Codeforces Round #188 (Div. 2) B. Strings of Power

B. Strings of Powertime limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 than10^{6 characters.}

Output

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

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams
or the %I64dspecifier.

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".

/*題目還是滿水的 不過TML了一次 WA了一次 寫個blog  紀念一下 漲點教訓 */#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;__int64 n,ans;int h[1000005],m[1000005],cnt1,cnt2,cnth,cntm;char s[1000005];void solve(){int i,j,len;cnt1=cnt2=0;len=strlen(s);for(i=0;i<=len-5;i++)    // 找出哪些位置出現了“heavy"和"metal" 並將首位置記錄在各自數組中    {        if(s[i]=='h')        {            if(s[i+1]=='e'&&s[i+2]=='a'&&s[i+3]=='v'&&s[i+4]=='y')            {                h[++cnt1]=i;            }        }        if(s[i]=='m')        {            if(s[i+1]=='e'&&s[i+2]=='t'&&s[i+3]=='a'&&s[i+4]=='l')            {                m[++cnt2]=i;            }        }    }    h[cnt1+1]=100000000;     // 類似貪心 要在最後加一個哨兵     m[cnt2+1]=100000005;    ans=0;    cnth=cntm=1;    while(cnth<=cnt1||cntm<=cnt2) // 開始用的二階迴圈 導致TMl 其實一階完全可以搞定    {        if(h[cnth]<m[cntm])  // 若不加哨兵 則cnth=cnt1後會一直執行語句一        {            cnth++;        }        else          // 同理m[]也要加哨兵  開始m[]沒加WA了一次        {            ans+=cnth-1;            cntm++;        }    }}int main(){int i,j,t;while(gets(s)!=NULL){solve();printf("%I64d\n",ans);}return 0;}