Happy 2009

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

很简单的字符串，主要是简单练下手，1A。

#include <iostream>
#include <string>
#include <map>

using namespace std;

map<string,int> smp;
int ans;

void init(){
	smp["h"] = 0;
	smp["ha"] = 0;
	smp["hap"] = 0;
	smp["happ"] = 0;
	ans = 0;
}

inline bool yes(char c){
	if (c == 'a' || c == 'h' || c == 'p' || c == 'y')return true;
	return false;
}

inline void add(char c){
	if (c == 'h'){
		++smp["h"];
	}
	else if (c == 'a' && smp["h"]>0){
		--smp["h"];
		++smp["ha"];
	}
	else if (c == 'p'){
		if (smp["hap"] > 0){
			--smp["hap"];
			++smp["happ"];
		}
		else if (smp["ha"] > 0){
			--smp["ha"];
			++smp["hap"];
		}

	}
	else if (c == 'y' && smp["happ"] > 0){
		++ans;
		--smp["happ"];
	}
}

void solve(string str){
	int len = str.length();

	for (int i = 0; i < len; ++i){
		if (yes(str[i])){
			add(str[i]);
		}
	}
}

int main(){
	string str;

	while (getline(cin, str)){
		init();
		solve(str);

		cout << ans << endl;
	}
	return 0;
}

HDU 2617 Happy 2009（字符串）,布布扣,bubuko.com