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

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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 than10⁶ 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".

/*题目还是满水的 不过TML了一次 WA了一次 写个blog  纪念一下 涨点教训 */#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++)    // 找出哪些位置出现了“heavy"和"metal" 并将首位置记录在各自数组中    {        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;     // 类似贪心 要在最后加一个哨兵     m[cnt2+1]=100000005;    ans=0;    cnth=cntm=1;    while(cnth<=cnt1||cntm<=cnt2) // 开始用的二阶循环 导致TMl 其实一阶完全可以搞定    {        if(h[cnth]<m[cntm])  // 若不加哨兵 则cnth=cnt1后会一直执行语句一        {            cnth++;        }        else          // 同理m[]也要加哨兵  开始m[]没加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;}