POJ1546 & HDU 1335 & ZOJ 1334 Basically Speaking（进制转换）

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-digital 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. 
  

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.

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.

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

    120
     78
   1765
    7CA
  ERROR
  11001
 12D687
   D071

Mid-Central USA 1995

#include <cstdio>
#include <cstring>
#include <cmath>
char s[117], ans[117];
int a, b;
int k;
int len;
int check( char c)
{
    if(c == 'A')
        return 10;
    else if(c == 'B')
        return 11;
    else if(c == 'C')
        return 12;
    else if(c == 'D')
        return 13;
    else if(c == 'E')
        return 14;
    else if(c == 'F')
        return 15;
    else
        return c-'0';
}
char check1(int c)
{
    if(c == 10)
        return 'A';
    else if(c == 11)
        return 'B';
    else if(c == 12)
        return 'C';
    else if(c == 13)
        return 'D';
    else if(c == 14)
        return 'E';
    else if(c == 15)
        return 'F';
    else
        return c+'0';
}
int POW(int a, int len)
{
    int ss = 1;
    for(int i = 0; i < len; i++)
    {
        ss*=a;
    }
    return ss;
}

void slove()
{
    int r, tt = 0;
    for(int i = 0; i < len; i++)
    {
        tt += check(s[i])*pow(a*1.0,len-1-i);
    }
    //printf("tt:%d\n",tt);
    k = 0;
    while(tt)
    {
       r = tt%b;
        ans[k++] = check1(r);
        tt/=b;
    }
}

int main()
{
    while(scanf("%s",s)!=EOF)
    {
        len = strlen(s);
        scanf("%d %d",&a,&b);
        slove();
        if(k > 7)
        {
            printf("  ERROR\n");
            continue;
        }
        for(int i = 1; i <= 7-k; i++)
            printf(" ");
        for(int i = k-1; i >= 0; i--)
            printf("%c",ans[i]);
        printf("\n");
    }
    return 0;
}