Print n digits from 1 to the maximum

The main consideration is the big number problem. N can easily exceed the range expressed by positive numbers, so it needs to be simulated using character arrays.

Add 1 first and then print. As follows:

# Include <stdio. h> # include <stdlib. h> // 1 is returned for overflow. Otherwise, 0int add (char * number) {int overflow = 0, I, current_value; int nLength = strlen (number), jinwei = 0; for (I = nLength-1; I> = 0; I --) {current_value = number [I]-'0' + jinwei; // Add carry if (I = nLength-1) for each bit // Add 1 current_value ++ for the second bit; if (current_value> = 10) {// when the value is equal to 10, carry if (I = 0) overflow = 1; // when the highest bit needs to be carried, it indicates that the maximum value of else {current_value-= 10 has been reached ;/ /If it is not the highest bit, the carry operation is required. First, the current BIT is subtracted from 10, and then the carry flag is set to use jinwei = 1 next time; number [I] = '0' + current_value ;}} else {// assign current_value to the number bit without carrying it. Then, the number [I] = '0' + current_value; break ;}} return overflow ;} // print this number and ignore the 0 void print_a_number (char * number) {int begin = 0, I; for (I = 0; I <strlen (number ); ++ I) {if (! Begin & number [I]! = '0') begin = 1; if (begin) printf ("% c", number [I]);} printf ("\ t ");} void print_all (int n) {char * number = (char *) malloc (n + 1) * sizeof (char); if (n <= 0) return; memset (number, '0', n); number [n] = '\ 0'; while (! Add (number) {print_a_number (number);} free (number) ;}void main () {print_all (2); getch ();}

Reference offoffoffer