[HDU 5512][2015ACM/ICPC亚洲区沈阳站] Pagodas 博弈论

Pagodas
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 506 Accepted Submission(s): 375

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 b.

Output
For each test case, output the winner (Yuwgna" orIaka”). 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）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5512

题意：给定集合，最初有两个数a,b建塔，２人可选(i,j有塔)i-j,j+i建塔，无法取的输；

思路：
gcd(a,b)的倍数为可选范围；
n/gcd(a,b)-2为剩余塔的个数；
奇偶判段先手胜负；

代码：

#include<iostream>#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;int n,a,b;int T;int main(){    scanf("%d",&T);    for(int i=1;i<=T;i++)    {        scanf("%d%d%d",&n,&a,&b);        int d=__gcd(a,b);        d=n/d-2;        if(d%2)        {            printf("Case #%d: Yuwgna\n",i);        }        else printf("Case #%d: Iaka\n",i);    }}