Letters比赛第六场1001 Basically Speaking解题报告

1001 Basically Speaking(POJ 1546)

基本思路：基本的进制转换，先转化成十进制再转化成目标进制即可。

代码如下：

 1 #include <cstdlib>
 2 #include <iostream>
 3 #include <cmath>
 4 
 5 using namespace std;
 6 
 7 int a = 0;
 8 int b = 0;
 9 char aNum[10];
10 char bNum[10];
11 
12 void solve()
13 {
14     int i = 7;
15     int j = 0;
16     int ans = 0;
17     int alen = strlen(aNum);
18 
19     for(j=0; j<alen; j++)
20     {
21         if(aNum[j] >= 'A')
22         {
23             aNum[j] = aNum[j] - 'A' + 10;
24         }
25         else
26         {
27             aNum[j] = aNum[j] - '0';
28         }
29     }
30     
31     for(i=0; i<alen; i++)
32     {
33         ans = a * ans + aNum[i];
34     }
35 
36     if(ans >= (int)pow((double)b, 7.0))
37     {
38         printf("  ERROR\n");
39     }
40     else
41     {
42         for(j=0; ans && j<7; j++)
43         {
44             bNum[j] = ans%b;
45             ans = ans/b;
46         }
47         i = 7-j;
48         while(i--)
49         {
50             printf(" ");
51         }
52         for(i=j-1; i>=0; i--)
53         {
54             printf("%0X", bNum[i]);
55         }
56         printf("\n");
57     }
58 }
59 
60 int main()
61 {
62 #ifdef MYLOCAL
63     freopen("input.txt", "r", stdin);
64     freopen("output.txt", "w", stdout);
65 #endif
66 
67     while(scanf("%s %d %d", aNum, &a, &b) != EOF)
68     {
69         solve();  
70     }
71 
72     return 0;
73 }