1 条题解
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0
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