#include<cstring>
#include<string.h>
#include<stdio.h>
#include<iostream>
#include<stdlib.h>
#include<math.h>
using namespace std;
void main( )
{ int n;
char c;
void sub11(),sub22(),sub33(),sub44(),sub55(),sub66(),sub77();
do
{ system("cls");
cout<<"\n\n\n" ;
cout<<"\n 1.数的合法输入并相加 " ;
cout<<"\n 2.线性插值之拉格朗日法" ;
cout<<"\n 3.二次插值（抛物线插值）之牛顿插值法" ;
cout<<"\n 4.n次插值之拉格朗日法 " ;
cout<<"\n 5.分段线性插值 " ;
cout<<"\n 6.最小二乘" ;
cout<<"\n 7.高斯消去 " ;
cout<<"\n 8. 退出" ;
cout<<"\n Please Input your choice!(1--8)\n" ;
scanf("%d",&n);
switch (n)
{
case 1: sub11( ); break;
case 2: sub22( ); break;
case 3: sub33( ); break;
case 4: sub44( ); break;
case 5: sub55( ); break;
case 6: sub66( ); break;
case 7: sub77( ); break;
case 8: goto end;
default : cout<<"\n输入有误\n" ;
}
cout<<"\n press anykey to continue\n" ;
cin>>c;
c=getchar( );
} while(n!=8);
end: ;
}
int is_number(const char *p)
{
int flag0=0;
int flag1=0;
int flag3=0;
while(*p)
{
if (*p<'0' || *p>'9')
{
if (*p!='.' && *p!='-') return -1;
if (*p=='.') flag0++;
if (*p=='-')
{
if (flag3>0) return -1;
else flag1++;
}
if (flag0 > 1 || flag1>1 ) return -1;
}
p++;
flag3++;
}
if (flag0==1)
return 1; //float
else
return 0 ;//integer
}
void sub11()
{
char s1[10],s2[10];
char *p1,*p2;
cout << "input first:\n";
cin >> s1;
cout << "input second:\n";
cin >>s2;
p1=s1;
p2=s2;
if (is_number(p1)>=0 && is_number(p2)>=0)
{ cout<<'\n'<<atof(p1)+atof(p2); }
else
{ cout<<s1<<s2; }
}
void sub22()
{
float x0,y0,x1,y1,xm,x,ym;float s;
cout<<"please input two points:(x0,y0),(x1,y1) and (x0!=x1)"<<endl;
cin>>x0>>y0>>x1>>y1;
ym=y0*((x-x1)/(x0-x1))+y1*((x-x0)/(x1-x0));
s=(y1-y0)/(x1-x0);
cout<<"please input xm:"<<endl;
cin>>xm;
cout<<"ym="<<(y0*((xm-x1)/(x0-x1))+y1*((xm-x0)/(x1-x0)))<<endl;
cout<<"方程是:y-"<<y1<<"="<<s<<"(x-"<<x1<<")"<<endl;
}
void sub33()
{
float x0,y0,x1,y1,x2,y2,xm,x,ym;float y01,y12,y012;
cout<<"please input two points:(x0,y0),(x1,y1),(x2,y2)and (x0!=x1!=x2)"<<endl;
cin>>x0>>y0>>x1>>y1>>x2>>y2;
y01=(y0-y1)/(x0-x1);y12=(y1-y2)/(x1-x2);y012=(y01-y12)/(x0-x2);
ym=y0+(x-x0)*y01+(x-x0)*(x-x1)*y012;
cout<<"please input xm:"<<endl;
cin>>xm;
cout<<"ym="<< y0+(xm-x0)*y01+(xm-x0)*(xm-x1)*y012<<endl;
}
void sub44()
{
int n,i,j;
double xm,ym=0;
double x[200],y[200], l[200];
cout << "input n=";
cin >> n;
cout << endl;
cout << "input xm=";
cin >> xm;
cout << endl;
cout << "你输入的xm="<<xm<<"那么 ym=?"<< endl;
for(i=0;i<=n;i++)
{
cout << "x[" << i << "]=";
cin >> x[i];
cout << endl;
}
for(i=0;i<=n;i++)
{
cout << "y[" << i << "]=";
cin >> y[i];
cout << endl;
}
for(i=0;i<=n;i++)
{
double b=1,c=1;
for(j=0;j<=n;j++)
