poj Magic Number（坑人的数学题）

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 integerx it satisfies that put y to the right of x, which will form a new integerz, 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 andn(m, n inclusively).

Sample Input

1 11 10

Sample Output

1

4

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>using namespace std;long long int a[100];int main(){    long long int n,m,i,j,k,sum;    a[0]=1;a[1]=2;a[2]=5,a[3]=25,a[4]=125;    k=5;sum=1;    for(i=1;i<=10;i++){a[k++]=sum*10;sum*=10;}sum=10;for(i=1;i<=10;i++){a[k++]=sum*2;sum*=10;}sum=5;for(i=1;i<=10;i++){a[k++]=sum*10;sum*=10;}sum=10;for(i=1;i<=10;i++){a[k++]=sum*25;sum*=10;}sum=10;for(i=1;i<=8;i++){a[k++]=sum*125;sum*=10;}    sort(a,a+k);    while(scanf("%lld%lld",&n,&m)!=EOF)    {int y=0;for(i=0;i<k;i++){if(a[i]>=n&&a[i]<=m)y++;}printf("%d\n",y);    }    return 0;}