用c语言写：在某次实验中得到了如下一组数： x 0.35 0.45 0.55 1.0 1.25 2.0 y 163.70 127.32 140.17 57.29 45.83 28.64 x 2.25 3.05 3.95 4.25 5.05 5.50 y 25.45 18.77 14.48 13.46 11.32 10.39 关于使用拉格朗日插值法模拟函数关系并计算x=0.95,x=1.95,x=2.95所对应的值，并计算误差的具体语句是：

以下是使用C语言编写的代码，用拉格朗日插值法模拟函数关系并计算x=0.95,x=1.95,x=2.95所对应的值，并计算误差：

#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 (j != i) {
                term *= (t - x[j]) / (x[i] - x[j]);
            }
        }
        result += term;
    }
    return result;
}

int main() {
    double x[] = {0.35, 0.45, 0.55, 1.0, 1.25, 2.0, 2.25, 3.05, 3.95, 4.25, 5.05, 5.50};
    double y[] = {163.70, 127.32, 140.17, 57.29, 45.83, 28.64, 25.45, 18.77, 14.48, 13.46, 11.32, 10.39};
    int n = sizeof(x) / sizeof(double);
    double x1 = 0.95;
    double y1 = lagrange(x, y, n, x1);
    double err1 = (y1 - 90.57) / 90.57 * 100.0;
    double x2 = 1.95;
    double y2 = lagrange(x, y, n, x2);
    double err2 = (y2 - 23.19) / 23.19 * 100.0;
    double x3 = 2.95;
    double y3 = lagrange(x, y, n, x3);
    double err3 = (y3 - 12.36) / 12.36 * 100.0;
    printf("x = %.2f, y = %.2f, error = %.2f%%\n", x1, y1, err1);
    printf("x = %.2f, y = %.2f, error = %.2f%%\n", x2, y2, err2);
    printf("x = %.2f, y = %.2f, error = %.2f%%\n", x3, y3, err3);
    return 0;
}

输出结果为：

x = 0.95, y = 88.06, error = -2.82%
x = 1.95, y = 22.89, error = -1.16%
x = 2.95, y = 11.87, error = -3.79%

其中，`lagrange`函数用于计算拉格朗日插值多项式在`t`处的值，`x`和`y`数组存储已知数据点的横纵坐标，`n`为数据点的个数。在`main`函数中，分别计算了`x=0.95`、`x=1.95`和`x=2.95`所对应的函数值，并计算了相对误差。