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，要求依次输出数据

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 xi);

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 xi[3] = {0.5, 0.7, 0.85};
    int n = 5;

    for (int i = 0; i < 3; i++) {
        double yi = lagrange(x, y, n, xi[i]);
        printf("x = %.2f, y = %.5f\n", xi[i], yi);
    }

    return 0;
}

double lagrange(double x[], double y[], int n, double xi) {
    double yi = 0;
    for (int i = 0; i < n; i++) {
        double li = 1;
        for (int j = 0; j < n; j++) {
            if (j != i) {
                li *= (xi - x[j]) / (x[i] - x[j]);
            }
        }
        yi += li * y[i];
    }
    return yi;
}

输出结果为：

x = 0.50, y = 0.49442
x = 0.70, y = 0.74002
x = 0.85, y = 0.96224