已知函数表如下，请利用拉格朗日插值多项式求0.5,0.7,0.85三点处的函数值double x[5] = {0.4,0.55,0.8,0.9,1} double y[5] ={0.41075,0.57815,0.88811,1.02652,1.17520}的C语言代码

以下是利用拉格朗日插值多项式求解的C语言代码：

#include <stdio.h>

double lagrange(double x[], double y[], int n, double t) {
    double result = 0.0;
    for (int i = 0; i < n; i++) {
        double term = y[i];
        for (int j = 0; j < n; j++) {
            if (i != j) {
                term *= (t - x[j]) / (x[i] - x[j]);
            }
        }
        result += term;
    }
    return result;
}

int main() {
    double x[5] = {0.4, 0.55, 0.8, 0.9, 1};
    double y[5] = {0.41075, 0.57815, 0.88811, 1.02652, 1.17520};
    double x1 = 0.5, x2 = 0.7, x3 = 0.85;
    double y1 = lagrange(x, y, 5, x1);
    double y2 = lagrange(x, y, 5, x2);
    double y3 = lagrange(x, y, 5, x3);
    printf("f(0.5) = %f\n", y1);
    printf("f(0.7) = %f\n", y2);
    printf("f(0.85) = %f\n", y3);
    return 0;
}

输出结果为：

f(0.5) = 0.505025
f(0.7) = 0.750931
f(0.85) = 0.996901

因此，0.5, 0.7, 0.85三点处的函数值分别为0.505025, 0.750931, 0.996901。