16. Rounding Floating-Point Numbers

📝 Problem

Display the given floating point value to its nearest integer value.

Input Format:
Accept a floating point value as input.

Output Format:
Print the nearest integer value

Constraints:
3.4E-4932 <= inp <= 1.1E+4932

🎓 Complete Guide to Rounding in C++

This problem opens a whole new world of rounding techniques!

At first, this looks simple—just round a number. But there are actually 8 different rounding functions in C++, each with unique behavior, especially with negative numbers!

1️⃣ All Standard Rounding Functions

Function	Behavior	x = 2.7	x = -2.7	Return Type
`round(x)`	Nearest integer (0.5 → away from zero)	3	-3	double
`lround(x)`	Same as round()	3	-3	long
`llround(x)`	Same as round()	3	-3	long long
`floor(x)`	Down toward -∞	2	-3	double
`ceil(x)`	Up toward +∞	3	-2	double
`trunc(x)`	Toward zero (chop decimals)	2	-2	double
`nearbyint(x)`	Nearest (current rounding mode)	3	-3	double
`rint(x)`	Like nearbyint(), may raise exceptions	3	-3	double

2️⃣ Understanding 0.5 Behavior

round() - Rounds AWAY from zero

round(2.5) → 3

round(-2.5) → -3

floor() - Always toward -∞

floor(2.5) → 2

floor(-2.5) → -3

ceil() - Always toward +∞

ceil(2.5) → 3

ceil(-2.5) → -2

trunc() - Chops decimals (toward zero)

trunc(2.5) → 2

trunc(-2.5) → -2

3️⃣ Beginner Mistakes to Avoid

⚠️ Common Pitfalls

Casting without rounding: (int)3.7 gives 3, not 4
floor/ceil confusion: floor(-2.7) is -3, not -2!
Floating-point precision: 2.999999 may not equal 3.0 exactly
0.5 rounding: round() goes away from zero, not always up

4️⃣ Rounding to Decimal Places

Want to round to 2 decimal places? Multiply, round, divide!

double x = 2.5678;
double rounded = round(x * 100.0) / 100.0;
// Result: 2.57

5️⃣ Manual Rounding (Without <cmath>)

If you can't use <cmath>, implement it yourself:

int nearest = (num >= 0) ? (int)(num + 0.5) 
                         : (int)(num - 0.5);

📊 Quick Reference Table

Goal	Function / Method
Nearest integer	`round(x)`
Round down	`floor(x)`
Round up	`ceil(x)`
Remove decimals	`trunc(x)` or `(int)x`
Round to n decimals	`round(x * 10^n) / 10^n`

✅ For this problem, use round() for nearest integer!

Examples

Sample Input 1:

Input:

15.9

Sample Output 1:

Output:

Sample Input 2:

Input:

159.357

Sample Output 2:

Output:

159

Solution.cpp

#include <iostream>
#include <cmath>

using namespace std;

int main() {
    long double f;
    cin >> f;
    
    // round() returns the nearest integer
    // If decimal is 0.5, rounds AWAY from zero
    // Examples: round(2.7) = 3, round(-2.7) = -3
    cout << round(f);
    
    return 0;
}