void SPL(int n, double *x, double *y, int ni, double *xi, double *yi)； 是你所要。
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已知 n 个点 x,y; x 必须已按顺序排好。要插值 ni 点，横坐标 xi[], 输出 yi[]。
程序里用double 型，保证计算精度。
SPL调用现成的程序。
现成的程序很多。端点处理方法不同，结果会有不同。想同matlab比较，你需 尝试 调用 spline（）函数 时，令 end1 为 1, 设 slope1 的值，令 end2 为 1 设 slope2 的值。
#include stdio.h
#include math.h
int spline (int n, int end1, int end2,
double slope1, double slope2,
double x[], double y[],
double b[], double c[], double d[],
int *iflag)
{
int nm1, ib, i, ascend;
double t;
nm1 = n - 1;
*iflag = 0;
if (n 2)
{ /* no possible interpolation */
*iflag = 1;
goto LeaveSpline;
}
ascend = 1;
for (i = 1; i n; ++i) if (x[i] = x[i-1]) ascend = 0;
if (!ascend)
{
*iflag = 2;
goto LeaveSpline;
}
if (n = 3)
{
d[0] = x[1] - x[0];
c[1] = (y[1] - y[0]) / d[0];
for (i = 1; i nm1; ++i)
{
d[i] = x[i+1] - x[i];
b[i] = 2.0 * (d[i-1] + d[i]);
c[i+1] = (y[i+1] - y[i]) / d[i];
c[i] = c[i+1] - c[i];
}
/* ---- Default End conditions */
b[0] = -d[0];
b[nm1] = -d[n-2];
c[0] = 0.0;
c[nm1] = 0.0;
if (n != 3)
{
c[0] = c[2] / (x[3] - x[1]) - c[1] / (x[2] - x[0]);
c[nm1] = c[n-2] / (x[nm1] - x[n-3]) - c[n-3] / (x[n-2] - x[n-4]);
c[0] = c[0] * d[0] * d[0] / (x[3] - x[0]);
c[nm1] = -c[nm1] * d[n-2] * d[n-2] / (x[nm1] - x[n-4]);
}
/* Alternative end conditions -- known slopes */
if (end1 == 1)
{
b[0] = 2.0 * (x[1] - x[0]);
c[0] = (y[1] - y[0]) / (x[1] - x[0]) - slope1;
}
if (end2 == 1)
{
b[nm1] = 2.0 * (x[nm1] - x[n-2]);
c[nm1] = slope2 - (y[nm1] - y[n-2]) / (x[nm1] - x[n-2]);
}
/* Forward elimination */
for (i = 1; i n; ++i)
{
t = d[i-1] / b[i-1];
b[i] = b[i] - t * d[i-1];
c[i] = c[i] - t * c[i-1];
}
/* Back substitution */
c[nm1] = c[nm1] / b[nm1];
for (ib = 0; ib nm1; ++ib)
{
i = n - ib - 2;
c[i] = (c[i] - d[i] * c[i+1]) / b[i];
}
b[nm1] = (y[nm1] - y[n-2]) / d[n-2] + d[n-2] * (c[n-2] + 2.0 * c[nm1]);
for (i = 0; i nm1; ++i)
{
b[i] = (y[i+1] - y[i]) / d[i] - d[i] * (c[i+1] + 2.0 * c[i]);
d[i] = (c[i+1] - c[i]) / d[i];
c[i] = 3.0 * c[i];
}
c[nm1] = 3.0 * c[nm1];
d[nm1] = d[n-2];
}
else
{
b[0] = (y[1] - y[0]) / (x[1] - x[0]);
c[0] = 0.0;
d[0] = 0.0;
b[1] = b[0];
c[1] = 0.0;
d[1] = 0.0;
}
LeaveSpline:
return 0;
}
double seval (int n, double u,
double x[], double y[],
double b[], double c[], double d[],
int *last)
{
int i, j, k;
double w;
i = *last;
if (i = n-1) i = 0;
if (i 0) i = 0;
if ((x[i] u) || (x[i+1] u))
{
i = 0;
j = n;
do
{
k = (i + j) / 2;
if (u x[k]) j = k;
if (u = x[k]) i = k;
}
while (j i+1);
}
*last = i;
w = u - x[i];
w = y[i] + w * (b[i] + w * (c[i] + w * d[i]));
return (w);
}
void SPL(int n, double *x, double *y, int ni, double *xi, double *yi)
{
double *b, *c, *d;
int iflag,last,i;
b = (double *) malloc(sizeof(double) * n);
c = (double *)malloc(sizeof(double) * n);
d = (double *)malloc(sizeof(double) * n);
if (!d) { printf("no enough memory for b,c,d\n");}
else {
spline (n,0,0,0,0,x,y,b,c,d,iflag);
