新手求帮助，我们要求用c语言写一个计算n次多项式的程序，但运行结果不准啊，求大佬看看有什么问题

[lanshanstar]

#include <stdio.h>
#include <math.h>

const double eps = 1e-20;


int main()
{


int n, y, k, i;
double m,p,u;
double j,sum;
printf("The function is like y = [(arr[n])x^n] + [(arr[n-1])x^(n-1)] + [(arr[n-2])x^(n-2)] + ... + [(arr[1])x] + arr[0]\n");
printf("Input n :");
scanf("%d",&n);
i = n;
double arr[n];
for(i=n; i>=0; i--)
{
printf("Input the arr[%d]: ",i);
scanf("%lf", &arr[i]);
}

double f(double x)
{

y = 0;
for(k=n; k>=0; k--)
{
y = y + arr[k] * pow(x,k);
}
return y;
}

printf("\nEnter the range[m,p]:");
scanf("%lf%lf",&m,&p);
if(f(m)*f(p)<0)
{
while(fabs(m-p)>eps)
{
i=(m+p)/2.0;
sum=f(i);
printf("[%lf %lf]\n",m,p);
if(fabs(sum)<eps)
{
break;
}
else
{if(f(i)*f(m)<0)
{
p=i;
}
else if(f(i)*f(p)<0)
{
m=i;
}
}
}
j=(m+p)/2;
printf("\nThe nearest result is:x=%lf\n",j);
}

else
{
printf("Wrong");
}
return 0;
}

[lanshanstar]

多项式求零点，比如(x+1)(x+2)(x+3) 这个函数, 我输入范围[-2.5,-4]，理论上的输出结果是-3，但它最后的输出结果是-3.25

[赵4老师]

#include <stdio.h>
#include <math.h>
const double eps = 1e-20;
double arr[10];
int n;
int n, k, i;
double m,p,u,y;
double j,sum,v;
double f(double x) {
    y = 0.0;
    for (k=n; k>=0; k--) {
        y = y + arr[k] * pow(x,(double)k);
    }
    return y;
}
int main() {
    printf("The function is like y = [(arr[n])x^n] + [(arr[n-1])x^(n-1)] + [(arr[n-2])x^(n-2)] + ... + [(arr[1])x] + arr[0]\n");
    printf("Input n :");
    scanf("%d",&n);
    if (n<1 || 10<=n) {
        printf("1<=n<10\n");
        return 1;
    }
    i = n;
    for (i=n; i>=0; i--) {
        printf("Input the arr[%d]: ",i);
        scanf("%lf", &arr[i]);
    }


    printf("\nEnter the range[m,p]:");
    scanf("%lf%lf",&m,&p);
    if (f(m)*f(p)<0.0) {
        while (fabs(m-p)>eps) {
            v=(m+p)/2.0;
            sum=f(v);
            printf("[%lf  %lf]\n",m,p);
            if (fabs(sum)<eps) {
                break;
            } else {
                if(f(v)*f(m)<0) {
                    p=v;
                } else if(f(v)*f(p)<0) {
                    m=v;
                }
            }
        }
        j=(m+p)/2.0;
        printf("\nThe nearest result is:x=%lf\n",j);
    } else {
        printf("Wrong");
    }
    return 0;
}

[lanshanstar]

@赵4老师 真的太感谢了，真的是大佬啊

[赵4老师]