【HDU5512 2015沈阳赛区D】【签到题 gcd博弈】Pagodas 取数x+y或x-y

Pagodas

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 72    Accepted Submission(s): 64

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

162 1 23 1 367 1 2100 1 28 6 89 6 810 6 811 6 812 6 813 6 814 6 815 6 816 6 81314 6 81994 1 131994 7 12

 

Sample Output

Case #1: IakaCase #2: YuwgnaCase #3: YuwgnaCase #4: IakaCase #5: IakaCase #6: IakaCase #7: YuwgnaCase #8: YuwgnaCase #9: IakaCase #10: IakaCase #11: YuwgnaCase #12: YuwgnaCase #13: IakaCase #14: YuwgnaCase #15: IakaCase #16: Iaka

 

Source

2015ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

 

#include<stdio.h>#include<string.h>#include<ctype.h>#include<math.h>#include<iostream>#include<string>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;template <class T> inline void gmax(T &a,T b){if(b>a)a=b;}template <class T> inline void gmin(T &a,T b){if(b<a)a=b;}const int N=0,M=0,Z=1e9+7,ms63=1061109567;int casenum,casei;int gcd(int x,int y){return y==0?x:gcd(y,x%y);}int main(){scanf("%d",&casenum);for(casei=1;casei<=casenum;casei++){int n,a,b;scanf("%d%d%d",&n,&a,&b);printf("Case #%d: %s\n",casei,(n/gcd(a,b))&1?"Yuwgna":"Iaka");}return 0;}/*【trick&&吐槽】这道题之所以给了这么多样例，就是为了让这题的结论更容易观察，让这题更算得上是签到题【题意】共有T（[1,500]）组数据。每组数据有1~n（[2,20000]）共n个正整数。一开始对于已有数字集合，除了a，b（1<=a,b<=n && a≠b），其他数字都是不存在的。对于每次成功的操作，我们是从已有数字集合之中，任选两个不同数x,y，得到z=x-y或者z=x+y，如果z是在[1,n]范围，而且z这个数字当前不存在，那我们就可以得到这个新数z，并把其放入已有数字集合之中。Yuwgna先手，Iaka后手，谁无法操作谁就输了。问你最后的winner是谁。【类型】签到 博弈 gcd【分析】首先应该想的问题是，数轴的范围是[1,n]，但是我们能选的数究竟有什么呢？样例中的{3 1 3}{8 6 8}这样组数据，前者可以取遍1,2,3，后者却只能取得2,4,6,8，让我们很快想到实际能选取的数字范围要由gcd判定。我们先令g=gcd(x,y)，那显然我们能取的数，必然是g的倍数。于是求出，接下来能取的数的个数num=n/g-2所以答案就是——【时间复杂度&&优化】O（Tlogn）【数据】162 1 23 1 367 1 2100 1 28 6 89 6 810 6 811 6 812 6 813 6 814 6 815 6 816 6 81314 6 81994 1 131994 7 12*/