ZOJ 3813 Pretty Poem (violence)

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 is many famous poets in the contemporary era. It is said a few ACM-ICPC contestants can even write poetic code. Some poems have a strict rhyme scheme like "Ababa" or "Ababcab". For example, "Niconiconi" was composed of a rhyme scheme "ababa" with a = "ni" and B = "CO".

More technically, we call a poem pretty if it can being decomposed into one of the following rhyme scheme: "Ababa" O R "Ababcab". The symbolA, B and C are different continuous non-empty substrings of the poem. By the punctuation characters should is ignored when considering the rhyme scheme.

You were given a line of poem, please determine whether it was pretty or not.

Input

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

There is a line of poem S (1 <= Length ( S ) <=). Would only contains alphabet characters S or punctuation Characters.

Output

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

Sample Input

3niconiconi~pettan,pettan,tsurupettanwafuwafu

Sample Output

Yesyesno

Title Link: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3818

Problem-solving ideas: string processing, direct violence enumeration A,b,c, directly determine whether to meet the two situations. Note A,b,c are different, I am two cases of classification open processing, the second case of AB as a substring directly judged, but also to judge AB can be decomposed into two different substrings, a, B and not the same as C.

The code is as follows:

#include <cstdio> #include <cstring> #include <cmath> #include <iostream> #include <string > #include <algorithm> #include <string>using namespace Std;//const int maxn=205;int main (void) {int t; scanf ("%d", &t), while (t--) {string s,a;cin>>s;int p=0,n=0;int len=s.length (); for (int i=0;i<len;i++) if (s [i]>= ' A ' &&s[i]<= ' Z ' | | S[i]>= ' a ' &&s[i]<= ' Z ') {a+=s[i];n++;}   for (int l1=1;l1<=n;l1++)//first case Ababa judgment {string a=a.substr (0,L1); Substring a always starts at the first position for the for (int l2=1;l2<=n;l2++) {if (l1+l2+l1+l2+l1>n)//length and is greater than the main string, without having to enumerate backwards break;string b=a.substr (L1,  L2); Substring B is always next to string AIF (A==A+B+A+B+A&AMP;&AMP;A!=B) {p=1;break;}} if (p) break;} if (p) {printf ("yes\n"); continue;}   for (int l1=2;l1<=n;l1++)//second case Ababcab judgment {string ab=a.substr (0,L1); Consider AB as a substring for (int l2=1;l2<=n;l2++) {if (l1+l1+l2+l1>n) break;string c=a.substr (2*L1,L2); if (A==AB+AB+C+AB)// Further judging when conditions are met the AB can be decomposed into different substrings A and b{int lab=ab.length (); for (int k=1;k<lab;k++) {string a=ab.subSTR (0,k); string B=ab.substr (K,lab-k); if (a!=c&&a!=b&&b!=c) {p=1;break;}} if (p) break;}} if (p) break;} if (p) printf ("yes\n"); elseprintf ("no\n");}}

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ZOJ 3813 Pretty Poem (violence)