HDU-4639 Hehe 简单DP

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4639

　　简单递推题，呵呵，不多说。。。

 //STATUS:C++_AC_15MS_272KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 char s[N];
 int f[N];
 int T;

 int main(){
  //   freopen("in.txt","r",stdin);
     int Ca=,i,j,cnt,ans,n,ok;
     f[]=,f[]=;
     for(i=;i<N;i++)
         f[i]=(f[i-]+f[i-])%MOD;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%s",s);
         n=strlen(s);
         ans=;ok=;
         for(i=;i<n;){
             if(s[i]!='h' || s[i+]!='e'){
                 i++;
                 continue;
             }
             for(cnt=;i<n- && s[i]=='h' && s[i+]=='e';i+=,cnt++);
             ans=(ans*f[cnt])%MOD;
         }

         printf("Case %d: %d\n",Ca++,ans);
     }
     return ;
 }