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 Description1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 Input2
13
10000
0 Sample OutputSecond win
Second win
First win SourceECJTU 2008 Autumn Contest Recommendlcy   |   We have carefully selected several similar problems for you:  2509 2512 1536 2510 1907  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[]; void init() {
     fs[] = ;
     fs[] = ;     for(int i =  ; i <=  ;i ++) {
         fs[i] = fs[i-] + fs[i-];
     }
 } bool find(long long n) {
     int l = ;
     int r = ;
     int mid = (l+r) >> ;  //   (l+r)/2
     while(l <= r) {
         if(fs[mid] == n) return true;
         else if(fs[mid] < n) l = mid + ;
         else  r = mid - ;
         mid = (l+r) >> ;
     }
     return false;
 } int main() {     init();
     long long n;
     while(~scanf("%lld",&n)&&n) {
         if(find(n))  puts("Second win");
         else    puts("First win");
     }
     return ;
 }