Shulou(Shulou.com)06/01 Report--

This article introduces the relevant knowledge of "how to achieve digital rounding on C++". In the operation of actual cases, many people will encounter such a dilemma. Then let the editor lead you to learn how to deal with these situations. I hope you can read it carefully and be able to achieve something!

Problem description

Write a program that accepts a positive floating-point value and outputs an approximate integer value of that value. If the value after the decimal point is greater than or equal to 5, round up; less than 5, round down.

Enter description

Enter a positive floating point value

Output description

Outputs the approximate integer value of this value

Example

Input

5.5

Output

six

Analysis

This question is taken as a whole, we should first know the principle of int type casting in C++, and then it will be more advantageous to help us to achieve this problem. The cast of the Int type retains only the integer part and discards all the decimal parts. According to this principle, we can easily take out the decimal part of the input floating point number and compare it with 0. 5 to determine whether to enter 1. 5.

In fact, this problem has a more convenient way, because int forced conversion only retains the integer part, while the decimal part of the upward 1 is greater than 0.5, if the decimal is called 0.5, then it meets the requirements of more than 0.5 upward 1, less than 0.5 do not carry, and then you can write a more simple code.

Code 1

# include

Using namespace std

Int main ()

{

Double number

Cin > > number

Int int_number

Int_number = number/1

Doublelast_number = number-int_number

If (last_number > = 0.5)

Int_number=int_number+1

Int output = int (int_number)

Cout

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