#include<stdio.h>/*要求的方程为: y=ax+b。N∑xy-∑x∑y _ _其中：a= --------------- b = y - axN∑(x^2)-(∑x)^2设：A=∑xy B=∑x C=∑y D=∑(x^2)NA-BC则： a = ------- b = C/N-aB/NND-BB注：N为要拟合的点数量*/#define POINT_NUM 30 //最多输入30个点进行线性拟和typedef struct POINT //点的结构{double x;double y;}POINTS;POINTS points[POINT_NUM]; //存放点数据/*参数说明：P[POINT_NUM]：传入要线性拟合的点数据（结构体数组）N：线性拟合的点的数量K0:直线斜率参数存放地址b0:直线截距参数存放地址*/void Least_square_method(POINTS P[],int N,double *K0,double *b0){int i=0;double K=0,b=0,A=0,B=0,C=0,D=0;for(i=0;i<N;i++){A+=P[i].x*P[i].y;B+=P[i].x;C+=P[i].y;D+=P[i].x*P[i].x;}K=(N*A-B*C)/(N*D-B*B);b=C/N-K*B/N;/*将计算得到的直线参数通过指针传递到函数外部*/*K0=K;*b0=b;/*printf("A=%f\nB=%f\nC=%f\nD=%f\n",A,B,C,D);if(b>0) printf("y=%fx+%f\n",K,b);else if(b==0)printf("y=%fx\n",K);else printf("y=%fx%f\n",K,b);*/}int main(){double k,b;points[0].x=123;points[1].x=130;points[2].x=141;points[3].x=150;points[4].x=165;points[0].y=150;points[1].y=160;points[2].y=175;points[3].y=190;points[4].y=205;Least_square_method(points,5,&k,&b);if(b>0) printf("函数为：y=%fx+%f\n",k,b);else if(b<0) printf("函数为：y=%fx%f\n",k,b);else printf("函数为：y=%fx\n",k);return 0;}/*在EXCEL中计算得到的拟合函数为 y = 1.3342x - 13.187程序运算结果： 函数为：y=1.334183x-13.187113*/
