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

B. Strings of Power

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/318/problem/B

Description

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.

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 %I64d specifier.

Sample Input

heavymetalisheavymetal

Sample Output

3

HINT

题意

一个字符串如果以heavy开头，metal结尾的话，就说明这是一个powerful句子，然后问你有多少个句子

题解：

对于metal，我们维护一个后缀和就好了

然后扫一遍就行啦……

代码：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 4000001
#define mod 10007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
//**************************************************************************************

string s;
//heavy
//metal
string s1="heavy";
string s2="metal";
int a[maxn];
int aa[maxn];
int b[maxn];
int bb[maxn];
int main()
{
    cin>>s;
    if(s.size()<)
    {
        cout<<"";
        return ;
    }
    for(int i=;i<s.size()-;i++)
    {
        for(int j=;j<;j++)
        {
            if(s[i+j]!=s1[j])
                break;
            if(j==)
                a[i]=;
        }
        for(int j=;j<;j++)
        {
            if(s[i+j]!=s2[j])
                break;
            if(j==)
                b[i]=;
        }
        aa[i]=a[i];
        bb[i]=b[i];
    }
    for(int i=;i<s.size();i++)
        a[i]+=a[i-];
    for(int i=s.size()-;i>=;i--)
        b[i]+=b[i+];

    ll ans=;
    for(int i=;i<s.size();i++)
    {
        if(aa[i]==)
        ans+=b[i+];
    }
    cout<<ans<<endl;
}