简单计算即可，考察的主要是 cmath 库的使用以及保留小数。

#include <bits/stdc++.h> using namespace std; int main(){ double a,b,c; cin>>a>>b; c=sqrt((aa)+(bb)); cout<<fixed<<setprecision(2)<<c<<endl; return 0; }

#include <iostream> #include <cmath> #include <iomanip> using namespace std; int main() { int a, b; cin >> a >> b; double c = sqrt(static_cast<double>(a * a + b * b)); cout << fixed << setprecision(2) << c << endl; return 0; }

#include<bits/stdc++.h> using namespace std; int main(){ double a,b; cin>>a>>b; printf("%.2lf",sqrt(aa+bb)); return 0; }

#include<cstring> #include<cmath> #include<bits/stdc++.h> using namespace std; int main(){ int a,b; cin>>a>>b; float c; c=sqrt(pow(a,2)+pow(b,2)); cout<<fixed<<setprecision(2)<<c; return 0; }

#include<iostream> #include<cmath> using namespace std; int main(){ int a,b; cin>>a>>b; double c; c=sqrt(aa+bb); printf("%.2lf",c); return 0; }

#include<bits/stdc++.h>; using namespace std; int main(){short a,b,c;cin>>a>>b;c=aa+bb;printf("%.2f",sqrt(c));return 0;}

#include <bits/stdc++.h> using namespace std; int main(){ int a,b; cin >> a >> b; printf("%.2f",sqrt(aa+bb)); return 0; }

#include <bits/stdc++.h> using namespace std; int main() { double a,b; cin>>a>>b; double c= pow(a,2)+pow(b,2); printf("%.2lf",sqrt(c)); return 0; }

#include<iostream> #include<iomanip> #include<cmath> using namespace std; int main(){ float a,b; cin>>a>>b; cout<<std::fixed<<setprecision(2)<<sqrt(aa+bb); return 0; }

#include <bits/stdc++.h> using namespace std; int main(){ int a,b; cin >> a >> b; printf("%.2f",sqrt(aa+bb)); return 0; }