hdu (2617) Happy 2009

Problem Description

No matter you know me or not. Bless you happy in 2009.

 

 

Input

The input contains multiple test cases.
Each test case included one string. There are made up of ‘a’-‘z’ or blank. The length of string will not large than 10000.

 

 

Output

For each test case tell me how many times “happy” can be constructed by using the string. Forbid to change the position of the characters in the string. The answer will small than 1000.

 

 

Sample Input

hopppayppy happy

happ acm y

hahappyppy

 

 

Sample Output

2

1

2

 

 

Author

yifenfei

 

 

Problem analysis:

The problem want  us  count the number of  "happy"  in a string.Of couse, "happy" may not a word, eg: "hoa plpy", in the string it includes one "happy";

But you should notice one case:

For example:  in the string "hy ppappa",it includes zero "happy",because it need a sequence of "h-a-p-p-y"; and the problem is mainly to solve this case;

At first,I have no ideas about this problem,and I have never solved this kind of problems before;

So here I post the code,and this represents a solution to some problems;

And the ideas come from others bolg;

#include<iostream>
#include<string>
using namespace std;
int main()
{
    char str[10055];
    while(gets(str))
    {
        int i;
        int h,a,p1,p2,y;
        h=a=p1=p2=y=0;
        for(i=0;str[i]!='\0';i++)
        {
            if(str[i]=='h')
                h++;
            else if(str[i]=='a')
            {
                if(h>0)
                {
                    h--;a++;
                }
            }
            else if(str[i]=='p')
            {
                if(p1>0)
                {
                    p1--;p2++;
                }
                else if(a>0)
                {
                    a--;p1++;
                }
            }
            else if(str[i]=='y')
            {
                if(p2>0)
                {
                    p2--;y++;
                }
            }
            
        }cout<<y<<endl;
    
    }    return 0;
}