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.)
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.
Sample Input
There is no input for this problem
Sample Output
2992 2993 2994 2995 2996 2997 2998 2999 ...
Source
汉化：求出满足分别以十进制、十二进制、十六进制表示时数字相加之和相等的四位数。
就是红果果的暴搜。
program ACRush;
var n:longint;
a:array[0..18]of longint;
procedure f(x:longint);
var temp:longint;
begin
temp:=n;
while temp<>0 do
begin
a[x]:=a[x]+temp mod x;
temp:=temp div x;
end;
end;
begin
for n:=2992 to 9999 do
begin
a[16]:=0;
a[12]:=0;
a[10]:=0;
f(12);
f(16);
f(10);
if (a[16]=a[10]) and (a[10]=a[12]) then writeln(n);
end;
end.