Multiply【暴力＋进制】

6*9 = 42" is not true for base 10, but is true for base 13. That is, 6(13) * 9(13) = 42(13) because 42(13) = 4 * 131 + 2 * 130 = 54(10).

You are to write a program which inputs three integers p, q, and r and determines the base B (2<=B<=16) for which p * q = r. If there are many candidates for B, output the smallest one. For example, let p = 11, q = 11, and r = 121. Then we have 11(3) * 11(3) = 121(3) because 11(3) = 1 * 31 + 1 * 30 = 4(10) and 121(3) = 1 * 32 + 2 * 31 + 1 * 30 = 16(10). For another base such as 10, we also have 11(10) * 11(10) = 121(10). In this case, your program should output 3 which is the smallest base. If there is no candidate for B, output 0.

Input

The input consists of T test cases. The number of test cases (T ) is given in the first line of the input file. Each test case consists of three integers p, q, and r in a line. All digits of p, q, and r are numeric digits and 1<=p,q, r<=1,000,000.

Output

Print exactly one line for each test case. The line should contain one integer which is the smallest base for which p * q = r. If there is no such base, your program should output 0.

Sample Input

3
6 9 42
11 11 121
2 2 2

Sample Output

13
3
0

问题简述：（略）

问题分析：

　　这是一个简单的进制转换问题。

　　输入的数据都是有数字（0-9）组成的。

　　由于如果有两个答案的话则输出小的值，所以需要从小到大试探进制。

　　开始时，先将数据按照10进制读入，然后再转换为指定的进制。

程序说明：

　　编写了共用的函数convert()用于转换数据。　

题记：

　　将共用功能用封装到函数中是一种好的做法。

参考链接：（略）

AC的C语言程序如下：

/* POJ1331 UVALive2526 Multiply */

#include 

#define BASE10 10
#define START 2
#define END 16

int convert(int val, int base)
{
    int ans, weight, r;

    ans = 0;
    weight = 1;
    while(val) {
        r = val % BASE10;
        val /= BASE10;

        if(r >= base)
            return -1;

        ans += weight * r;

        weight *= base;
    }

    return ans;
}

int main(void)
{
    int t, p, q, r, i;
    int p2, q2, r2;

    scanf("%d", &t);
    while(t--) {
        scanf("%d%d%d", &p, &q, &r);

        for(i=START; i<=END; i++) {
            p2 = convert(p, i);
            if(p2 < 0)
                continue;

            q2 = convert(q, i);
            if(q2 < 0)
                continue;

            r2 = convert(r, i);
            if(r2 < 0)
                continue;

            if(p2 * q2 == r2)
                break;
        }

        printf("%d\n", i <= END ? i : 0);
    }

    return 0;
}