• Self Numbers[HDU1128]


    Self Numbers

    Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 5392    Accepted Submission(s): 2388

     

    Problem Description
    In 1949 the Indian mathematician D.R. Kaprekar discovered a class of numbers called self-numbers. For any positive integer n, define d(n) to be n plus the sum of the digits of n. (The d stands for digitadition, a term coined by Kaprekar.) For example, d(75) = 75 + 7 + 5 = 87. Given any positive integer n as a starting point, you can construct the infinite increasing sequence of integers n, d(n), d(d(n)), d(d(d(n))), .... For example, if you start with 33, the next number is 33 + 3 + 3 = 39, the next is 39 + 3 + 9 = 51, the next is 51 + 5 + 1 = 57, and so you generate the sequence
    33, 39, 51, 57, 69, 84, 96, 111, 114, 120, 123, 129, 141, ...

     

    The number n is called a generator of d(n). In the sequence above, 33 is a generator of 39, 39 is a generator of 51, 51 is a generator of 57, and so on. Some numbers have more than one generator: for example, 101 has two generators, 91 and 100. A number with no generators is a self-number. There are thirteen self-numbers less than 100: 1, 3, 5, 7, 9, 20, 31, 42, 53, 64, 75, 86, and 97.

     


    Write a program to output all positive self-numbers less than or equal 1000000 in increasing order, one per line.

     

     

     

    Sample Output
    1
    3
    5
    7
    9
    20
    31
    42
    53
    64
    |
    | <-- a lot more numbers
    |
    9903
    9914
    9925
    9927
    9938
    9949
    9960
    9971
    9982
    9993
    |
    |
    |
     

     

    Source
    Mid-Central USA 1998
     

     

    Recommend
    Eddy

    #include<stdio.h>
    #include<string.h>
    bool f[1000100];
    int main()
    {
    	int i;
    	memset(f,false,sizeof(f));
    	for (i=1;i<=1000000;i++)
    	{
    		int tmp=i,M=i;
    		while (tmp)
    		{
    			M+=tmp%10;
    			tmp/=10;
    		}
    		f[M]=true;
    	}
    	for (i=1;i<=1000000;i++)
    		if (!f[i]) printf("%d
    ",i);
    	return 0;
    }
    

     

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  • 原文地址:https://www.cnblogs.com/dramstadt/p/3260874.html
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