sicily 1014. Specialized FourDig

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 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.

 

从2992开始，输出所有10进制、12进制、16进制下各位数之和相等的四位数

先用最笨的方法写了一个答案，成功AC，加了注释

 

#include<stdio.h>

int main()
{
    int num;
    int a, b, c, d;        /* 10进制的各位数字 */ 
    int e, f, g, h;        /* 16进制的各位数字 */
    int i, j, k, l;        /* 12进制的各位数字 */
    
    for ( num = 2992; num <= 9999; num++ )
    {
        /* 用求余得到10进制的各位数字 */
        a = num % 10;
        b = ( num - a )% 100  / 10 ;
        c = ( num - a - b ) % 1000 / 100;
        d = (num - a - b - c) / 1000 ;
        
        /* 求16进制的各位数字 */
        e = num % 16;
        f = ( num - e )% 256  / 16  ;
        g = ( num - e - f ) % 4096 / 256;
        h = (num - e - f - g) / 4096 ;
        
        /* 求12进制的各位数字 */
        i = num % 12;
        j = ( num - i )% 144  / 12 ;
        k = ( num - i - j ) % 1728 / 144;
        l = (num - i - j - k) / 1728 ;
        
        /* 如果三个进制的各位数字之和相等，输出 */
        if ((a+b+c+d == e+f+g+h) && (a+b+c+d == i+j+k+l))
           printf("%d\n", num);
    }
    
    return 0;
}

写的时候觉得其实可以不用这么麻烦，因为int型做除法的时候会自动取整，但是因为是实验课上随便写写的，也没有认真优化。

下课之后上网参考了一下前辈的答案，改成了下面这样，简洁了不少，也AC了

#include<stdio.h>

int main()
{
    int num;
    int dectotal;        /* 10进制的各位数字 */ 
    int doztotal;        /* 12进制的各位数字 */
    int hextotal;        /* 16进制的各位数字 */
    
    for ( num = 2992; num <= 9999; num++ )
    {
        /* 计算各进制的各位数字之和 */
        dectotal = (num % 10) + (num % 100)/10 + (num % 1000)/100 + num / 1000;
        doztotal = (num % 12) + (num % 144)/12 + (num % 1728)/144 + num / 1728;
        hextotal = (num % 16) + (num % 256)/16 + (num % 4096)/256 + num / 4096;
        
        /* 输出三个和相等的数 */ 
        if( dectotal == doztotal && doztotal == hextotal )
        printf("%d\n", num);
    }
    
    return 0;
}

另外发现一个问题，实验室电脑用的是DEVCPP，懒得设置了所以直接每个程序都用system("pause")避免一闪而过，课上提交的时候西西里不堪重负，都是waiting，回来一看发现RF了，还好程序比较简单，主体内容只是简单的数学运算，很明显那个RF就是system("pause") 才知道原来OJ是不许用这个函数的，看来以后在实验室调试完提交还得先删掉这行和相应的#include<stdlib.h>