（HDU）1335 -- Basically Speaking（基础和交流）

题目链接：http://vjudge.net/problem/HDU-1335

思路：把A进制的数字符串处理编程十进制形式，再转换成B进制。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <string>
#include <cstdlib>

using namespace std;

int main()  
{  
    char s[100];  
    int num[100],a,b;  
    while(~scanf("%s%d%d",s,&a,&b))  
    {  
        int len = strlen(s);  
        memset(num,0,sizeof(num));  
        for(int i = 0; i < len; i++)  
        {  
            if(s[i]>='0' && s[i] <= '9')  
                num[i] = s[i] - '0';  
            else  
                num[i] = s[i] - 'A' + 10;  
        }  
        int A,NumA;  
        NumA = 0,A = 1;  
        for(int i = len-1; i >= 0; i--)  
        {  
            NumA += (num[i]*A);  
            A *= a;  
        }  
        char ans[100];  
        int k = 0;  
        memset(ans,0,sizeof(ans));  
        while(NumA)  
        {  
            char temp = NumA % b + '0';  
            if(temp >='0' && temp<='9')  
                ans[k++] = temp;  
            else  
                ans[k++] = NumA%b+'A'-10;  
            NumA /= b;  
        }  
        ans[k] = '';  
        if(k > 7)  
            printf("  ERROR
");  
        else  
        {  
            for(int i = 0;i < 7-k;i++)  
                printf(" ");  
            for(int i = k-1; i >= 0; i--)  
                printf("%c",ans[i]);  
            printf("
");  
        }  
    }  
  
    return 0;  
}