zoj 2406 Specialized Four-Digit Numbers

Specialized Four-Digit Numbers
Time Limit: 2 Seconds     
Memory Limit: 65536 KB
Find and list all four-digit numbers in decimal notation that have the property that the sum of its four digits equals the sum of its digits when represented in hexadecimal (base 16) notation and also equals the sum of its digits when represented in duodecimal (base 12) notation.
For example, the number 2991 has the sum of (decimal) digits 2+9+9+1 = 21. Since 2991 = 1*1728 + 8*144 + 9*12 + 3, its duodecimal representation is 189312, and these digits also sum up to 21. But in hexadecimal 2991 is BAF16, and 11+10+15 = 36, so 2991 should be rejected by your program.
The next number (2992), however, has digits that sum to 22 in all three representations (including BB016), so 2992 should be on the listed output. (We don't want decimal numbers with fewer than four digits - excluding leading zeroes - so that 2992 is the first correct answer.)
 
Input
There is no input for this problem.
 
Output
Your output is to be 2992 and all larger four-digit numbers that satisfy the requirements (in strictly increasing order), each on a separate line with no leading or trailing blanks, ending with a new-line character. There are to be no blank lines in the output. The first few lines of the output are shown below.
 
Sample Input
There is no input for this problem.
 
Sample Output
2992 2993 2994 2995 2996 2997 2998 2999
分析:
(1)問題の意味は簡単で、いろいろな進数の下で数字の和が等しい4桁を求めることです.16進法と12進法の解法は,因子に分解して計算し,これらの数字を構成できる因子のすべての場合を先に求め,各因子の係数を組み合わせて加算した和を返す.

#include <stdio.h>

int i;
int base_duo[4] = {1,12,144,1728};
int base_hex[4] = {1,16,256,4096};

//  12       
int get_sum_duo(int num)
{
    int ex[4]={0},sum=0;
    for(i=3;i>=0;i--)
    {

        if(num>=base_duo[i])
        {
            num -= base_duo[i];
            ex[i]++;
            i++;
        }
    }
    for(i=0;i<4;i++)
    sum+=ex[i];
    return sum;
}

//  16       
int get_sum_hex(int num)
{
    int ex[4]={0},sum=0;
    for(i=3;i>=0;i--)
    {
        if(num>=base_hex[i])
        {
            num -= base_hex[i];
            ex[i]++;
            i++;
        }
    }
    for(i=0;i<4;i++)
    sum+=ex[i];
    return sum;
}
//  10    
int get_sum(int num)
{
    int sum=0;
    while(num>0)
    {
        sum += num%10;
        num /=10;
    }
    return sum;
}

int main()
{
    int j,tem;
    for(j=2992;j<10000;j++)
    {
        tem = get_sum(j);
        if(get_sum_duo(j)==tem&&get_sum_hex(j)==tem)
        printf("%d
",j);
    }
    return 0;
}