if (iflag==0) printf("I got coef b,c,d now\n"); else printf("x not in order or other error\n");
for (i=0;ini;i++) yi[i] = seval(ni,xi[i],x,y,b,c,d,last);
free(b);free(c);free(d);
};
}
main(){
double x[6]={0.,1.,2.,3.,4.,5};
double y[6]={0.,0.5,2.0,1.6,0.5,0.0};
double u[8]={0.5,1,1.5,2,2.5,3,3.5,4};
double s[8];
int i;
SPL(6, x,y, 8, u, s);
for (i=0;i8;i++) printf("%lf %lf \n",u[i],s[i]);
return 0;
}
#includestdio.h
#includemath.h
double Lagrange1(double *x, double *y, double xx) //拉格郎日插值
{
int i,j;
double *a,yy=0.000;
a=new double[6];
for(i=0;i 6;i++)
{
a[i]=y[i];
for(j=0;j 6;j++)
if(j!=i)
a[i]*=(xx-x[j])/(x[i]-x[j]);
yy+=a[i];
}
delete a;
return yy;
}
double Lagrange2(double *x, double *y, double input) //分段线性插值
{
double output;
int i;
for (i=0;i5;i++)
{
if (x[i] = input x[i+1] = input)
{
output=y[i] +(y[i+1]-y[i])*(input-x[i])/(x[i+1]-x[i]);
break;
}
}
return output;
}
double Lagrange3(double *x,double *y,double u) //分段二次插值
{
int i,k=0;
double v;
for(i=0;i6;i++)
{
if(ux[1])
{
k=0;
v=y[k]*(u-x[k+1])*(u-x[k+2])/((x[k]-x[k+1])*(x[k]-x[k+2]))+y[k+1]*(u-x[k])*(u-x[k+2])/((x[k+1]-x[k])*(x[k+1]-x[k+2]))+y[k+2]*(u-x[k])*(u-x[k+1])/((x[k+2]-x[k])*(x[k+2]-x[k+1]));
}
if((x[i]uu=x[i+1])(fabs(u-x[i])=fabs(u-x[i+1])))
{
k=i-1;
v=y[k]*(u-x[k+1])*(u-x[k+2])/((x[k]-x[k+1])*(x[k]-x[k+2]))+y[k+1]*(u-x[k])*(u-x[k+2])/((x[k+1]-x[k])*(x[k+1]-x[k+2]))+y[k+2]*(u-x[k])*(u-x[k+1])/((x[k+2]-x[k])*(x[k+2]-x[k+1]));
}
if ((x[i]uu=x[i+1])fabs(u-x[i])fabs(u-x[i+1]))
{
k=i;
v=y[k]*(u-x[k+1])*(u-x[k+2])/((x[k]-x[k+1])*(x[k]-x[k+2]))+y[k+1]*(u-x[k])*(u-x[k+2])/((x[k+1]-x[k])*(x[k+1]-x[k+2]))+y[k+2]*(u-x[k])*(u-x[k+1])/((x[k+2]-x[k])*(x[k+2]-x[k+1]));
}
if(ux[4])
{
k=3;
v=y[k]*(u-x[k+1])*(u-x[k+2])/((x[k]-x[k+1])*(x[k]-x[k+2]))+y[k+1]*(u-x[k])*(u-x[k+2])/((x[k+1]-x[k])*(x[k+1]-x[k+2]))+y[k+2]*(u-x[k])*(u-x[k+1])/((x[k+2]-x[k])*(x[k+2]-x[k+1]));
}
}
return v;
}
void main()
{
double x[6] = {0.0, 0.1, 0.195, 0.3, 0.401, 0.5},y[6] = {0.39894,0.39695,0.39142,0.38138,0.36812,0.35206};
double u;
scanf("%lf",u);
printf("%f\n",Lagrange1(x,y,u)); //拉格郎日插值
printf("%f\n",Lagrange2(x,y,u)); //分段线性插值
printf("%f\n",Lagrange3(x,y,u)); //分段二次插值
}
#include
int main()
{
int x,y;
scanf("%d",x);
if(0xx10) y=3*x+2;
else
{if(x=0) y=0;
else
{if (x0) y=x*x;
else printf("go die\n");
}
}
printf("%d",y);
return 0;
}该程序的分段函数如下:
f(x)=3x+2 (0x10)
f(x)=1 (x=0)
f(x) = x*x (x0)
#include stdio.h
#include math.h
void main()
{
float x;
double y;
printf("Please input the value of x:");
scanf("%f",x);
if(x=-10x=4)
{
y=fabs(x-2);
printf("y=%.2f\n",y);
}
else if(x=5x=7)
{
y=x+10;
printf("y=%.2f\n",y);
}
else if(x=8x=12)
{
y=pow(x,4);
printf("y=%.2f\n",y);
}
else
printf("No answer\n");
}
我这里有2个程序,第一个用了2个函数
第二个用了1个函数,感觉误差小些
说下用法吧
先输入数据 两个两个的输,中间用空格隔开,比如
please input data1: 11 11.08然后回车
依次输入完五组数据
完了你想查某个温度的溶解度,他要求你输入温度你输入11.5
它就输出对应的溶解度,然后他提示你是否继续查溶解度,是就输入y,想结束程序
就输入n.
