HDU-5512 Pagodas(GCD)(2015ACM/ICPC亚洲区沈阳站-重现赛)

Pagodas

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

Total Submission(s): 921    Accepted Submission(s): 654


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.

 

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:有n个塔,编号为1~n,其中编号为a,b的塔已经是完美的,如果一个塔的编号k是某2个完美塔的编号相加或相减:k=i-j或k=i+j,那么这个塔就可以重建为完美塔,被重建为完美塔的塔不可以再次被重建.

现在Yuwgna和Iaka轮换重建这些塔,直到一个人不能重建塔则另一个人算获胜,问谁会获胜

题解:设gcd(a,b)=c,那么a=kc,b=pc,可以知道任何可以重建的塔的编号都会是c的倍数,因此只要找出n中有多少个数是c的倍数,再判断奇偶性就可以了

#include<cstdio>
int gcd(int a,int b){
return b==0?a:gcd(b,a%b);
}
int main(){
int T,n,a,b;
// freopen("in.txt","r",stdin);
scanf("%d",&T);
for(int cas=1;cas<=T;cas++){
printf("Case #%d: ",cas);
scanf("%d%d%d",&n,&a,&b);
int k=gcd(a,b);
printf("%s\n",(n/k)%2==1?"Yuwgna":"Iaka");
}
return 0;
}