01: Numeric conversion, 01 numeric Conversion

01: Numeric conversion, 01 numeric Conversion
01: Numeric Conversion

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Total time limit:

1000 ms

 

Memory limit:

65536kB

Description

Returns the conversion of any two non-negative integers of different hexadecimal notation (decimal order ~ 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

The input contains 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.

Output

The output contains a row. This row has an integer that is the converted 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

Source

2005 ~ 2006 Final Examination of Introduction to computing in the Medical Department

1 # include <iostream> 2 # include <cstdio> 3 # include <cmath> 4 # include <cstring> 5 using namespace std; 6 char a [10001]; 7 int B [10001]; 8 int now; 9 int tot; 10 int n; // unconverted hexadecimal 11 int m; // The hexadecimal 12 int da [10001]; 13 char w [20] = {'A', 'B', 'C', 'D ', 'E', 'F'}; 14 int main () 15 {16 cin> n; 17 scanf ("% s", & ); 18 19 cin> m; 20 int l = strlen (a); 21 if (a [0] = '0' & l = 1) 22 {23 cout <"0"; 24 return 0; 25} 26 for (int I = 0; I <l; I ++) 27 {28 if (a [I]> = 97 & a [I] <= 122) 29 {30 a [I] = a [I]-32; 31} 32 if (a [I]> = 49 & a [I] <= 57) 33 {34 B [I] = a [I]-48; 35} 36 else if (a [I]> = 65 & a [I] <= 90) 37 {38 B [I] = a [I]-55; 39} 40} 41 int ans = 0; 42 int now = 0; 43 for (int I = L-1; i> = 0; I --) 44 {45 ans = ans + B [I] * (pow (n, now); 46 now ++; 47} 48 int now2 = 1; 49 while (ans! = 0) 50 {51 int r = ans % m; 52 B [now2] = r; 53 ans = ans/m; 54 now2 ++; 55} 56 for (int I = now2-1; I> = 1; I --) 57 {58 if (B [I] <10) 59 cout <B [I]; 60 else 61 {62 int k = B [I]-10; 63 cout <w [k]; 64} 65} 66/* for (int I = now2; i> = 0; I --) 67 cout <da [I]; */68 return 0; 69}

Note that a is 0.