C ++ Programming Practice Guide 1.15 find out the implementation of the rewriting requirements of the number of input files, and program design practice 1.15

C ++ Programming Practice Guide 1.15 find out the implementation of the rewriting requirements of the number of input files, and program design practice 1.15

Rewrite requirement 1: using a single-chain table

#include <cstdlib>#include <iostream>using namespace std;struct LinkNode{       int data;       LinkNode *next;};class PALINDROME{      int low,up;      int a[100];      int count;      public:             PALINDROME(int t1,int t2);             int IsPalin(int x);             LinkNode* IsPalinAndStore();             void OutputResults(LinkNode* Head);};PALINDROME::PALINDROME(int t1,int t2){   count=0;   low=t1;   up=t2;}LinkNode* PALINDROME::IsPalinAndStore(){     LinkNode* Head=new LinkNode;     Head->next=NULL;     LinkNode* p=Head;     for(int i=low;i<=up;i++)     {             int x=i*i;             if(IsPalin(x))             {                LinkNode* newLinkNode=new LinkNode;                newLinkNode->next=NULL;                newLinkNode->data=i;                p->next=newLinkNode;                p=newLinkNode;             }     }     return Head;}void PALINDROME::OutputResults(LinkNode* Head){     LinkNode* p=Head->next;     cout<<"count="<<count<<endl;     cout<<"x"<<'\t'<<"x*x"<<endl;     while(p)     {        cout<<p->data<<'\t'<<p->data*p->data<<endl;        p=p->next;     }}int PALINDROME::IsPalin(int x){    int i=0,j,n;    int a[100];    while(x)    {       a[i]=x%10;       x=x/10;       i++;    }    n=i;    for(i=0,j=n-1;i<=j;i++,j--)       if(a[i]!=a[j])          return 0;       return 1;}int main(int argc, char *argv[]){    LinkNode* Head=new LinkNode;    PALINDROME p(100,1000);    Head=p.IsPalinAndStore();    p.OutputResults(Head);    system("PAUSE");    return EXIT_SUCCESS;}