【acm1197】 Specialized Four-Digit Numbers

题目：
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 = 11728 + 8144 + 912 + 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***

代码：

#include<iostream>
using namespace std;
int main(){
	for(int i=1000;i<9999;i++){
		int sum1=0,sum2=0,sum3=0;
		int j=i;
		int k=i;
		int l=i;
		while(l){
			sum1+=l%10;
			l=l/10; 
		
		}
	
		while(j){
			sum2+=j%16;
			j=j/16;
		}
		while(k){
			sum3+=k%12;
			k=k/12;
		}
		//	cout<<sum1<<" "<<sum2<<" "<<sum3<<endl;
		if(sum1==sum2&&sum2==sum3){
			cout<<i<<endl;
		}
	
		 
			}
	return 0;
}

分析：这一题还是有点东西的，题目问的意思是输出所有的四位数，这个四位数的十进制、十二进制、十六进制他们的各个位数相加和必须相等。那么我们其实要知道的是一个十进制的数如何转换成十六进制（或者十二进制）的。就是用这个十进制的数不断的除以16，所得的余数倒序就能得到。那我们用这个十进制数不断对十六取余所得到的和与十进制各个数得到的和相比，若相等则证明符合要求。整理要注意判断三个数相等的格式。