12 条题解

  • 1
    @ 2026-3-15 21:27:29

    这才叫粗暴AC（初学者专用）

    
#include<bits/stdc++.h>
using namespace std;

int main(){
	double a,b,c,x1,x2;
	cin>>a>>b>>c;
	x1=(-b+sqrt(b*b-4*a*c))*10000/(2*a);
	x2=(-b-sqrt(b*b-4*a*c))*10000/(2*a);
	if(a==0||b*b-4*a*c<=0){
		cout<<"No answer!";
		return 0;
	}
	if(x1<x2){
		cout<<"x1=";
		cout<<fixed<<setprecision(5)<<x1/10000;
		cout<<";x2=";
		cout<<fixed<<setprecision(5)<<x2/10000;
		return 0;
	}
	if(x1>x2){
		cout<<"x1=";
		cout<<fixed<<setprecision(5)<<x2/10000;
		cout<<";x2=";
		cout<<fixed<<setprecision(5)<<x1/10000;
		return 0;
	}
	if(x1==x2){
		cout<<"x1=x2=";
		cout<<fixed<<setprecision(5)<<x2/10000;
		return 0;
	}
}
    • 0
      @ 2025-9-14 16:18:47

      我是 高手 初学者，能AC就行~

      #include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int main()
{
	double a,b,c,x1,x2;
	cin >> a >> b >> c;
	if(a == 0 || b*b - 4*a*c < 0)
		cout << "No answer!";
	else
	{
		x1 = (-b + sqrt(b * b - 4 * a * c)) / (2 * a);
		x2 = (-b - sqrt(b * b - 4 * a * c)) / (2 * a);
		if(x1 == x2)
			printf("x1=x2=%.5f",x1);
		else
		{
			if(x1 > x2)
				swap(x1,x2);
			printf("x1=%.5f;x2=%.5f",x1,x2);
		}
	}
	
	return 0;
}
    • 0
      @ 2025-7-4 18:47:59 
      #include <stdio.h>
#include <math.h>
#include <iostream>
using namespace std;
int main()
{
	double x1,x2;
	double a,b,c;
    cin >> a >> b >> c;
    if ((b * b - 4 * a * c) < 0)
    {
        cout<<"No answer!";
    }
    else
    {
        if((b * b - 4 * a * c) == 0)
        {
        	x1 = -1 * (b / (2 * a));
            printf("X1=X2=%.5lf\n",x1);
        }
        else
        {
        	x1 = (-b + sqrt(b*b - 4*a*c))/(2*a);
        	x2 = (-b - sqrt(b*b - 4*a*c))/(2*a);
            if(x1 > x2)
            {
                swap(x1,x2);
            }
            printf("x1=%.5lf;x2=%.5lf\n",x1,x2);
        }
    }

}
      • 0
        @ 2024-12-1 19:01:57 
        #include<bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
int const N=1e5+10;
double a,b,c;
int main(){
	cin>>a>>b>>c;
	double x1=(-b+sqrt(b*b-4*a*c))/(2*a),x2=(-b+-sqrt(b*b-4*a*c))/(2*a);
	if(a==0){
		cout<<"No answer!";
		return 0;
	}else if(b*b-4*a*c<0){
		cout<<"No answer!";
		return 0;
	}
	else if(x1==x2){
		cout<<"x1=x2="<<fixed<<setprecision(5)<<x1;
	}else if(x1<x2){
		cout<<"x1="<<fixed<<setprecision(5)<<x1<<";x2="<<fixed<<setprecision(5)<<x2;
	}else if(x1>x2){
		cout<<"x1="<<fixed<<setprecision(5)<<x2<<";x2="<<fixed<<setprecision(5)<<x1;
	}
	return 0;
}
        • 0
          @ 2023-3-26 20:25:11

          876AC代码，之前的题解都是错的

          //需注意：当b*b-4*a*c<0时,没有答案
#include<iostream>
#include<iomanip>
#include<stdio.h>
#include<math.h>
using namespace std;
int main(){
	double a,b,c,x,y;
	cin>>a>>b>>c;
	if(a==0||b*b-4*a*c<0){
		cout<<"No answer!";
		return 0;
	}
	x=(-b+sqrt(b*b-4*a*c))/(2*a);
	y=(-b-sqrt(b*b-4*a*c))/(2*a);
	if(fabs(x-y)<1e-6){
		printf("x1=x2=%.5lf",x);
		return 0;
	} 
	if(x>y)swap(x,y);
	printf("x1=%.5lf;x2=%.5lf",x,y);
	return 0;
}
          • @ 2026-3-15 21:25:00

