1．鏈接地址

?????https://vjudge.net/problem/POJ-1316

2．問題描述

?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.?

輸入樣例

?None

輸出樣例

1 3 5 7 9 20 31 42 53 64|| <-- a lot more numbers| 9903 9914 9925 9927 9938 9949 9960 9971 9982 9993

3．解題思路

?采取了打表后再判斷的方法，不過在另一篇題解發(fā)現(xiàn)了一種數(shù)學解的方法，還沒看太懂先po上來再研究看看https://blog.csdn.net/xinghongduo/article/details/5805240

4．算法實現(xiàn)源代碼

#include<iostream> #include<cstring> using namespace std;int main() {int num[10001];int i,j,k,m;memset(num,0,sizeof(num));for( i= 0;i < 10; i++){for(j = 0;j < 10; j++){for(k = 0;k < 10; k++){for(m = 0;m < 10; m++){if((i+j+k+m+i*1000+j*100+k*10+m) < 10000)num[i+j+k+m+i*1000+j*100+k*10+m] = 1;}}}}for(i = 1;i < 10000;i++)if(num[i] == 0)printf("%d\n",i); }

?

轉(zhuǎn)載于:https://www.cnblogs.com/KasenBob/p/11163601.html

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