Codeforces Round #136 (Div. 2) / 221B Little Elephant and Numbers (数论)

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本文介绍了一道Codeforces上的编程题，任务是找出给定正整数x的所有除数中，有多少个除数在十进制表示下至少与x共享一个相同的数字。文章提供了完整的C++代码实现，并详细解释了算法思路。

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 x andd 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

1. 题目让你求d的个数！

2. i : 1~sqrt(x)，判断i和x/i

3. 数组开外面最好。。

完整代码：

/*30ms,0KB*/

#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>

char x[10];
bool has[10];///我错了，以后哪怕数组再小，数组声明一定要放在外面！

int main()
{
	scanf("%s", x);
	int len = strlen(x), xx = atoi(x);
	if (xx == 1)
		putchar('1');
	else
	{
		for (int i = 0; i < len; ++i)
			has[x[i] - '0'] = true;
		int count = 0, sqrtx = (int)sqrt(xx);
		for (int i = 1; i <= sqrtx; ++i)
			if (xx % i == 0)
			{
				memset(x, 0, sizeof(x));
				sprintf(x, "%d", i); //int转string：把i写到x中
				len = strlen(x);
				for (int j = 0; j < len; ++j)
					if (has[x[j] - '0'])
					{
						++count;
						break;
					}
				if (i * i != xx)
				{
					memset(x, 0, sizeof(x));
					sprintf(x, "%d", xx / i);
					len = strlen(x);
					for (int j = 0; j < len; ++j)
						if (has[x[j] - '0'])
						{
							++count;
							break;
						}
				}
			}
		printf("%d", count);
	}
	return 0;
}