P1024 一元三次方程求解(二分答案)

思路:

  

求这个根,然后有一个关键的条件|x1-x2|>=1,然后就是从-100,枚举到+100,每次二分(i, i+1)注意如果f(i)*f(i+1)>0则不进行二分,如果,你觉得这样的值不行的话就把每次 i++ 变成 i+=0.5;就好了。反正数据范围很小..

#include<iostream>
#include<cstdio>
using namespace std;

double a, b, c, d;

double f(double x){ return a*x*x*x + b*x*x + c*x + d; }

void half(double l, double r){
    if (r - l <= 0.001){ printf("%.2lf ", l); return; }
    double mid = (l + r) / 2;
    double ans_l, ans_r;
    ans_l = f(l)*f(mid);    ans_r = f(r)*f(mid);
    if (f(mid) == 0)printf("%,2lf ", mid);
    if (f(r) == 0)printf("%.2lf ", r);
    if (ans_l < 0)half(l, mid);
    else if (ans_r < 0)half(mid, r);
}

int main(){
    cin >> a >> b >> c >> d;
    for (double i = -100; i <= 99; ++i){
        if (f(i)*f(i + 1) <= 0){ half(i, i + 1.0); }
    }
    cout << endl;
}

猜你喜欢

转载自www.cnblogs.com/ALINGMAOMAO/p/10458753.html