Codeforces Gym 100513I I. Sale in GameStore 暴力-优快云博客

在游戏店进行限时促销活动时，如何最大化地利用朋友的赠款策略来获取尽可能多的游戏？本文通过贪心算法解决这个问题，详细阐述了实现过程和优化思路。

I. Sale in GameStore

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/gym/100513/problem/I

Description

A well-known Berland online games store has announced a great sale! Buy any game today, and you can download more games for free! The only constraint is that the total price of the games downloaded for free can't exceed the price of the bought game.

When Polycarp found out about the sale, he remembered that his friends promised him to cover any single purchase in GameStore. They presented their promise as a gift for Polycarp's birthday.

There are n games in GameStore, the price of the i-th game is pi. What is the maximum number of games Polycarp can get today, if his friends agree to cover the expenses for any single purchase in GameStore?

Input

The first line of the input contains a single integer number n (1 ≤ n ≤ 2000) — the number of games in GameStore. The second line contains n integer numbers p1, p2, ..., pn (1 ≤ pi ≤ 105), where pi is the price of the i-th game.

Output

Print the maximum number of games Polycarp can get today.

Sample Input

5
5 3 1 5 6

Sample Output

HINT

题意

给你一堆游戏，告诉你可以买一个游戏，并且可以赠送总价值不超过这个游戏价值的东西

问你最后在只买一个游戏的情况下，能够拥有多少个游戏

题解：

贪心一下，很显然这个人买最贵的，然后拿最便宜的几个

代码

#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 test freopen("test.txt","r",stdin)
const int maxn=202501;
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
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;
}
//**************************************************************************************

int n;
int a[maxn];
int sum[maxn];
int main()
{
    int n=read();
    for(int i=1;i<=n;i++)
        a[i]=read();
    sort(a+1,a+1+n);
    for(int i=1;i<=n;i++)
        sum[i]=sum[i-1]+a[i];
    for(int i=n-1;i>=1;i--)
        if(sum[i]<=a[n])
        {
            cout<<i+1<<endl;
            return 0;
        }
    cout<<"1"<<endl;
}