poj 1316 Self Numbers

最新推荐文章于 2022-03-22 19:03:50 发布

Ambition_Bit

最新推荐文章于 2022-03-22 19:03:50 发布

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本文介绍Self-Numbers的概念及其生成算法。通过印度数学家D.R. Kaprekar的定义，Self-Numbers是指那些不能通过将一个正整数与其各位数字之和相加而得到的数。文章提供了一个简单的C++程序，用于输出小于10000的所有Self-Numbers。

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Self Numbers

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 24659		Accepted: 13803

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

题目解析：

这个题目就是让求由自身加上其各位和得不到的数，直接暴力，1A

代码：

#include <iostream>
using namespace std;
const int maxn = 1e6;
int a[maxn];
int solve(int n)
{
    int ans = 0;
    while(n)
    {
        ans+=(n%10);
        n/=10;
    }
    return ans;
}

int main()
{
    for(int i=1;i<=10000;i++)
        a[i+solve(i)]=1;
    for(int i=1;i<=10000;i++)
        if(a[i]==0)
        cout<<i<<endl;
    return 0;
}