[选拔赛]Strings of Power

Strings of Power
Volodya likes listening to heavy metal and (occasionally) reading. No wonder Volodya is especially interested in texts concerning his favourite music style.


Volodya calls a string powerful if it starts with "heavy" and ends with "metal". Finding all powerful substrings (by substring Volodya means a subsequence of consecutive characters in a string) in a given text makes our hero especially joyful. Recently he felt an enormous fit of energy while reading a certain text. So Volodya decided to count all powerful substrings in this text and brag about it all day long. Help him in this difficult task. Two substrings are considered different if they appear at the different positions in the text.


For simplicity, let us assume that Volodya's text can be represented as a single string.


Input
Input contains a single non-empty string consisting of the lowercase Latin alphabet letters. Length of this string will not be greater than 106 characters.


Output
Print exactly one number — the number of powerful substrings of the given string.


Sample test(s)
input
heavymetalisheavymetal
output
3
input
heavymetalismetal
output
2
input
trueheavymetalissotruewellitisalsosoheavythatyoucanalmostfeeltheweightofmetalonyou
output
3
Note
In the first sample the string "heavymetalisheavymetal" contains powerful substring "heavymetal" twice, also the whole string "heavymetalisheavymetal" is certainly powerful.

In the second sample the string "heavymetalismetal" contains two powerful substrings: "heavymetal" and "heavymetalismetal".

解题报告

此题属于字符串匹配问题，当出现一个heavy时，如果后面有几个metal，那么匹配数就是几，如果有多个heavy那么需要考虑先后问题。于是先找到匹配的heavy，计数rec++，碰到metal就用rec+ans（初始值为0），于是匹配数就是用rec+ans最后的结果。代码如下：

#include<stdio.h>#include<string.h>#define ll __int64char a[2000000];int main(){    int len,i;    while(scanf("%s",a)!=EOF)    {        ll rec=0,ans=0;        len=strlen(a);        for(i=0;i<len;i++)        {            if(a[i]=='h'&&a[i+1]=='e'&&a[i+2]=='a'&&a[i+3]=='v'&&a[i+4]=='y')            {                i+=4;                rec++;            }            else if(a[i]=='m'&&a[i+1]=='e'&&a[i+2]=='t'&&a[i+3]=='a'&&a[i+4]=='l')            {                i+=4;                ans=ans+rec;            }        }        printf("%I64d\n",ans);    }    return 0;}