Number conversion (9 degree oj Question 44)

Question [html] Question Description: Calculate the conversion of any two non-negative integers in different hexadecimal notation (2-decimal ~ Hexadecimal), the given integer is within the range that long can express. Different hexadecimal notation is ,..., 9, a, B ,..., f) or (0, 1 ,..., 9, A, B ,..., F ). Input: enter only one row, which contains three integers a, n, and B. A indicates that the n after it is an a-hexadecimal integer, and B indicates that you want to convert a-hexadecimal integer n to a-hexadecimal integer. A and B are decimal integers, 2 = <a, B <= 16. Data may contain leading zeros. Output: There may be multiple groups of test data. For each group of data, the output contains one row, and the row has an integer that is converted to the hexadecimal number. Uppercase letters are all used for output, that is, (0, 1,..., 9, A, B,..., F ). Sample input: 15 Aab3 7 sample output: 210306 tip: Different hexadecimal integers can be expressed using strings. The concept of this question is a digital conversion problem, examining the conversion algorithms between different hexadecimal systems. The conversion between different hexadecimal formats can be converted to decimal by using decimal as the intermediary R: multiplied by the Weight Sum Method: decimal to R: divided by the base to obtain the remainder method, the sequence stack is used as the data structure to help pay attention to the leading 0 problem. Therefore, strings cannot be output directly because a = B, the AC code [cpp] # include <stdio. h> # include <stdlib. h> # include <string. h> # define MAX 200 struct stack {char data [MAX]; int top ;}; long int switchToTen (int a, char * s); int main () {long int a, B, n; char str [MAX]; while (scanf ("% ld % s % ld", & a, str, & B )! = EOF) {n = switchToTen (a, str); if (n = 0) {printf ("0");} else {int temp; struct stack * s = (struct stack *) malloc (sizeof (struct stack); s-> top = 0; while (n) {temp = n % B; if (temp> = 0 & temp <= 9) {s-> data [s-> top ++] = temp + '0 ';} else {s-> data [s-> top ++] = temp-10 + 'a';} n/= B;} while (s-> top) {printf ("% c", s-> data [-- s-> top]) ;}} printf ("\ n");} return 0 ;} long int switchToTen (int a, char * s) {int I, B; long int n, c; for (n = 0, c = 1, I = strlen (s) -1; I> = 0; I --) {if (s [I]> = 'A' & s [I] <= 'Z ') B = s [I]-55; else if (s [I]> = 'A' & s [I] <= 'Z ') B = s [I]-87; else B = s [I]-'0'; n + = B * c; c * = a;} return n ;}