SGU 169 numbers 数学

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，规律是通过计算，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···ak-1

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：

#include"bits/stdc++.h"
using namespace std;
int k;
int main()
{
    while(scanf("%d",&k)!=EOF)
    {
        k--;
        int ans=1;
        if(!k) puts("8");
        else
        {
            if(k%3==0) ans+=2;
            if(k%6==0) ans++;
            pi(ans);
        }
    }
    return 0;
}