HDU 1335 Basically Speaking

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

 

Basically Speaking

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K(Java/Others)
Total Submission(s): 962 Accepted Submission(s):437

Problem Description

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

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

It will support bases 2 through 16.

 

Input

The input for your prototype programwill consist of one base conversion per line. There will be threenumbers per line. The first number will be the number in the baseyou are converting from. The second number is the base you areconverting from. The third number is the base you are convertingto. There will be one or more blanks surrounding (on either sideof) the numbers. There are several lines of input and your programshould continue to read until the end of file isreached.

 

Output

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

 

Sample Input

1111000 210 1111000 2 16 2102101 3 10 2102101 3 15 12312 4 2 1A 15 2 123456710 16 ABCD 16 15

 

Sample Output

120 78 17657CA ERROR 11001 12D687 D071

 

Source

Mid-Central USA 1995

 

Recommend

Ignatius.L

 

题目分析：前面是一个7位的a进制数，转换成后面的进制输出，如果超过7位输出ERROR

分析：进制转换。

代码如下：
#include<iostream>
#include<string>
#include<math.h>
using namespace std;
int fun1(int,int);  //将一个10进制的数转换成要求的进制
intfun2(int);      //将一个已知进制的数转换成10进制
chars[8];          //保存输入的数据
intx[8];          //保存输出的数据
int main()
{
 int m,n,len; 
 int t,l,i;
 while(cin>>s)
 {
  len=strlen(s);
  cin>>m>>n;
  t=fun2(m);
  //printf("s=%s m=%d n=%dt=%d\n",s,m,n,t);
  if(fun1(t,n)==-1)cout<<" ERROR"<<endl;
  else
  {
   l=fun1(t,n);
   for(i=7-l;i>0;i--)
    cout<<'';
   for(i=l-1;i>=0;i--)
   {
    if(x[i]>=10)
     printf("%c",x[i]-10+'A');         //这个地方还是用printf好一点。。
    else
     printf("%d",x[i]);
   }
   cout<<'\n';
  }
  memset(s,'\0',sizeof(s));
  memset(x,'\0',sizeof(x));
 }
 return 0;
}
int fun1(int t,int m)
{
 int i=0;
 while(t)
 {
  x[i++]=t%m;
  t=t/m;
  if(i==8) return -1;
 }
 return i;
}
int fun2(int m)
{
 int len=strlen(s);
 int i,t=0;
 for(i=len-1;i>=0;i--)
 {
  if(s[i]>='A')
   s[i]=s[i]-'A'+10;
  else
   s[i]=s[i]-'0';
  t+=s[i]*pow(m,len-1-i);
 }
 return t;
}