6 solutions

  • 1
    @ 2026-1-11 17:54:58
    #include<stdio.h>
    #include<math.h>
    int main()
    {
        double a,b,c,D,x_1,x_2;
        scanf("%lf %lf %lf",&a,&b,&c);
        D=pow(b,2)-4.0*a*c;
        x_1=(-b+sqrt(D))/(2.0*a);
        x_2=(-b-sqrt(D))/(2.0*a);
        if(D<0)
        printf("No answer!");
        else if(x_1==x_2)
        printf("x1=x2=%lf",x_1);
             else if(x_1<x_2)
             printf("x1=%.5lf;x2=%.5lf",x_1,x_2);
                  else
                  printf("x1=%.5lf;x2=%.5lf",x_2,x_1);
    
        return 0;
    }
    
    • 1
      @ 2023-9-30 20:02:29
      #include <stdio.h>
      #include <math.h>
      int main()
      {
      	double a,b,c;
      	scanf("%lf %lf %lf",&a,&b,&c);
      	double t=b*b-4*a*c;
      	double x1=(-b+sqrt(t))/(2*a);
      	double x2=(-b-sqrt(t))/(2*a);
      	while(t>=0){
      		if(t==0)
      			printf("x1=x2=%.5f",x1);
      		else//t>0
      		{
      			if(x1<x2)
      				printf("x1=%.5f;x2=%.5f",x1,x2);
      			else//x2<x1
      				printf("x1=%.5f;x2=%.5f",x2,x1);
      		}
      		return 0;
      	}	
      	printf("No answer!");
      	return 0;
      }
      
      • 0
        @ 2024-7-28 21:20:47
        #include<iostream>
        #include<cmath>
        using namespace std;
        int main()
        {
            double a,b,c,x1,x2;
            scanf("%lf %lf %lf",&a,&b,&c);
            x1 = (-b + sqrt(b * b - 4 * a * c)) / (2 * a);
            x2 = (-b - sqrt(b * b - 4 * a * c)) / (2 * a);
            double res = b * b - 4 * a * c;
            if(res < 0){
                printf("No answer!");
            }else 
                if(res == 0){
                    printf("x1=x2=%.5lf",x1);
                }else
                    if(res > 0){
                        if(x1 > x2){
                            swap(x1,x2);
                        }
                        printf("x1=%.5lf;x2=%.5lf",x1,x2);
                    }else printf("No answer!");
            return 0;
        }
        
        • 0
          @ 2023-11-17 16:34:53
          #include <stdio.h>
          #include <math.h> //要用到开平方和平方
          int main()
          {
              double a, b, c;
              scanf("%lf %lf %lf", &a, &b, &c);/*现在只能说是一元二次型方程, 因为a的值未知*/
              double d = pow(b, 2) - 4 * a * c; /*d就是那个德尔塔但是没开根号*/
              if (d < 0 || a == 0) /*如果a等于零那么就是一元一次方程*/
              {
                  printf("No answer!");
              }
              else if (d >= 0)
              {
                  double x1 = (0 - b + sqrt(d)) / (2 * a);
                  double x2 = (0 - b - sqrt(d)) / (2 * a);
                  if (x1 < x2)
                  {
                      printf("x1=%.5f;x2=%.5f", x1, x2);
                  }
                  else if (x1 > x2)
                  {
                      printf("x1=%.5f;x2=%.5f", x2, x1);
                  }
                  else if (x1 == x2)
                  {
                      printf("x1=x2=%.5f", x1);
                  }
              }
              return 0;
          }
          
          • 0
            @ 2023-10-18 10:54:33
            #include <stdio.h>
            #include <math.h>
            int main()
            {
                double a,b,c,d,x1,x2,max,min;
                scanf("%lf %lf %lf",&a,&b,&c);
                d=b*b-4*a*c;x1=(-b+sqrt(d))/(2*a);x2=(-b-sqrt(d))/(2*a);
                if(x1==x2){printf("x1=x2=%.5f",x1);}else{
                if(x1<x2){max=x2;min=x1;}else{max=x1;min=x2;};
                if(d<0){printf("No answer!");};
                if(d>0){printf("x1=%.5f;x2=%.5f",min,max);};}
                return 0;
            }
            
            • 0
              @ 2023-10-6 11:48:09
              #include<math.h>
              int main()
              {
                  double a,b,c,d,x1,x2,e;
                  scanf("%lf %lf %lf",&a,&b,&c);
                  d=b*b-4*a*c;
                  if(d<0){
                      printf("No answer!");
                  }else{
                      if(d==0){
                          x1=x2=(-b)/(2*a);
                          printf("x1=x2=%.5f",x1);
                      }else{
                          e=sqrt(d);
                          x1=(-b+e)/(2*a);
                          x2=(-b-e)/(2*a);
                         if(x1<x2){
                          printf("x1=%.5f;x2=%.5f",x1,x2);
                          }else{
                              printf("x1=%.5f;x2=%.5f",x2,x1);
                              }
                      }
                  }return 0;
              }
              
              • 1

              Information

              ID
              6735
              Time
              1000ms
              Memory
              128MiB
              Difficulty
              9
              Tags
              (None)
              # Submissions
              839
              Accepted
              93
              Uploaded By