每25个数字统一

Question

所以我有这个：

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

这很好，但是我做了一个购买功能，所以你可以买1,10,25的物品。我想减少其计数通过的每25个数字的bought.timeToComeplete by 2，所以例如我有1来自那个项目，然后我决定购买25，然后我会有26个。我可以用很多if statements来做，但我不想整天写if(bought.count >= 25 && bought.count <= 50) ..等等p>

Answer 1

只需添加一个额外的值即可保存已升级的频率，并检查您是否获得了下一个＆＃34;级别＆＃34;。 对于此解决方案，我假设bought.count和bought.needForUpgrade的类型为整数：

int boughtLevel = 0;

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

否则你必须在分裂之前施放它们：

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
}