hdu2617

Happy 2009

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3290    Accepted Submission(s): 1077

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 happyhapp acm yhahappyppy

 

Sample Output

21
2
题意分析：这题要注意happy的顺序不能变,所以要最先产生h然后a然后p,p,y,给每个字母都对应一个数组的数,要产生a时必须有h,要产生p时必须有a,要产生happy时,必须有一个h,一个a,两个p,一个y;而且要保证所有的happy都没颠倒次序,所以每一个字母的个数不能超过前面那个字母的个数( 例如hayppahppy,这种只有一个 )
AC代码：
#include<stdio.h>#include<string.h>int n,num[5];char str2[100005];//happyvoid cal(  ){     int k = 0;     while( str2[k] )     {            if( str2[k] == 'h' )                ++num[0];            if( str2[k] == 'a' && num[0] && num[1] < num[0]  )                num[1]++;            if( str2[k] == 'p' && num[1] && ( num[2] < 2 * num[1] ) )                num[2]++;            if( str2[k] == 'y' && num[2] >= 2 && num[1] && num[0] )            {                ++n;                num[0] -= 1;                num[1] -= 1;                num[2] -= 2;            }            ++k;            } }int main( ){    while( gets( str2 ) )    {           memset( num,0,sizeof( num ) );           n = 0;           cal( );           printf( "%d\n",n );           }    return 0;}