ZOJ1180 Self Numbers 开始刷水题模式

最新推荐文章于 2019-08-31 17:24:15 发布

KBeam

最新推荐文章于 2019-08-31 17:24:15 发布

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本文链接：https://blog.youkuaiyun.com/oShenJingBing12/article/details/50965774

本文介绍了一种用于生成自数（self-numbers）的算法，并提供了一个C++实现示例。自数是由印度数学家D.R. Kaprekar于1949年发现的一类特殊整数，对于任意正整数n，定义d(n)为n加上其各位数字之和。如果一个数不能通过这种方式由小于它的数生成，则该数被称为自数。

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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.

就是对每个数取模判断这是个水题贴上代码

#include <iostream>
using namespace std;
int a[1000001]={0};
int d(int n)
{
	int sum=n;
	while(n)
	{
		sum+=n%10;
		n/=10;
	}
	return sum;
}
int main()
{
	int i,t;
	for(int i=1;i<=1000000;i++)
	{
		t=d(i);
		if (t <= 1000000)
		{
		a[t]=1;
		}
	}
	for(int i=1;i<=1000000;i++)
	{
		if(!a[i])
		{
			cout << i<<endl;
		}
	}
	return 0;
}