Solve equation

Description

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

A=k*B+d

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.

Input

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.

Output

For each test case, output the solution “(k,d)” to the equation in Base 10.

Sample Input

32bc 33f 16

123 100 10

1 1 2

Sample Output

(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>