hdu 2516 取石子游戏 （FIB博弈）

取石子游戏

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4223 Accepted Submission(s): 2531

Problem Description
1堆石子有n个,两人轮流取.先取者第1次可以取任意多个，但不能全部取完.以后每次取的石子数不能超过上次取子数的2倍。取完者胜.先取者负输出”Second win”.先取者胜输出”First win”.

Input
输入有多组.每组第1行是2<=n<2^31. n=0退出.

Output
先取者负输出”Second win”. 先取者胜输出”First win”.
参看Sample Output.

Sample Input

2
13
10000
0

Sample Output

Second win
Second win
First win

思路：经典的FIB博弈模型题.

ac代码：

/* ***********************************************Author       : AnICoo1Created Time : 2016-07-30-08.39 SaturdayFile Name    : D:\MyCode\2016-7月\2016-7-30.cppLANGUAGE     : C++Copyright  2016 clh All Rights Reserved************************************************ */#include<stdio.h>#include<math.h>#include<string.h>#include<stack>#include<set>#include<map>#include<queue>#include<vector>#include<iostream>#include<algorithm>#define MAXN 1010000#define LL long long#define ll __int64#define INF 0xfffffff#define mem(x) memset(x,0,sizeof(x))#define PI acos(-1)#define eps 1e-8using namespace std;ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}ll lcm(ll a,ll b){return a/gcd(a,b)*b;}ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}//headll fib[110];int main(){    fib[1]=1;fib[2]=2;    for(int i=3;i<=50;i++) fib[i]=fib[i-1]+fib[i-2];    int n;    while(scanf("%d",&n)!=EOF,n)    {        int bz=0;        for(int i=1;i<=50;i++)            if(fib[i]==n)            bz=1;        if(bz) printf("Second win\n");        else printf("First win\n");    }    return 0;}