5
15 2 1 5 3

Sample Output

HINT

题意

一堆人排队做事情，如果这个人等了ts还没轮到他，他就会走了

问最多能服务多少人

题解：

傻逼题

这题也配扔到D题难度？

cf药丸

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
/*

inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
*/
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************


int a[maxn];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    scanf("%d",&a[i]);
    sort(a,a+n);
    int miss=0,ans=0;
    for(int i=0;i<n;i++)
    {
        if(a[i]>=miss)
        {
            miss+=a[i];
            ans++;
        }
    }
    cout<<ans<<endl;
}