Vanya and Food Processor

题意翻译

瓦尼亚在一个垂直的食品处理器中粉碎土豆。 你可以把它想象成一个圆柱体，从上面塞入，从下面粉碎后吐出。 每个土豆可以视为条状。 处理器中的土豆高度不超过$h$(否则会满出来)，处理器每秒粉碎$k$厘米的土豆。如果处理器里剩不到$k$厘米土豆，则粉碎所有剩余的土豆。 瓦尼亚有$n$条土豆，第$i$块的长度等于$a_i$。他把它们按顺序从$1$号到$n$号放进食品处理器，从$1$号开始，到$n$号结束。 每秒会发生如下事件： 1.如果还有至少一条土豆没放进去，瓦尼亚将它们逐一放入处理器，直到没有足够的空间放置下一片，即塞到塞不进为止。 2.处理器粉碎了$k$厘米或剩下全部的土豆。 ## 输入输出格式 ### 输入格式： 输入的第一行包含整数$n$、$h$和$k$ （$1≤n≤100 000,1≤k≤h≤109$），含义如上。 第二行包含$n$个整数$a_i$（1≤a_i≤h），即土豆长度。 ### 输出格式： 最短需要粉碎所有土豆的时间。

题目描述

Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed $ h $ and the processor smashes $ k $ centimeters of potato each second. If there are less than $ k $ centimeters remaining, than during this second processor smashes all the remaining potato. Vanya has $ n $ pieces of potato, the height of the $ i $ -th piece is equal to $ a_{i} $ . He puts them in the food processor one by one starting from the piece number $ 1 $ and finishing with piece number $ n $ . Formally, each second the following happens: 1. If there is at least one piece of potato remaining, Vanya puts them in the processor one by one, until there is not enough space for the next piece. 2. Processor smashes $ k $ centimeters of potato (or just everything that is inside). Provided the information about the parameter of the food processor and the size of each potato in a row, compute how long will it take for all the potato to become smashed.

输入输出格式

输入格式

The first line of the input contains integers $ n $ , $ h $ and $ k $ ( $ 1<=n<=100000,1<=k<=h<=10^{9} $ ) — the number of pieces of potato, the height of the food processor and the amount of potato being smashed each second, respectively. The second line contains $ n $ integers $ a_{i} $ ( $ 1<=a_{i}<=h $ ) — the heights of the pieces.

输出格式

Print a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.

输入输出样例

输入样例 #1

5 6 3
5 4 3 2 1

输出样例 #1

5

输入样例 #2

5 6 3
5 5 5 5 5

输出样例 #2

10

输入样例 #3

5 6 3
1 2 1 1 1

输出样例 #3

2

说明

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.