本文主要是介绍杭电OJ 1197. Specialized Four-Digit Numbers，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目描述：
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
... ...
思路：
题目大意，从10进制数2992开始，依次输出符合以下要求的十进制四位数：
该四位数在10进制、12进制和16进制的表示下，四位数之和均相同。
以十进制2992为例：
10进制 2992 四位数之和2+9+9+2=22
12进制 1894 四位数之和1+8+9+4=22
16进制 0bb0 四位数之和0+11+11+0=22
因此2992是我们要的十进制四位数。
题目重点考察进制的转换，用for循环取余得到转换后数字的每位数的数值大小，参考最基本的2进制转换。
实现（C++）：
#include <iostream>
using namespace std;int main(){for(int i=2992; i<10000; i++){int temp_dec=i;int temp_hex=i;int temp_12=i;int sum_dec=0;int sum_hex=0;int sum_12=0;for(int j=0; j<4; j++){sum_dec+=temp_dec%10;temp_dec/=10;}for(int k=0; k<4; k++){sum_hex+=temp_hex%16;temp_hex/=16;}if(sum_dec==sum_hex){for(int l=0; l<4; l++){sum_12+=temp_12%12;temp_12/=12;}if(sum_hex==sum_12)cout<<i<<endl;}}
}
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