Problem Description
1堆石子有n個,兩人輪流取.先取者第1次可以取任意多個,但不能全部取完.以後每次取的石子數不能超過上次取子數的2倍。取完者勝.先取者負輸出"Second win".先取者勝輸出"First win".
Input
輸入有多組.每組第1行是2<=n<2^31. n=0退出.
Output
先取者負輸出"Second win". 先取者勝輸出"First win".
參看Sample Output.
Sample Input
2
13
10000
0
Sample Output
Second win
Second win
First win
Source
ECJTU 2008 Autumn Contest
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解析:
斐波那契博弈:當n爲斐波那契數時,先手必敗,否則先手必勝
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll f[55];
int n;
int main()
{
f[0]=f[1]=1;
for(int i=2;i<55;i++) f[i]=f[i-1]+f[i-2];
while(~scanf("%d",&n)&&n)
{
int flag=0;
for(int i=0;i<55;i++)
{
if(f[i]==n)
{
cout<<"Second win"<<endl;
flag=1;
break;
}
if(f[i]>n) break;
}
if(!flag) cout<<"First win"<<endl;
}
}