HDU 5512 Pagodas (2015沈阳现场赛，找规律+gcd)

Pagodas

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 70    Accepted Submission(s): 62

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

Source

2015ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

【题意】：

选已存在的i j来建新塔(i+j or i-j)

【解题思路】：

沈阳最水的题，仍然挂了2发。。。

一开始以为除了a b均是偶数的情况均能出现所有数（被样例带歪了）

多写几个例子就发现可能出现的数字是gcd(a,b)的倍数，答案即为判断 n/gcd(a,b)-2 奇偶~

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #define eps 1e-8
 #define zero(x)(((x)>0?(x):-(x))<eps)
 #define PI acos(-1.0)
 #define LL long long
 #define maxn 100100
 #define IN freopen("in.txt","r",stdin);
 using namespace std;

 int gcd(int a,int b)
 {
     return !b? a:gcd(b,a%b);
 }

 int main(int argc, char const *argv[])
 {
     //IN;

     int t,ca=;scanf("%d",&t);
     while(t--)
     {
         int n,a,b;
         scanf("%d %d %d",&n,&a,&b);

         int cnt = n/gcd(a,b) - ;

         if(cnt%==) printf("Case #%d: Iaka\n",ca++);
         else printf("Case #%d: Yuwgna\n", ca++);
     }

     return ;
 }