Codeforces Beta Round #98 (Div. 2) / 137A Postcards and photos (模拟)

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Zemax光学设计——单透镜设计（1）在镜头数据编辑器中面2那一栏中的“曲率半径”栏右侧的方格中单击，弹出“在面2上的曲率解”对话框（2）将对话框中的求解类型设为“F数”（3）将“F/#”设置为10设置完成后，其中的一些数据会发生变化在初始结构中，透镜的曲率半径和厚度未知，，这些参数需要软件自动优化得到，但可以使用透镜后表面上边缘厚度解得到近轴焦平面的位置在透镜后表面的厚度上单击鼠标右键，选择边缘光线求解类型，它表示近轴边缘光线会自动在下一个表面上聚焦并确定距离值。阅读详情

Polycarpus has postcards and photos hung in a row on the wall. He decided to put them away to the closet and hang on the wall a famous painter's picture. Polycarpus does it like that: he goes from the left to the right and removes the objects consecutively. As Polycarpus doesn't want any mix-ups to happen, he will not carry in his hands objects of two different types. In other words, Polycarpus can't carry both postcards and photos simultaneously. Sometimes he goes to the closet and puts the objects there, thus leaving his hands free. Polycarpus must put all the postcards and photos to the closet. He cannot skip objects. What minimum number of times he should visit the closet if he cannot carry more than 5 items?

Input

The only line of the input data contains a non-empty string consisting of letters "С" and "P" whose length does not exceed 100characters. If the i-th character in the string is the letter "С", that means that the i-th object (the numbering goes from the left to the right) on Polycarpus' wall is a postcard. And if the i-th character is the letter "P", than the i-th object on the wall is a photo.

Output

Print the only number — the minimum number of times Polycarpus has to visit the closet.

   Sample test(s)
 
     input
   
CPCPCPC

     output
   
7

     input
   
CCCCCCPPPPPP

     output
   
4

     input
   
CCCCCCPPCPPPPPPPPPP

     output
   
6

     input
   
CCCCCCCCCC

     output
   
2

Note

In the first sample Polycarpus needs to take one item to the closet 7 times.

In the second sample Polycarpus can first take 3 postcards to the closet; then 3 more. He can take the 6 photos that are left in the similar way, going to the closet twice.

In the third sample Polycarpus can visit the closet twice, both times carrying 3 postcards. Then he can take there 2 photos at once, then one postcard and finally, he can carry the last 10 photos if he visits the closet twice.

In the fourth sample Polycarpus can visit the closet twice and take there all 10 postcards (5 items during each go).

以now为参考标准进行统计~

完整代码：

/*30ms,0KB*/

#include<cstdio>
#include<cstring>
#include<cmath>

char str[101];

int main()
{
	int count = 0, i = 0;
	scanf("%s", str);
	int len = strlen(str);
	char now = str[0];
	while (i < len)
	{
		int nowcount = 1;
		while (++i < len && str[i] == now)
			++nowcount;
		count += ceil((double)nowcount / 5);
		now = str[i];
	}
	printf("%d", count);
	return 0;
}