4 solutions
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2
#include<bits/stdc++.h> using namespace std; int main() { double a,b,c,x1/*根1*/,x2/*根2*/,n; cin>>a>>b>>c; n = b*b-4*a*c; if (n>=0)//有解 { x1 = (-b+sqrt(n))/(2*a);//根1 x2 = (-b-sqrt(n))/(2*a);//根2 if (n==0) cout<<fixed<<setprecision(5)<<"x1=x2="<<x1;//根相等 else if (x1<x2) cout<<fixed<<setprecision(5)<<"x1="<<x1<<";x2="<<x2; else cout<<fixed<<setprecision(5)<<"x1="<<x2<<";x2="<<x1; } else cout<<"No answer!";//无解 return 0; }
Information
- ID
- 865
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 3
- Tags
- (None)
- # Submissions
- 125
- Accepted
- 31
- Uploaded By