中国DOS联盟

Slightly simplified the code

1. @echo off &SetLocal EnableDelayedExpansion
   
2. set /p x=Please enter the dividend:
   
3. set /p y=Please enter the divisor:
   
4. set /p n=Please enter the number of decimal places to retain:
   
5. set /a count=0
   
6. set /a bjsh=10
   
7. :loop
   
8. if %count% geq %n% goto start
   
9. set /a bjsh*=10
   
10. set /a count+=1
    
11. goto loop
    
12. :start
    
13. set /a rest=%x%*%bjsh%/%y%-%x%*%bjsh%/10/%y%*10
    
14. if "%n%"=="0" (set /a last=%x%/%y%) else (
    
15. set /a last=%x%*%bjsh%/10/%y%-%x%*%bjsh%/100/%y%*10
    
16. )
    
17. if %rest% geq 5 set /a last+=1
    
18. if "%n%"=="0" echo The result is: %last% & pause & goto exit
    
19. if "%n%"=="1" set /a t=%x%/%y% && echo The result is: !t!.%last% & pause & goto exit
    
20. set /a result=%x%*%bjsh%/100/%y%
    
21. set /a end=%n%-1
    
22. call echo The result is: %%result:~0,-%end%%%.%%result:~-%end%%%%%last%% && pause
    
23. :exit
    

BJSH posted on: 2007-04-01 14:43

Brother flyinspace's words I need to think carefully

Last edited by bjsh on 2007-4-1 at 02:53 PM ]

Let's change the algorithm..

1. Get the result of dividend / divisor.
2. Get the remainder of dividend / divisor.
3. The dividend minus the remainder is the actual integer divisor.
3. If the remainder is 0, then the number is divisible. Prompt for divisibility, and then output several 0s.
4. If not 0, then actual integer divisor / divisor is the value of the actual integer part (this value must be accurate).
5. Get the number of digits of the divisor. Suppose it is 2 digits. The output decimal point is 7 digits. (Neither value can be greater than 10)
6. Use a loop to multiply the remainder by 10 (10 to the 9th power).
7. The result of 6 divided by the divisor (then use the rounding algorithm) (give a prompt) to output the decimal part.
8. Output the result: %integer part%.%decimal part%

In this way, the precision is also satisfied. Your requirement is also met.

How do you think, bjsh?

Brother flyinspace indeed gave me a reminder;

The system can only accurately calculate the ninth digit exactly;

That is to say, the maximum accurate value is eight digits;

So my previous algorithm is absolutely no problem for smaller numbers in terms of accuracy;

For example, 18/7

But for large numbers like 100000/7, the accuracy is smaller than the former; it is also related to the system's maximum ability to calculate up to the ninth digit;

So in this regard, I adopted Brother flyinspace's suggestion;

First find the remainder; convert the large number into a small number; then use the remainder to continue dividing

Hehe, if you think it's good, give me more points : )

Am I quite cheeky

I need to correct my previous statement.

System

The digital precision limit is 32 bits.

So the ninth digit is not guaranteed to be calculated;

For example, set /a p=4000000000/7
C:\>set /a p=4000000000/7
-42138185

So the eighth digit is the one calculated precisely;

Therefore, the final precision should be 7 digits

It's just a problem with a long int. The precision is right here.

set /a p=400000000 % 7
set /a a=400000000 - % p%
set /a integer_part=(400000000 - %a%) / 7
The above must be divisible. Why is it not accurate for 9 digits...

New code with precision 7

1. @echo off &SetLocal EnableDelayedExpansion
   
2. set /p x=Please enter the dividend:
   
3. set /p y=Please enter the divisor:
   
4. :n
   
5. set /p n=Please enter the number of decimal places to retain:
   
6. if %n% gtr 7 echo The maximum precision is 7 digits, please re-enter && goto n
   
7. set /a count=0
   
8. set /a bjsh=10
   
9. set /a system_result=%x%/%y%
   
10. set /a system_rest=%x%%%%y%
    
11. set x=%system_rest%
    
12. :loop
    
13. if %count% geq %n% goto start
    
14. set /a bjsh*=10
    
15. set /a count+=1
    
16. goto loop
    
17. :start
    
18. set /a rest=%x%*%bjsh%/%y%-%x%*%bjsh%/10/%y%*10
    
19. if "%n%"=="0" (set /a last=%rest%) else (
    
20. set /a last=%x%*%bjsh%/10/%y%-%x%*%bjsh%/100/%y%*10
    
21. )
    
22. if "%n%"=="0" if %rest% geq 5 set /a last=%system_result%+1 && echo Result is: !last! & pause & goto exit
    
23. if %rest% geq 5 set /a last+=1
    
24. if "%n%"=="1" echo Result is: %system_result%.%last% & pause & goto exit
    
25. set /a result=%x%*%bjsh%/100/%y%
    
26. set /a end=%n%-1
    
27. call echo Result is: %%system_result%%.%%result:~-%end%%%%%last%% && pause
    
28. :exit
    

Posted by BJSH on: 2007-04-01 15:44

Brother flyinspace,What you said makes sense!Through cyclic remainder;It may be of arbitrary precision

Final code;
Precise to any digit

1. @echo off
   
2. set /p x=Please enter the dividend:
   
3. set /p y=Please enter the divisor:
   
4. set /p n=Please enter the number of decimal places to retain:
   
5. set /a end=%n%-1
   
6. set /a count=0
   
7. set /a system_result=%x%/%y%
   
8. set /a rest=%x%%%%y%
   
9. set result=%system_result%.
   
