統一爲每25號

Question

所以我有這個：

if (bought.count == bought.needForUpgrade) 
     { 
      bought.timeToComplete /= 2; 
      bought.needForUpgrade += 25; 
     } 

這是很好的，但我犯了一個團購功能，這樣你就可以通過1,10,25購買項目，我想減少bought.timeToComeplete by 2爲每25號的數字通過，所以例如我有1從該項目，然後我決定購買25，然後我會有26.我可以做很多if statements但我不想整天寫它if(bought.count >= 25 && bought.count <= 50) ..等

Answer 1

只需添加一個額外的值，以節省您已經升級的頻率，並檢查你是否有下一個「級」。 對於這個解決方案，我認爲bought.count和bought.needForUpgrade是整數類型：

int boughtLevel = 0; 

if (bought.count/bought.needForUpgrade > boughtLevel) 
{ 
    // Do your stuff here 
    boughtLevel = bought.count/bought.needForUpgrade; 
} 

否則你就必須將之前丟掉。

int boughtLevel = 0; 

if ((int)(bought.count/bought.needForUpgrade) > boughtLevel) 
{ 
    // Do your stuff here 
    boughtLevel = (int)(bought.count/bought.needForUpgrade); // Not necessary because boughtLevel is of type integer 
}