Codeforces Round #136 (Div. 2) B. Little Elephant and Numbers_b litle elephant and numbers-优快云博客

本文链接：https://blog.youkuaiyun.com/xzf19930910/article/details/21639953

本文探讨了一个关于数字x的特殊性质问题：找到所有能被x整除且与x至少共享一个数字的正整数d的数量。通过输入x（范围为1到10^9），读者将学习一种有效的方法来计算符合条件的d的数量，巧妙地避免了直接循环可能带来的超时问题。

The Little Elephant loves numbers.

He has a positive integer x. The Little Elephant wants to find the number of positive integers d, such that d is the divisor of x, and xand d have at least one common (the same) digit in their decimal representations.

Help the Little Elephant to find the described number.

Input

A single line contains a single integer x (1 ≤ x ≤ 109).

Output

In a single line print an integer — the answer to the problem.

Sample test(s)

input
1

output
1

input
10

output
2

题意：给你一个数n，求n一共有多少个其中至少包含一个他本身的数字的因子（一共有三行英文，我愣是看了半个小时还没看懂，惭愧啊！！！）。

另外直接循环会超时，可以同时遍历i和i/n，止痒就可以过啦~

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
int s[10];
int f(int x)
  {
      while(x)
      {
         if(s[x%10]==1)
            return 1;
          x=x/10;
        }
   return 0;
  }
int main()
{
  int n,i,ans;
  while(~scanf("%d",&n))
   {
       int x=n;
     memset(s,0,sizeof(s));
     ans=0;
     while(x)
      {
          s[x%10]=1;
          x=x/10;

      }
     for(i=1;i*i<=n;i++)
       if(n%i==0)
        {
            ans+=f(i);
         if(n/i!=i)
          ans+=f(n/i);
        }
    printf("%d\n",ans);
   }
	return 0;
}