Hihocoder #1095 : HIHO Drinking Game （微软笔试题）

最新推荐文章于 2018-06-13 00:43:25 发布

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深入解析水杯挑战游戏的数学原理，揭示如何通过预测和策略运用，实现小Ho在游戏中获得胜利的最低饮料倒入量T。

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题目1 : HIHO Drinking Game

时间限制: 10000ms

单点时限: 1000ms

内存限制: 256MB

描述

Little Hi and Little Ho are playing a drinking game called HIHO. The game comprises N rounds. Each round, Little Hi pours T milliliter of water into Little Ho's cup then Little Ho rolls a K-faces dice to get a random number d among 1 to K. If the remaining water in Little Ho's cup is less than or equal to d milliliter Little Hi gets one score and Little Ho drinks up the remaining water, otherwise Little Ho gets one score and Little Ho drinks exactly d milliliter of water from his cup. After N rounds who has the most scores wins.

Here comes the problem. If Little Ho can predict the number d of N rounds in the game what is the minimum value of T that makes Little Ho the winner? You may assume that no matter how much water is added, Little Ho's cup would never be full.

输入

The first line contains N(1 <= N <= 100000, N is odd) and K(1 <= K <= 100000).
The second line contains N numbers, Little Ho's predicted number d of N rounds.

输出

Output the minimum value of T that makes Little Ho the winner.

样例输入

5 6
3 6 6 2 1

样例输出

题意分析

小Hi和小Ho正在玩这样一个游戏，在每局游戏的开始，小Hi手持一瓶可以认为是无穷无尽的饮料，而小Ho手中有一个空杯子。

一局游戏分为N轮，在每轮行动中，小Hi先向小Ho手中的杯子倒入T个单位的饮料（倒入的数量在一局游戏开始之前约定好且在整局游戏中固定），然后小Ho掷出一个均匀的K面骰子得到一个1..K之间的数d，如果杯中饮料的单位数小于等于d，则小Hi记一分，且小Ho将杯中剩余饮料一饮而尽，否则小Ho记一分，小Ho喝掉杯中d个单位的饮料。在N轮结束后，分高者获胜。

那么问题来了，如果小Ho能够预测这局中每轮自己所掷出的点数，那么最小的能使得小Ho获胜的T（每轮小Hi倒入小Ho杯子的饮料的单位数）是多少？

杯子的饮料的单位数和你选着的T有关。

应为T越大Ho的分数就越大，所以可以用二分

中间题意没看懂，一直是WA。

#include
  
   
using namespace std;

const int N = 100001;
int a[N];
int n, m;

int check(int k)
{
    int cnt = 0;
    int res = 0;
    for(int i = 0; i < n; i++)
    {
        res += k;
        if(res > a[i])
        {
            cnt++;
            res -= a[i];
        }
        else
            res = 0;
    }
    return cnt;
}

int main(void)
{
    while(cin >> n >> m)
    {
        int i, j;
        for(i = 0; i < n; i++) cin >> a[i];
        i = 0;
        j = m+1;
        while(i < j)
        {
            int mid = (i+j)/2;
            if(check(mid) > n/2)
            {
                j = mid-1;
            }
            else
            {
                i = mid+1;
            }
        }
        if(check(i) <= n/2) i++;
        cout << i << endl;
    }
    return 0;
}