HDU5512Pagodas

Pagodas

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1493    Accepted Submission(s): 1033

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

题意:

有n个庙，只有两个是完好的，其他n-2个都可以修。两个和尚轮流修庙，规则是任意两个完好的庙的序号的和或差，谁无法继续修就会输掉。

分析：

这个题是个简单的博弈，想想之后发现就是求出符合题目要求的可以修的庙的总数目，如果是奇数则先修的赢，偶数则后修的赢。关键是求一共有多少个可以修的庙，多次尝试后发现可以用a和b的最大公约数来求，这个最大公约数就是可以修的两个塔之间的距离。

代码：

#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
int gcd(int a, int b){
	return b == 0? a : gcd(b, a % b);
}
int main()
{
	int t, n, a, b;
	int cas = 0;
	scanf("%d", &t);
	while(t--){
		scanf("%d%d%d", &n, &a, &b);
		if(a < b)
			swap(a, b);
		int x = gcd(a, b);
		int min = a % x;
		int cnt = n / x;
		if(cnt * x + min <= n)
			cnt ++;
		printf("Case #%d: ", ++cas);
		if(cnt % 2)
			printf("Iaka\n");
		else
			printf("Yuwgna\n");
	}
	return 0;
}

题目比较水，但还是想了好久，还得加油。