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最新推荐文章于 2019-10-06 08:14:11 发布

原创最新推荐文章于 2019-10-06 08:14:11 发布 · 478 阅读

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#each #numbers #input #output

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本文介绍了一个关于两个玩家轮流给出数字的游戏，通过设定的规则来判断最终赢家的算法实现。输入包括目标数字N和每轮可增加的最大值k，输出为胜利者的名字。

2699: 16 and 18

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	1s	65536K	285	65	Standard

16 and 18 are good friends. They are playing a game.

The rules are as following:

(1) 16 and 18 will give numbers in turn. At first 16, then 18, and then 16...
(2) If the previous one give a number A, the number the next one gives, namely B, must satisfy that
1 <= B - A <= k.
(3) Who first gives the number larger than or equal to N will lose the game.
(4) The game always starts with 16, and he can only give a number from [1, k] in the first round.
(5) Suppose that both of them will try their best to win.

Input

In each line, there will be two integers N (0 < N <= 100, 000, 000) and k (0 < k <= 100). Input terminates when N = k = 0

Output

For each case, print the winner's name in each line.

Sample Input

Sample Output

18
16
18

Problem Source: homeboy

博弈问题详细情况可以看我转载的博弈问题总结

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
int n,k;
while(scanf("%d%d",&n,&k),n&&k)
{
if((n-1)%(k+1)==0)cout<<18<<endl;
else cout<<16<<endl;
}
return 0;
}