Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3159 Accepted Submission(s): 2110
Total Submission(s): 3159 Accepted Submission(s): 2110
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 aand 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.
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.
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
Recommend
找规律
1 #include<iostream> 2 #include<cstdio> 3 using namespace std; 4 5 int t,n,a,b,sum; 6 7 int main() 8 { 9 scanf("%d",&t); 10 for(int T=1;T<=t;T++) 11 { 12 scanf("%d%d%d",&n,&a,&b); 13 if(a>b) swap(a,b); 14 while(b-a>0&&b-a!=a) 15 { 16 b=b-a; 17 if(a>b) swap(a,b); 18 } 19 sum=(n-b)/a; 20 if(sum&1) printf("Case #%d: Yuwgna ",T); 21 else printf("Case #%d: Iaka ",T); 22 } 23 return 0; 24 }