題目1 : HIHODrinking Game

時間限制:10000ms

單點時限:1000ms

記憶體限制:256MB

        Little Hi and Little Ho are playing adrinking game called HIHO. The game comprises N rounds. Each round, Little Hipours T milliliter of water into Little Ho's cup then Little Ho rolls a K-facesdice to get a random number d among 1 to K. If the remaining water in LittleHo's cup is less than or equal to d milliliter Little Hi gets one score andLittle Ho drinks up the remaining water, otherwise Little Ho gets one score andLittle Ho drinks exactly d milliliter of water from his cup. After N rounds whohas the most scores wins.Here comes the problem. If Little Ho can predict thenumber d of N rounds in the game what is the minimum value of T that makesLittle Ho the winner? You may assume that no matter how much water is added, LittleHo's cup would never be full.

    輸入The firstline 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 Nrounds.

輸出Output theminimum value of T that makes Little Ho the winner.

樣例輸入

5 6

3 6 6 2 1

樣例輸出

4

解題思路：

該題的目的是要求取出最小的T，並且K的序列以及得分的規則是已知的。因而可以很容易寫出每一個輪迴中

因此，可以假設：

score = f(T)

而要證明f(T)

我們設s[i]，r[i]表示第i輪開始，還沒有新增 T 單位飲料時，小Ho的得分和剩餘飲料的體積；s'[i]，r'[i]表示第i輪開始，還沒有新增 T' 單位飲料時，小Ho的得分和剩餘飲料的體積。

我們要證明：對於i = 1..N+1，都有s[i] ≤ s'[i]且r[i] ≤ r'[i]。

利用數學歸納法，i = 1 時，s[i] = s'[i] = 0，r[i] = r'[i] = 0，結論成立。

假設i = n 時結論成立，那麼當i = n+1 時：r[n+1] = max(r[n]+T-d, 0),r'[n+1] = max(r'[n]+T'-d, 0)。

由於r[n] ≤ r'[n]，T < T'，所以r[n]+T < r'[n]+T'

• 當d < r[n]+T < r'[n]+T時，r[n+1] = r[n]+T-d, r'[n+1] = r'[n]+T'-d, s[n+1] = s[n]+1,s'[n+1] = s'[n]+1，易知結論成立；
• 當r[n]+T ≤ d < r'[n]+T時，r[n+1] = 0, r'[n+1] = r'[n]+T'-d > 0, s[n+1] = s[n],s'[n+1] = s'[n]+1，易知結論成立；
• 當r[n]+T < r'[n]+T ≤ d時，r[n+1] = r'[n] = 0, s[n+1] = s[n], s'[n+1] = s'[n]，易知結論成立。

綜上所述，對於i = 1..N+1，都有s[i] ≤ s'[i]且r[i] ≤ r'[i]。而f(T) = s[N+1] ≤ s'[N+1] = f(T')，所以函式f(T)是單調遞增的。

<span style="color:#333333;">#include<iostream>
using namespace std;

int main(){
	int N;
	int K;
	int T;
	int flag=0;
	cin>>N>>K;
	int d[100000]={0};
	int min=100000;
	int max=0;
	int mid=0;
	
	for(int i=0;i<N;i++){
		cin>>d[i];
		if(min>d[i])
			min=d[i];
		if(max<d[i])
			max=d[i];
	}
	if(min==max){               //</span><span style="color:#ff0000;">此處需要注意，當出現min與max相同時，是不會進入while迴圈的，此時T</span><span style="color:#333333;">
		max++;             //</span><span style="color:#ff0000;">應為max+1</span><span style="color:#333333;">
		T=max;
	}
	else{
		max++;
		while(min+1<max){
			mid=(min+max)/2;
			T=mid;
			int score_O=0;
			int height=0;
			for(int i=0;i<N;i++){
				height+=T;
				if(height<=d[i]){
					height=0;
				}
				else{
					score_O++;
					height-=d[i];
				}
			}
			if(score_O<=N/2){
				min=mid;
			}
			else{
				max=mid;
			}
		}
	}
	T=max;
	cout<<T;
	return 0;
	//delete []d;
}</span>