Red packet【二分】

最新推荐文章于 2018-01-27 15:05:06 发布

Somethingwll

最新推荐文章于 2018-01-27 15:05:06 发布

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分类专栏：大菜鸟小历程二分查找

本文链接：https://blog.youkuaiyun.com/wang907553141/article/details/52047303

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本文介绍了一个有趣的红包分配问题：如何确保成为最幸运的人。通过输入总金额、人数等参数，算法计算出最少需要获得多少金额才能确保自己在分配中是最幸运的那个。

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Red packet

Time Limit:1000MS Memory Limit:65535KB 64bit IO Format:

Submit Status

Description

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

Sample Output

Impossible
4
6

以下纯属渣渣胡扯大神勿喷

#include<cstdio>   
#include<algorithm>   
using namespace std;
int main()  
{  
    int t,n,m,k,i,j,a;  
    scanf("%d",&t);  
    while(t--)  
    {  
        int sum=0;  
        int mm=0;  
        scanf("%d %d %d",&n,&m,&k);  
        for(i=0;i<k;i++)  
        {  
            scanf("%d",&a);  
            sum+=a;  
            mm=max(mm,a);  
        }
        int ss=m-sum-(n-k-1);  // 能得到的最大钱数 （让后面的每人都分 1 元） 
        int ma=mm+1;    // 至少要得到的钱数 
        if(ss<=mm)    
            printf("Impossible\n");  
        else  
        {  
            int left=ma,right=ss;
            int mid,ans=0;  
            while(left<=right)  
            {  
                mid=(left+right)/2;
                if(mid>ss-mid+1)   // 除前面的 k 人外，要比 剩下的所有人中 钱最多的那个人 多 ，之前ss是让后面每个人都得 1 元，这里让ss加 1 ，是为了保证让后面有一个人钱数不是 1 元，这样才有可比性，要不然后面全都是 1 元，得到的就不是最佳答案了
				{
					ans=mid;
                    right=mid-1;  
				}
				else
					left=mid+1;
            }  
            printf("%d\n",ans);  
        }  
    }  
    return 0;  
}