[ZOJ 3622] Magic Number

Magic Number

---

Time Limit: 2 Seconds      Memory Limit: 32768 KB

---

A positive number y is called magic number if for every positive integer x it satisfies that put y to the right of x, which will form a new integer z, z mod y = 0.

Input

The input has multiple cases, each case contains two positve integers m, n(1 <= m <= n <= 2^31-1), proceed to the end of file.

Output

For each case, output the total number of magic numbers between m and n(m, n inclusively).

Sample Input

1 1
1 10

Sample Output

1
4

分析：
设y有k位数，则需要满足 (x*10^k+y)%y==0
--> (x*10^k%y+0)%y==0
--> x*10^k%y==0
--> 由于x是任意的，假设x为质数，则需要满足(10^k)%y==0
直接判断会超时，先打表。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <cmath>
#include <string>
using namespace std;
#define N 100010

int n,m;
int len;
int num[N]=
{,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
};

int main()
{
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        if(m>n) swap(m,n);
        int cnt=;
        for(int i=;i<;i++)
        {
            if(m<=num[i] && num[i]<=n) cnt++;
        }
        cout<<cnt<<endl;
    }
    return ;
}