12--Prints 1 to the largest n for the number.

/* Title: Print 1 to the largest n for the number. For example: Enter 3, print 1 to 999. The problem of solving ideas, this is a large number of questions. The self-addition operation of large number. But pay attention to 2 points. (1): Judging whether the cross-border, in other words, is to reach the maximum value of the discrimination. The strcmp can be selected, but the complexity is O (n), because only when rounding is possible, and only 99 ... 999 will not cross when rounding up, so I am judged if I is 0. (2): output, note that 0 is removed, 0098, into 98.*/#include <stdio.h> #include <iostream>using namespace std;//determine whether the maximum value, exit output. Only when the highest one implements a carry, is it overflow. BOOL Isoverflow (char *number) {int length = strlen (number)-1;//string The last one is ' \ s ', here is a minus 1 operation. printf ("Length%d\n", length), int ntakeover = 0;//carry flag, determines whether to carry bool Isoverflow = false;//is not the maximum value for (int i = length; I &G t;= 0; i--)//string from left to right, but number is from right to left. {//The first loop is the number that gets the lowest bit, and the future loop is used for carry. int nSum = number[i]-' 0 ' + ntakeover;if (i = = length)//Only the lowest bit does plus 1, which is the first time. Nsum++;if (NSum >= 10)//if greater than or equal to 10, it is necessary to carry {if (i = = 0)//Because here from 1 to length-1, it is already the length of the number, if 0, is out of bounds. {isoverflow = True;break;} Ntakeover = 1;nsum-= 10;number[i] = ' 0 ' + nSum;} else//not greater than 10, complex value, and on exit loop {number[i] = ' 0 ' + Nsum;break;}} Return isoverflow;//returns whether the symbol is out of bounds. }//printing numbers, note that the front of the string is 0 non-printing, such as 098, printing only 98. void Printnumber (char *number) {int length = strlen (number); bool boolSingalprint = false;for (int i = 0; i < length; i++) {if (Boolsingalprint = = False && number[i]! = ' 0 ') Boolsinga Lprint = true;if (Boolsingalprint = = True) cout << number[i];} cout << ' t ';} void formonetonum (int num) {if (num <= 0) {cout << re-enter << Endl;return;} The end of the string is added to, so the length plus 1char *number = new Char[num + 1];memset (number, ' 0 ', num), number[num] = '!isoverflow '; R) {printnumber (number);} Delete[]number;} int main () {cout << Please enter a number length. \ n '; int numbers = 0;cin >> number;formonetonum; return 0;}

　　

12--Prints 1 to the largest n for the number.