HDU1128 UVA640 UVALive5326 ZOJ1180 Self Numbers【水题】

Self Numbers

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8398    Accepted Submission(s): 3683

Problem 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. 


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

 

Sample Output

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

 

Source

Mid-Central USA 1998

Regionals 1998 >> North America - Mid-Central USA

问题链接：HDU1128 UVA640 UVALive5326 ZOJ1180 Self Numbers

问题简述：（略）

问题分析：

　　这个问题的关键是设计一个标记数组，是与不是用逻辑真假区分开，根据标记输出就可以了。需要注意的是有可能重复标记，另外标记时下标有可能越界，需要加以限定。

程序说明：（略）

题记：

　　共同的功能封装到函数，想快的话加上修饰inline就可以了。

参考链接：（略）

AC的C++语言程序如下：

/* HDU1128 UVA640 UVALive5326 ZOJ1180 Self Numbers */#include <iostream>#include <string.h>#include <stdio.h>using namespace std;const int BASE = 10;const int N = 1000000;bool flag[N + 1];inline int getsumdigit(int n){    int sum = 0;    while(n) {        sum += n % BASE;        n /= BASE;    }    return sum;}int main(){    int k;    memset(flag, false, sizeof(flag));    for(int i=1; i<=N; i++) {        k = i + getsumdigit(i);        if(k <= N)            flag[k] = true;        if(!flag[i])            printf("%d\n", i);    }    return 0;}