最小二乘法c语言求长度,计算方法中最小二乘法如何用C语言编程

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yydc1108gd

2014.05.17

采纳率：44%    等级：9

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#include

#include

#define epsilon 1e-6

void nihe1(int n,int m,float sum_x,float sum_y,float sum_xy,float x2);

void nihe2(int n,int m,float sum_x,float sum_y,float sum_xy,float x2,float x2y,float x3,float x4);

int main(){

float x[100]={0.0};

float y[100]={0.0};

int n,i,flag=1;

float sum_y=0.0,sum_x=0,x2=0,sum_xy=0.0,x3=0,x4=0,x2y=0.0;

printf("请你输入需要测试的数据(先输入x[]，后输入y[])的个数:");

scanf("%d",&n);

for(i = 0; i < n; i++){

scanf("%f",&x[i]);}

for(i = 0; i < n; i++){

scanf("%f",&y[i]);}

for(i = 0; i < n; i++){

sum_x += x[i];

sum_y += y[i];

sum_xy += x[i]*y[i];

x2 += x[i]*x[i];

x2y += x[i]*x[i]*y[i];

x3 += x[i]*x[i]*x[i];

x4 += x[i]*x[i]*x[i]*x[i];}

printf("---------------请你输入的要拟合的函数------------------

");

printf(" 1、拟合一次函数

");

printf(" 2、拟合二次函数

");

scanf("%d",&flag);

switch(flag){

case 1:

nihe1(n,flag+1,sum_x,sum_y,sum_xy,x2); break;

case 2:

nihe2(n,flag+1,sum_x,sum_y,sum_xy,x2,x2y,x3,x4); break;

default:

printf("ERROR

");}

return 0;}

void nihe1(int n,int m,float sum_x,float sum_y,float sum_xy,float x2){

int i,k,j;

float t,s=0;

float a[2][3] = {{(float)n,sum_x,sum_y},{sum_x,x2,sum_xy}};

n=m;

//if(m == 3)

// a[3][4] = {{n,sum_x,sum_y},{sum_x,x2,x3,sum_xy},{x2,x3,x4,x2y}};

for(k=0;k

for(i=k+1;i

if( abs((int)a[i][k]) > abs((int)a[k][k]) )

for(j=k;j

t=a[k][j];

a[k][j]=a[i][j];

a[i][j]=t; }

if( abs((int)a[k][k]) < epsilon) {

printf("

Error,主元消去法 cann't be durable,break at %d!

",k+1);

return; }

for(i=k+1;i

a[i][k]=a[i][k] / a[k][k];

for(j=k+1;j

a[i][j]=a[i][j]-a[i][k] * a[k][j]; }}

a[n-1][n]=a[n-1][n] / a[n-1][n-1];

for(k=n-2;k>=0;k--) {

s=0;

for(j=k+1;j

s+=a[k][j]*a[j][n];

a[k][n]=( a[k][n]-s ) / a[k][k]; }

printf("

*****The Result*****

");

for(i=0;i

printf(" x[%d]=%.4f

",i+1,a[i][n]);

printf("函数为：p(x) = %.4f + (%.4f)*x

",a[0][n],a[1][n]);

getchar();}

void nihe2(int n,int m,float sum_x,float sum_y,float sum_xy,float x2,float x2y,float x3,float x4){

int i,k,j;

float t,s=0;

float a[3][4]=

{{(float)n,sum_x,x2,sum_y},{sum_x,x2,x3,sum_xy},{x2,x3,x4,x2y}};

n=m;

for(k=0;k

for(i=k+1;i

if( abs((int)a[i][k]) > abs((int)a[k][k]) )

for(j=k;j

t=a[k][j];

a[k][j]=a[i][j];

a[i][j]=t; }

if( abs((int)a[k][k]) < epsilon) {

printf("

Error,主元消去法 cann't be durable,break at %d!

",k+1);

return; }

for(i=k+1;i

a[i][k]=a[i][k] / a[k][k];

for(j=k+1;j

a[i][j]=a[i][j]-a[i][k] * a[k][j]; } }

a[n-1][n]=a[n-1][n] / a[n-1][n-1];

for(k=n-2;k>=0;k--) {

s=0;

for(j=k+1;j

s+=a[k][j]*a[j][n];

a[k][n]=( a[k][n]-s ) / a[k][k]; }

printf("

*****The Result*****

");

for(i=0;i

printf(" x[%d]=%.4f

",i+1,a[i][n]);

printf("函数为：p(x) = %.4f + (%.4f)*x + (%.4f)*x*x

",a[0][n],a[1][n],a[2][n]);

getchar();}

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