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; }