001:NUMBER BASE CONVERSION

Messj

总时间限制: 1000ms 内存限制: 65536kB

描述
Write a program to convert numbers in one base to numbers in a second base. There are 62 different digits:
{ 0-9,A-Z,a-z }
HINT: If you make a sequence of base conversions using the output of one conversion as the input to the next, when you get back to the original base, you should get the original number.

输入
The first line of input contains a single positive integer. This is the number of lines that follow. Each of the following lines will have a (decimal) input base followed by a (decimal) output base followed by a number expressed in the input base. Both the input base and the output base will be in the range from 2 to 62. That is (in decimal) A = 10, B = 11, ..., Z = 35, a = 36, b = 37, ..., z = 61 (0-9 have their usual meanings).

输出
The output of the program should consist of three lines of output for each base conversion performed. The first line should be the input base in decimal followed by a space then the input number (as given expressed in the input base). The second output line should be the output base followed by a space then the input number (as expressed in the output base). The third output line is blank.

样例输入

1. 8
   
2. 62 2 abcdefghiz
   
3. 10 16 1234567890123456789012345678901234567890
   
4. 16 35 3A0C92075C0DBF3B8ACBC5F96CE3F0AD2
   
5. 35 23 333YMHOUE8JPLT7OX6K9FYCQ8A
   
6. 23 49 946B9AA02MI37E3D3MMJ4G7BL2F05
   
7. 49 61 1VbDkSIMJL3JjRgAdlUfcaWj
   
8. 61 5 dl9MDSWqwHjDnToKcsWE1S
   
9. 5 10 42104444441001414401221302402201233340311104212022133030

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样例输出

1. 62 abcdefghiz
   
2. 2 11011100000100010111110010010110011111001001100011010010001
   
3. 
   
4. 10 1234567890123456789012345678901234567890
   
5. 16 3A0C92075C0DBF3B8ACBC5F96CE3F0AD2
   
6. 
   
7. 16 3A0C92075C0DBF3B8ACBC5F96CE3F0AD2
   
8. 35 333YMHOUE8JPLT7OX6K9FYCQ8A
   
9. 
   
10. 35 333YMHOUE8JPLT7OX6K9FYCQ8A
    
11. 23 946B9AA02MI37E3D3MMJ4G7BL2F05
    
12. 
    
13. 23 946B9AA02MI37E3D3MMJ4G7BL2F05
    
14. 49 1VbDkSIMJL3JjRgAdlUfcaWj
    
15. 
    
16. 49 1VbDkSIMJL3JjRgAdlUfcaWj
    
17. 61 dl9MDSWqwHjDnToKcsWE1S
    
18. 
    
19. 61 dl9MDSWqwHjDnToKcsWE1S
    
20. 5 42104444441001414401221302402201233340311104212022133030
    
21. 
    
22. 5 42104444441001414401221302402201233340311104212022133030
    
23. 10 1234567890123456789012345678901234567890

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Messj

1. #include<cstdio>
   
2. #include<iostream>
   
3. #include<cstring>
   
4. 
   
5. const int maxn=1000;
   
6. using namespace std;
   
7. 
   
8. int n,m;
   
9. char str1[maxn],str2[maxn];
   
10. int t[maxn],A[maxn];
    
11. 
    
12. void solve(char *str1,char *str2,int n,int m)
    
13. {
    
14.         int len,i,k;
    
15.         len = strlen(str1);
    
16.         for(i=len;i>=0;i--)
    
17.         {
    
18.                 t[len-1-i]=str1[i]-(str1[i]<58?48:str1[i]<91?55:61);    //ascii '0'-48 'A'-65 'a'-97
    
19.         }
    
20.         for(i=0;len;)
    
21.         {
    
22.                 for(i=len;i>0;i--)
    
23.                 {
    
24.                         t[i-1]=t[i]%m*n;
    
25.                         t[i]/=m;
    
26.                 }
    
27.                 A[k++]=t[0]%m;
    
28.                 t[0]/=m;
    
29.                 while(len>0&&!t[len-1])
    
30.                 {
    
31.                         len--;
    
32.                 }
    
33.         }
    
34.         str2[k]=NULL;
    
35.         for(i=0;i<k;i++)
    
36.         {
    
37.                 str2[k-1-i] = A[i]+(A[i]<10?48:A[i]<36?55:61);
    
38.         }
    
39. }
    
40. int main()
    
41. {
    
42.         int i,T;
    
43.         cin>>T;
    
44.         for(i=0;i<T;i++)
    
45.         {
    
46.                 cin>>n>>m>>str1;
    
47.                 solve(str1,str2,n,m);
    
48.                 cout<<n<<" "<<str1<<endl;
    
49.                 cout<<m<<" "<<str2<<endl;
    
50.                 if(i!=T-1)
    
51.                         cout<<endl;
    
52.         }
    
53. }

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