中国DOS联盟

Fame and fortune are external things; what matters is to acquire knowledge.

Marked as (Essence) for easier finding. ^_^

Originally posted by namejm at 2006-11-17 16:27:
It might be that some moderator operated improperly and removed the essence, and now it is re-added as essence. However, there is a little regret that the original poster of this post was zouzhxi, but when returning to the post list interface to look...

It's not that the essence was removed, but that the PID was changed when merging topics. Oh, my.

Let's attack the "castle" again. First, release my calculation results and see if the answers are correct first?!

Question 19, 10 digits; this should be no problem, 9 sixes and one 2.

Question 20, 14, 10, 35 (the problem of the least common multiple)

Question 21, 1.8 degrees

Looking forward to the building owner's reply.

@echo off

setlocal enabledelayedexpansion

set a=3333333333

set b=9999999999

set a1=%a%0000000000

set/a jg=1%a1:~-9%-%a:~-9%

set a1=%a1:~0,-9%

set/a jg1=%a1:~-9%-1-%a:~0,-9%

set jg=%a1:~0,-9%%jg1%%jg%

for /l %%a in (0,1,19) do for /l %%b in (0,2,8) do (set tm=!jg:~%%a,1!

if !tm!==%%b set/a gs+=1)

echo %a%*%b%=%jg%

echo 共有%gs%个偶数  

pause 

### 第19题：

@echo off

setlocal enabledelayedexpansion

set a=3333333333

set b=9999999999

set a1=%a%0000000000

set/a jg=1%a1:~-9%-%a:~-9%

set a1=%a1:~0,-9%

set/a jg1=%a1:~-9%-1-%a:~0,-9%

set jg=%a1:~0,-9%%jg1%%jg%

for /l %%a in (0,1,19) do for /l %%b in (0,2,8) do (set tm=!jg:~%%a,1!

if !tm!==%%b set/a gs+=1)

echo %a%*%b%=%jg%

echo 共有%gs%个偶数  

pause 

@echo off

set a=5

set b=7

set C=2

set/a ji=a*b*c

set/a a1=ji/a

set/a b1=ji/b

set/a c1=ji/c

echo 这三个齿轮齿数最少应分别是%a1%,%b1%,%c1%齿!

pause

**第20题：**

@echo off

set a=5

set b=7

set C=2

set/a ji=a*b*c

set/a a1=ji/a

set/a b1=ji/b

set/a c1=ji/c

echo The number of teeth of these three gears should be at least %a1%, %b1%, %c1% teeth!

pause

The first two questions are correct. As for the last question? Since it's not the question I made, I'm also calculating it, and I don't know if the result is correct!!!

Question 21 write the corresponding code, I can't, but I can talk about my thinking:
1>First of all, it should be clear that the mass of a cup of water can be ignored compared to a basin of water.
Suppose the temperature of the hot water in the cup drops by X degrees at the very beginning, then it drops by X-3 degrees for the second time
According to the physics formula: Q=cm△t, we can know:
3/5=(X-3)/X
Solve for X=7.5
2>Let the temperature of the water in the basin rise by a degree when heating water for the third time, then:
a/3=(4.5-a)/4.5
Solve for a=1.8

This is the thinking of the above problem, but unfortunately, I can't make it automatically complete with a computer.

????
I haven't written it out yet...
!!!

The calculation method on floor 68 should be wrong, right? Where does 4.5 come from?

I still can't write it with BAT... I don't know if any expert can post the code for reference!!!

The 4.5 on the 68th floor is based on 7.5 - 3, that's because for the first time, there was a problem with the hot water, and it was cooled down by 3 degrees the first time.

The standard answer is to rise another 2 degrees.

According to physics, it's impossible!

Could it really be a so-called primary school problem, and we've artificially made it complicated!

It's a junior high school problem. This problem doesn't seem suitable for solving with a program.