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=10100;
const int INF=0x3f3f3f3f;
const int MOD=10007,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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=1,i,j,cnt,ans,n,ok;
    f[1]=1,f[2]=2;
    for(i=3;i<N;i++)
        f[i]=(f[i-1]+f[i-2])%MOD;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",s);
        n=strlen(s);
        ans=1;ok=1;
        for(i=0;i<n;){
            if(s[i]!='h' || s[i+1]!='e'){
                i++;
                continue;
            }
            for(cnt=0;i<n-1 && s[i]=='h' && s[i+1]=='e';i+=2,cnt++);
            ans=(ans*f[cnt])%MOD;
        }

        printf("Case %d: %d
",Ca++,ans);
    }
    return 0;
}