HDU 5512 Pagodas(GCD)

庙宇重建博弈问题

最新推荐文章于 2021-07-13 14:54:15 发布

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最新推荐文章于 2021-07-13 14:54:15 发布

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分类专栏： ACMICPC网络赛

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本文介绍了一个关于庙宇重建的博弈问题，两个和尚通过特定规则重建庙宇，探讨了获胜策略及其实现方法。

Pagodas(传送门)

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/65536K (Java/Other)
Total Submission(s) : 22 Accepted Submission(s) : 15

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 $1\le a\neq b\le n$ ) 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 $i~(i\not\in\{a,b\}~and~1\le i\le n)$ 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 $t~(1\le t\le 500)$ which is the number of test cases. For each test case, the first line provides the positive integer $n~(2\le n\le 20000)$ 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）

题意

有 $n$ 个庙经过长时间风吹雨打需要修补，只有两座（被标记为 $a$ ， $b$ ）完好无损不需要修补，有两个和尚轮流去修补这 $n-2$ 个庙，每个和尚每次只能修补一个庙标记为 $i$ ，并要求 $i$ 满足 $i=j+k$ 或者 $i=j-k$ ，每个庙只能被修建一次；
其中 $j$ 和 $k$ 代表已经修建好的庙， $Yuwgna$ 先开始，问最后谁不能修建谁输

解题思路

假设接下来我要进行修复的寺庙为 $s$ ,那么 $s = ax + by$ ,其中 $x,y$ 倍数, $a,b$ 为初始时完好的寺庙编号。

几个必须知道的点：

如果编号为 $1$ 的寺庙出现了，那么所有的寺庙都可以被修复
编号为 $1$ 的寺庙出现的前提是， $s = 1$ 即 $gcd(a,b) = s = 1$
如果没有出现编号为 $1$ 的寺庙，则是直接求解 $[1,n]$ 中有多少个 $gcd(a, b)$ 的倍数

代码

/*除去冗长的头文件*/

int t;
LL a, b, n;
bool vis[MAXN];



LL gcd(LL a, LL b) {
    return b ? gcd(b, a % b) : a;
}

LL lcm(LL a, LL b) {
    return  a * b / gcd(a, b);
}
int main() {
#ifndef ONLINE_JUDGE
    FIN;
    //FOUT;
#endif
    IO_Init();
    int cas = 1;
    scanf("%d", &t);
    while(t --) {
        scanf("%lld %lld %lld", &n, &a, &b);
        LL sum = n / gcd(a, b);
        printf("Case #%d: ", cas ++);
        if(~ sum & 1) printf("Iaka\n");
        else printf("Yuwgna\n");
    }
    return 0;
}