UVALive3199 Specialized Four-Digit Numbers【进制】

Find and list numbers in decimal notation that have the property that the sum of its digits equals thesum of its digits when represented in hexadecimal (base 16) notation and also equals the sum of itsdigits 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 upto 21. But in hexadecimal 2991 is BAF16, and 11 + 10 + 15 = 36, so 2991 should be rejected by yourprogram.

  The next number (2992), however, has digits that sum to 22 in all three representations (includingBB016), so 2992 should be on the listed output. (We don’t want decimal numbers with fewer than fourdigits — excluding leading zeroes — so that 2992 is the first correct answer.)

Input

  The input file will contain several possible cases. Each test case consists of a line containing a pair ofpositive integers, separated by a space.

  Input will terminate with ‘0 0’ in a line byself. This line should not be processed.

Output

  For each input case, your output is to be all the numbers between the couple of integers (both included),that satisfy the requirements (in strictly increasing order), each on a separate line with no leading ortrailing blanks, ending with a new-line character.

  Separate the output for consecutive cases by a single blank line.

Sample Input

2900 3000

0 0

Sample Output

2992

2993

2994

2995

2996

2997

2998

2999

Regionals 2004 >> North America - Pacific Northwest

问题链接：UVALive3199 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。需要注意的是输入数据格式和输出数据格式。

程序说明：（略）

题记：（略）

参考链接：（略）

AC的C语言程序如下：

/* UVALive3199 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 start, end, d10, flag, i;    flag = 0;    while(~scanf("%d%d", &start, &end) && (start || end)) {        if(flag)            printf("\n");        flag = 1;        for(i=start; i<=end; i++) {            d10 = sum_digits(i, 10);            if(sum_digits(i, 12) == d10 && sum_digits(i, 16) == d10)                printf("%d\n", i);        }    }    return 0;}