Red packet 二分算法

最新推荐文章于 2024-03-29 09:50:14 发布

Ethan__Han

最新推荐文章于 2024-03-29 09:50:14 发布

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分类专栏：二分

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本文介绍了一个抢红包问题的算法解决方案，使用二分查找法来确定成为最幸运接收者的最小金额阈值。通过输入红包总数、参与人数及已知的红包分配情况，算法能够计算出最少需要多少金额才能确保自己是最幸运的人。

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Red packet
Time Limit:1000MS Memory Limit:65535KB 64bit IO Format:%I64d & %I64u

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
3
3 5 2
2 1
4 10 2
2 3
4 15 2
3 5
Sample Output
Impossible
4
6

这个题就是抢红包呗，二分直接算，但是注意在二分之前的各种判断。
代码如下：

#include<cstdio>
#include<algorithm>
using namespace std;
int n,m,k,sum,ans,a[100000+11],ri,le,mid,max0;
bool f(int x)
{
    if(x <= max0)
        return 0;
    if(x > (m-sum-x-(n-k-2)))
        return 1;
    return 0;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        sum=0;
        max0=0;
        scanf("%d %d %d",&n,&m,&k);
        for(int i = 0 ; i < k ; i++)
        {
            scanf("%d",&a[i]);
            max0=max(max0,a[i]);
            sum+=a[i];
        }
        if((m-sum-(n-k-1))<= max0)
        {
            printf("Impossible\n");
            continue;
        }
        le=max0;
        ri=m-sum-(n-k-1);
        while(le <= ri)
        {
            mid=( le + ri)/2;
            if(f(mid))
            {
                ans=mid;
                ri=mid-1;
            }
            else
            {
                le=mid+1;
            }
        }
        printf("%d\n",ans);
    }
}