Hehe（hdu4639，斐波拉契数+排列组合）

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分类专栏：组合计数 hdu 2013年7月

本文链接：https://blog.youkuaiyun.com/JHC23/article/details/10011471

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本文介绍了一道名为“笑笑问题”的编程竞赛题目，该题涉及字符串处理与组合数学中的斐波那契数列。通过分析字符串中特定子串出现的不同意义，利用斐波那契数列求解所有可能的意义数量。

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

Hehe

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

Total Submission(s): 224 Accepted Submission(s): 135

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

wanshangniyoukongme

womenyiqichuqukanxingxingba

bulehehewohaiyoushi

eheheheh

Sample Output

Case 1: 1

Case 2: 1

Case 3: 2

Case 4: 3

Source

2013 Multi-University Training Contest 4

Recommend

zhuyuanchen520

解析：

题意：“hehe”有两种意思，给出字符串，问这个字符串有几种含义

思路：斐波拉契数+排列组合，”he“的个数满足斐波拉契数，然后把每段连续含有"he "的个数的斐波拉契数相乘

Accepted 15MS 272K 691 B C++

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxn=10100+10;
const int mod=10007;
int f[maxn];
char s[maxn];
void init()
{
f[1]=f[0]=1;
for(int i=2;i<maxn;i++)
f[i]=(f[i-1]+f[i-2])%mod;
}
int main()
{
int T,i,c=0;
init();
scanf("%d",&T);
while(T--)
 {
 	scanf("%s",s);
 	int len=strlen(s);
 	int ans=1,t=0;
 	for(i=0;i<len;i++)
 	 {
 	 	if(s[i]=='h'&&s[i+1]=='e')
 	 	{
 	 	 t=1;
 	 	 i+=2;
 	 	 while(s[i]=='h'&&s[i+1]=='e'&&(i+1)<len)
 	 	 {
 	 	 t++;
 	 	 i+=2;
 	 	 }
 	 	 ans=(ans*f[t])%mod;
 	 	}
 	 }
 	 printf("Case %d: %d\n",++c,ans);
 }
 return 0;
}