【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;

}