spoj MCUR98（输出数列题）

MCUR98 - Self Numbers

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Background

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 ben 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.

Problem

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

Input

There is no input.

Output

All positive self-numbers less than 1000000 in increasing order, one per line.

题解：数字题，按条件把是self numbers的标记，最后把不是得输出就好了，

代码：

#include<bits/stdc++.h>#define MAXN 1000005int vis[MAXN];int f(int x){    int ans=x;    while(x)    {        ans+=x%10;        x/=10;    }    return ans;}int main(){    int mark=f(1);vis[mark]=1;    for(int i=2;i<=1000000;i++)    {        if(i%10==0)        {            mark=f(i);            vis[mark]=1;        }        else        {        mark+=2;        vis[mark]=1;}    }    for(int i=1;i<1000000;i++)    {        if(!vis[i])            printf("%d ",i);    }    return 0;}