F Solve equation（FZU 2102）

                 F Solve equation

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 16123 100 101 1 2

Sample Output

(0,700)(1,23)(1,0)

题意分析：

第一行输入一个整数T，接下来有T组数据

每行输入三个数a,b,c;

其中c代表a,b的进制，例如c=3，那a和b就是三进制数。

我们要先把a,b化为10进制数。

然后输出（k,d）中k为a/b，d为a%b。

代码如下：

#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;int main(){    int t;    scanf("%d",&t);    while(t--)    {        char a[105],b[105];        int n;        int x=0,y=0,c=0,d=0;        scanf("%s%s%d",&a,&b,&n);        int len1=strlen(a);        int len2=strlen(b);        int k=1,l=1;        for(int i=len1-1; i>=0; i--)   //求a的十进制数        {            if(a[i]>='a')                a[i]=a[i]-'a'+10;            else                a[i]=a[i]-'0';            x+=a[i]*k;            k*=n;        }        for(int i=len2-1; i>=0; i--)    //求b的十进制数        {            if(b[i]>='a')                b[i]=b[i]-'a'+10;            else                b[i]=b[i]-'0';            y+=b[i]*l;            l*=n;        }        c=x/y;        d=x%y;        printf("(%d,%d)\n",c,d);    }    return 0;}