Pretty Poem

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



#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<cmath>
#include<vector>
#include<limits.h>
#include<algorithm>
#define LL long long
#define mod 1e9 + 7

using namespace std;

const int M = 55;

char a[M];
char s[M];
char A[M];
char B[M];
char C[M];

int main()
{
    int t;
    cin >> t;
    getchar();
    while( t-- )
    {
        gets(s);
        int temp = 0;
        memset(a,'',sizeof(a));
        for(int i = 0; i < strlen(s); ++i)
            if((s[i] >= 'a' && s[i] <= 'z') || (s[i] >= 'A' && s[i] <= 'Z'))
                a[temp++] = s[i];
        int len = strlen(a);
        int ans = 0;
        for(int i = 1; i < 17; ++i)
        {
            for(int j = 1; j < 25; ++j)
            {
                if(3 * i + j * 2 != len)
                    continue;
                memset(A,'',sizeof(A));
                memset(B,'',sizeof(B));
                memset(C,'',sizeof(C));
                strncpy(A,a,i);
                strncpy(B,a + i + j,i);
                strncpy(C,a + 2 * i + 2 * j,i);
                if(strcmp(A,B) == 0 && strcmp(A,C) == 0 && strcmp(B,C) == 0)
                {
                    memset(A,'',sizeof(A));
                    memset(B,'',sizeof(B));
                    strncpy(A,a + i,j);
                    strncpy(B,a + 2 * i + j,j);
                    if(strcmp(A,B) == 0 && strcmp(A,C) != 0)
                    {
                        ans = 1;
                        break;
                    }
                }
            }
            if(ans)
                break;
        }
        for(int i = 2; i < 17; ++i)
        {
            if(3 * i >= len)
                break;
            memset(A,'',sizeof(A));
            memset(B,'',sizeof(B));
            memset(C,'',sizeof(C));
            strncpy(A,a,i);
            strncpy(B,a + i,i);
            strncpy(C,a + len - i,i);
            if(strcmp(A,B) == 0 && strcmp(A,C) == 0 && strcmp(B,C) == 0)
            {
                for(int j = 1; j < i; ++j)
                {
                    memset(A,'',sizeof(A));
                    memset(B,'',sizeof(B));
                    memset(C,'',sizeof(C));
                    strncpy(A,a,j);
                    strncpy(B,a + j,i - j);
                    strncpy(C,a + 2 * i,len - 3 * i);
                    if(strcmp(A,B) != 0 && strcmp(A,C) != 0 && strcmp(B,C) != 0)
                    {
                        ans = 1;
                        break;
                    }
                }
            }
            if(ans)
                break;
        }
        if(ans)
            puts("Yes");
        else
            puts("No");
    }
    return 0;
}
View Code
原文地址:https://www.cnblogs.com/767355675hutaishi/p/3960552.html