Coin Game
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1627 Accepted Submission(s): 936
Problem Description
After hh has learned how to play Nim game, he begins to try another coin game which seems much easier.
The game goes like this:
Two players start the game with a circle of n coins.
They take coins from the circle in turn and every time they could take 1~K continuous coins.
(imagining that ten coins numbered from 1 to 10 and K equal to 3, since 1 and 10 are continuous, you could take away the continuous 10 , 1 , 2 , but if 2 was taken away, you couldn‘t take 1, 3, 4, because 1 and 3 aren‘t continuous)
The player who takes the last coin wins the game.
Suppose that those two players always take the best moves and never make mistakes.
Your job is to find out who will definitely win the game.
Input
The first line is a number T(1<=T<=100), represents the number of case. The next T blocks follow each indicates a case.
Each case contains two integers N(3<=N<=109,1<=K<=10).
Output
For each case, output the number of case and the winner "first" or "second".(as shown in the sample output)
Sample Input
2 3 1 3 2
Sample Output
Case 1: first Case 2: second
Author
NotOnlySuccess
Source
2011
Alibaba Programming Contest
题意：n枚银币构成一个环，每次可以去1~k之间任意个连续的硬币(取完不合并- -)；
分析：一个简单的博弈，注意取完一段之后环是不会合并起来的，只要考虑下对称性就好了。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 1e9;
const int MOD = 1e9+7;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define lson (i<<1)
#define rson ((i<<1)|1)
#define N 50010
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int main()
{
int T,n,k,cas=1;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&k);
printf("Case %d: ",cas++);
if(n%2==1&&k==1) cout<<"first"<<endl;
else if(k>=n) cout<<"first"<<endl;
else cout<<"second"<<endl;
}
return 0;
}