            需注意：当bb-4a*c=0时,也没有答案

        • -1
          @ 2022-8-16 11:30:35 
          #include <stdio.h>
#include <math.h>
#include <iostream>
using namespace std;
int main()
{
	double a,b,c;
    cin >> a >> b >> c;
    if (a==0)
    {
        cout<<"No answer!";
    }
    else
    {
        double x1,x2;
        x1 = (-b + sqrt(b*b - 4*a*c))/(2*a);
        x2 = (-b - sqrt(b*b - 4*a*c))/(2*a);
        if( fabs(x1-x2)< 1e-6)
        {
            printf("X1=X2=%.5lf\n",x1);
        }
        else
        {
            if(x1 > x2)
            {
                swap(x1,x2);
            }
            printf("x1=%.5lf;x2=%.5lf\n",x1,x2);
        }
    }

}
          • -1
            @ 2022-1-7 10:57:41 
            #include <stdio.h>
#include <math.h>
#include <iostream>
using namespace std;
int main()
{
	double a,b,c;
    cin >> a >> b >> c;
    if (a==0)
    {
        cout<<"No answer!";
    }
    else
    {
        double x1,x2;
        x1 = (-b + sqrt(b*b - 4*a*c))/(2*a);
        x2 = (-b - sqrt(b*b - 4*a*c))/(2*a);
        if( fabs(x1-x2)< 1e-6)
        {
            printf("X1=X2=%.5lf\n",x1);
        }
        else
        {
            if(x1 > x2)
            {
                swap(x1,x2);
            }
            printf("x1=%.5lf;x2=%.5lf\n",x1,x2);
        }
    }

}
            • -1
              @ 2022-1-7 10:46:55 
              #include <iostream>
#include <math.h>
#include<iomanip>
using namespace std;
int main()
{
    double x,y,a,b,c;
    cin>>a>>b>>c;
    if (a==0)
    {
        cout<<"No answer!";
        return 0;
    }
    x=(sqrt(b*b-4*a*c)-b)/(2*a);
    y=(-sqrt(b*b-4*a*c)-b)/(2*a);
    if (x>y){
        swap(x,y);
    }
    if (fabs(x-y)<1e-6)
    {
        cout<<"x1=x2="<<fixed<<setprecision(5)<<x<<endl;
    }
    else
    {
        cout<<"x1="<<fixed<<setprecision(5)<<x<<";x2="<<fixed<<setprecision(5)<<y<<endl;
    }
}
              • -2
                @ 2023-3-26 22:01:02 
                #include <stdio.h>
#include <math.h>
#include <iostream>
using namespace std;
int main()
{
	double a,b,c;
    cin >> a >> b >> c;
    if (a==0)
    {
        cout<<"No answer!";
    }
    else
    {
        double x1,x2;
        x1 = (-b + sqrt(b*b - 4*a*c))/(2*a);
        x2 = (-b - sqrt(b*b - 4*a*c))/(2*a);
        if( fabs(x1-x2)< 1e-6)
        {
            printf("X1=X2=%.5lf\n",x1);
        }
        else
        {
            if(x1 > x2)
            {
                swap(x1,x2);
            }
            printf("x1=%.5lf;x2=%.5lf\n",x1,x2);
        }
    }

}
                • -2
                  @ 2023-3-10 20:00:03

                  真题解

                  https://jingyan.baidu.com/article/ca00d56c2e0952e99eebcf86.html

                  不会一元二次的人一定要看

                  完整代码如下：

                  #include <stdio.h>
#include <math.h>
#include <iostream>
using namespace std;
int main()
{
	double x1,x2;
	double a,b,c;
    cin >> a >> b >> c;
    if ((b * b - 4 * a * c) < 0)
    {
        cout<<"No answer!";
    }
    else
    {
        if((b * b - 4 * a * c) == 0)
        {
        	x1 = -1 * (b / (2 * a));
            printf("X1=X2=%.5lf\n",x1);
        }
        else
        {
        	x1 = (-b + sqrt(b*b - 4*a*c))/(2*a);
        	x2 = (-b - sqrt(b*b - 4*a*c))/(2*a);
            if(x1 > x2)
            {
                swap(x1,x2);
            }
            printf("x1=%.5lf;x2=%.5lf\n",x1,x2);
        }
    }

}
                  • -2
                    @ 2022-2-9 11:14:53

                    #include #include <math.h> #include using namespace std; int main() { double x,y,a,b,c; cin>>a>>b>>c; if (a==0) { cout<<"No answer!"; return 0; } x=(sqrt(bb-4ac)-b)/(2a); y=(-sqrt(bb-4ac)-b)/(2a); if (x>y){ swap(x,y); } if (fabs(x-y)<1e-6) { cout<<"x1=x2="<<fixed<<setprecision(5)<<x<<endl; } else { cout<<"x1="<<fixed<<setprecision(5)<<x<<";x2="<<fixed<<setprecision(5)<<y<<endl; } }

                    • -4
                      @ 2024-6-15 20:13:28

                      #include<bits/stdc++.h> using namespace std; int main(){ for(long long i=0;i<=514514;i++){ cout<<"1111111111111111111111111111111111111111111111111"; } return 0; }

                      • 1

                      求一元二次方程

                      信息

                      ID
                      876
                      时间
                      1000ms
                      内存
                      256MiB
                      难度
                      6
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