15 条题解
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1曾致瑾 (zengzhijin) LV 6 @ 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; }
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12022-11-12 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); }
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12022-11-4 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; }
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02024-8-6 16:19:41@
#include<iostream> #include<iomanip> #include<cmath> 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; }
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02024-8-6 16:19:27@
#include<iostream> #include<iomanip> #include<cmath> 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; }
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02024-8-1 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; }
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02024-8-1 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; }
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02024-8-1 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; }
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02024-8-1 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; }
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02024-8-1 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; }
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02024-6-4 21:27:58@
#include<iostream>
#include<cmath>
#include<iomanip>
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;
}
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02024-2-3 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))); }
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02023-3-12 19:46:36@
#include <iostream> #include <stdio.h> #include <string.h> #include <queue> #include <math.h> #include <vector> #include <algorithm> #include <iomanip> #include <stack> #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;
}
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-42022-2-8 17:31:35@
#include <stdio.h> #include <iostream> #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);
}
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-42022-1-17 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; }
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信息
- ID
- 822
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- 6
- 标签
- 递交数
- 989
- 已通过
- 339
- 上传者