2011大纽约区域赛试题 Decoding EDSAC Data 解题报告

　　2011大纽约程序设计竞赛，D题 Decoding EDSAC Data（题目链接）。

D ⋅ Decoding EDSAC Data

The world's first full-scale, stored-program, electronic, digital computer was the EDSAC (Electronic Delay Storage Automatic Calculator). The EDSAC had an accumulator-based instruction set, operating on 17-bit words(and 35-bit double words), and used a 5-bit teletypewriter code for input and output.

The EDSAC was programmed using a very simple assembly language: a single letter opcode followed by an unsigned decimal address, followed by the the letter 'F' (for full word) or 'D' (for double word). For example, the instruction "A 128 F" would mean "add the full word at location 128 to the accumulator", and would be assembled into the 17-bit binary value, 11100000100000000, consisting of a 5-bit opcode (11100="add"), an 11-bit operand (00010000000 = 128), and a single 0 bit denoting a full word operation (a 1 bit would indicate a double word operation).

Although arithmetic on the EDSAC was fixed point two's complement binary, it was not mere intger arithmetic (as is common with modern machines). The EDSAC hardware assumed a binary point between the leftmost bit and its immediate successor. Thus the hardware could handle only values in the range -1.0 ≤ x < 1.0. For example:

Value	Binary Representation
-1.0	10000000000000000
½	01000000000000000
¾	01100000000000000
-½	11000000000000000

As you can see, the largest possible positive value was:

01111111111111111 = 0.9999847412109375

and the smallest possible positive value was:

00000000000000001 = 2^-16 = 0.0000152587890625

(This also happens to be the increment between successive values on the EDSAC).

By a curious coincidence(or an elegant design decision), the opcode for the add operation(11100) was the same as the teleprinter code for the letter 'A'. The opcode for subtract was the same as the teleprinter code for 'S'(01100), and so on. This simplified the programming for the assembler (which, incidentally, was a mere 31 instructions long). The EDSAC teleprinter alphabet was "PQWERTYUIOJ#SZK*?F@D!HNM&LXGABCV" (with 'P'=00000, 'Q'=00001, and so on, up to 'V'=11111)

Unfortunately, the EDSAC assembler had no special directives for data values. On the other hand, there was no reason that ordinary instructions couldn't be used for this, thus, an EDSAC programmer desiring to reserve space for the constant ¾ (represented as 01100000000000000) would use the instruction "S O F" and for ⅓ (which is approximately represented as 00101010101010101) "T 682 D", and so on.

Your job is to write a program that will translate EDSAC instructions into the appropriate decimal fractions.

Input

The first line of input contains a single integer P ( 1 ≤ P ≤ 1000 ) which is the number of data sets that follow. Each data set is a single line that contains N (the dataset number), followed by a space, followed by an EDSAC instruction of the form: c□d□s, where c is a single character in the EDSAC alphabet, d is an unsigned decimal number (0 ≤ d < 2¹¹), and s is either a 'D' or 'F'. Note: □ represents a single space.

Output

For each data set there is one line of output. It contains the data set number (N) followed by a single space, followed by the exact decimal fraction represented by the by the EDSAC instruction, including a minus sign (for negative values). The format for the decimal fraction is: sb.ddd..., where s is an optional minus sign, b is either a 1 or 0, and d is any decimal digit (0-9). There must be at least 1 and at most 16 digits after the decimal point. Trailing zeros in the fraction must be suppressed.

Sample Input	Sample Output
13 1 P 0 F 2 I 0 F 3 & 0 F 4 ? 0 F 5 Q 1228 D 6 P 0 D 7 V 2047 D 8 * 2047 D 9 ? 0 D 10 P 256 F 11 V 1536 F 12 T 682 D 13 T 54 F	1 0.0 2 0.5 3 -0.5 4 -1.0 5 0.0999908447265625 6 0.0000152587890625 7 -0.0000152587890625 8 0.9999847412109375 9 -0.9999847412109375 10 0.0078125 11 -0.015625 12 0.3333282470703125 13 0.31414794921875

　　这一题，我是直接用机器中的二进制来做的，一位一位地判断。前五位（细分成第一位，接下来的4位），中间11位，后一位，分开来处理。另外，正负也分别处理。

　　C语言源代码如下：

#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <string.h>

const char * code = "PQWERTYUIOJ#SZK*?F@D!HNM&LXGABCV";

int main (void)
{
    int testcases;
    int testcase;
    char firstpart;
    int secondpart;
    char thirdpart;
    int i;
    unsigned short number;
    long long result;
    int sign;
    long long current;
    char final[100];;
    scanf( "%d", &testcases );
    while ( testcases -- )
    {
        scanf( "%d %c %d %c", &testcase, &firstpart, &secondpart, &thirdpart );
        for ( i = 0; code[i] != '\0' ; i ++ )
        {
            if ( code[i] == firstpart )
                break;
        }
        assert( code[i] != '\0' );
        if ( i >= 0x10 )
        {
            number = i - 0x10 ;
            sign = -1;
        }
        else
        {
            number = i;
            sign = 1;
        }
        number <<= 12;
        number += (secondpart << 1);
        if ( thirdpart == 'D' )
            number ++;
        result = 0;
        current = 152587890625;
        if ( sign > 0 )
        {
            for ( i = 1 ; i <= 16 ; i ++ )
            {
                if ( (number & 1 ) == 1 )
                {
                    result += current;
                }
                number >>= 1;
                current <<= 1;
            }
#ifdef _WIN32
            sprintf( final, "0.%016I64d", result );
#else
            sprintf( final, "0.%016lld", result );
#endif
        }
        else
        {
            number --;
            for ( i = 1 ; i <= 16 ; i ++ )
            {
                if ( (number & 1 ) == 0 )
                {
                    result += current;
                }
                number >>= 1;
                current <<= 1;
            }
#ifdef _WIN32
            sprintf( final, "-0.%016I64d", result );
#else
            sprintf( final, "-0.%016lld", result );
#endif
        }
        if ( !strcmp(final, "-0.0000000000000000" ) )
            strcpy( final, "-1.0" );
        else
        {
            if ( sign > 0 )
            {
                for ( i = 17; i > 2 ; i -- )
                {
                    if ( final[i] == '0' )
                        final[i] = '\0';
                    else
                        break;
                }
            }
            else
            {
                for ( i = 18; i > 3 ; i -- )
                {
                    if ( final[i] == '0' )
                        final[i] = '\0';
                    else
                        break;
                }
            }
        }
        printf( "%d %s\n", testcase, final );
    }
    return EXIT_SUCCESS;
}