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
1 /*
2 许杰浩的题解：
3 AC代码：
4 */
5 #include <iostream>
6 #include <cstdio>
7 #include <string.h>
8 #include <cstring>
9 using namespace std;
10 char str[55],s1[55];
11 char s2[55],s3[55],s4[55];
12 int cnt;
13 void getap(){
14 cnt=1;
15 int l=strlen(str);
16 for(int i=0;i<l;i++){
17 if(str[i]<=‘z‘&&str[i]>=‘a‘){
18 s1[cnt++]=str[i];
19 }
20 else if(str[i]<=‘Z‘&&str[i]>=‘A‘){
21 s1[cnt++]=str[i];
22 }
23 }
24 }
25 void getstr(int i,int j,char *s){
26 int p=0;
27 for(int k=0;k<55;k++)s[k]=‘\0‘;
28 for(int k=i;k<=j;k++){
29 s[p++]=s1[k];
30 }
31 }
32 void print(int l,int k){
33 for(int i=l;i<=k;i++){
34 printf("%c",s1[i]);
35 }
36 cout<<endl;
37 }
38 bool solve1(){
39 int len=cnt-1;
40 if(len<5)return false;
41 for(int i=1;i<=len;i++){
42 if(s1[i]!=s1[len])continue;
43 if(len-i*3<2)break;
44 if((len-i*3)%2==1)continue;
45 int l=(len-i*3)/2;
46 bool flag=1;
47 for(int j=1;j<=i;j++){
48 if(s1[j]!=s1[j+l+i]){flag=0;break;} //ababa 1a 2a
49 if(s1[j]!=s1[len-i+j]){flag=0;break;} //ababa 1a 3a
50 }
51 //print(1,i);
52 //print(i+1,i+l);
53 if(!flag)continue;
54 if(l==i){ //ababa a b
55 bool ok1=0;
56 for(int j=1;j<=i;j++){
57 if(s1[j]!=s1[j+i]){ok1=1;break;}
58 }
59 if(!ok1)flag=0;
60 }
61 if(!flag)continue;
62 for(int j=1;j<=l;j++){ //ababa 1b 2b
63 if(s1[j+i]!=s1[j+i*2+l]){flag=0;break;}
64 }
65 if(flag)return true;
66 }
67 return false;
68 }
69 bool solve2(){
70 int len=cnt-1;
71 if(len<7)return false;
72 for(int i=2;i<=len;i++){
73 if(s1[i]!=s1[len])continue;
74 if(len-i*3<1)break;
75 bool flag=1;
76 //print(1,i);
77 for(int j=1;j<=i;j++){ //ababcab
78 if(s1[j]!=s1[len-i+j]){flag=0;break;} //ababcab 1ab 3ab
79 if(s1[j]!=s1[i+j]){flag=0;break;} //ababcab 1ab 2ab
80 }
81 if(!flag)continue;
82 getstr(2*i+1,len-i,s4);
83 //print(2*i+1,len-i);
84 for(int j=1;j<i;j++){
85 getstr(1,j,s2);
86 getstr(j+1,i,s3);
87 if(strcmp(s2,s3)!=0&&strcmp(s4,s2)!=0&&strcmp(s3,s4)!=0){ //a!=b c!=a b!=c
88 //print(1,j);print(j+1,i);
89 return true;
90 }
91 }
92 }
93 return false;
94 }
95
96 int main(){
97 int t;
98 scanf("%d",&t);
99 while(t--){
100 scanf("%s",str);
101 getap();
102 if(solve1()||solve2())printf("Yes\n");
103 else printf("No\n");
104 }
105 }