ZOJ 3818 (use of substr function)

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 symbol A, 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 punctuatio n 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

Main topic:

This topic is said, give you a length of not more than 55 of the article, asked you to the similar with Ababa and Ababcab format string to find out, if you can find such a string, output Yes,

Problem Solving Ideas:

This question is my own study of substr (i,j) function, substr (I,J) refers to a string of length J, starting with Str[i]. Just fine, we enumerate substr (0,i) as a

SUBSTR (i,j) as B, then the length of a is the length of the I,b is J, the two strings are stitched together to see if we can get a string to satisfy the form.

For the enumeration of C, we can use len-(i+j) to get the length of C and then pass substr (I+j), len-(I+J).

Code:

# include<cstdio># include<iostream># include<cstring># include<string>using namespacestd;# define MAX -CharStr[max];strings;intMainvoid){    intT scanf"%d",&t);  while(t--) {scanf ("%s", str); intLen =strlen (str);  for(inti =0; I < len;i++ )        {            if(Islower (str[i]) | |Isupper (Str[i])) s+=Str[i]; }        //cout<<s<<endl;Len =s.size (); intFlag =0;  for(inti =1; I < len&&flag==0; i++ )        {             for(intj =1; J < len&&flag==0; j + + )            {                stringA = S.substr (0, i); stringB =s.substr (I,J); if(a==B)Continue; if(a+b+a+b+a==s) {flag=1;  Break; }                if(len-(I+J) *3>0 )                {                    stringAB = A +C; stringC = S.substr ((i+j) *2, len-(I+J) *3); if(a==c| | b==C)Continue; if(ab+ab+c+ab==s) {flag=1;  Break; }                }            }        }        if(flag) puts ("Yes"); Elseputs ("No");    S.clear (); }    return 0;}

　　

ZOJ 3818 (use of substr function)