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[]={,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, , , ,, , , , ,, , ,, ,, };
         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=;
           for(i=;i<;i++)
           {
               if(a[i]>=m&&a[i]<=n)
                 t++;
               if(a[i]>n)
                   break;
           }
           printf("%lld\n",t);
       }
    return ;
 }

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=,m,k;
    while(scanf("%I64d%I64d",&x,&y)!=EOF)
    {
        int i,j;
        k=;
        for(i=;i<pow(,);i++)
        {
            t=;
            m=i;
            while(m)
            {
              t++;m/=;
            }
            int p=;
            for(j=;j<=i;j++)
            {
                if((j*(__int64)pow(,t))%i)
                {
                    p=;
                    break;
                }
            }
            if(p)
            {
                printf("%I64d  ",i);
            }
        }
        printf("\n");
    }
    return ;
}