第一次网络赛之许杰浩题解

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

/*
许杰浩的题解：
AC代码：
*/
#include <iostream>
#include <cstdio>
#include <string.h>
#include <cstring>
using namespace std;
char str[55],s1[55];
char s2[55],s3[55],s4[55];
int cnt;
void getap(){
    cnt=1;
    int l=strlen(str);
    for(int i=0;i<l;i++){
        if(str[i]<='z'&&str[i]>='a'){
            s1[cnt++]=str[i];
        }
        else if(str[i]<='Z'&&str[i]>='A'){
            s1[cnt++]=str[i];
        }
    }
}
void getstr(int i,int j,char *s){
    int p=0;
    for(int k=0;k<55;k++)s[k]='\0';
    for(int k=i;k<=j;k++){
        s[p++]=s1[k];
    }
}
void print(int l,int k){
    for(int i=l;i<=k;i++){
        printf("%c",s1[i]);
    }
    cout<<endl;
}
bool solve1(){
    int len=cnt-1;
    if(len<5)return false;
    for(int i=1;i<=len;i++){
        if(s1[i]!=s1[len])continue;
        if(len-i*3<2)break;
        if((len-i*3)%2==1)continue;
        int l=(len-i*3)/2;
        bool flag=1;
        for(int j=1;j<=i;j++){
            if(s1[j]!=s1[j+l+i]){flag=0;break;}                        //ababa    1a 2a
            if(s1[j]!=s1[len-i+j]){flag=0;break;}                    //ababa    1a 3a
        }
        //print(1,i);
        //print(i+1,i+l);
        if(!flag)continue;
        if(l==i){                                                    //ababa    a b
            bool ok1=0;
            for(int j=1;j<=i;j++){
                if(s1[j]!=s1[j+i]){ok1=1;break;}
            }
            if(!ok1)flag=0;
        }
        if(!flag)continue;
        for(int j=1;j<=l;j++){                                        //ababa        1b 2b
            if(s1[j+i]!=s1[j+i*2+l]){flag=0;break;}
        }
        if(flag)return true;
    }
    return false;
}
bool solve2(){
    int len=cnt-1;
    if(len<7)return false;
    for(int i=2;i<=len;i++){
        if(s1[i]!=s1[len])continue;
        if(len-i*3<1)break;
        bool flag=1;
        //print(1,i);
        for(int j=1;j<=i;j++){                                                //ababcab
            if(s1[j]!=s1[len-i+j]){flag=0;break;}                            //ababcab  1ab 3ab
            if(s1[j]!=s1[i+j]){flag=0;break;}                                //ababcab  1ab 2ab
        }
        if(!flag)continue;
        getstr(2*i+1,len-i,s4);
        //print(2*i+1,len-i);
        for(int j=1;j<i;j++){
            getstr(1,j,s2);
            getstr(j+1,i,s3);
            if(strcmp(s2,s3)!=0&&strcmp(s4,s2)!=0&&strcmp(s3,s4)!=0){        //a!=b c!=a b!=c
                //print(1,j);print(j+1,i);
                return true;
            }
        }
    }
    return false;
}

int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%s",str);
        getap();
        if(solve1()||solve2())printf("Yes\n");
        else printf("No\n");
    }
}