POJ2196 HDU1197 ZOJ2405 Specialized Four-Digit Numbers【进制＋水题】

Specialized Four-Digit Numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8413		Accepted: 6126

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.
Sample Input

There is no input for this problem

Sample Output

2992
2993
2994
2995
2996
2997
2998
2999
...

Source
Pacific Northwest 2004

问题链接：POJ2196 HDU1197 ZOJ2405 Specialized Four-Digit Numbers


问题分析：四位数的１０进制数字之和，１２进制数字之和，１６进制数字之和都相同的输出。同时题目中给出了其值的范围。

程序说明：（略）

题记：（略）

参考链接：（略）

AC的C语言程序如下：

/* POJ2196 HDU1197 ZOJ2405 Specialized Four-Digit Numbers */

#include <stdio.h>

int sum_digits(int n, int base)
{
int sum = 0;

while(n) {
sum += n % base;
n /= base;
}

return sum;
}

int main(void)
{
int d10, i;

for(i=2292; i<=9999; i++) {
d10 = sum_digits(i, 10);
if(sum_digits(i, 12) == d10 && sum_digits(i, 16) == d10)
printf("%d\n", i);
}

return 0;
}