#includestdio.h
#includestdlib.h
#define NUMBER 5
typedef struct
{
double x;
double y;
}Point;
double * parabola(Point*, Point*, Point*);
double calculate(double* , double );
int main()
{
double x = 0, y = 0;
double* f = NULL;
int i = 0, n = 0;//n为要插的中间那个点的位置
char c = 'y';
Point p[NUMBER];
for(i = 0; i NUMBER; i++)
{
printf("Please input the point%d:", i+1);
scanf("%lf%lf", p[i].x, p[i].y);
while(10 != getchar())
{
continue;
}
}
while('n' != c)
{
printf("Please input the temperature:");
scanf("%lf", x);
while(10 != getchar())
{
continue;
}
//计算插值点的位子
if(2 * x = (p[1].x + p[2].x)) n = 1;
else if(2 * x = (p[NUMBER-2].x + p[NUMBER-1].x)) n = NUMBER - 2;
else n = (int)NUMBER / 2;
printf("%d\n", n);
f = parabola(p[n-1],p[n],p[n+1]);//计算抛物线方程
y = calculate(f,x);//计算溶解度
printf("The solubility at this temperature is %lf\n", y);
printf("Continue?(y/n) ");
scanf("%c", c);
while(10 != getchar())
{
continue;
}
//释放内存
free(f);
}
return 0;
}
//下面为计算抛物线方程的函数
//设抛物线函数为y = a * x^2 + b * x + c
//以下f[0]为a,f[1]为b,f[2]为c
double * parabola(Point* p1, Point* p2, Point* p3)
{
double temp1 = 0, temp2 = 0;
double * f = NULL;
f = (double*)calloc(3,sizeof(double));
if(NULL == f)
{
printf("Calloc failed!\n");
return NULL;
}
temp1 = (p2-y - p1-y)/(p2-x - p1-x);
temp2 = (p3-y - p2-y)/(p3-x - p2-x);
f[0] = (temp2 - temp1)/(p3-x - p1-x);
f[1] = temp1 - f[0] * (p1-x + p2-x);
f[2] = p1-y - p1-x * (p1-x * f[0] + f[1]);
return f;
}
//根据温度计算溶解度
double calculate(double* f, double x)
{
double y;
y = x * (f[0] * x + f[1]) + f[2];
return y;
}
//这里是第二个程序
#includestdio.h
#includestdlib.h
#define NUMBER 5
typedef struct
{
double x;
double y;
}Point;
double parabola(Point*, Point*, Point*, double);
int main()
{
double x = 0, y = 0;
int i = 0, n = 0;//n为要插的中间那个点的位置
char c = 'y';
Point p[NUMBER];
for(i = 0; i NUMBER; i++)
{
printf("Please input the data%d:", i+1);
scanf("%lf%lf", p[i].x, p[i].y);
while(10 != getchar())
{
continue;
}
}
while('n' != c)
{
printf("Please input the temperature:");
scanf("%lf", x);
while(10 != getchar())
{
continue;
}
//计算插值点的位子
if(2 * x (p[1].x + p[2].x)) n = 1;
else if(2 * x (p[NUMBER-2].x + p[NUMBER-1].x)) n = NUMBER - 2;
else n = (int)NUMBER / 2;
//printf("%d\n", n);
y = parabola(p[n-1],p[n],p[n+1],x);
printf("The solubility at this temperature is %lf\n", y);
printf("Continue?(y/n) ");
scanf("%c", c);
while(10 != getchar())
{
continue;
}
}
return 0;
}
//直接套用公式计算溶解度
double parabola(Point* p1, Point* p2, Point* p3, double x)
{
double temp1 = 0, temp2 = 0, temp3 = 0, y = 0;
double a = 0, b = 0, c = 0;
temp1 = (p2-y - p1-y) * (x - p1-x);
temp2 = (p3-y - p1-y) * (p2-x - p1-x) - (p2-y - p1-y) * (p3-x - p1-x);
temp3 = (p3-x - p1-x) * (p3-x - p1-x) * (p2-x - p1-x);
y = p1-y + temp1 / (p2-x - p1-x) + temp2 * (x - p1-x) * (x - p2-x) / temp3;
return y;
}
该程序误差绝对小 但结果和你不是完全符合
你的数据好象有点问题 14度的溶解度不太正常
