使用Python仅使用已知数字来获取数值。

huangapple
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英文:

using python to get to a value with only known numbers

问题

我有2个已知的数字，1500和1000，我想弄清楚需要添加多少个每个数字才能达到4000。编写最有效的代码的方法是什么？

英文:

let's say I have only have 2 known numbers, 1500 & 1000 and I want to figure out how many of each I need to add to get to 4000. what would be the most efficient way to code that?

答案1

得分: 2

floor division，但只有两个选项，当然可能还会有余数。

  1. goal = 4000
  3. a = goal // 1500
  4. goal -= a * 1500
  5. b = goal // 1000
  6. goal -= b * 1000
  8. results = {'balance': goal, '1500': a, '1000': b}
  9. for k, v in results.items():
  10.     print(k, ':', v)
英文:

floor division, but with only two options, of course there still might be a remainder.

  1. goal = 4000
  3. a = goal // 1500
  4. goal -= a * 1500
  5. b = goal // 1000
  6. goal -= b * 1000
  8. results = {'balance': goal, '1500': a, '1000': b}
  9. for k, v in results.items():
  10.     print(k, ':', v)

答案2

得分: 0

这是一种你可以做的方法。

  1. # 设置变量
  2. N_1 = 1000
  3. N_2 = 1500
  4. N_3 = 4000
  6. # 计算数字2可以整除数字3的次数
  7. Answer = N_3 / N_2
  8. print(Answer)
  10. # 或者你可以这样计算数字1可以整除数字3的次数
  11. answer = N_3 / N_1
  12. print(answer)
  14. # 要更改这些数字以查看它们在4000中整除的次数，只需设置变量为不同的值。

这将告诉你数字在4000中整除的次数，只需将变量设置为不同的值即可。

英文:

Here is one way you could do that.

  1. #set variabes
  2. N_1 = 1000
  3. N_2 = 1500
  4. N_3 = 4000
  5. #set answer to how many times number 2 will go into number 3
  6. Answer = N_3 / N_2
  7. print (Answer)
  8. # or you could do this to get number 1 will go into number 3
  9. answer = N_3 / N_1
  10. print (answer)

This will tell you how many times the numbers go into 4000 to change the numbers just set the variable to different values.

答案3

得分: 0

根据我的情况解决了这个问题... 这是对我有效的方法，但可能是基于我使用的两个数字的幸运情况。

  1. 目标 = 4000
  2. a = 目标 // 1500
  3. 目标 -= a * 1500
  4. b = 目标 // 1000
  5. 目标 -= b * 1000
  7. if 目标 == 0:
  8.     threetall = a
  9.     twotall = b
  11. elif 目标 == 500:
  12.     目标 += 1500
  13.     b = 目标 // 1000
  15.     threetall = a - 1
  16.     twotall = b
英文:

Solved this based on my situation... this is what worked for me but properly was luck based on the 2 numbers I was using.

  1. goal = 4000
  2. a = goal // 1500
  3. goal -= a * 1500
  4. b = goal // 1000
  5. goal -= b * 1000
  7. if goal == 0:
  8. 	threetall = a
  9. 	twotall = b
  11. elif goal == 500:
  12. 	goal += 1500
  13. 	b = goal // 1000
  15. 	threetall = a - 1
  16. 	twotall = b

huangapple
  • 本文由 发表于 2023年6月9日 07:54:53
  • 转载请务必保留本文链接：https://go.coder-hub.com/76436385.html
  • math
  • python
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