HDU 1335 Basically Speaking(进制转换问题)

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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

120
78
1765
7CA
ERROR
11001
12D687
D071
题目大意：输入一个 数（字符串） 输入一个进制a（原数的进制）,再输入一个进制b 要求把这个数以进制b输出。如果答案超过7个单位，输出ERROR,右对齐.思路：就是把这个字符串转化为数，再转化为字符串。这道题，我自己写的没过，也不知为什么感觉自己测试数据都过了（就是一直WA，WA15次之久）但我看见一个自认为很优秀的代码。上代码。

#include <iostream>
#include <stdio.h>
#include <cstdio>
#include <cstring>
using namespace std;

int main()
{
int se,re;
char x;
char s[20],r[20];
while(scanf("%s%d%d",s,&se,&re)>0)
{
int l=strlen(s),k=0,perc;
long long sum=0;
for(int i=0; i<l; i++)
{
if(s[i]>='0'&&s[i]<='9')        //把进制转化为数这点很巧妙
sum=sum*se+s[i]-'0';
else if(s[i]>='A'&&s[i]<='Z')
sum=sum*se+s[i]-'A'+10;
}
while(sum>0)
{
perc=sum%re;
if(perc>=10)
r[k++]=perc-10+'A';
else
r[k++]=perc+'0';
sum=sum/re;
}
for(int i=0;i<k/2;i++)
{
x=r[i];
r[i]=r[k-i-1];
r[k-i-1]=x;
}
if(k>7)
cout<<"  ERROR"<<endl;
else
printf("%7s\n",r);
memset(s,0,sizeof(s));
memset(r,0,sizeof(r));
/*这个事是作者写的 ，我修改了一下。
{
for(int j=0; j<7-k; j++)
cout<<" ";
for(int j=k-1; j>=0; j--)
cout<<r[j];
cout<<endl;
}
*/
}
return 0;
}

但我的代码一直WA，其实思路一样的，只是实现方式问题。如果有路过的，虚心求教！

#include<stdio.h>
#include<string.h>
#include<math.h>
int main()
{
long int n,d;
char s[200],len;
char c[200],x;
int t;
long int i,j,k;
long int sum,ss;
memset(s,0,sizeof(s));
memset(c,0,sizeof(c));
while(scanf("%s%ld%ld",s,&n,&d)>0)
{
len=strlen(s);
j=0;
sum=0;
for(i=len-1;i>=0;i--)
{
if('0'<=s[i]&&s[i]<='9')
sum=sum+(s[i]-'0')*pow(n,j++);
else
{
switch(s[i])
{
case 'A':sum=sum+10*(long int)pow(n,j++);break;
case 'B':sum=sum+11*(long int)pow(n,j++);break;
case 'C':sum=sum+12*(long int)pow(n,j++);break;
case 'D':sum=sum+13*(long int)pow(n,j++);break;
case 'E':sum=sum+14*(long int)pow(n,j++);break;
case 'F':sum=sum+15*(long int)pow(n,j++);break;
}
}
}
if(sum==0) {
printf("      0\n");  memset(s,0,sizeof(s));
memset(c,0,sizeof(c));continue;
}
ss=sum;
k=0;
while(ss)
{
t=ss%d;
if(0<=t&&t<=9)
c[k++]=t+'0';
if(10<=t&&t<=15)
c[k++]=55+t;
ss=ss/d;
}
for(i=0;i<k/2;i++)
{
x=c[i];
c[i]=c[k-i-1];
c[k-i-1]=x;
}
/*
if(k<8)
{
for(j=0;j<7-k;j++)
printf(" ");
k=k-1;
for(;k>=0;k--)
printf("%c",c[k]);
}
*/
if(k<8)
printf("%7s",c);
else
printf("  ERROR");
printf("\n");
memset(s,0,sizeof(s));
memset(c,0,sizeof(c));
}
return 0;
}