ZOJ - 3818 Pretty Poem

Description

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

题意：求满足"ABABA" or "ABABCAB"组合的字符串
思路：暴力求A和B，推出C
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;

int main() {
	int t;
	char ch[60];
	scanf("%d", &t);
	while (t--) {
		string str;
		scanf("%s", ch);
		int len = strlen(ch);
		for (int i = 0; i < len; i++)
			if (isalpha(ch[i]))
				str += ch[i];

		len = str.length();
		int flag = 0;
		for (int i = 1; i < len/2 && !flag; i++) 
			for (int j = 1; j < len/2 && !flag; j++) {
				string A = str.substr(0, i);
				string B = str.substr(i, j);
				if (A == B)
					continue;
				if (A + B + A + B + A == str) {
					flag = 1;
					continue;
				}
				if (len - (i + j) * 3 > 0) {
					string AB = A + B;
					string C = str.substr(2*(i+j), len-3*(i+j));
					if (A == C || B == C)
						continue;
					if (AB + AB + C + AB == str) {
						flag = 1;
						continue;
					}
				}
			}
		if (flag)
			printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}