关于HDOJ 1197题(进制转换) 的理解(C/C++)

题目描述

                                              Specialized Four-Digit Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10438 Accepted Submission(s): 7566

Problem 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 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



Source
Pacific Northwest 2004

原题链接

More info:Question

Accepted代码

#include<iostream>
using namespace std;
/*input 10进制数a，
*进制b
*output  b进制下该数的各项和*/
int count_(int a, int b) {
    int x=a,res = 0;
    while ( x >= b ) {
        res += (x%b);
        x /= b;
    }
    res += x;
    return res;
}
int main(){
    for (int i = 2992; i < 10000; i++)
        if (count_(i, 10) == count_(i, 12) && count_(i, 12) == count_(i, 16) )
            cout << i << endl;
    return 0;
}

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参考博客 weirdo_coder