Hehe

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 355    Accepted Submission(s): 202

Problem Description

As we all know, Fat Brother likes MeiZi every much, he always find some topic to talk with her. But as Fat Brother is so low profile that no one knows he is a rich-two-generation expect the author, MeiZi always rejects him by typing “hehe” (wqnmlgb). You have to believe that there is still some idealized person just like Fat Brother. They think that the meaning of “hehe” is just “hehe”, such like “hihi”, “haha” and so on. But indeed sometimes “hehe” may really means “hehe”. Now you are given a sentence, every “hehe” in this sentence can replace by “wqnmlgb” or just “hehe”, please calculate that how many different meaning of this sentence may be. Note that “wqnmlgb” means “我去年买了个表” in Chinese.

 

 

Input

The first line contains only one integer T, which is the number of test cases.Each test case contains a string means the given sentence. Note that the given sentence just consists of lowercase letters.

T<=100

The length of each sentence <= 10086

 

 

Output

For each test case, output the case number first, and then output the number of the different meaning of this sentence may be. Since this number may be quite large, you should output the answer modulo 10007.

 

 

Sample Input

4 wanshangniyoukongme womenyiqichuqukanxingxingba bulehehewohaiyoushi eheheheh

 

 

Sample Output

Case 1: 1 Case 2: 1 Case 3: 2 Case 4: 3

 

 

Source

 

 

Recommend

zhuyuanchen520

 

 我用的是组合的方法：

#include
     
      #include
      
       #include
       
        #include
        
         using namespace std;const int mod=10007;char str[12000];int C[1000][1000];void init(){    for(int i=0;i<1000;i++){        C[i][i]=C[i][0]=1;        for(int j=1;j
         
          =2){                     for(int i=1;i<=cnt/2;i++){                    tmp=(tmp+C[i+cnt-i*2][i])%mod;                }            }            ans=(ans*tmp)%mod;        }        printf("Case %d: %d\n",++cases,ans);    }    return 0;}
         
        
       
      
     

 

另一种思路：

思路： 斐波那契数列

#include 
     
        #include 
      
         #include 
       
          #include 
        
           #include 
         
            #include 
          
            using namespace std; //斐波那契数列 const int V = 5050 + 50; const int MaxN = 500 + 5; const int mod = 10000 + 7; int T, ans, num[V]; char ch[10100]; int main() { int i, j, k; scanf("%d", &T); num[0] = num[1] = 1; for(i = 2; i <= V; ++i) num[i] = (num[i - 1] + num[i - 2]) % mod; for(i = 1; i <= T; ++i) { ans = 1; int sum = 0; scanf("%s", &ch); int len = strlen(ch); for(j = 1; j < len; ++j) { if(ch[j] == 'e' && ch[j - 1] == 'h') { sum++; j++; } else { ans = (ans * num[sum]) % mod; sum = 0; } } ans = (ans * num[sum]) % mod; printf("Case %d: %d\n", i, ans); } }