Red packet 【模拟】

最新推荐文章于 2024-03-29 09:50:14 发布

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最新推荐文章于 2024-03-29 09:50:14 发布

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本文介绍了一个关于新年红包分配的算法问题。该问题涉及到如何将一定数量的钱公平地分配给多人，确保至少有一人能成为“幸运儿”，即获得比其他人更多的红包金额。文章通过具体的输入输出示例解释了算法的设计思路，并提供了实现该算法的C++代码。

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New Year is coming! Our big boss Wine93 will distribute some “Red Package”, just like Alipay and Wechat.

Wine93 has m yuan, he decides to distribute them to n people and everyone can get some money(0 yuan is not allowed and everyone’s money is an integer), Now k people has gotten money, it’s your turn to get “Red Package”, you want to know, at least how much money to give you, then you can must become the “lucky man”. and the m yuan must be used out.

Noting that if someone’s money is strictly much than others’, than he is “lucky man”.

Input
Input starts with an integer T (T <= 50) denoting the number of test case.
For each test case, three integers n, m, k (1 <= k < n <= 100000, 0< m <= 100000000) will be given.
Next line contains k integers, denoting the money that k people get. You can assume that the k integers’ summation is no more than m.
Output
Ouput the least money that you need to become the “lucky man”, if it is impossible, output “Impossible” (no quote).
Sample Input
3
3 5 2
2 1
4 10 2
2 3
4 15 2
3 5
Sample Output
Impossible
4
6
Hint
无

果然自己菜啊，就这个模拟，还wa还几次。看来以后要多写点模拟，要多锻炼一下严谨的思考方式。
代码

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#define LL long long
using namespace std;
const int MAXN = 500+10;
const int MAXM = 3000000;
/*------------------------------*/
int main(){
    int t;cin>>t;
    while(t--){
        int n,m,k;
        scanf("%d%d%d",&n,&m,&k);
        int max_sum=0;
        int sum=0;
        for(int i=0;i<k;i++){
            int a;scanf("%d",&a);
            sum+=a;
            max_sum=max_sum>a?max_sum:a;
        }
        m=m-sum;   n=n-k;
        if(m<=n) puts("Impossible");
        else {
            m=m-n; // 剩下的每个人首先得到一个
            if(m&1){
                int val=(m-1)/2+1+1;//剩下的人中最多得到这么多
                if(val>=max_sum+1) printf("%d\n",val);
                else {
                    m=m+1-max_sum-1;
                    if(m<0) puts("Impossible");
                    else printf("%d\n",max_sum+1);
                }
            }else {
                int val=m/2+1+1;
                if(val>=max_sum+1) printf("%d\n",val);
                else {
                    m=m+1-max_sum-1;
                    if(m<0) puts("Impossible");
                    else printf("%d\n",max_sum+1);
            }
        }
    }
    }
    return 0;
}