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       : AnICoo1
Created Time : 2016-07-30-08.39 Saturday
File Name    : D:\MyCode\2016-7月\2016-7-30.cpp
LANGUAGE     : 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-8
using 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;}
//head
ll 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;
}