CodeForce 677B Vanya and Food Processor

Consider the first sample.

1. First Vanya puts the piece of potato of height 5 into processor. At the end of the second there is only amount of height 2 remaining inside.
2. Now Vanya puts the piece of potato of height 4. At the end of the second there is amount of height 3 remaining.
3. Vanya puts the piece of height 3 inside and again there are only 3 centimeters remaining at the end of this second.
4. Vanya finally puts the pieces of height 2 and 1 inside. At the end of the second the height of potato in the processor is equal to 3.
5. During this second processor finally smashes all the remaining potato and the process finishes.

In the second sample, Vanya puts the piece of height 5 inside and waits for 2 seconds while it is completely smashed. Then he repeats the same process for 4 other pieces. The total time is equal to 2·5 = 10 seconds.

In the third sample, Vanya simply puts all the potato inside the processor and waits 2 seconds.

题意：给出n个土豆，机器容纳最高h，每秒碎 k，求研碎时间，放土豆的时候不能高于h

思路：重点在于判断下一个是否超过了最该高度

ＡＣ代码：

# include <iostream>
# include <algorithm>
using namespace std;
typedef long long int ll;
int main()
{
	ll n, h, k;
	while(cin >> n >> h >> k)
	{
		ll t = 0, m = 0;
		for(int i = 0; i < n; i++)
		{
			ll tep;
			cin >> tep;
			if(m + tep <= h)
			{
				m += tep;
			}
			else
			{
				m = tep;
				t++;
			}
			t += m / k;  
			m %= k; 
			// cout << m << " " << t << endl;
		}
		t += m / k;  
		m %= k;  
		if (m > 0)  
		{  
			t++;  
		} 
		cout << t << endl;
	}


	return 0;
}