方程的解

Description

求ax2+bx+c=0方程的实根。a,b,c由键盘输入. 解方程要考虑系数a等于零的情况，且解x1、x2必须是float型。a等于零有两种情况(b==0,b!=0)，a不等于零有三种情况（delta>0、==0、<0），先计算得到x1、x2，再printf输出

Input

输入三个数a,b,c

Output

输出方程的实根，如果方程有实根，则输出根；如果方程有2个不等实根，则分2行输出，第一行输出较大根，第二行输出较小根。其余情况（如无实根等）则输出No

Sample Input

1 -3 2

Sample Output

2.000000
1.000000

HINT

#include <stdio.h>
#include <stdlib.h>
 
 
#include <math.h>
int main(void)
{
float a,b,c;
float data;
float x1,x2;
scanf("%f%f%f",&a,&b,&c);
 
if(a ==  0)
{
if(b == 0)
printf("No\n");
else
{
x1 = -c / b;
printf("%f\n",x1);
}
}
else
{
   data = b * b - 4 * a * c;
x1 = (-b + sqrt(data)) / (2 * a);
x2 = (-b - sqrt(data)) / (2 * a);
if(data < 0)
   {
printf("No\n");
}
else
{
if(data == 0)
{
printf("%f\n",x1);
}
else
{
if(x1 > x2)
        {
       printf("%f\n",x1);
       printf("%f\n",x2);
         }
        else
         {
        printf("%f\n",x2);
        printf("%f\n",x1);
         }
}
}
}
return 0;
}

#include<stdio.h>
double a,b,c,d,x[3];
double f(double x){
	return (a*x*x*x + b*x*x + c*x + d);
} 
int main(){
	double f1,f2,f3,f4,high,mid,low;
	int i,j = 0;
	scanf("%lf %lf %lf %lf",&a,&b,&c,&d);
	f1 = f(-100);f2=f(-99);
	for(int i=-99;i <= 100;i ++){
		double ans=f1*f2;
		//printf("%.2lf %.2lf\n",f1,f2);
		if(ans==0){
			if(f1==0){
				x[j ++] =i-1;
			}
		}else if(ans<0){
			high = i+1; low = i;
			do{
				mid = (high + low) / 2;
				f4 = f(low); f2 = f(high);
				f3 = f(mid);
				if(f4 * f3 <= 0) high = mid;
				if(f2 * f3 <= 0) low = mid;
			}while((high - low) < 0.0001);
			x[j ++] = high;
		}
		f1=f(i);
		f2=f(i+1);
	}
	printf("%.2lf %.2lf %.2lf\n",x[0],x[1],x[2]);
	return 0;
}

https://www.luogu.org/problemnew/show/P1024

题解：

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
double a,b,c,d,x[3];
double f(double x){
	return (a*x*x*x + b*x*x + c*x + d);
}
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    double f1,f2,f3,f4,high,mid,low;
	int j = 0;
	scanf("%lf %lf %lf %lf",&a,&b,&c,&d);
	f1 = f(-100);f2=f(-99.5);
	for(double i=-99.5;i <= 100;i +=0.5){
		double ans=f1*f2;
		if(ans==0){
			if(f1==0){
				x[j ++] =i-0.5;
			}
		}else if(ans<0){
			high = i; low = i-0.5;
			do{
				mid = (high + low) / 2;
				f2 = f(high);
				f3 = f(mid);
				f4 = f(low);
				//printf("%.2f %.2f %.2f\n",f2,f3,f4);
				if(f4 * f3 <= 0) high = mid;
				if(f2 * f3 <= 0) low = mid;

			}while((high - low) > 0.0001);
			x[j ++] = high;
		}
		f1=f(i);
		f2=f(i+0.5);
	}
	printf("%.2f %.2f %.2f\n",x[0],x[1],x[2]);
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}