【博弈】【HDU】取石子游戏

取石子游戏

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7192    Accepted Submission(s): 4313

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

 

Source

ECJTU 2008 Autumn Contest

 

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寻找规律可以知道，必败的情况下都是斐波那契数，所以做一匹配就可以了，使用离线和二分法搜索来提高效率。

 

#include<bits/stdc++.h>
using namespace std;

long long  fs[55];

void init() {
    fs[0] = 1;
    fs[1] = 2;
    
    for(int i = 2 ; i <= 50 ;i ++) {
        fs[i] = fs[i-1] + fs[i-2];
    }
}

bool find(long long n) {
    int l = 0; 
    int r = 50;
    int mid = (l+r) >> 1;  //   (l+r)/2
    while(l <= r) {
        if(fs[mid] == n) return true;
        else if(fs[mid] < n) l = mid + 1;
        else  r = mid - 1;
        mid = (l+r) >> 1;
    }
    return false;
}

int main() {
    
    init();
    long long n;
    while(~scanf("%lld",&n)&&n) {
        if(find(n))  puts("Second win");
        else    puts("First win");    
    }
    return 0;
}