hdu5512 Pagodas_c - pagodas hdu - 5512-优快云博客

本文介绍了一个关于两位和尚轮流重建古塔的游戏问题。通过分析古塔编号间的数学关系，利用最大公约数等数学原理，确定了在游戏中获胜的策略，并提供了解决方案的代码实现。

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Pagodas

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2605 Accepted Submission(s): 1761

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

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）

题解：orz。。。

 细想就能知道：只要j或者k中有一个1，说明1~n(j,k除外)都能修，只要j或者k互质，那么肯定能修塔1（exgcd可证明），又回到了1状态，此时1~n（j,k除外）都能修。

 还有一种情况就是j,k不互质，此时只要是gcd(j,k)的倍数都能修，求出n数内gcd(j,k)的倍数的个数（用到鸽笼原理/容斥原理）,n/(gcd(j,k))判断奇偶性即可。

代码：

#include<bits/stdc++.h>
using namespace std;
int gcd(int n,int m){
	if(!m)return n;
	return gcd(m,n%m);
}
int main(){
	int t1,ii,n,j,k;
	scanf("%d",&t1);
	for(ii=1;ii<=t1;ii++){
		printf("Case #%d: ",ii);
	scanf("%d%d%d",&n,&j,&k);
	if(gcd(j,k)==1)n&1?printf("Yuwgna\n"):printf("Iaka\n");
	 else n/gcd(j,k)&1?printf("Yuwgna\n"):printf("Iaka\n");
}
}