{
if(i!=j){b=b*(xm-x[j]);}
}
for(j=0;j<=n;j++)
{
if(i!=j){c=c*(x[i]-x[j]);}
}
l[i]=b/c;
}
for(i=0;i<=n;i++)
{
ym=ym+y[i]*l[i];
}
cout << ym<<endl;
}
void sub55()
{
int i,n;
double xm,ym;
cout << "input xm=";
cin >> xm;
cout << endl;
cout << "你输入的xm="<<xm<<"那么 ym=?"<< endl;
cout << "你现在拥有几个点的坐标？至少得有2个点";
cin >> n;
cout << endl;
cout << "那么你现在一一输入这些坐标\n";
double x[200],y[200];
for(i=0;i<n;i++)
{
cout << "x[" << i << "]=";
cin >> x[i];
cout << endl;
}
for(i=0;i<n;i++)
{
cout << "y[" << i << "]=";
cin >> y[i];
cout << endl;
}
for (i = 0; i<n; i++)
{
if (x[i] <= xm && xm<=x[i+1] )
ym=(y[i]*((xm-x[i+1])/(x[i]-x[i+1]))+y[i+1]*((xm-x[i])/(x[i+1]-x[i])));
cout<<"ym="<<ym<<endl;
double s=(y[i+1]-y[i])/(x[i+1]-x[i]);
cout<<"该段对应的方程是:y-"<<y[i+1]<<"="<<s<<"(x-"<<x[i+1]<<")"<<endl;
cout<<"其中x的区间为【"<<x[i]<<"，"<<x[i+1]<<"】"<<"显然xm="<<xm<<"在这一区间内"<<endl;
break;
}
}
void sub66()
{
int n,i;
float a,b,xsum=0,ysum=0,squaresum=0,xy=0,x[25],y[25],D;
cout<<"请输入点数n:\n";
cin>>n;
cout<<"请输入所有点坐标:\n";
for(i=0;i<n;i++)
{
cin>>x[i]>>y[i];
xsum+=x[i];
ysum+=y[i];
squaresum+=x[i]*x[i];
xy+=x[i]*y[i];
}
D=n*squaresum-xsum*xsum;
if(D==0)
cout<<"原方程无解或有多组解!\n";
else
{
a=(squaresum*ysum-xsum*xy)/D;
b=(n*xy-xsum*ysum)/D;
}
cout<<"y="<<a<<"+"<<b<<"x\n";
}
void sub77()
{
cout<<"测试例子："<<endl;
cout<<"a[1][1] = 4 a[1][2] = 1 a[1][3] = 1 a[1][4] = 1"<<endl;
cout<<"a[2][1] = 1 a[2][2] = -1 a[2][3] = 2 a[2][4] = -2"<<endl;
cout<<"a[3][1] = 2 a[3][2] = 2 a[3][3] = -1 a[3][4] = 2"<<endl;
cout<<"此方程组的解为 ："<<endl;
cout<<"x[1] = 1 x[2] = -1 x[3] = -2 "<<endl;
cout<<endl;
int n;
cout << "你有几行方程? 不限规模";
cin >> n;
cout << endl;
cout << "请输入" <<n<<"行"<<n+1<<" 列数据:"<<endl;
cout <<"[注：其中每一行的最后一个值代表b1b2...b"<<n<<"]"<<endl;
float **a ;
a=new float*[n+1];
int i,j;
for(i=1;i<=n;i++)
{
a[i]=new float[n+1];
for(j=1;j<=n+1;j++)
{
cout << "a[" << i << "]["<<j<<"]= ";
cin >> a[i][j];
}
cout << endl;
}//输入数据
for(i=1;i<=n;i++)
{
if(a[i][i]==0)
{
cout<<"无解！"<<endl;
goto end;
}
}
float t;
for(i=1;i<=n;i++)
{
t=1/a[i][i];
for(int l=1;l<=n+1;l++)
a[i][l]=t * a[i][l];
for(int m=i+1;m<n+1;m++)
{
t=a[m][i];
for(j=1;j<=n+1;j++)
a[m][j]=t * a[i][j] - a[m][j];
}
}//转化过程
cout<<endl<<"转化为上三角矩阵："<<endl;
for( i=1;i<=n;i++)
{
for(j=1;j<=n+1;j++)
cout<<"a["<<i<<"]["<<j<<"] = "<<a[i][j]<<'\t';
cout<<endl;
}//输出上三角
cout<<endl<<"此方程组的解为 ："<<endl;
float *x;
x = new float [n];
x[n] = a[n][n+1]/a[n][n];
for( i=n-1;i>0;i--)
{
for(j=n;j>i;j--)
a[i][n+1] = a[i][n+1] - a[i][j] * x[j];
x[i] = a[i][n+1] / a[i][i];
}//计算解
for( i=1;i<=n;i++)
cout<<"x["<<i<<"] = "<<x[i]<<"\t";
cout<<endl;//输出解
end:;
}
