______________________________贴一个进制转换的题吧________________________1197________

Specialized Four-Digit Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5291    Accepted Submission(s): 3823


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
 

英语比较难,实际上意思就是你把四位数中  10进制 12进制 16进制,他们各自的各个位数相加,如果他们的和相等的话,那么就输出这个数字...just so so...下面附上正确的代码

#include<stdio.h>
int main()
{
    int i,j,shi,shier,shiliu,x1,x2,x3,x4;
    for(i=2992;i<10000;i++)
    {
        shi=shier=shiliu=0;
        x1=i%10;
        x2=(i%100-x1)/10;
        x3=(i%1000-x2-x1)/100;
        x4=i/1000;
        shi=x1+x2+x3+x4;
        j=i;
        while(j>0)
        {
            shier=shier+j%12;
            j=j/12;
        }
        j=i;
        while(j>0)
        {
            shiliu=shiliu+j%16;
            j=j/16;
        }
        if(shi==shier&&shier==shiliu)
            printf("%d
",i);
    }
    return 0;
}