169.Numbers
Let us call P(n) - the product of all digits of number n (in decimal notation).
For example, P(1243)=1*2*4*3=24; P(198501243)=0.
Let us call n to be a good number, if (p(n)<>0) and (n mod P(n)=0).
Let us call n to be a perfect number, if both n and n+1 are good numbers.
You are to write a program, which, given the number K, counts all such
numbers n that n is perfect and n contains exactly K digits in decimal notation.
Input
Only one number K (1<=K<=1000000) is written in input.
Output
Output the total number of perfect k-digit numbers.
Sample test(s)
Input
1
Output
8
题意：一道很有意思的题，看到这个题的时候就很容易想到这道题是有规律的，但最后也是查找题解才找到的
规律就是通过计算，得到k位数除个位数之外，所有的非个位数都为1，所以只需看最后一位的情况与前面组成的数能构成多少perfect number
假设n的各个位数为a1，a2···ak，那么n+1的各个位数为a1，a2···ak+1
因为要求n mod P（n）=0，所以有n=s1*a1*a2···*ak，n+1=s2*a1*a2···*（ak+1）
在此处s1与s2因为n mod P（n）=0所以肯定是整数（因为n一定是P（n）的m倍），所以有1=[s2*(ak+1)-s1*ak]*a1*a2···
由此可得出a1，a2···肯定为1
设个位数为x
x=1, perfect numbers
x=2, 当6|(k-1)时是perfect numbers
x=3，不是
x=4，不是
x=5，当3|(k-1)时是perfect numbers
x=6，当6|(k-1)时是perfect numbers
x=7，不是
x=8，不是
代码：
1 //#include"bits/stdc++.h"
2 #include<sstream>
3 #include<iomanip>
4 #include"cstdio"
5 #include"map"
6 #include"set"
7 #include"cmath"
8 #include"queue"
9 #include"vector"
10 #include"string"
11 #include"cstring"
12 #include"time.h"
13 #include"iostream"
14 #include"stdlib.h"
15 #include"algorithm"
16 #define db double
17 #define ll long long
18 #define vec vectr<ll>
19 #define mt vectr<vec>
20 #define ci(x) scanf("%d",&x)
21 #define cd(x) scanf("%lf",&x)
22 #define cl(x) scanf("%lld",&x)
23 #define pi(x) printf("%d\n",x)
24 #define pd(x) printf("%f\n",x)
25 #define pl(x) printf("%lld\n",x)
26 //#define rep(i, x, y) for(int i=x;i<=y;i++)
27 #define rep(i, n) for(int i=0;i<n;i++)
28 const int N = 1e4+ 5;
29 const int mod = 1e9 + 7;
30 const int MOD = mod - 1;
31 const int inf = 0x3f3f3f3f;
32 const db PI = acos(-1.0);
33 const db eps = 1e-10;
34 using namespace std;
35 int k;
36 int main()
37 {
38 while(scanf("%d",&k)!=EOF)
39 {
40 k--;
41 int ans=1;
42 if(!k) puts("8");
43 else
44 {
45 if(k%3==0) ans+=2;
46 if(k%6==0) ans++;
47 pi(ans);
48 }
49 }
50 return 0;
51 }
原文地址：https://www.cnblogs.com/mj-liylho/p/8962320.html