Pagodas HDU - 5512(找规律)

Pagodas

HDU - 5512
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

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

题意：给你三个数：n,a,b，一开始一个集合里面有两个数：a和b，然后两个人轮流往这个集合里面增加数字，增加的这个数字的原则是，这个集合里面任选两个数的和或差，集合里面的数字不能重复，同时这个数字不能大于 n 。
思路：把数字的加减都模拟一遍之后发现，无论怎样，最后得到的这个集合里面的数列，其实是一个等差数列，并且会发现这个等差数列有一个特点，因为这个等差数列是我们根据a，b相加减生成的，中间并没有加入其他的数字，因此这个等差数列首项其实就是公差也就是他的公式an = nd，所以由一开得到的 a 和 b 来相减，然后不断地取最小的两个数相减，然后得到等差数列的差，这个差同时也是等差数列一开始的那个数字，也可以直接取ab的最大公因数，然后拿 n 除以这个差就是在 n 的范围内可以得到的数字的个数了，然后因为分先手和后手，所以最后只要判断一下个数的奇偶数就可以得到答案了。
code：

#include 
#include 
#include 
using namespace std;

int main(){
    int cas = 0;
    int t,n,a,b;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d",&n,&a,&b);
        if(a < b) swap(a,b);
        int cnt = 0;
        while(1){
            cnt = a - b;
            if(b == cnt) break;
            a = max(cnt,b);
            b = min(cnt,b);
        }
        int ans = n / cnt;
        printf("Case #%d: ",++cas);
        if(ans % 2 == 0) printf("Iaka\n");
        else printf("Yuwgna\n");
    }
    return 0;
}