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

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

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;
}

2011大纽约区域赛试题 Decoding EDSAC Data 解题报告相关推荐

【2020大数据应用赛试题】Spark分析处理
文章目录 2020大数据应用赛试题任务一.Spark技术栈有哪些组件?简述其功能,及应用场景. 任务二.本题目使用spark进行数据分析数据说明题目题目一题目二题目三题目四 2020大数 ...
武汉工程大学第一届程序设计女生赛(牛客contest 4746)解题报告 Apare_xzc
武汉工程大学第一届程序设计女生赛解题报告 xzc 2020.3.8 比赛链接:武汉工程大学第一届程序设计女生赛 A. Multiplication (101/861) 分析: 问x平方几次后就会> ...
[CF/AT]各大网站网赛体验部部长第一季度工作报告
文章目录 CodeForces #712 (Div. 1)--1503 A. Balance the Bits B. 3-Coloring C. Travelling Salesman Problem ...
HDOJ 4239 - Decoding EDSAC Data 模拟
题意: 给了一系列的用操作数转换为17位二进制数的关系..又给出了17位二进制数转化成十进制小数的关系...现在给出操作数..请退出其对应十进制小数. 题解: 这个不恶心..简单题...控制小数位.. ...
22行代码AC_试题历届试题油漆面积【解题报告】
励志用更少的代码做更高效的表达 X星球的一批考古机器人正在一片废墟上考古. 该区域的地面坚硬如石.平整如镜. 管理人员为方便,建立了标准的直角坐标系. 每个机器人都各有特长.身怀绝技.它们感兴趣的内容 ...
hihoCoder 1114 小Hi小Ho的惊天大作战：扫雷·一最详细的解题报告
题目来源:小Hi小Ho的惊天大作战:扫雷·一解题思路:因为只要确定了第一个是否有地雷就可以推算出后面是否有地雷(要么为0,要么为1,如果不是这两个值就说明这个方案行不通),如果两种可能中有一种成功, ...
ACM/ICPC2016沈阳网络赛（不完全）解题报告
比赛地址: http://acm.hdu.edu.cn/contests/contest_show.php?cid=724 1003.hannnnah_j's Biological Test 题目大意 ...
2020蓝桥杯国赛Java大学B组解题报告
文章目录试题 A: 美丽的 2 试题 B: 扩散试题 C: 阶乘约数试题 D: 本质上升序列试题 E: 玩具蛇试题 F: 蓝肽子序列试题 H: 画廊试题 A: 美丽的 2 问题描述小蓝 ...
【WZOI第二次NOIP模拟赛Day1T2】世界末日解题报告
[WZOI第二次NOIP模拟赛Day1T2]世界末日 Problem 2 世界末日 (doomsday.pas/c/cpp) 背景话说CWQ大牛终于打开了那扇神秘大门,但迎接他的不是什么神秘的东西, ...

2011大纽约区域赛试题 Decoding EDSAC Data 解题报告

D ⋅ Decoding EDSAC Data

2011大纽约区域赛试题 Decoding EDSAC Data 解题报告相关推荐

最新文章

热门文章