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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.
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.
Output the minimum value of T that makes Little Ho the winner.
5 6 3 6 6 2 1
4
思路：求最值时考虑使用二分搜索。
#include <iostream> using namespace std; const int MAXN=100005; const int INF=0x3f3f3f3f; int n,k; int d[MAXN]; bool test(int T) { int cup=0; int hi_score=0; int ho_score=0; for(int i=0;i<n;i++) { cup+=T; if(cup<=d[i]) { hi_score++; cup=0; } else { ho_score++; cup-=d[i]; } } return ho_score>hi_score; } int main() { cin>>n>>k; for(int i=0;i<n;i++) { cin>>d[i]; } int l=0,r=INF; while(r-l>1) { int mid=(l+r)>>1; if(test(mid)) { r=mid; } else { l=mid; } } cout<<r<<endl; return 0; }
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原文地址：http://www.cnblogs.com/program-ccc/p/5729375.html
