HDU 3951 Coin Game（博弈水题）

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<=10^9,1<=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

大致题意：n个硬币围成一圈，两个人轮流拿，每次只能拿连续的硬币，数量最多不超过k个，谁先无法操作谁输。问先手赢还是后手赢。

思路：先手赢的情况只有两种，1.n<=k，即先手一次性全那完，2.k==1，且n为奇数。其他情况下，不论先手怎么操作，后手都能将其变成对称的局面，这样的话先手必输。

代码如下

//#include<bits/stdc++.h>#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;  int main(){      int T;    scanf("%d",&T);    for(int cas=1;cas<=T;cas++)    {        int flag=0;        int n,k;        scanf("%d%d",&n,&k);        printf("Case %d: ",cas);        if(n<=k)            printf("first\n");        else if(k==1&n%2)            printf("first\n");        else             printf("second\n");    }    return 0;}