POJ 2196 Specialized Four-Digit Numbers

Specialized Four-Digit Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7310   Accepted: 5343 Description 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 1893 12, and these digits also sum up to 21. But in hexadecimal 2991 is BAF 16, 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 BB0 16), 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 ... Source Pacific Northwest 2004

题目大意：

列出所有满足这样特性的四位数：其四位数字的和等于这个数以十六进制表示时四位数字的和，也等于这个数以十二进制表示时四位数字的和。

#include <iostream>
using namespace std;
int main()
{
	int n;
	for(n=2992;n<=9999;n++)
	{
		int sum1=0;
		int a1,a2,a3,a4;
		a1=n/1000;
		a2=n/100%10;	
		a3=n/10%10;
		a4=n%10;
		sum1=a1+a2+a3+a4;
		int sum2=0,m;
		m=n;
		while(m)
		{
			sum2=sum2+m%12;
			m=m/12;
		}
		int sum3=0;
		m=n;
		while(m)
		{
			sum3=sum3+m%16;
			m=m/16;
		}
		if(sum1==sum2&&sum2==sum3)
			cout<<n<<endl;
	}
	return 0;
}

附0MS代码

#include <cstdio>

int sum(int a, int b)
{
    int ret = 0;
    while (a)
    {
        ret += a % b;
        a /= b;
    }
    return ret;
}

int main()
{
    for (int i = 2992; i <= 9999; i++)
        if (sum(i, 10) == sum(i, 12) && sum(i, 10) == sum(i, 16))
            printf("%d\n", i);
    return 0;
}