又双叒叕是一道水题

直接看代码吧：

#include<iostream>
#include<cmath>
#define QwQ return 0;
using namespace std;
double n;
int main(){
	cin>>n;
	n=ceil(n);
	cout<<(n<=5?12:12+(n-5)*2);
	QwQ
}

#include<bits/stdc++.h>
#define int double
using namespace std;
signed main(){
	int n;
	cin>>n;
	n=ceil(n);
	if(n<=5)cout<<12;
	else cout<<12+(n-5)*2;
}

别忘了要向上取整。

#include<bits/stdc++.h>
//#define int long long
#define INF 0x3f3f3f
using namespace std;
double x;
signed main(){
	cin>>x;
	if(x!=int(x))x=int(x)+1;
	if(x<=5)cout<<"12";
	else cout<<12+(x-5)*2;
	return 0;
}

玩把大的：不用 if。

#include<bits/stdc++.h>
using namespace std;
int main()
{
	double x;
	cin>>x;
	x=ceil(x);
	int a=12+(x-5)*2;
	cout<<a;
	return 0;
}

C++ :

#include <bits/stdc++.h>
using namespace std;
int main() {
	double x;
	cin >> x;
	int b = (int)(x - 5.0);
	if (x <= 5) {
		cout<<"12";
	}
	else {
		int ans = 0;
		if (b < x) {
			ans = 12 + (b+1)*2;
		}
		else {
			ans = 12 + b * 2;
		}
		if (ans == 24) ans = 22;
		cout<<ans;
	}
	return 0;
}

Python :

# coding=utf-8
a=float(input())
s=0
if a<=5:
    s=12
else:
    if a%1==0:
        s=12+(a//1-5)*2
    else:
        s=12+(a//1-4)*2
print(int(s))