H - Specialized Four-Digit Numbers
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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
题意分析：
查找并列出十进制表示法中所有四位数的数字，这些数字的四位数之和等于十六进制（以16为基数）表示时的位数之和，以及以双十进制（以12为基数）表示时的位数之和。
例如，数字2991的总和是（十进制）数字2+9+9+1=21。由于2991=1*1728+8*144+9*12+3，它的双字母表示是1893（12），这些数字的总和也是21。但是在十六进制中2991是BAF16，11+10+15=36，所以2991应该被程序拒绝。
然而，下一个数字（2992）在所有三种表示法（包括BB016）中的数字总和为22，因此2992应该在列出的输出中。（我们不希望十进制数少于四位数（不包括前导零），因此2992是第一个正确答案。）
最后输出符合要求且大于2992的数。
解题思路：
这道题本身是一个考察进制转换的题。要求的是得到一个数对应的10进制、12进制、16进制的每位数的加和。本身对于三种进制计算和的方法是通用的。计算出加和之后比较是否一直，如果一致则输出。
编码：
#include <bits/stdc++.h> using namespace std; int sum10(int a) { int ans = 0; while (a > 0) { ans += a % 10; a = a / 10; } return ans; } int sum12(int a) { int ans = 0; while (a > 0) { ans += a % 12; a = a / 12; } return ans; } int sum16(int a) { int ans = 0; while (a > 0) { ans += a % 16; a = a / 16; } return ans; } int main() { for(int i = 2992; i < 10000; i++) { int num10 = sum10(i); int num12 = sum12(i); int num16 = sum16(i); if (num10 == num12 && num12 == num16) { cout << i << endl; } } return 0; }