HDU 5512 Pagodas（思维数学）_pagodas csdn-优快云博客

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本文介绍了一个关于两位和尚轮流重建古塔的游戏问题。通过数学方法确定了每种情况下获胜者，利用最大公约数来判断剩余可操作次数的奇偶性以决定胜负。

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Pagodas

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2555 Accepted Submission(s): 1725

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

Output

For each test case, output the winner (``Yuwgna" or ``Iaka"). Both of them will make the best possible decision each time.

Sample Input

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

如题求最后动不了的就输，那么求可以移动的步数即可以10 6 8为例只能修复的房子为 4 6 8 10 2 即可知道6 8的步数等于10/gcd（6,8）即可

#include <stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main(int argc, char *argv[])
{
    int t;
    int case1=1;
    scanf("%d",&t);
    while(t--)
    {
        int all,s,e;
        scanf("%d %d %d",&all,&s,&e);
        int gcd=__gcd(s,e);
        if((all/gcd)%2==0)
        {
            printf("Case #%d: Iaka\n",case1++);
        }
        else
        {
            printf("Case #%d: Yuwgna\n",case1++);
        }
    }
    return 0;
}