Happy 2009
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2394 Accepted Submission(s): 802
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.
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;
}