1 条题解

  • 0
    @ 2025-4-7 21:28:46

    C :

    #include<stdio.h>
    int main(){int a,b,c;
    	int disc;
    	double p1,p2;
    	scanf("%d %d %d",&a,&b,&c);
    	if(a!=0){
    		disc=b*b-4*a*c;
    		p1=-b/(2.0*a);
    			p2=sqrt(disc)/(2.0*a);
    			printf("%.4lf %.4lf\n",p1-p2,p1+p2);
    		}
    return 0;}
    

    C++ :

    #include <stdio.h>
    #include <math.h>
    int main() {
    	int a, b, c, delta;
    	double ans0, ans1, sqrt_delta;
    	scanf("%d %d %d", &a, &b, &c);
    	/* 计算delta的值(b*b-4*a*c) */
    	delta = b * b - 4 * a * c;
    	sqrt_delta = sqrt(delta);
    	ans0 = (-b - sqrt_delta) / (2.0 * a);
    	ans1 = (-b + sqrt_delta) / (2.0 * a);
    	printf("%.4f %.4f\n", ans0, ans1);
    	return 0;
    }
    
    

    Pascal :

    var 
      a,b,c,d:double;
    
    begin
      readln(a,b,c);
      d:=b*b-4*a*c;
      writeln( (-b-sqrt(d))/(2*a):0:4 ,' ', (-b+sqrt(d))/(2*a):0:4 );
    end.
    
    

    Java :

    import java.util.*;
    public class Main {
    	public static void main(String args[]) {
    		Scanner cin = new Scanner(System.in);
    		double a,b,c;
    		double d;
    		double x1,x2;
    		a=cin.nextInt();
    		b=cin.nextInt();
    		c=cin.nextInt();
    		d=Math.pow(b, 2)-4*a*c;
    		if(d==0){
    			x1=-b/(2*a);
    			System.out.printf("%.4f\n",x1);
    		}
    		else if(d>0){
    			x1=(-b+Math.sqrt(d))/(2*a);
    			x2=(-b-Math.sqrt(d))/(2*a);
    			System.out.printf("%.4f %.4f\n",x2,x1);
    		}
    		else {
    			System.out.println("no answer");
    		}
    	}
    }
    

    Python :

    from math import sqrt
    a,b,c = [float(x) for x in raw_input().split()]
    print "%.4f %.4f" %((-b-sqrt(b*b-4*a*c))/(2*a), (-b+sqrt(b*b-4*a*c))/(2*a))
    
    • 1

    信息

    ID
    1249
    时间
    1000ms
    内存
    32MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者