Specialized Four-Digit Numbers

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本文探讨了如何找出所有四位数，这些数在十进制、十二进制和十六进制下数字之和相等的问题。通过编写简单的函数来计算不同进制下数字的和，并验证这些和是否相等。

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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

题解：写几个简单函数就行了，水题一道；

代码：

#include <iostream>

using namespace std;

int ten(int m)
{
    int sum=0;
    while(m>0)
    {
        sum+=m%10;
        m/=10;
    }
    return sum;
}//求十进制所有位数和;

int twelve(int m)
{
    int sum=0;
    while(m>0)
    {
        sum+=m%12;
        m/=12;
    }
    return sum;
}//求十二进制所有位数和;

int sixteen(int m)
{
    int sum=0;
    while(m>0)
    {
        sum+=m%16;
        m/=16;
    }
    return sum;
}//求十六进制所有位数和;

int equal(int a,int b,int c)
{
    if(a==b && b==c) return 1;
    else return 0;
}//判断三个数是否相等;

int main()
{
    for(int i=2992;i<=9999;i++)
    {
        if(equal(ten(i),twelve(i),sixteen(i))) cout<<i<<endl;
    }
    return 0;
}