zoj 1180 Self Numbers（大数，灵活题）

最新推荐文章于 2018-05-26 17:06:15 发布

ccyy-

最新推荐文章于 2018-05-26 17:06:15 发布

阅读量652

点赞数

CC 4.0 BY-SA版权

分类专栏：灵活题稍难

本文链接：https://blog.youkuaiyun.com/cacyth/article/details/45114965

灵活题同时被 2 个专栏收录

91 篇文章

订阅专栏

稍难

65 篇文章

订阅专栏

本文介绍了一种特殊的数——自数，由印度数学家D.R. Kaprekar在1949年发现。自数是一类无法通过其他数字与其各位数字之和（如abc+a+b+c）构成的数。文中提供了寻找1000000以内所有自数的两种算法实现，并展示了部分结果。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=180

找出1000000以内所有不能由其他数字（abc+a+b+c）组成的数

因为a+b+c+...最大也就是9+9+9+9+9+9=54，所以直接暴力即可

#include<iostream>
#include<algorithm>
#include<string>
#include<map>
#include<cmath>
#include<string.h>
#include<stdlib.h>
#include<cstdio>
#define ll long long
using namespace std;
int main(){
	for(int i=1;i<=1000000;++i){
		int u=0;
		for(int j=1;j<=54&&i-j>0;++j){
			int p=i-j,s=0;
			while(p>0){
				s+=p%10;
				p/=10;
			}
			if(s==j){
				u=1;
				break;
			}
		}
	  	if(u==0)
			cout<<i<<endl;
	}
	return 0;
}

不过上面的代码效率没有下面的高：

#include <cstring>
#include <cstdlib>
#include <cstdio>
using namespace std;
int hash[1000005];
void deal(int n){
    int rec = n;
    while( n > 0 ){
        int c = n % 10;
        n /= 10;
        rec += c;
    }
    hash[rec] = 1;
}
int main(){
    for(int i=1;i<=1000000;++i){
        if(!hash[i])
            printf("%d\n",i);
        deal(i);
    }
    return 0;
}