CodeForces 677B Vanya and Food Processor（打土豆，模拟）

http://codeforces.com/problemset/problem/677/B

B. Vanya and Food Processor

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed hand 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 ai. 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.

Input

The first line of the input contains integers n, h and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ h ≤ 109) — 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 ai (1 ≤ ai ≤ h) — the heights of the pieces.

Output

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

Examples

input

5 6 35 4 3 2 1

output

5

input

5 6 35 5 5 5 5

output

10

input

5 6 31 2 1 1 1

output

2

Note

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：

用机器打土豆，要求：

没达到机器额定容量 h 可以持续往里面放，每秒钟打下 k，问最少的时间。

输入数据提示：

给定 n h k ，分别是土豆数量，机器限制的土豆高度，每秒钟打下的高度。

思路：

看注释模拟的部分，问题就是很尴尬。我明明输出的是 5 ，这bug。。。。。

未通过代码Code：

#include<cstdio>#include<cstring>const int MYDD=1103+1e5;int pot[MYDD];int main() {int n,h,k;while(scanf("%d%d%d",&n,&h,&k)!=EOF) {for(int j=1; j<=n; j++)scanf("%d",&pot[j]);int ans=0,sum=0;pot[0]=0;for(int j=0; j<=n; ) {//j++while(sum<=h) {//一直装填j++;sum+=pot[j];}sum-=pot[j];//去掉本轮次装的最后一个ans+=sum/k;//耗时sum%=k;//加工后还能添加的高度if(sum+pot[j]>h) {//下一个土豆不能够加入ans++;sum=0;}j--;//为了 while()第一个pot[j]累加到 sum}printf("%d\n",ans);}return 0;}/*By: Shyazhut*/