hdu 1335 Basically Speaking

题目地址：

http://acm.hdu.edu.cn/showproblem.php?pid=1335

题目描述：

Basically Speaking

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2427    Accepted Submission(s): 922

Problem Description

The Really Neato Calculator Company, Inc. has recently hired your team to help design their Super Neato Model I calculator. As a computer scientist you suggested to the company that it would be neato if this new calculator could convert among number bases. The company thought this was a stupendous idea and has asked your team to come up with the prototype program for doing base conversion. The project manager of the Super Neato Model I calculator has informed you that the calculator will have the following neato features: 
It will have a 7-digit display.

Its buttons will include the capital letters A through F in addition to the digits 0 through 9.

It will support bases 2 through 16. 

 

Input

The input for your prototype program will consist of one base conversion per line. There will be three numbers per line. The first number will be the number in the base you are converting from. The second number is the base you are converting from. The third number is the base you are converting to. There will be one or more blanks surrounding (on either side of) the numbers. There are several lines of input and your program should continue to read until the end of file is reached.

 

Output

The output will only be the converted number as it would appear on the display of the calculator. The number should be right justified in the 7-digit display. If the number is to large to appear on the display, then print "ERROR'' (without the quotes) right justified in the display. 

 

Sample Input

   
   
   
   

    
    
    
    1111000  2 10
1111000  2 16
2102101  3 10
2102101  3 15
  12312  4  2
     1A 15  2
1234567 10 16
   ABCD 16 15
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
        120
     78
   1765
    7CA
  ERROR
  11001
 12D687
   D071
   
   
   
   

题意：

任意进制间的转换。

题解：

就是吧m进制转化为10进制（辗转相乘，连乘，注意连乘技巧，叠加乘，而不是直接取够着m的几次方），然后再吧10进制转化为n进制（辗转相除）。注意个别的位上是字母的情况。

代码：

#include<stdio.h>
#include<string.h>
#define SWAP(x,y) x^=y;y^=x;x^=y;
char m_str[100],n_str[100];
int m_base=0,n_base=0;//base number
//m base convert into the n base
int m_to_n(int mb,char *ms,int nb,char *ns)
{
    int len_m=strlen(ms),len_n=0;
    int tens=0,i=0,j=0;
    for(i=0;i<=len_m-1;i++) tens=ms[i]<='9' ? tens*mb+ms[i]-'0' :tens*mb+ms[i]-'A'+10 ;//convert to the 10 base, low string bit map to high number bit
    while(tens>0) {ns[len_n++]=tens%nb>=10 ? tens%nb-10+'A' : tens%nb+'0';tens/=nb;}
    i=0;j=len_n-1;n_str[len_n]='\0';
    while(i<j){SWAP(n_str[i],n_str[j]);i++;j--;}
    return(0);
}
int main()
{
    while(scanf("%s%d%d",m_str,&m_base,&n_base)!=EOF)
    {
        m_to_n(m_base,m_str,n_base,n_str);
        if(strlen(n_str)<=7) printf("%7s\n",n_str);
        else printf("%7s\n","ERROR");
    }
    return(0);
}