【codeforces 20B】Equation

【题目链接】:http://codeforces.com/contest/20/problem/B

【题意】



给你一个方程,让你输出这个方程的解的情况.

【题解】



a==0,b==0,c==0时,为恒等式,无穷解;

a==0,b==0,c!=0时,为恒不等式,无解;

a==0,b!=0,为一次方程,有唯一解-c/b

a!=0的时候,按照正常的二次方程求解;

x1和x2的关系可能会因为a的正负改变的,不能直接输出,要判断一下大小再控制输出;



【Number Of WA】



4



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)
#define Open() freopen("D:\\rush.txt","r",stdin)
#define Close() ios::sync_with_stdio(0),cin.tie(0)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 1e6+100;
const int MOD = 1e9+7;
LL a,b,c;
double temp;
int main(){
    //Open();
    Close();
    cin >> a >> b >> c;
    if (a==0 && b==0 && c==0){
        cout <<-1<<endl;
        return 0;
    }
    if (a==0 && b==0 && c!=0){
        cout <<0<<endl;
        return 0;
    }
    if (a==0 && b!=0){
        cout <<1<<endl;
        double ans = (-1.0)*c/b;
        cout << fixed << setprecision(10) << ans << endl;
        return 0;
    }
    temp = b*b-4*a*c;
    if (temp<0){
        cout <<0<<endl;
    }else if (temp==0){
        cout <<1<<endl;
        double ans = -1.0*b/(2*a);
        cout << fixed << setprecision(10) << ans << endl;
    }else if (temp>0){
        cout <<2<<endl;
        temp = sqrt(temp);
        double ans1 = (-1.0*b-temp)/(1.0*2*a),ans2 = (-1.0*b+temp)/(1.0*2*a);
        if (ans1>ans2) swap(ans1,ans2);
        cout << fixed << setprecision(10) << ans1 <<endl<<ans2<<endl;
    }
    return 0;
}