开发者社区> 谙忆> 正文
阿里云
为了无法计算的价值
打开APP
阿里云APP内打开

HDOJ 1197 Specialized Four-Digit Numbers

简介: Problem Description Find and list all four-digit numbers in decimal notation that have the property that the sum of its four digits equals...
+关注继续查看

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

题意:输入一个十进制数,看这个十进制的数的各个位数上的和,和它的12进制和16进制上的各个位数上的和是不是相等,相等就输出。
(十进制的数范围[2992-9999])

public class Main{
    public static void main(String[] args) {
        for(int i=2992;i<=9999;i++){
            int a = i/1000+(i/100)%10+(i/10)%10+i%10;
            String str = Integer.toString(i, 12);
            //System.out.println(str);
            int b = 0;
            for(int j=0;j<str.length();j++){
                switch(str.charAt(j)){
                case 'a':b=b+10;break;
                case 'b':b=b+11;break;
                default:b=b+(new Integer(str.charAt(j)+""));
                }
            }
            if(a!=b){
                continue;
            }
            int c=0;
            str = Integer.toString(i, 16);
            //System.out.println(str);
            for(int j=0;j<str.length();j++){
                switch(str.charAt(j)){
                case 'a':c=c+10;break;
                case 'b':c=c+11;break;
                case 'c':c=c+12;break;
                case 'd':c=c+13;break;
                case 'e':c=c+14;break;
                case 'f':c=c+15;break;
                default:c=c+(new Integer(str.charAt(j)+""));
                }
            }
            if(a!=c){
                continue;
            }
            System.out.println(i);
        }

    }

}

版权声明:本文内容由阿里云实名注册用户自发贡献,版权归原作者所有,阿里云开发者社区不拥有其著作权,亦不承担相应法律责任。具体规则请查看《阿里云开发者社区用户服务协议》和《阿里云开发者社区知识产权保护指引》。如果您发现本社区中有涉嫌抄袭的内容,填写侵权投诉表单进行举报,一经查实,本社区将立刻删除涉嫌侵权内容。

相关文章
HDOJ 1197 Specialized Four-Digit Numbers
HDOJ 1197 Specialized Four-Digit Numbers
0 0
HDOJ 1061 Rightmost Digit(循环问题)
HDOJ 1061 Rightmost Digit(循环问题)
0 0
HDOJ 2058 The sum problem
HDOJ 2058 The sum problem
0 0
HDOJ 1001Sum Problem
HDOJ 1001Sum Problem
0 0
HDOJ(HDU) 1720 A+B Coming(进制)
HDOJ(HDU) 1720 A+B Coming(进制)
0 0
HDOJ 2029 Palindromes _easy version(回文串)
HDOJ 2029 Palindromes _easy version(回文串)
0 0
HDOJ 2035 人见人爱A^B
HDOJ 2035 人见人爱A^B
0 0
HDOJ-1001 Sum Problem
Problem Description Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge).
910 0
+关注
谙忆
GitHub: https://github.com/chenhaoxiang
文章
问答
文章排行榜
最热
最新
相关电子书
更多
低代码开发师(初级)实战教程
立即下载
阿里巴巴DevOps 最佳实践手册
立即下载
冬季实战营第三期:MySQL数据库进阶实战
立即下载