poj magic number

                          Problem H

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 24   Accepted Submission(s) : 8

Problem Description

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
怎样打表
这些数都是有规律的从100开始后面都只是加了个0而已，所以可以很快找出来。pow（2,31）=2147483648；

#include<stdio.h>
#include<math.h>
int main()
{
      long long int a[55]={1,2,5,10,20,25,50,100,125,200,250,500,1000,1250,2000,2500,5000,10000,12500,20000,25000,50000,100000,125000,200000,250000,500000,1000000,1250000,2000000,2500000,5000000,10000000,12500000,20000000, 25000000, 50000000,100000000 ,125000000, 200000000, 250000000, 500000000,1000000000 ,1250000000, 2000000000, 2500000000,5000000000,10000000000 ,12500000000, 20000000000};
        long long m,n,i,t;//不能用I64，wa了用long long，而且vc编译器是非法的。
      while(~scanf("%lld%lld",&m,&n))
      {
          t=0;
          for(i=0;i<55;i++)
          {
              if(a[i]>=m&&a[i]<=n)
                t++;
              if(a[i]>n)
                  break;
          }
          printf("%lld
",t);
      }
   return 0;
}

把数组的那些数是怎样弄出来的代码：

#include<stdio.h>
#include<math.h>
int main()
{
    __int64 x,y,t=0,m,k;
    while(scanf("%I64d%I64d",&x,&y)!=EOF)
    {
        int i,j;
        k=0;
        for(i=1;i<pow(2,31);i++)
        {
            t=0;
            m=i;
            while(m)
            {
              t++;m/=10;
            }
            int p=1;
            for(j=1;j<=i;j++)
            {
                if((j*(__int64)pow(10,t))%i)
                {
                    p=0;
                    break;
                }
            }
            if(p)
            {
                printf("%I64d  ",i);
            }
        }
        printf("
");
    }
    return 0;
}