#include <stdio.h>
float f(float x);
float a3, a2, a1, a0;
int main()
{
float a, b;
scanf("%f %f %f %f", &a3, &a2, &a1, &a0);
scanf("%f %f", &a, &b);
float left, mid, right;
left = a;
right = b;
while (left <= right - 0.001 && f(left) * f(right) <= 0)
{
if (f(left) == 0)
{
printf("%.2f", &left);
return 0;
}
if (f(right) == 0)
{
printf("%.2f", right);
return 0;
}
mid = (left + right) / 2;
if (f(mid) * f(left) > 0)
{
left = mid;
}
else
{
right = mid;
}
}
printf("%.2f", mid);
return 0;
}
float f(float x)
{
float result;
result = a3 * x * x * x + a2 * x * x + a1 * x + a0;
return result;
}
基础编程题目集 7-18 二分法求多项式单根 (20分)
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转载自blog.csdn.net/qq_44458489/article/details/105399965
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