Just a Numble模拟除法的运算

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Just a Numble

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1526 Accepted Submission(s): 675

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 2 5 7 123 123

Sample Output

5 0 8

开始是没有读懂题，那叫一个晕呐，后来结合例子分析，原来是求1/n小数点后面第m个数字并输出，如果没有的话，输出0

其实挺简单的，自己却郁闷了这么久才做出来；哎，要加油了！！！

模拟除法的运算

#include<stdio.h>int main(){int n,m,temp,s;while(scanf("%d%d",&n,&m)!=EOF){   if(n==1) printf("0\n");   else   {           temp=10;        while(m--)        {           s=temp/n;           temp=temp%n*10;//temp%10==0时可能中断；        }        printf("%d\n",s);   }}return 0;}

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