Pagodas

时间：2015-10-31 15:45:16 收藏：0 阅读：193

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" or ``Iaka"). Both of them will make the best possible decision each time.

Sample Input

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

题意：给定一个区间（1-n）还有2个数a，b 这2个数的位置上都有一根柱子。 2个人在这个区间内建柱子（同一个位置只能建一次）

建柱子的条件是建在2根已经存在的柱子的差值或和值（必须在区间内）。

分析：找到这个区间内能放多少根柱子就A

#include<cstdio>
#include<cstring>
#include<stack>
#include<vector>
#include<queue>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
const int oo = 1e9+7;
const int maxn = 1e6+7;
int main()
{
    int T, n, a, b, ans, cas=1, t;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d %d %d", &n, &a, &b);
        if(a < b) swap(a, b);
        printf("Case #%d: ", cas++);
        while(b)
        {
            t = a;
            a = b;
            b = t%b;
        }
        ans = n/a;
        if(ans%2 == 1) printf("Yuwgna\n");
        else printf("Iaka\n");
    }
    return 0;
}

原文：http://www.cnblogs.com/PersistFaith/p/4925595.html