SRM150 Div.2 Easy WidgetRepairs

この頃なんか親から色々言われて

結構耳が痛いです(笑)

分けて考える

PROBLEM STATEMENT

When a widget breaks, it is sent to the widget repair 
shop, which is capable of repairing at most numPerDay 
widgets per day.
Given a record of the number of widgets that arrive at the 
shop each morning, your task is to determine how many days 
the shop must
operate to repair all the widgets, not counting any days 
the shop spends entirely idle.



For example, suppose the shop is capable of repairing at 
most 8 widgets per day, and over a stretch of 5 days, it 
receives 10, 0, 0, 4, and 20 widgets, respectively.  The 
shop would operate on days 1 and 2, sit idle on day 3, and 
operate again on days 4 through 7.  In total, the shop 
would operate for 6 days to repair all the widgets.



Create a class WidgetRepairs containing a method days that 
takes a sequence of arrival counts arrivals (of type
vector ) and an int numPerDay, and calculates the 
number of days of operation.


DEFINITION
Class:WidgetRepairs
Method:days
Parameters:vector , int
Returns:int
Method signature:int days(vector  arrivals, int 
numPerDay)


CONSTRAINTS

arrivals contains between 1 and 20 elements, inclusive.
Each element of arrivals is between 0 and 100, inclusive.
numPerDay is between 1 and 50, inclusive.

一日の仕事量が決まっていて、日毎に入ってくる仕事の数与えられるからそれを何日で終えられるか？ただし仕事がない日はカウントしない。
という問題。
仕事が増えていく(配列の長さ分)と、余った仕事を片付けるという二つのループで考えるとうまくいきます。

class WidgetRepairs {
   public:
   int days(vector <int> arrivals, int numPerDay)
  {
	int count = 0, sum = 0, i = 0;
	while (i<arrivals.size()) {
		sum += arrivals[i];
		if (sum == 0) {
			i++;
			continue;
		} else {
			sum -= numPerDay;
		}
		if (sum < 0) sum = 0;
		count++; i++;
	}
	while (sum > 0) {
		sum -= numPerDay;
		count++;
	}
	return count;
  }
};