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

---

Author: QU, Zhe
Contest: ZOJ Monthly, July 2012
//下午月赛的第一题

//开始少了2个数，郁闷了

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <cmath>
#include <stack>
using namespace std;
int dp[100];
int nu;
void set()
{   dp[nu++]=1;
    int r[5]={1,2,4,5,8};
    int i,j;
    for(i=10;i<=1000000000;i*=10)
    {
        for(j=0;j<5;j++)
         if(i%r[j]==0)
          dp[nu++]=i/r[j];
    }
}
int main()
{
    set();
    int i,k;
    sort(dp,dp+nu);

    dp[nu++]=1250000000;
    dp[nu++]=2000000000;
    int n,m;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        for(k=i=0;i<nu;i++)
          if(m<=dp[i]&&n>=dp[i])
            k++;
        printf("%d\n",k);
    }
    return 0;
}