Use iterative method to get the maximum number of 1 to n bits

This is the Huang Teacher Sword Point of offer above the 12th question, this question first notice cannot use the integer int type as the operand, because N is large when obviously overflows. This big data is generally used as a string to represent.

The direct method is: 1. For the addition of strings, it involves cyclic carrying and jumping out of judgment.

2. Print out the string, note that printing 01 is legal, otherwise you need to add additional judgment.

Another idea is that the output of these numbers is actually an n-bit full array. Each bit of the number has 0~9, which is possible in 10, so that every time the n bit is set, the n+1 bit of the iteration is set.

voidPrintArray (Char*ptr,intlength) {    BOOLFlag =false;  for(size_t i=0; i<length; ++i) {if(Ptr[i]! ='0'||flag) {Flag=true; cout<<Ptr[i]; }} cout<<Endl;}voidExhaustiverecursive (Char*ptr,intLengthintN) {    if(n = = Length-1) {PrintArray (ptr, length); return ; }     for(size_t i=0; i<Ten; ++i) {ptr[n+1] = i+'0'; Exhaustiverecursive (ptr, length, n+1); }}voidPrintnmaxrecursive (intN) {    if(n = =0)    {        return; }    Char*number_array =New Char[n+1];  for(size_t i =0; i<Ten; i++) {number_array[0] = i+'0'; Exhaustiverecursive (Number_array, N,0); }    Delete[] number_array;}

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The above is the iterative method of their own attempt, not through a lot of use case testing, for reference only.

Use iterative method to get the maximum number of 1 to n bits