There are two routes to solve this problem.

One is to read the input as a string, then convert the substring before . (the decimal point) to the integer, and then increment if the character right after . or the third last character is \(5\) or more.

Another is, depending on programming language, to use a nearest-whole library equipped to each language. If the behavior for the boundary, that is, when the decimal part is \(0.5\), is undefined or inappropriate, or not known, then you may avoid it by adding a sufficiently small positive number (like \(0.0005\) in this case).
In Python, without this tweak the operation against \(0.5\) will fail, resulting in WA.

Sample code in C++ (with the first route):

#include <bits/stdc++.h>

using namespace std;

#define rep(i, n) for(int i = 0; i < n; ++i)

int main(void) {
	string X;
	int a;
	int ans = 0;

	cin >> X;
	int n = X.size();
	rep(i, n - 4) {
		a = (int)(X[i] - '0');
		ans = 10 * ans + a;
	}
	if (X[n - 3] >= '5')ans++;
	cout << ans << endl;
	return 0;
}

Sample code in C++ (with the second route):

#include <bits/stdc++.h>

using namespace std;

int main(void) {
	double a;
	cin >> a;
	cout << (int)round(a+0.0005) << endl;
        //cout << (int)round(a) << endl; でも良い
	return 0;
}

Sample code in C++ (with the third route):

x=float(input())
print(int(round(x+0.0005,0)))