[ZOJ 3622] Magic Number

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


分析：
设y有k位数，则需要满足 (x*10^k+y)%y==0
--> (x*10^k%y+0)%y==0
--> x*10^k%y==0
--> 由于x是任意的，假设x为质数，则需要满足(10^k)%y==0
直接判断会超时，先打表。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <cmath>
#include <string>
using namespace std;
#define N 100010

int n,m;
int len;
int num[N]=
{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};

int main()
{
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        if(m>n) swap(m,n);
        int cnt=0;
        for(int i=0;i<45;i++)
        {
            if(m<=num[i] && num[i]<=n) cnt++;
        }
        cout<<cnt<<endl;
    }
    return 0;
}