44:正整數的任意進制轉換

6

18,2345678A123,18

15,23456,18

12,2345678,20

16,12345678,23

25,3456AB,21

18,AB1234567,22

 

2345678A123

114E0

22B7A4

21A976L

7C2136

22JF0G367

 1 #include "stdlib.h"
 2 #include <iostream>
 3 #include<cstring>
 4 #include <stdio.h>
 5 using namespace std;
 6 int main()
 7 {
 8     int n;
 9     cin >> n;
10     for (int i = 0; i < n; i++)
11     {
 
12         char num[800];//設置一個數組儲存q進制數
13         int p, q;
14         cin >> p;
15         cin.ignore();
16         cin.getline(num, 800, ‘,‘);
17         cin >> q;
18         char res[800]; int count = 0;
19         while (1)
20         {
21             int x = 0; //歸零
22             int len = 0;
23
 
             for (int i = 0; i < strlen(num);)//對當前數字進行除q法
24             {
25                 while (x < q&&i<strlen(num))//x小於q的時候，直接將它作為余數*p給下一個位數  當i達到最後時直接將其作為余數存進另一個數組
26                 {
27                     if (num[i] >= ‘A‘&&num[i] <= ‘Z‘)
28                         x = x*p + num[i] + 10 - ‘A‘;
29                     else x = x*p +num[i]- ‘0‘;
30                     num[i] = ‘0‘;
31                     i++;
32                 }
33                 if (x >= q)//x比q大的時候，進行除法
34                 {
35                     if (x / q > 9) num[i - 1] = x / q + ‘A‘ - 10;
36                     else num[i - 1] = x / q + ‘0‘;
37                     x %= q;
38                 }
39             }
40             if (x > 9)//余數進行轉換以後存入數組res
41                 res[count++] = x - 10 + ‘A‘;
42             else
43                 res[count++] = x + ‘0‘;
44             len = -1;
45             for(int i=0;i<strlen(num)&&num[i]==‘0‘;i++)
46             {
47                 len = i;
48             }
49             if (strlen(num) - len < 2)
50                 break;
51         }
52         for (int i = count - 1; i >= 0; i--)
53             cout << res[i];
54         cout << endl;
55     }
56 }

這個真的花了蠻長時間的

並不知道除k取余法是個什麽鬼

感謝poj提問區的大佬們的點撥

POJ44:正整數的任意進制轉換