Solve equation

You are given two positive integers A and B in Base C. For the equation:

We know there always existing many non-negative pairs (k, d) that satisfy the equation above. Now in this problem, we want to maximize k.

For example, A="123" and B="100", C=10. So both A and B are in Base 10. Then we have:

(1) A=0*B+123

(2) A=1*B+23

As we want to maximize k, we finally get one solution: (1, 23)

The range of C is between 2 and 16, and we use ‘a‘, ‘b‘, ‘c‘, ‘d‘, ‘e‘, ‘f‘ to represent 10, 11, 12, 13, 14, 15, respectively.

The first line of the input contains an integer T (T≤10), indicating the number of test cases.

Then T cases, for any case, only 3 positive integers A, B and C (2≤C≤16) in a single line. You can assume that in Base 10, both A and B is less than 2^31.

(0,700)

(1,23)

(1,0) 

题意：属于进制转换，比较简单的一道题，首先将A、B两个数转换为10进制，然后直接输出A/B，A%B；

<span style="font-size:14px;"># include <cstdio>
# include <cstring>
# include <iostream>
using namespace std;

char a[1000],b[1000];
int c,A,B;

void exchange(char *a,char *b,int c)
{
    A=B=0;
    int i;
    for(i=0;i<strlen(a);i++)
    {
        A*=c;
        switch(a[i])
        {
            case '1':
            case '2':
            case '3':
            case '4':
            case '5':
            case '6':
            case '7':
            case '8':
            case '9':
            case '0':
                A+=a[i]-48;    break;
            default:
                A+=a[i]-87;
        }
    }
    for(i=0;i<strlen(b);i++)
    {
        B*=c;
        switch(b[i])
        {
            case '1':
            case '2':
            case '3':
            case '4':
            case '5':
            case '6':
            case '7':
            case '8':
            case '9':
            case '0':
                B+=b[i]-48;    break;
            default:
                B+=b[i]-87;
        }
    }
}

int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int k;
        scanf("%s%s%d",&a,&b,&c);
        exchange(a,b,c);
        printf("(%d,%d)\n",A/B,A%B);
    }
    return 0;
}
</span>