Pagodas
Time Limit:
2000/
1000 MS (Java/Others) Memory Limit:
65536/
65536 K (Java/Others)
Total Submission(s):
1395 Accepted Submission(s):
974
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
Case
Case
Case
Case
Case
Case
Case
Case
Case
Case
Case
Case
Case
Case
Case
Source
2015ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）
不难发现 最终可以建
n/gcd(a,b)−2
个塔，判断一下奇偶性即可
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<ctime>
#include <string.h>
#include<math.h>
using namespace std;
#define ll long long
#define pii pair<int,int>
const int inf=
1e9+
7;
const int N =
2e3+
5;
int gcd(
int a,
int b){
return b?gcd(b,a%b):a;
}
int main(
int argc,
char *argv[])
{
int T;
scanf(
"%d",&T);
for(
int t=
1;t<=T;++t){
int n,a,b;
scanf(
"%d%d%d",&n,&a,&b);
bool flag=(n/gcd(a,b))%
2;
printf(
"Case #%d: %s\n",t,flag?
"Yuwgna":
"Iaka");
}
return 0;
}
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