直接上代码，有空更新题解

#include <iostream>
#include <stdio.h>
using namespace std;

int main()
{
    double x,y;

    cin>>x;
    if(0<=x&&x<5){
        y=-x+2.5;
    }
    else if(5<=x&&x<10){
        y=2-1.5*(x-3)*(x-3);
    }
    else {
    y=x/2-1.5;
    }
    printf("%.3f",y) ;
    return 0;
}

#include<iomanip> 
#include<cmath> 
using namespace std; 
double x,y;
int main() 
{
	cin>>x; 
	if(0<=x&&x<5) 
	{ 
		y=x*-1+2.5; 
	} 
	else if(5<=x&&x<10)
	{ 
 		y=2-1.5*(x-3)*(x-3);
	} 
  	else if(10<=x&&x<20) 
	{ 
  		y=x/2-1.5; 
	} 
  	cout<<fixed<<setprecision(3)<<y;
	return 0;

}

虽然不是最简解，很多地方还可以优化，但凡请各位大佬不要差评，毕竟我才学c++不到10 年 天，仅供参考，保证能AC，就当一个乐子吧，谢谢大佬们的观看！

#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
	double x;
	cin >> x;
	if (0 <= x && x < 5)
		cout << fixed << setprecision(3) << (-x + 2.5);
	else if (5 <= x && x < 10)
		cout << fixed << setprecision(3) << (2 - 1.5 * (x - 3) * (x - 3));
	else if (10 <= x && x < 20)
		cout << fixed << setprecision(3) << (x / 2 - 1.5);
	
    return 0;
}

//冒泡排序
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
int main(){
	double x,sum;cin>>x;if(x>=0&&x<=5) sum=-x+2.5;
	else if(x>=5&&x<=10){
		sum=2-1.5*(x-3)*(x-3);
	}
	else if(x>=10&&x<=20)
		sum=x*0.5-1.5;
	cout<<fixed<<setprecision(3)<<sum<<endl;
	return 0;
}

#include<iomanip>
using namespace std;
int main()
{
	double w,h;
	cin>>w;
	if(0<=w&&w<5)
	{
	h=-w+2.5;
	}
	else if(5<=w&&w<10)
	{
	h=2-1.5*(w-3)*(w-3);
	}
	else if(10<=w&&w<20)
	{
	h=w/2-1.5;
	}
	cout<<fixed<<setprecision(3)<<h;
}

#include <iostream>
#include <stdio.h>
using namespace std;

int main()
{
    double x,y;

    cin>>x;
    if(0<=x&&x<5){
        y=-x+2.5;
    }
    else if(5<=x&&x<10){
        y=2-1.5*(x-3)*(x-3);
    }
    else {
    y=x/2-1.5;
    }
    printf("%.3f",y) ;
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
const int N=1e7+10;
const int INF=0x3f3f3f3f;
int main()
{
	double x,y;
	cin>>x;
	if(0<=x&&x<5)
	{
		y=-x+2.5;
	}
	else if(5<=x&&x<10)
	{
		y=2-1.5*(x-3)*(x-3);
	}
	else if(10<=x&&x<20)
	{
		y=x/2-1.5;
	}
	cout<<fixed<<setprecision(3)<<y;
	return 0;
}

#include #include using namespace std; int main(){ double n; cin>>n; if(0<=n&&n<5) n=-n+2.5; else if(5<=n&&n<10) n=2-1.5*(n-3)*(n-3); else n=n/2-1.5; cout<<fixed<<setprecision(3)<<n<<endl; return 0; }

#include #include using namespace std; int main(){ double n; cin>>n; if(0<=n&&n<5) n=-n+2.5; else if(5<=n&&n<10) n=2-1.5*(n-3)*(n-3); else n=n/2-1.5; cout<<fixed<<setprecision(3)<<n<<endl; return 0; }

#include #include using namespace std; int main(){ double n; cin>>n; if(0<=n&&n<5) n=-n+2.5; else if(5<=n&&n<10) n=2-1.5*(n-3)*(n-3); else n=n/2-1.5; cout<<fixed<<setprecision(3)<<n<<endl; return 0; }

#include<bits/stdc++.h>
#include <iostream>
#include <iomanip>
using namespace std;
int main(){
    double n;
    cin>>n;
    if(0<=n&&n<5)
        n=-n+2.5;
    else if(5<=n&&n<10)
        n=2-1.5*(n-3)*(n-3);
    else
        n=n/2-1.5;
    printf("%.3lf",n);
    return 0;
}

#include <iostream>
#include <iomanip>
using namespace std;
int main(){
    double n;
    cin>>n;
    if(0<=n&&n<5)
        n=-n+2.5;
    else if(5<=n&&n<10)
        n=2-1.5*(n-3)*(n-3);
    else
        n=n/2-1.5;
    cout<<fixed<<setprecision(3)<<n<<endl;
    return 0;
}

ggf nj