拉格朗日-
#include <iostream>
using namespace std;
int main()
{
    float x,y,z,X[50],Y[50];
    int j,m,n;
    cout<<"请输入要计算的x,x=";
        cin>>x;
    cout<<"请输入插值节点个数n,n=";
        cin>>n;
    cout<<"请输入"<<n+1<<"组X,Y的值"<<endl;
    for(j=0;j<=n;j++)
        cin>>X[j]>>Y[j];
    y=0;
    for(m=0;m<=n;m++)
    {
        z=1;
        for(j=0;j<=n;j++)
            if(j!=m)
                z=z*(x-X[j])/(X[m]-X[j]);
            y=y+z*Y[m];
    }
    cout<<"拉格朗日插值的结果为:"<<endl;
           cout<<"当x="<<x<<"时,  y="<<y<<endl;
        return 0;
}


欧拉-

#include<iostream.h>
int main()
{
    double x0,y0,a,b,h,x,y,Y;
    int n,i;
    cout<<"请输入初始值"<<endl;
    cin>>x0>>y0;
    cout<<"请输入取值范围"<<endl;
    cin>>a>>b;
    cout<<"请输入步长"<<endl;
    cin>>h;
    x=x0,y=y0;
    n=(b-a)/h;
    cout<<"结果为："<<endl;
    for(i=1;i<n+1;i++)
    {
        Y=y+h*(y-2*(x+(i-1)*h)/y);
            cout<<i*h<<"      ";
            cout<<Y<<endl;
        y=Y;
    }
    return 0;
}


改进
的欧拉-
#include<iostream>
using namespace std;
float f(float x,float y);
int main()
{
 float x0,y0,a,b,x1,y1,yp,yc,h;
 int i,n;
 cout<<"请输入已知的一组x0,y0的值:";
 cin>>x0>>y0;
 cout<<"请输入步长h, h=";
 cin>>h;
 cout<<"请输入定义域（两端的数值）:";
  cin>>a>>b;
  n=(b-a)/h;
  cout<<" n"<<"   Xn"<<"    Yn"<<endl;
  for(i=1;i<=n+1;i++)
  {
   yp=y0+h*f(x0,y0);
   x1=x0+h;
   yc=y0+h*f(x1,yp);
   y1=(yp+yc)/2;
   cout.width(2);
   cout<<i;
   cout.width(6);
   cout<<x1;
   cout.width(10);
   cout<<y1;
   cout<<endl;
   x0=x1;
   y0=y1;
  }
 return 0;
}
float f(float x,float y)
{
 float z;
 z=y-2*x/y;
 return z;
}


牛顿插-
#include<iostream.h>
int main()
{
    int i,n,j;
    float X[100],Y[100],a,b[100]={0},x,y;
    cout<<"请输入节点总数n："<<endl;
    cin>>n;
    cout<<"请输入要求的x的值："<<endl;
    cin>>x;
    for(i=0;i<=n;i++)
    {
        cout<<"请输入第"<<i+1<<"组x、y的值："<<endl;
        cin>>X[i]>>Y[i];
    }
    y=Y[0];
    for(i=0;i<=n;i++)
    {
        for(j=0;j<=n;j++)
        {
            a=1.0;
            if(j!=i)
            {
                a=a*(X[i]-X[j]);
            }
        }
            b[i+1]=b[i]+Y[i]/a;
    }
    for(i=0;i<n;i++)
    {
        y=y+b[i+1]*(x-X[i]);
        
    }
    cout<<"所求的函数值为："<<y<<endl;
    return 0;
}



二分-
#include<iostream.h>
float f(float X);
int main()
{
    float a,b,x;
    cout<<"请输入估计实根所在的区间。"<<endl;
    cin>>a>>b;
    while ((a-b)*(a-b)>=0.000025)
    {
        x=(a+b)/2;
        if (f(x)*f(a)>0)
            a=x;
        else b=x;
    }
    cout<<x;
    return 0;
}
float f(float X)
{
    float z;
    z=X*X*X-X-1;
    return z;
}


四阶-
#include<iostream.h>
int main()
{
    double x0,y0,a,b,k1,k2,k3,k4,xn,yn,h,Y;
    int i,n;
    cout<<"请输入初值："<<endl;
    cin>>x0>>y0;
    cout<<"请输入范围："<<endl;
    cin>>a>>b;
    cout<<"请输入步长："<<endl;
    cin>>h;
    n=(b-a)/h;
    xn=x0;
    yn=y0;
    cout<<"结果为："<<endl;
    for(i=1;i<n+1;i++)
    {
        k1=yn-2*xn/yn;
        k2=yn+h*k1/2-(2*xn+h)/(yn+h*k1/2);
        k3=yn+h*k2/2-(2*xn+h)/(yn+h*k2/2);
        k4=yn+h*k3-2*(xn+h)/(yn+h*k3);
        Y=yn+h*(k1+2*k2+2*k3+k4)/6;
        cout<<i*h<<"    ";
        cout<<Y<<endl;
        xn=i*h;
        yn=Y;
    }
    return 0;
}



迭代
#include<iostream.h>
#include<math.h>
double f(double X);
int main()
{
    double a,b,c;
    cout<<"请输入估计实根。"<<endl;
    cin>>a;
    b=f(a);
    while ((a-b)*(a-b)>=0.000000000001)
    {
        b=f(b);
        c=b;
        a=b;
    }
    cout<<c;
    return 0;
}
double f(double X)
{
    double z;
    z=pow(X+1,1.0/3);
    return z;
}

呵呵 不错

数值分析，哈哈 终于成为历史了...