zoukankan      html  css  js  c++  java
  • HDU 4639 Hehe

    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<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    
    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<i;j++)
                C[i][j]=(C[i-1][j-1]+C[i-1][j])%mod;
        }
    }
    
    int main(){
    
        //freopen("input.txt","r",stdin);
    
        int t,cases=0;
        scanf("%d",&t);
        init();
        while(t--){
            scanf("%s",str);
            int cnt=0,ans=1;
            int len=strlen(str);
            for(int i=0;i<len;i++){
                cnt=0;
                int tmp=1;
                while(i+1<len && str[i]=='h' && str[i+1]=='e'){     //找出每次连续在一起的he的个数
                    i+=2;
                    cnt++;
                }
                if(cnt>=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
    ",++cases,ans);
        }
        return 0;
    }

    另一种思路:

    思路: 斐波那契数列

    #include <cstdio>  
    #include <cstring>  
    #include <algorithm>  
    #include <iostream>  
    #include <cmath>  
    #include <vector>  
    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
    ", i, ans);  
        }  
    } 
  • 相关阅读:
    DNS正向解析实现
    基于bind软件部署DNS服务器
    DNS解析工具使用案例
    DNS服务工作原理
    3
    .deb/.whl/.tgz/.tar/gz软件包安装方式(Ubuntu)
    博客系列目录
    Databricks 第3篇:pyspark.sql 通过JDBC连接数据库
    Databricks 第2篇:pyspark.sql 简介
    数据预处理 第4篇:数据预处理(sklearn 插补缺失值)
  • 原文地址:https://www.cnblogs.com/jackge/p/3231561.html
Copyright © 2011-2022 走看看