zoukankan      html  css  js  c++  java
  • Specialized Four-Digit Numbers

    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 189312, 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
    ...
    
    
    
    #include <iostream>
    #include <cmath>
    using namespace std;
    int main()
    {int n;
    	for(n=2992;n<9999;n++)
    	{int a=0,b=0,c=0,a1,a2;
    	a1=a2=n;
    	a=n/1000+n/100-n/1000*10+n%100/10+n%10;
    	while(a1>=12)
    	{b=b+a1%12;
    	a1=a1/12;}
    	b=b+a1;
    	while(a2>=16)
    	{c=c+a2%16;a2=a2/16;}
    	c=c+a2;
    	
    	if(a==b&&b==c)
    		cout<<n<<endl;
    	
    	
    	
    	
    	
    	
    	}
    return 0;
    
    
    }

  • 相关阅读:
    javascript功能插件大集合
    基于Swiper插件的简单轮播图的实现
    LeetCode24. 两两交换链表中的节点
    530. 二叉搜索树的最小绝对差
    416. 分割等和子集
    GAN ——Generative Adversarial Network 理解与推导(一)
    面试题 02.08. 环路检测(返回环路开头节点)
    141.环形链表-LeetCode
    357. 计算各个位数不同的数字个数 ——LeetCode
    LSTM的理解
  • 原文地址:https://www.cnblogs.com/lengxia/p/4387852.html
Copyright © 2011-2022 走看看