Just a Numble（求小数的第N位数）

Just a Numble

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2410    Accepted Submission(s): 1160

Problem Description

Now give you two integers n m, you just tell me the m-th number after radix point in 1/n,for example n=4,the first numble after point is 2,the second is 5,and all 0 followed

 

Input

Each line of input will contain a pair of integers for n and m(1<=n<=10^7,1<=m<=10^5)

 

Output

For each line of input, your program should print a numble on a line,according to the above rules

 

Sample Input

4 25 7123 123

 

Sample Output

508

 

#include<stdio.h>int main(){int n,m,i,j,sum;int k;while(scanf("%d%d",&n,&m)!=EOF){sum=1;for(i=0;i<m;i++){sum=sum*10;//这个题的意思是让每一位都X10，比如求得第一位k=sum/n;//k即为第一位小数sum=sum%n;//求得余数，还得让余数继续除以n也很好理解！}printf("%d\n",k%10);}}

也很好理解的，不过最好还是借用一下画画来更加清晰地认识！