POJ-2196 Specialized Four-Digit Numbers-10,12,16進位制的各個位數相加彼此相等的數

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.) 

There is no input for this problem

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.

There is no input for this problem

2992
2993
2994
2995
2996
2997
2998
2999
...

Pacific Northwest 2004

//輸出從2992-9999之前滿足10,12,16進位制的各個位數相加彼此相等的數
#include <iostream>
#include <stdio.h>
#include <cmath>
using namespace std;

int cal(int n,int b)//引數分別表示要進行計算的數和進位制，這個函式用來計算各個位數之和
{
    int ans=0,a;
    a=n;
    while(a>0)
    {
        ans+=(a%b);
        a/=b;
    }
    return ans;
}

int main()
{
    int i;
    for(i=2992; i<=9999; ++i)
        if(cal(i,10)==cal(i,12)&&cal(i,12)==cal(i,16)&&cal(i,10)==cal(i,16))
            cout<<i<<endl;
    return 0;
}