#include <stdio.h> #include <stdlib.h> #include <string.h> #include <math.h> #define NR(x) sizeof(x)/sizeof(x[0]) //最小2乘法實現(xiàn) void lsqe(const double *x, const double *y, int n, double *b1, double *b0) { int i; double sumx,sumy,sumx2,sumxy; sumx = 0.0; sumy = 0.0; sumx2 = 0.0; sumxy = 0.0; //計算N次 for (i = 0; i < n; i++) { //將橫坐標(biāo)方向的x值進行累加 sumx = sumx + x[i]; //將縱坐標(biāo)方向的y值進行累加 sumy = sumy + y[i]; sumx2 = sumx2 + pow(x[i], 2.0); sumxy = sumxy + (x[i] * y[i]); } //根據(jù)公式求解b1和b0的值 *b1 = (sumxy - ((sumx * sumy)/(double)n)) / (sumx2-(pow(sumx,2.0)/(double)n)); *b0 = (sumy - ((*b1) * sumx)) / (double)n; return; } int main(void) { double x[] = {1.1 , 1.2 , 1.3 , 1.4 , 1.5 ,1.6} ; double y[] = {4.1 , 4.2 , 4.3 , 4.4 , 4.5 , 4.6} ; double b0 , b1 ; lsqe(x,y,NR(x),&b0,&b1); printf("%lf,%lf\n",b0,b1); return 0 ; }