Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

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There is a rectangular pool table ABCD with side lengths m and n, where m and n are integers with m Assume A,B,C,D is the lower left corner,upper left corner, upper right corner, lower right corner respectively. The length of AB is m, and the length of AD is n.

B---------C
| |
| |
A---------D

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Input

Input consist of mutiple cases. Each case include two integers m and n, both of which fits in 32-bit signed integer.

Output

For each case, you should find out which pocket (A,B,C,D) the ball first hit and how many reflections the ball make .

Sample Input

6 10

Sample Output

C 6

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假设球撞到边界时不会反弹，而是继续进入下一个与原图一样大小的边界，那么，由于出发的角度是45度，经过了m，n的公倍数的时候就会撞击一次洞，第一次撞击自然就是最小公倍数了（这样的话其实30度60度也是同样的思路）。见下图：

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上图已经显示得很清晰了，这样，我们只需计算出最小公倍数，然后计算撞击横边和纵边的次数，最后再根据撞击次数确定洞就行：

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#include 
  
   

int gcd(int a, int b) {//求最大公约数不用递归会快
    int c = a % b;
    while (c != 0) {
        a = b;
        b = c;
        c = a % b;
    }
    return b;
}

int main() {
    int n, m, lcm, hit_horizontal_size, hit_vertical_size;
    while (~scanf("%d%d", &m, &n)) {
        lcm = n * m / gcd(n, m);
        hit_horizontal_size = lcm / m - 1;//注意由于最后到洞的时候是不算撞边的，所以都要减去一
        hit_vertical_size = lcm / n - 1;
        //1 & 1 = 1, 0 & 1 = 0，判断奇偶这样快
        if (hit_horizontal_size & 1) {//如果撞击水平边的次数是奇数次，那么可能的点是AD
            if (hit_vertical_size & 1) {//如果撞击竖直边的次数是奇数次，那么可能的点是AB
                printf("A ");
            } else {
                printf("D ");
            }
        } else {
            if (hit_vertical_size & 1) {
                printf("B ");
            } else {
                printf("C ");
            }
        }
        printf("%d\n", hit_horizontal_size + hit_vertical_size);
    }
    return 0;
}

  

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