每日一题之 hiho 1095 HIHO Drinking Game

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
4

题意：

两个人A，B玩游戏，一共N轮，每一轮A会往B的杯子里倒入T毫升水，然后B掷一个K面的骰子范围为1到K，如果掷出的点数小于等于d毫升，那么A加一分，B把杯子里剩下的水喝掉，否则A喝掉杯子中的d毫升水，并且B加一分。最后得分高的胜出。现在假设B能预知每次掷骰子的点数，求最小的T使得B一定获胜。

思路：

二分求T啦，每次判断mid是否满足条件。直接上框架就好。

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

const int maxn = 1e5+5;

int n,k;
int A[maxn];

bool judge(int x)
{
    int Hi = 0, Ho = 0;
    long long remain = 0;
    for (int i = 0; i < n; ++i) {
        remain += x;
        if (remain <= A[i]) {
            ++Hi;
            remain = 0;
        }
        else {
            ++Ho;
            remain -= A[i];
        }
    }
    if (Hi < Ho) return true;
    else 
        return false;
}

int Search(int low,int high)
{
    while(low < high) {
        int mid = (low + high) / 2;
        if (!judge(mid)) low = mid + 1;
        else 
            high = mid ;
    }

    return low;
}


int main()
{
    cin >> n >> k;

    for (int i = 0; i < n; ++i)
        cin >> A[i];

    int res = Search(1,1e9);
    cout << res << endl;

    return 0;
}