#include<stdlib.h>
#include<math.h>
int main ()
{
//Inputs
double f(double x, double y, double z)
double d2f(double x, double y, double z)
double d3f(double x, double y, double z)
double a, b, tol
int iter, maxIter, n, i
float ya, yb, dy//重定义a，b
printf("Enter the initial (a) and final (b) values of x:")
scanf("%f%f",&a,&b)
printf("Enter boundary values of y at x=a (ya) and x=b (yb):")
scanf("%f%f", &ya,&yb)
printf("Enter number of subdivision n:")
scanf("%d", &n)//定义是整形。存的时候又用f??
float A[3][n+1], y[n+1]// tridiagonal. We do not use A(.,0)
double h = (b-a)/(n+1)
double alpha = y[0]
double beta = y[n+1]
const int SUBD = 0
const int DIAG = 1
const int SUPD = 2
float B[n+1]
B[0] = 0
// Initial guess at solution
for(i=1i<=ni++)
y[i] = alpha + i*h*(beta - alpha)/(b-a)
// Newton loop
for(iter=0iter <maxIteriter++) {//maxIter的值呢？？？？ 没初始化怎么使用？
后面的太乱了。。。只要你关系没有错，应该没问题的- -
先看看一元线性回归函数代码：
// 求线性回归方程：Y = a + bx
// dada[rows*2]数组：X, Y；rows：数据行数；a, b：返回回归系数
// SquarePoor[4]：返回方差分析指标: 回归平方和，剩余平方和，回归平方差，剩余平方差
// 返回值：0求解成功，-1错误
int LinearRegression(double *data, int rows, double *a, double *b, double *SquarePoor){
int m
double *p, Lxx = 0.0, Lxy = 0.0, xa = 0.0, ya = 0.0
if (data == 0 || a == 0 || b == 0 || rows < 1)
return -1
for (p = data, m = 0 m < rows m ++)
{
xa += *p ++
ya += *p ++
}
xa /= rows // X平均值
ya /= rows // Y平均值
for (p = data, m = 0 m < rows m ++, p += 2)
{
Lxx += ((*p - xa) * (*p - xa)) // Lxx = Sum((X - Xa)平方)
Lxy += ((*p - xa) * (*(p + 1) - ya)) // Lxy = Sum((X - Xa)(Y - Ya))
}
*b = Lxy / Lxx // b = Lxy / Lxx
*a = ya - *b * xa // a = Ya - b*Xa
if (SquarePoor == 0)
return 0
// 方差分析
SquarePoor[0] = SquarePoor[1] = 0.0
for (p = data, m = 0 m < rows m ++, p ++)
{
Lxy = *a + *b * *p ++
SquarePoor[0] += ((Lxy - ya) * (Lxy - ya)) // U(回归平方和)
SquarePoor[1] += ((*p - Lxy) * (*p - Lxy)) // Q(剩余平方和)
}
SquarePoor[2] = SquarePoor[0] // 回归方差
SquarePoor[3] = SquarePoor[1] / (rows - 2) // 剩余方差
return 0
}
实例计算：
double data1[12][2] = {// X Y
{187.1, 25.4},
{179.5, 22.8},
{157.0, 20.6},
{197.0, 21.8},
{239.4, 32.4},
{217.8, 24.4},
{227.1, 29.3},
{233.4, 27.9},
{242.0, 27.8},
{251.9, 34.2},
{230.0, 29.2},
{271.8, 30.0}
}
void Display(double *dat, double *Answer, double *SquarePoor, int rows, int cols)
{
double v, *p
int i, j
printf("回归方程式: Y = %.5lf", Answer[0])
for (i = 1 i < cols i ++)
printf(" + %.5lf*X%d", Answer[i], i)
printf(" ")
printf("回归显著性检验: ")
printf("回归平方和：%12.4lf 回归方差：%12.4lf ", SquarePoor[0], SquarePoor[2])
printf("剩余平方和：%12.4lf 剩余方差：%12.4lf ", SquarePoor[1], SquarePoor[3])
printf("离差平方和：%12.4lf 标准误差：%12.4lf ", SquarePoor[0] + SquarePoor[1], sqrt(SquarePoor[3]))
printf("F 检 验：%12.4lf 相关系数：%12.4lf ", SquarePoor[2] /SquarePoor[3],
sqrt(SquarePoor[0] / (SquarePoor[0] + SquarePoor[1])))
printf("剩余分析: ")
printf(" 观察值 估计值 剩余值 剩余平方 ")
for (i = 0, p = dat i < rows i ++, p ++)
{
v = Answer[0]
for (j = 1 j < cols j ++, p ++)
v += *p * Answer[j]
printf("%12.2lf%12.2lf%12.2lf%12.2lf ", *p, v, *p - v, (*p - v) * (*p - v))
}
system("pause")
}
int main()
{
double Answer[2], SquarePoor[4]
if (LinearRegression((double*)data1, 12, &Answer[0], &Answer[1], SquarePoor) == 0)
Display((double*)data1, Answer, SquarePoor, 12, 2)
return 0
}
#include <stdio.h>#define N 10
int main(){
int a[N*N],x=0,y=0,m=0,i
for(i=0i<N*Ni++){
a[x+y*N]=i
x+=((m+1)&1)*(1-m)
y+=((m+0)&1)*(2-m)
if(x==y||x+y==N-1){
m=++m&3
if(!m)x=++y
}
}
for(i=0i<N*Ni++){
printf("%02d%c",a[i],(i%N+1)/N*10)
}
getchar()
}
