ZOJ 3818

       这道题没什么固定算法，关键就是思维是否够缜密。

       我的方法是先找出DDE模式（其中D的长度必须大于2），再根据具体的情况进行下一步处理。这个具体情况就是当D>E时，E就是A；当D<E时，把E拆成CD，也就是CAB。大的方向说完了，就是细节问题，当D>E时，E能否和D的前半段匹配，如果可以匹配那么D拆成AB后AB是否符合题目要求，当D<E时，D能否和E的后半段匹配，同样E拆成CAB后是否符合题目要求。

       最后感谢一下Kewowlo博客里的数据。

代码（C++）：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>

#define MAX 55
using namespace std;

char s[MAX];

int main()
{
    //freopen("in.txt","r",stdin);

    int t,i,j,k,id,len;
    bool flag;
    string str,str1,str2,str3;

    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",s);
        str="";
        for(i=0;i<strlen(s);i++)    //去除标点符号
        {
            if(s[i]>='a'&&s[i]<='z'||s[i]>='A'&&s[i]<='Z') str+=s[i];
        }
        flag=false;
        len=str.length();
        for(i=1;i<len;i++)
        {
            if(str[0]==str[i])
            {
                if(i<2) continue;   //如果D的长度小于2，直接跳过
                if(i*2-1<len&&str.substr(0,i)==str.substr(i,i))   //判断是否存在DD的情况
                {
                    id=2*i;
                    if(len-id>i){   //针对ABABCAB模式
                        for(j=i-1,k=0;j>=0;j--,k++)     //判断E的后半段是否有D
                        {
                            if(str[len-1-k]!=str[j]) break;
                        }
                        if(j<0){
                            str3=str.substr(id,len-i-id);   //判断是否能分成符合条件的ABC，即判断ABC是否不相同且不为空
                            for(j=1;j<i;j++)    //枚举A的长度
                            {
                                str1=str.substr(0,j);
                                str2=str.substr(j,i-j);

                                if(str1!=str2&&str1!=str3&&str2!=str3) break;
                            }
                            if(j<i){
                               flag=true;
                               break;
                            }
                        }
                    }else if(len-id<i&&len>id){  //针对ABABA模式
                        for(j=0;j<len-id;j++)   //判断E是否为D的前半段
                        {
                            if(str[j]!=str[id+j]) break;
                        }
                        if(j==len-id&&str.substr(0,j)!=str.substr(j,i-j)){  //此时E为A，判断是否D所包含的AB不相同
                            flag=true;
                            break;
                        }
                    }
                }
            }
        }
        if(flag) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}

题目（ http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5350）：

Pretty Poem

---

Time Limit: 2 Seconds      Memory Limit: 65536 KB

---

Poetry is a form of literature that uses aesthetic and rhythmic qualities of language. There are many famous poets in the contemporary era. It is said that a few ACM-ICPC contestants can even write poetic code. Some poems has a strict rhyme scheme like "ABABA" or "ABABCAB". For example, "niconiconi" is composed of a rhyme scheme "ABABA" with A = "ni" and B = "co".

More technically, we call a poem pretty if it can be decomposed into one of the following rhyme scheme: "ABABA" or "ABABCAB". The symbol A, B and C are different continuous non-empty substrings of the poem. By the way, punctuation characters should be ignored when considering the rhyme scheme.

You are given a line of poem, please determine whether it is pretty or not.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

There is a line of poem S (1 <= length(S) <= 50). S will only contains alphabet characters or punctuation characters.

Output

For each test case, output "Yes" if the poem is pretty, or "No" if not.

Sample Input

3
niconiconi~
pettan,pettan,tsurupettan
wafuwafu

Sample Output

Yes
Yes
No