Algorithm title: Print 1 to the maximum n digits

Note : This article is only for learning Exchange, reprint please indicate the source, welcome reprint! today saw the sword refers to the offer on the 12th topic, such as the following:Enter the number n. The number of n bits 10 digits from 1 to the maximum is printed sequentially.

For example, if you enter 3, the,..., 999 is printed.

when I see this topic. The first feeling is to use recursion, why? First we have to start with one of our actual figures, for example 123. The number plus 1 is actually divided into two steps, such as the following: Step 1: The lowest bit plus 1; Step 2: If rounding occurs, the effect of the carry is propagated to a higher level ( this is where recursion is located ). before this algorithm problem is completed. I want to insert a " detail to determine success or failure!" "We write a program, in fact the general idea can be thought of." And often ignore the details of the consideration, such as the question to be considered, such as the following: 1. How to represent an n-digit? ( with character array ) 2. Each plus 1 is in the lowest position; 3. Add 1 if rounding occurs. The carry is propagated to the high-level number (the loop can be used here as well as recursion ); 4. If the highest bit occurs, then overflow, the overflow can be used as a sign of the last number printed; 5. When printing a number, it is only possible to start printing from a number that is not 0 of the highest digit. ( this conforms to the normal representation of the numbers )        recursive implementations such as the following:

 #include <iostream> #include <cstring>using namespace Std;bool Increment ( char *str,int length)//string plus 1, assuming that an overflow occurs, returns True, otherwise returns FALSE{IF (Str==null && length<1)//Assuming an overflow occurred. then returns False{return false;} int sum=str[length-1]-' 0 ' +1;if (sum<10)//Assuming no rounding occurs {Str[length-1]+=1;return true;} Else{if (length-1==0)//If overflow is assumed, returns False{return false;} str[length-1]= ' 0 ';//Take carry, leave the remaining task to the previous length-1 character return Increment (str,length-1);//recursive expression}}void print (char *str)//print character, When printing, remove the previous 0{bool isbegin=false;//identification to be able to start output int i;for (I=0;i<strlen (str); i++) {if (!isbegin && str[i]!= ' 0 ')/ /Assuming that the first non-0 character is found, the identity can start output {isbegin=true;} if (Isbegin) {cout<<str[i];}} Cout<<endl;} void Tomaxn (int n)//prints from 1 to the maximum number of n digits {if (n>=1) {char *str=new char[n+1];memset (str, ' 0 ', n); str[n]= ' + '; while (Increment ( STR,N)//Assuming a smooth growth of {print (str);} delete []str;}} int main () {int N;while (cin>>n) {if (n>=1 && n<=5) {TOMAXN (n);}} return 0;} 

      non-recursive implementations such as the following:

 #include <iostream> #include <cstring>using namespace Std;bool Increment ( char * str)//used to add 1 to the corresponding number of the current string, which returns true to indicate the addition of {int len=strlen (str); int current=str[len-1]-' 0 ' +1;//the digit digit plus 1, Current represents a value of 1 after int i=len-1;//to propagate the chain carry reaction after the single plus 1 while (i>=0)///with I for the present bit {if (CURRENT<10)//If plus 1 does not carry {str[i]=str[i]+ 1;break;} ELSE//assumes that the current bit takes place in the carry {if (i==0)//Assuming that the rounding is the highest bit. The overflow {return false is directly occurred;} ELSE//assumes that the carry is not the highest level, here can guarantee the i!=0, because there is a branch of 0 to deal with {str[i]= ' 0 ';//First the standard zero i=i-1;//start processing the standard of the previous current=str[i]-' 0 ' +1;}} return true;} void print (char *str)//Displays the number {bool Begin=false;int i;int Len=strlen (str); for (i=0;i<len;i++) {if (!begin && str[i]!= ' 0 ') {begin=true;} if (begin)//Assume that the first high number {Cout<<str[i] that is not 0 is found;}} Cout<<endl;} void Tomaxn (int n)//client called the function {char *str=new char[n+1];memset (str, ' 0 ', n);//Note the initial value is in the middle, not the third parameter str[n]= ' n '; Increment (str)) {print (str);} delete []str;} int main () {int N;while (cin>>n) {if (n>=1 && n<=5) {TOMAXN (n);}} return 0;} 

        test results such as the following:       References :"The sword means offer"

Algorithm title: Print 1 to the maximum n digits