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 思路：求最值时考虑使用二分搜索。
#include <iostream>
using namespace std;
const int MAXN=;
const int INF=0x3f3f3f3f;
int n,k;
int d[MAXN];
bool test(int T)
{
int cup=;
int hi_score=;
int ho_score=;
for(int i=;i<n;i++)
{
cup+=T;
if(cup<=d[i])
{
hi_score++;
cup=;
}
else
{
ho_score++;
cup-=d[i];
}
}
return ho_score>hi_score;
}
int main()
{
cin>>n>>k;
for(int i=;i<n;i++)
{
cin>>d[i];
}
int l=,r=INF;
while(r-l>)
{
int mid=(l+r)>>;
if(test(mid))
{
r=mid;
}
else
{
l=mid;
}
}
cout<<r<<endl;
return ;
}