输出1到最大的N位数,用递归和非递归实现，同时满足大整数的要求，很巧妙

题目：输入数字n，按顺序输出从1最大的n位10进制数。比如输入3，则输出1、2、3一直到最大的3位数即999

分析：这是一道很有意思的题目。看起来很简单，其实里面却有不少的玄机。

 

//============================================================================ // Name : 100题之输出1到最大的N位数.cpp // Author : // Version : // Copyright : Your copyright notice // Description : Hello World in C++, Ansi-style //============================================================================ #include <iostream> using namespace std; int n; void printN(int n) { char *a=new char[n+1]; int *answer=new int[n+1]; for(int i=0;i<n;i++) { a[i]='0'; answer[i]=0; } a[n]='/0'; bool temp; while(true) { int k=n-1; temp=false; for(int i=0;i<n;i++) { //cout<<"temp "<<temp<<"　i "<<i<<" answer[i]　"<<answer[i]<<endl; if(temp&&answer[i]==0) { //cout<<"(temp&&answer[i] "<<bool(temp&&answer[i])<<endl; cout<<answer[i]<<" "; } else if(answer[i]!=0) { cout<<answer[i]<<" "; temp=true; } } if(!temp&&answer[n-1]==0) cout<<answer[n-1]; cout<<endl; while(k>=0) { if(answer[k]<9) { answer[k]++; break; } else { answer[k]=0; k--; } } if(k==0) break; } } void printRecurisong(char *a,int *answer,int index,int n) { if(index==n) { int temp=false; for(int i=0;i<n;i++) { if(temp&&answer[i]==0) { cout<<answer[i]<<" "; } else if(answer[i]!=0) { cout<<answer[i]<<" "; temp=true; } } if(!temp&&answer[n-1]==0) cout<<answer[n-1]; cout<<endl; return; } for(answer[index]=0;answer[index]<=9;answer[index]++) { printRecurisong(a,answer,index+1,n); } } void printNRecurision(int n) { char *a=new char[n+1]; int *answer=new int[n+1]; for(int i=0;i<n;i++) { a[i]='0'; answer[i]=0; } a[n]='/0'; printRecurisong(a,answer,0, n); } int main() { printN(4); cout<<"recurision"<<endl; printNRecurision(4); return 0; }