计算卡车货运能力在一场比赛中 [英] Calculating truck cargo capacity in a game

问题描述

这更多的是一个数学/算法的问题不是一个编程问题，但我希望你们能帮忙呢。

情景1：

  

播放机1 40 箱在他的库存。

     

播放机1 2 车，

     

1个小（容量：8箱）

     

1个中（容量：16箱）

---

  

由于容量：

     

A 小卡车能装8箱

     

A 中卡车能装16箱

     

A 大卡车能装30箱

多少车没有播放器1需要能够采取所有40箱？

情景2，如果有货物已在卡车发生什么呢？

  

玩家1有40个包装箱及2卡车上面的场景。

     

如果小已经有2箱在其负荷，给他8-2 = 6的空间

     

如果中已经有4个箱子中的负载，让他16-4 = 8的空间

     

多少车没有播放器1需要采取所有40个箱子？什么算法呢？

情景3：无车

  

播放机1 0卡车在所有。有多少车他需要采取所有40箱？此外，什么是你可以使用algoritm？

情景＃4：有太多的车

  

播放机1 10辆卡车，全部大的能力。有多少车没有考虑到船舶上的所有40箱？

我的想法。

情景1，

  2辆卡车，1小= 8和1个中等大小= 16
8 + 16 = 24箱
40  -  24 = 16辆卡车？ //这看起来是错误的。
 

卡车的成本都在早些时候完成（你先买）。

我觉得我的算法是错误的。我需要一个基数来划分呢？难道我把它用卡车？

在这个任何帮助将是非常有益的。

解决方案

我建议以下算法（伪code）

 做卡车= 1，number_trucks
  current_capacity（卡车）= total_capacity（卡车） -  loaded_crates（卡车）
ENDDO
根据current_capacity排序卡车（大前）
remaining_crates = 40
办卡= 1，number_trucks
  如果（remaining_crates  -  current_capacity（卡车）大于0）
    载货车全
    remaining_crates  -  = current_capacity（卡车）
  其他
    如果（卡车！=上车）
      如果（remaining_crates  -  current_capacity（卡车+ 1）0）
        与remaining_crates载货车
        remaining_crates = 0
      ENDIF
    其他
      载货车全
      remaining_crates  -  = current_capacity（卡车）
    ENDIF
  ENDIF
ENDDO
根据total_capacity（truck_class）（最大的开始）按truck_class
DO truck_class = 1，number_truck_classes
  而（remaining_crates  -  total_capacity（truck_class）大于0）
    买truck_class卡车
    remaining_crates  -  = total_capacity（truck_class）
  端，而
  如果（truck_class == number_truck_classes＆安培;＆安培; remaining_crates大于0）
    买truck_class卡车
  ENDIF
ENDDO
 

This is more of a math/algorithm question than a programming question, but I hope you guys can help anyway.

Scenario #1:

Player 1 has 40 crates in his inventory.

Player 1 has 2 trucks,

1x small (capacity: 8 crates)

1x medium (capacity: 16 crates)

---

Given capacity:

A small truck can hold 8 crates

A medium truck can hold 16 crates

A large truck can hold 30 crates

How many trucks does player 1 need to be able to take all 40 crates?

Scenario #2, what happens if there is cargo already in trucks?

Player 1 has 40 crates and 2 trucks as above scenario.

If small already has 2 crates in its load, giving him 8-2 = 6 space

If medium already has 4 crates in its load, giving him 16-4 = 8 space

How many trucks does player 1 need to take all 40 crates? What would the algorithm be?

Scenario #3: No trucks

Player 1 has 0 trucks at all. How many trucks does he need to take all 40 crates? Again, what is the algoritm you would use?

Scenario #4: Too many trucks

Player 1 has 10 trucks, all at large capacity. How many trucks does it take to ship all 40 crates?

I'm thinking.

Scenario 1,

2 trucks, 1 small = 8 and 1 medium = 16
8+16 = 24 crates
40 - 24 = 16 trucks?? // This looks wrong.

Costs of trucks are done earlier on (you buy them first).

I think my algorithm is wrong. Do I need to divide it by a base number? Do I divide it by trucks?

Any help on this would be very helpful.

解决方案

I suggest the following algorithm (in pseudocode)

do truck = 1,number_trucks
  current_capacity(truck) = total_capacity(truck) - loaded_crates(truck)
enddo
sort trucks according to current_capacity (biggest first)
remaining_crates = 40
do truck = 1,number_trucks
  if(remaining_crates - current_capacity(truck) > 0)
    load truck full
    remaining_crates -= current_capacity(truck)
  else
    if(truck != last truck)
      if(remaining_crates - current_capacity(truck+1) > 0)
        load truck with remaining_crates
        remaining_crates = 0
      endif
    else
      load truck full
      remaining_crates -= current_capacity(truck)
    endif
  endif
enddo
sort truck_class according to total_capacity(truck_class) (biggest first)
do truck_class = 1,number_truck_classes
  while(remaining_crates - total_capacity(truck_class) > 0)
    buy truck of truck_class
    remaining_crates -= total_capacity(truck_class)
  end while
  if(truck_class == number_truck_classes && remaining_crates > 0)
    buy truck of truck_class
  endif
enddo

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