hdu5512 Pagodas 规律题 gcd

Pagodas

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1296    Accepted Submission(s): 913

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

 

算出gcd = GCD(a, b)，发现规律，只能取gcd的正整数倍的数字，然后判断下n里面有多少个可以取的数字，ans = n / gcd，判断下奇偶，就可以输出答案了

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int T, n, a, b, gcd, ans, kase;

int GCD(int a, int b) {
	if (!b) return a;
	return GCD(b, a % b);
}

int main()
{
	cin >> T;
	kase = 0;
	while (T--) {
		scanf("%d%d%d", &n, &a, &b);
		printf("Case #%d: ", ++kase);
		gcd = GCD(a, b);
		ans = n / gcd;
		if (ans & 1) {
			puts("Yuwgna");
		}
		else {
			puts("Iaka");
		}
	}
	return 0;
}