Magic Number - ZOJ 3622 水题

Magic Number

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Time Limit: 2 Seconds      Memory Limit: 32768 KB

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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

思路：直接打表就可以得到结果，水题。

AC代码如下;

#include<cstdio>
#include<cstring>
using namespace std;
long long num[100],m,n;
int main()
{ int i,j,k;
  num[1]=1;
  num[2]=2;
  num[3]=5;
  num[4]=10;
  num[5]=20;
  num[6]=25;
  num[7]=50;
  num[8]=100;
  num[9]=125;
  num[10]=200;
  num[11]=250;
  num[12]=500;
  for(i=13;i<=52;i++)
   num[i]=num[i-5]*10;
  while(~scanf("%lld%lld",&m,&n))
  { k=0;
    for(i=1;i<=52;i++)
     if(m<=num[i] && num[i]<=n)
      k++;
    printf("%d\n",k);
  }
}