18 条题解

  • 1
    @ 2024-6-4 21:27:58

    #include

    #include

    #include

    using namespace std;

    int main()

    {

    double a,b,c,d,p;

    cin>>a>>b>>c;

    p=(a+b+c)/2;

    d=p*(p-a)(p-b)(p-c);

    cout << setiosflags (ios::fixed<<setprecision(2);

    cout <<sqrt(d);

    return 0;

    }

    • 1
      @ 2023-1-23 16:02:33 
      #include <iostream>
#include <math.h>
using namespace std;
int main(){
	double a,b,c;
	cin>>a>>b>>c;
	double p=(a+b+c)/2;
	double S=sqrt(p*(p-a)*(p-b)*(p-c));
	printf("%.2f",S);
	return 0; 
}
      • 0
        @ 2025-10-17 20:49:41

        #include<bits/stdc++.h> using namespace std; int main() { double a, b, c, p, m;

        cin >> a >> b >> c;

p = (a + b + c) / 2;

m = p * (p - a) * (p - b) * (p - c);

cout << fixed << setprecision(2) << sqrt(m);



return 0;

        }

        • 0
          @ 2025-3-2 20:42:21

          #include<bits/stdc++.h> using namespace std; int main() { float a,b,c,p; cin>>a>>b>>c; p=(a+b+c)/2; cout<<fixed<<setprecision(2)<<sqrt(p*(p-a)(p-b)(p-c)); }

          • 0
            @ 2022-11-12 12:01:49 
            #include<bits/stdc++.h>
using namespace std;
int main()
{
    double a,b,c;
    cin >> a >> b >> c;
    double p = (a + b + c)/2;
    double sum = p * (p-a)*(p-b)*(p-c);
    sum = sqrt(sum);
    printf("%.2lf\n",sum);
}
            • 0
              @ 2022-11-4 21:58:25 
              #include <bits/stdc++.h>

using namespace std;

int main()
{
	double a, b, c, p, m;

	cin >> a >> b >> c;

	p = (a + b + c) / 2;

	m = p * (p - a) * (p - b) * (p - c);

	cout << fixed << setprecision(2) << sqrt(m);

	return 0;
}
              • -1
                @ 2025-3-2 20:41:59

                #include<bits/stdc++.h> using namespace std; int main() { float a,b,c,p; cin>>a>>b>>c; p=(a+b+c)/2; cout<<fixed<<setprecision(2)<<sqrt(p*(p-a)(p-b)(p-c)); }

                • -1
                  @ 2024-8-6 16:19:41

                  #include #include #include using namespace std; double a, b, c, p, m; int main() { cin >> a >> b >> c; p = (a + b + c) / 2; m = p * (p - a) * (p - b) * (p - c); cout << fixed << setprecision(2) << sqrt(m); return 0; }

                  • -1
                    @ 2024-8-6 16:19:27

                    #include #include #include using namespace std; double a, b, c, p, m; int main() { cin >> a >> b >> c; p = (a + b + c) / 2; m = p * (p - a) * (p - b) * (p - c); cout << fixed << setprecision(2) << sqrt(m); return 0; }

                    • -1
                      @ 2024-8-1 10:58:32 
                      using namespace std;
double f(double a,double b,double c)
{
	double p=(a+b+c)/2.0;
	p=sqrt(p*(p-a)*(p-b)*(p-c));
	return p;
}
int main()
{
	double a,b,c;
	cin>>a>>b>>c;
	cout<<fixed<<setprecision(2)<<f(a,b,c)<<endl;
    return 0;
}
                      • -1
                        @ 2024-8-1 10:56:52

                        #include<bits/stdc++.h> using namespace std;

                        double f(double a,double b,double c) { double p=0; p=(a+b+c)/2; p=sqrt(p*(p-a)(p-b)(p-c)); return p; } int main() { double a,b,c; cin>>a>>b>>c; cout<<fixed<<setprecision(2)<<f(a,b,c); return 0; }

                        • -1
                          @ 2024-8-1 10:56:49

                          函数

                          #include<bits/stdc++.h>
using namespace std;
double p,k;
double san(double a,double b,double c){
	p=(a+b+c)/2.0;
	k=sqrt(p*(p-a)*(p-b)*(p-c));
	return k;
}
int main(){
	double a,b,c;
	cin>>a>>b>>c;
	cout<<fixed<<setprecision(2)<<san(a,b,c);
	return 0; 
}
                          • -1
                            @ 2024-8-1 10:45:27

                            函数：

                            #include<iostream>
#include<iomanip>
#include<cmath>
using namespace std;
double f(double a, double b, double c)
{
	double p = (a + b + c) / 2;
	return sqrt(p * (p - a) * (p - b) * (p - c));
}
int main()
{
	double a, b, c;
	cin >> a >> b >> c;
	cout << fixed << setprecision(2) << f(a, b, c);
	return 0;
}
                            • -1
                              @ 2024-2-3 17:06:27 
                              #include<iostream>
#include<cstdio>
#include<string>
#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int INF=0x3f3f3f3f;
const int N=2e5+10;
double a,b,c;
int main(){
	cin>>a>>b>>c;
	double p=(a+b+c)/2;
	printf("%.2f",sqrt(p*(p-a)*(p-b)*(p-c)));
}
                              • -1
                                @ 2023-3-12 19:46:36

                                #include #include <stdio.h> #include <string.h> #include #include <math.h> #include #include #include #include #include <bits/stdc++.h> using namespace std; int main(){ double a,b,c,p,m; cin>>a>>b>>c; p=(a+b+c)/2; double sum; m= p * (p - a) * (p - b) * (p - c); cout << fixed << setprecision(2) << sqrt(m);

                                return 0;

                                }

                                • -2
                                  @ 2024-8-1 10:57:36 
                                  #include<bits/stdc++.h>
using namespace std;

double f(double a,double b,double c)
{
	double p=0;
	p=(a+b+c)/2;
	p=sqrt(p*(p-a)*(p-b)*(p-c));
	return p;
}
int main()
{
	double a,b,c;
	cin>>a>>b>>c;
	cout<<fixed<<setprecision(2)<<f(a,b,c);
	return 0;
}
                                  • -5
                                    @ 2022-2-8 17:31:35

                                    #include <stdio.h> #include #include <math.h> using namespace std; int main() { double a , b , c , p; cin >> a >> b >> c; p = (a + b + c)/2; double num = p*(p - a)(p - b)(p - c); double s = sqrt ( num ); printf("%.2lf",s);

                                    }

                                    • -5
                                      @ 2022-1-17 22:21:56 
                                      #include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <math.h>
#include <vector>
#include <algorithm>
#include <iomanip>
#include <stack>

using namespace std;

#define LL long long
const int N =1e5+10;
const int INF =0x3f3f3f3f;

int main(){
    double a,b,c,p;
    cin>>a>>b>>c;
    p=(a+b+c)/2;
    cout<<fixed<<setprecision(2)<<sqrt(p*(p-a)*(p-b)*(p-c))<<endl;
    return 0;
}
                                      • 1

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                                      信息

                                      ID
                                      822
                                      时间
                                      1000ms
                                      内存
                                      128MiB
                                      难度
                                      6
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                                      递交数
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                                      已通过
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