poj1316 Self Numbers

Self Numbers

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 21683 Accepted: 12209

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

135792031425364 | |       <-- a lot more numbers |9903991499259927993899499960997199829993

Source

Mid-Central USA 1998

题意：让你求1-10000之间的自数，自数就是不能由一个数的本身加上这个数各个数位上的数字的和生成。

做法：先建立一个ss函数用来求本身以及各个数位上的和，然后用一个1-10000的循环算出各自的ss（），并把这些所产生的数组标记为1，然后再用一个1-10000的循环输出标记为0的数。

#include <iostream>#include <cstdio>#include <climits>#include <cstring>#include <cstdlib>#include <cmath>#include <vector>#include <queue>#include<map>#include <algorithm>#include<ctime>#define esp 1e-6#define LL long long#define inf 0x0f0f0f0fusing namespace std;int ss(int x){    int sum=0;    sum=x;    while(x!=0)    {        sum+=x%10;        x=x/10;    }    return sum;}int main(){    int num[20005];    int i;    memset(num,0,sizeof(num));    for(i=1;i<=10000;i++)    {        num[ss(i)]=1;    }    for(i=1;i<=10000;i++)        if(num[i]==0)        printf("%d\n",i);    return 0;}