YT 督促训练 fzoj 2012Solve equation 各种进制转换为10进制



A - Solve equation
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d
& %I64u
Submit Status Practice FZU
2102

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

3

2bc 33f 16

123 100 10

1 1 2

Sample Output

(0,700)

(1,23)

(1,0)

#include <iostream>
#include<string.h>
#include<cstdio>
#include<cmath>
#define ll long long
using namespace std;
int zhuanhuan(char a[],int c)///将任意进制转化为10进制
{
int i=0;
int sum=0;
int b[1000]={0};
int len=strlen(a);
for(i=0;i<len;i++)
{
/*
switch(a[i])
{
case 'a':b[i]=10;break;
case 'b':b[i]=11;break;
case 'c':b[i]=12;break;
case 'd':b[i]=13;break;
case 'e':b[i]=14;break;
case 'f':b[i]=15;break;
default:b[i]=a[i]-'0';break;
}
*/
if(a[i]>=0&&a[i]<='9')
b[i]=a[i]-'0';
else
b[i]=a[i]-'a'+10;

}
for(int j=0;j<len;j++)
sum+=b[j]*pow(c,len-j-1);
return sum;
}
int main()
{
char x[1000],y[1000];
int t,a,b,c;
scanf("%d",&t);
while(t--)
{
scanf("%s%s%d",&x,&y,&c);
a=zhuanhuan(x,c);
b=zhuanhuan(y,c);
if(a<b)
printf("(0,%d)\n",a);
else
{
printf("(%d,%d)\n",a/b,a%b);
}
}
return 0;
}

再次注意字符转化为数字