Self Numbers
Time Limit: 1000MS |
Memory Limit: 10000K |
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Total Submissions: 20966 |
Accepted: 11775 |
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.
Input
No input for this problem.
Output
Write a program to output all positive self-numbers less than 10000 in increasing order, one per line.
Sample Input
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
题意:让一个数加上其各位上的和,组成一个新数。输出不能够被这样组成的数。
原谅我在回家等火车的时候闲来无事切切水题………………
AC代码如下:
#include<cstdio>
#include<cstring>
using namespace std;
int num[20010];
int main()
{ int i,j,k;
for(i=1;i<=10000;i++)
{ num[i/1000+i/100%10+i/10%10+i%10+i]=1;
if(!num[i])
printf("%d\n",i);
}
}
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