幾道數值計算題目的c語言實現

引用請註明出處：http://blog.csdn.net/int64ago/article/details/7084211

注：個人習慣原因，帶了點c++語句

一、 用牛頓法求在附近的實根，取四位有效數字。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>

using namespace std;

const double eps=1e-4;
const double x0=2;

double FunOfSet(double x)
{
	return pow(x,3.0)-3*x-1;
}

void ProGetRes()
{
	double Begin=x0-1,End=x0+1;
	double Mid;
	while(fabs(FunOfSet(Begin)-FunOfSet(End))>eps)
	{
		Mid=(Begin+End)/2;
		if(FunOfSet(Begin)*FunOfSet(Mid)<0)
			End=Mid;
		else Begin=Mid;
	}
	printf("The root of fuction beside 2 is: %.4f\n",Mid) ;
}
int main()
{
	ProGetRes();
	return 0;
}

二、 用高斯消去法求解下列方程組

    

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>

using namespace std;

const int LenOfMatrix=3;

double a[LenOfMatrix+1][LenOfMatrix+2]={{0,0,0,0,0},{0,2,-1,-1,4},{0,3,4,-2,11},{0,3,-2,4,11}};
double Res[LenOfMatrix+1];
void GussPro()
{
	int i,j,k,MaxPos,n=3;
    double mmax;
    for(k=1;k<n-1;k++)
    {
		mmax=-1;
        for(i=k;i<=n;i++)
            if(fabs(a[i][k])>mmax)
            {
                mmax=a[i][k];
                MaxPos=i;
            }
		double temp;
		for(j=1;j<=n+1;j++)
		{
		    temp=a[k][j];
            a[k][j]=a[MaxPos][j];
            a[MaxPos][j]=temp;
		}
		for(i=k+1;i<=n;i++)
            for(j=k+1;j<=n+1;j++)
				a[i][j]=a[i][j]-(a[i][k])*(a[k][j])/(a[k][k]);
	}
    Res[n]=(a[n][n+1])/(a[n][n]);
	double sum;
    for(k=n-1;k>=1;k--)
    {
		sum=0;
        for(j=k+1;j<=n;j++)
            sum+=(a[k][j])*(Res[j]);
        Res[k]=(a[k][n+1]-sum)/(a[k][k]);
    }
	cout<<"For the follow matrix :"<<endl;
	cout<<"2x1-1x2-1x3=4"<<endl;
	cout<<"3x1+4x2-2x3=11"<<endl;
	cout<<"3x1-2x2+4x3=11"<<endl;
	cout<<"The answer is:	 ";
    for(i=1;i<=n;i++)
        printf("x[%d]=%.4f  ",i,Res[i]);
	cout<<endl;
}
int main()
{
	GussPro();
    return 0;
}

三、已知函數表

x     1.1275	1.1503	1.1735	1.1972
f(x)  0.1191	0.13954	0.15932	0.17903

應用拉格朗日插值公式計算f(1.1300)的近似值。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>

using namespace std;

const int LenOfNum=4;
double x[]={1.1275,1.1503,1.1735,1.1972};
double y[]={0.1191,0.13954,0.15932,0.17903};
double Li[LenOfNum];
int i,j;
void GetLi(double x0)
{
	for(i=0;i<LenOfNum;i++)
	{
		Li[i]=1;
		for(j=0;j<LenOfNum;j++)
			if(i!=j)Li[i]*=((x0-x[j])/(x[i]-x[j]));
	}
} 

void ProLag()
{
	double x0=1.1300;
	double res=0;
	GetLi(x0);
	for(i=0;i<LenOfNum;i++)
		res+=(y[i]*Li[i]);
	printf("f(1.1300)= %f\n",res);
}
int main()
{
	ProLag();
}