HDU Basiclly Speaking

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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 101111000  2 162102101  3 102102101  3 15  12312  4  2     1A 15  21234567 10 16   ABCD 16 15

Sample Output

    120     78   1765    7CA  ERROR  11001 12D687   D071
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/*基本思路：就是将数据先转换为十进制，然后再装换为，别的进制，刚开始，想的太麻烦，需要注意的是，当数字大于10的时候，如何转换为A,B....D所有的进制转换，转换为十进制，之后在进行转换，是最简单的；*/#include<stdio.h>#include<string.h>#include<math.h>int main(){char a[1001],c[1001];int d1,d2,tt,i,j,k,le,len,res,n;while(scanf("%s%d%d",a,&d1,&d2)!=EOF){len = strlen(a);res = 0;for(i=0,j=len-1;i<len;i++,j--){if(a[i]>='0'&&a[i]<='9')tt = a[i] - '0';elsett = a[i] -'A' + 10;res += tt*(int)pow(d1,j);}n=0;while(res){k = res % d2;if(k>=0&&k<=9)c[n++] = k+48;else if(k>=10&&k<=15)c[n++] = k+55;// 55 = 65 - 10;res /= d2;}c[n]='\0';//在没有对字符数组c清零的情况下，要在末尾加上结束标记；le=strlen(c);if(n>7)printf("  ERROR\n");else{for(j=1;j<=7-le;j++)printf(" ");for(i=n-1;i>=0;i--)printf("%c",c[i]);printf("\n");}}return 0;}

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