Pagodas(传送门)
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/65536K (Java/Other)
Total Submission(s) : 22 Accepted Submission(s) : 15
Problem Description
n
pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the Yuntai Mountain, labelled from
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
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
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
个庙经过长时间风吹雨打需要修补,只有两座(被标记为
其中
j
和
解题思路
假设接下来我要进行修复的寺庙为
s
,那么
几个必须知道的点:
- 如果编号为 1 的寺庙出现了,那么所有的寺庙都可以被修复
- 编号为
1 的寺庙出现的前提是, s=1 即 gcd(a,b)=s=1 - 如果没有出现编号为
1
的寺庙,则是直接求解
[1,n] 中有多少个 gcd(a,b) 的倍数
代码
/*除去冗长的头文件*/
int t;
LL a, b, n;
bool vis[MAXN];
LL gcd(LL a, LL b) {
return b ? gcd(b, a % b) : a;
}
LL lcm(LL a, LL b) {
return a * b / gcd(a, b);
}
int main() {
#ifndef ONLINE_JUDGE
FIN;
//FOUT;
#endif
IO_Init();
int cas = 1;
scanf("%d", &t);
while(t --) {
scanf("%lld %lld %lld", &n, &a, &b);
LL sum = n / gcd(a, b);
printf("Case #%d: ", cas ++);
if(~ sum & 1) printf("Iaka\n");
else printf("Yuwgna\n");
}
return 0;
}