e-olymp Problem11 Big accuracy

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Big accuracy

The rational fraction m/n is given. Write it in the decimal notation with k digits after the decimal point.

Input

Three numbers m, n, k are written in one line. 0 < m, n ≤ 100, 0 ≤ k ≤ 1000.

Output

Print k digits after the decimal point of the given number.

Time limit 1 second

Memory limit 64 MiB

Input example #1

1 2 3

Output example #1

0.500

题意：给你N，M，K，求N/M ，保留K位。
     本题有点小坑的地方是，K等于0的时候是取整数部分，保留K位是直接扔掉后面的，不需要四舍五入。
JAVA代码：

import java.text.DecimalFormat;
import java.util.*;
import java.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.InputMismatchException;

public class Main
{
  public static void main (String[] argv)
  {
      new Main();
  }
  boolean test = false;
  public Main(){
      FastReader in = new FastReader(new BufferedReader(new InputStreamReader(System.in)));
      //FastReader in = new FastReader(new BufferedReader(new FileReader("Main.in")));
      Scanner input = new Scanner(System.in);
      BigDecimal x = input.nextBigDecimal();
      BigDecimal y = input.nextBigDecimal();
      int k = input.nextInt();
      if(k == 0){
          BigDecimal ans = x.divide(y,1,BigDecimal.ROUND_DOWN);
          int answer = ans.intValue();
          System.out.println(answer);
      }//如果只需要整数部分，就直接取int
      else{
          BigDecimal ans = x.divide(y,k,BigDecimal.ROUND_DOWN);
          System.out.println(ans);
      }//保留K位，并且不需要四舍五入
      input.close();
  }
  static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;
        public FastReader(BufferedReader in)
        {
            br = in;
        }

        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    String line = br.readLine();
                    if (line == null || line.length() == 0) return "";
                    st = new StringTokenizer(line);
                }
                catch (IOException  e)
                {
                    return "";
                    //e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {
            return Long.parseLong(next());
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                return "";
                //e.printStackTrace();
            }
            return str;
        }
    }
}