HDU 5512 - Pagodas - [gcd解决博弈]

题目链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=5512

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Problem Description
n pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the Yuntai Mountain, labelled from 1 to n. However, only two of them (labelled a and b, where 1≤a≠b≤n) withstood the test of time.

Two monks, Yuwgna and Iaka, decide to make glories great again. They take turns to build pagodas and Yuwgna takes first. For each turn, one can rebuild a new pagodas labelled i (i∉{a,b} and 1≤i≤n) if there exist two pagodas standing erect, labelled j and k respectively, such that i=j+k or i=j−k. Each pagoda can not be rebuilt twice.

This is a game for them. The monk who can not rebuild a new pagoda will lose the game.

Input
The first line contains an integer t (1≤t≤500) which is the number of test cases.
For each test case, the first line provides the positive integer n (2≤n≤20000) and two different integers a and b.

Output
For each test case, output the winner (``Yuwgna" or ``Iaka"). Both of them will make the best possible decision each time.

 

 

Sample Input

16

2 1 2

3 1 3

67 1 2

100 1 2

8 6 8

9 6 8

10 6 8

11 6 8

12 6 8

13 6 8

14 6 8

15 6 8

16 6 8

1314 6 8

1994 1 13

1994 7 12

 

Sample Output

Case #1: Iaka

Case #2: Yuwgna

Case #3: Yuwgna

Case #4: Iaka

Case #5: Iaka

Case #6: Iaka

Case #7: Yuwgna

Case #8: Yuwgna

Case #9: Iaka

Case #10: Iaka

Case #11: Yuwgna

Case #12: Yuwgna

Case #13: Iaka

Case #14: Yuwgna

Case #15: Iaka

Case #16: Iaka

 

题意：

给出N(N>=2)座佛塔，其中只有两座经历了时间的考验，存留了下来，他们的编号记为a,b；

现在两个僧人Yuwgna和Iaka要重建这N座佛塔；

但是重建有规则：只能重建佛塔k，满足k = i + j或者k = i - j，并且 i 和 j 是目前情况下已经建好的佛塔；

知道不能重建佛塔为止，Yuwgna先开始重建；

问两个人谁是最后一个建佛塔的；

 

题解：

给出a,b后，不难发现所有可以重建的宝塔的编号是一个等差数列；

并且等差数列的差d=gcd(a,b)；

这就很好办了，我们可以立即求出在1~n的范围内到底可以重建多少佛塔，记为cnt；

然后两人轮流来么，只要判断一下cnt是奇是偶即可；

 

AC代码：

 #include<cstdio>
 int gcd(int m,int n){return n?gcd(n,m%n):m;}
 int n,a,b;
 int main()
 {
     int t;
     scanf("%d",&t);
     for(int kase=;kase<=t;kase++)
     {
         scanf("%d%d%d",&n,&a,&b);
         int cnt=,d=gcd(a,b);
         for(int i=gcd(a,b);i<=n;i+=d) cnt++;
         printf("Case #%d: %s\n",kase,(cnt%)?"Yuwgna":"Iaka");
     }
 }