10. :loop
    
11. set /a rest*=10
    
12. set /a last=%rest%/%y%
    
13. set /a count+=1
    
14. if %count% geq %n% set /a rest=%rest%%%%y%*10/%y% && goto jump_loop
    
15. set result=%result%%last%
    
16. set /a rest=%rest%%%%y%
    
17. goto loop
    
18. :jump_loop
    
19. if %n%==0 if %last% geq 5 set /a result+=1 && echo The result is: !result!&& goto exit
    
20. if %rest% geq 5 set /a last+=1
    
21. echo %result%%last% & pause
    

BJSH posted on: 2007-04-01 18:28

Last edited by bjsh on 2007-4-1 at 06:40 PM ]

The idea of this code is attributed to flyinspace. I originally wanted to give him ten points; unfortunately, the forum restricts it to only two points..

May all those who have read it fulfill my wish for me

The previous ones broke the lower limit;
The following one after modification, which cannot calculate higher than 10^9 times, broke both the upper and lower limits;
Can calculate any number of digits; and can be precise to any number of digits; (it seems that for something too large like 1 million digits, it will prompt that the output line is too long)

1. @echo off & setlocal enabledelayedexpansion
   
2. set /p x=Please enter the dividend:
   
3. set /p y=Please enter the divisor:
   
4. if %y% gtr 2000000000 echo The divisor is too large && goto exit
   
5. set /p n=Please enter the number of decimal places to keep:
   
6. if %x% gtr 2000000000 goto big
   
7. :start
   
8. set /a count=0
   
9. set /a system_result=%x%/%y%
   
10. set /a rest=%x%%%%y%
    
11. set result=%system_result%.
    
12. :loop
    
13. set /a rest*=10
    
14. set /a last=%rest%/%y%
    
15. set /a count+=1
    
16. if %count% geq %n% set /a rest=%rest%%%%y%*10/%y% && goto jump_loop
    
17. set result=%result%%last%
    
18. set /a rest=%rest%%%%y%
    
19. goto loop
    
20. :jump_loop
    
21. if %n%==0 if %last% geq 5 set /a result+=1 && echo The result is: !result!&& goto exit
    
22. if %rest% geq 5 set /a last+=1
    
23. echo The result is: %result%%last% && goto exit
    
24. :big
    
25. set /a l=0
    
26. set result=
    
27. :big_loop
    
28. set p=%x:~0,1%
    
29. set /a m=%l%*10+%p%
    
30. set /a s=%m%/%y%
    
31. set result=%result%%s%
    
32. if "%result%"=="0" set result=
    
33. set /a l=%m%%%%y%
    
34. set x=%x:~1%
    
35. if not defined x goto big_end
    
36. goto big_loop
    
37. :big_end
    
38. set result=%result%.
    
39. set /a count=0
    
40. set rest=%l%
    
41. goto loop
    
42. :exit
    
43. pause
    

BJSH posted on: 2007-04-01 18:56

Last edited by bjsh on 2007-11-22 at 10:30 PM ]

Unable to solve the problem where the divisor is greater than 10^9

Last edited by bjsh on 2007-4-1 at 08:53 PM ]

Originally posted by bjsh at 2007-4-1 07:01 PM:
Unable to solve the problem where the dividend is greater than 10^9

In dividend / divisor = result.

The dividend can be greater than 10^9.
But the divisor cannot.

set dividend=9999999999999999999999999999999
Assume this 9 has 20 digits. Of course, the length can be arbitrary.

Use set new=%dividend:~0,9% to take out 9 digits.

Of course, at the application level, we will use a for loop to solve the problem of continuous values.

Here, only one situation is explained. You can handle other situations by yourself :）

First, judge whether new is greater than the divisor
Yes --》If it is greater, start doing the remainder operation...

Assume the divisor is 111111110.

Then, after obtaining the precise value according to the above method...

Obtain the number of digits of the remainder.

In this example, it is one digit... and the value is 9.

Then, use set to fill in 8 digits later.

In this way, loop...

Take out all 20 digits of the number.
Then the code for the integer part is done..
You can also handle the decimal part by yourself..

I made a mistake above;

It's impossible to solve the problem where the divisor is greater than 10^9;

As for the dividend, the idea used in my last code is the same as flyinspace's;

That code is applicable to any number of digits of the dividend

Last edited by bjsh on 2007-4-1 at 08:51 PM ]

;)Large number. I use cyclic interception to get numbers, but there are still problems. For example, 123456789123456. First take the first nine digits for calculation, then add zeros at the end, and then add the results. There is a problem with adding very large numbers, and the result is not very correct. There is a decimal problem. When magnified, there is a big difference in numbers. It seems that cmd is not suitable for calculation. I can only do this much~~~
----The previous ones are all experts. I will study hard and work harder: )

Last edited by nanhui112 on 2008-1-28 at 04:43 PM ]