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.
InputThe 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.OutputFor 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
思路：
即使是数学水题，对我来说都是一项巨大的挑战，数学这么差，真的是玩不下去啦。
看了一下题解，才勉强弄懂了写法。
首先，这个题意让我琢磨不清，按照我的做法，应该是Iaka先走才对。
假设有n为正无穷，a=15,b=10,那么这两位老和尚可以建造的塔就是5，10，15，20，25，30。。。。
不难发现全部都是gcd(a,b)的倍数，所以当n不是正无穷时，可以建造的塔的数量也就只有n/gcd(a，b)种了。
而且，这n/gcd(a,b)个数一定会全部出现，所以把会出现的数字的个数再判断一下奇偶就行啦。
代码
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#define fuck(x) cout<<#x<<" = "<<x<<endl;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100086;
const int inf = 2.1e9;
const ll INF = 999999999999999;
const double eps = 1e-6;
int gcd(int a,int b)
{
return b?gcd(b,a%b):a;
}
int main()
{
// ios::sync_with_stdio(false);
// freopen("in.txt","r",stdin);
int T;
scanf("%d",&T);
int n,a,b;
int cases = 0;
while(T--){
cases++;
scanf("%d%d%d",&n,&a,&b);
int ans =n/gcd(a,b);
fuck(ans)
printf("Case #%d: ",cases);
if(ans&1){printf("Yuwgna\n");}
else printf("Iaka\n");
}
return 0;
}
原文地址：https://www.cnblogs.com/ZGQblogs/p/9743289.html