本文主要是介绍二分法求多项式单根，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
输出格式：
在一行中输出该多项式在该区间内的根，精确到小数点后2位。
输入样例：
3 -1 -3 1
-0.5 0.5
输出样例：
0.33
idea
- 精确到小数点后两位
=>阈值为0.001
solution1
#include <stdio.h>
#include <math.h>
double f(double a[], double x);
int main(){double arr[4], a, b;for(int i = 0; i < 4; i++)scanf("%lf", arr + i);scanf("%lf%lf", &a, &b);while(b - a > 0.001 && f(arr, b)*f(arr, a) <= 0){if(f(arr, b) == 0){printf("%.2f", b);return 0;}else if(f(arr, a) == 0){printf("%.2f", a);return 0;}else{double mid = (a + b) / 2;if(f(arr, a)*f(arr, mid) < 0) b = mid;else a = mid;}}printf("%.2f", (a + b) / 2);return 0;
}double f(double a[], double x){double ans = 0, power = 1;for(int i = 3; i >= 0; i--){ans += a[i]*power;power *= x;}return ans;
}
solution2
#include <stdio.h>
#include <math.h>
double f(double a[], double x);
int main(){double arr[4], a, b, mid;for(int i = 0; i < 4; i++)scanf("%lf", arr + i);scanf("%lf%lf", &a, &b);while(b - a > 0.001){mid = (a + b) / 2;if(f(arr, mid) == 0) break;else if(f(arr, a)*f(arr, mid) < 0) b = mid;else a = mid;}printf("%.2f", (a + b) / 2);return 0;
}double f(double a[], double x){double ans = 0, power = 1;for(int i = 3; i >= 0; i--){ans += a[i]*power;power *= x;}return ans;
}
solution3
#include <stdio.h>
double f(double a[], double x);
int main(){double arr[4], a, b, mid;for(int i = 0; i < 4; i++)scanf("%lf", arr + i);scanf("%lf%lf", &a, &b);while(b - a > 0.001){mid = (a + b) / 2;if(f(arr, mid) == 0) break;else if(f(arr, a)*f(arr, mid) < 0) b = mid;else a = mid;}printf("%.2f", (a + b) / 2);return 0;
}double f(double a[], double x){return a[0]*x*x*x + a[1]*x*x + a[2]*x + a[3];
}
这篇关于二分法求多项式单根的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
