zoj 3818 Pretty Poem（暴力處理字串）2014年牡丹江賽區網路賽，zoj3818

zoj 3818 Pretty Poem（暴力處理字串）2014年牡丹江賽區網路賽，zoj3818

Pretty PoemTime 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

3niconiconi~pettan,pettan,tsurupettanwafuwafu

Sample Output

YesYesNo

題意：給出一個長度不超過50的字串，判斷它是不是pretty串（忽略標點符號）。pretty串的定義如下：如果一個串可以寫成ABABA或者ABABCAB的形式，且A、B、C兩兩不相等，那麼這個串就是pretty串。

分析：因為長度不超過50，所以可以直接枚舉A、B的長度，然後暴力求解。比賽時我枚舉的是AB整體的長度，不知道為什麼一直WA。取子串時，可以用substr取代迴圈。

#include<iostream>#include<string>using namespace std;bool is_pretty1(string s){    if(s.length() < 5) return false;    for(int lenA = 1; lenA <= (s.length() - 2) / 3; lenA++) {        string A = s.substr(0, lenA);        for(int lenB = 1; lenA * 3 + lenB * 2 <= s.length(); lenB++) {            if(lenA * 3 + lenB * 2 != s.length()) continue;            string B = s.substr(lenA, lenB);            if(A != B) {                string AA = s.substr(lenA + lenB, lenA);                string BB = s.substr(lenA * 2 + lenB, lenB);                string AAA = s.substr(lenA * 2 + lenB * 2, lenA);                if(A == AA && A == AAA && B == BB)                    return true;            }        }    }    return false;}bool is_pretty2(string s){    if(s.length() < 7) return false;    for(int lenA = 1; lenA <= (s.length() - 4) / 3; lenA++) {        string A = s.substr(0, lenA);        for(int lenB = 1; 3 * lenA + 3 * lenB < s.length(); lenB++) {            string B = s.substr(lenA, lenB);            if(A != B) {                int lenC = s.length() - 3 * lenA - 3 * lenB;                string AA = s.substr(lenA + lenB, lenA);                string BB = s.substr(lenA * 2 + lenB, lenB);                string C = s.substr(lenA * 2 + lenB * 2, lenC);                string AAA = s.substr(lenA * 2 + lenB * 2 + lenC, lenA);                string BBB = s.substr(lenA * 3 + lenB * 2 + lenC, lenB);                if(A == AA && A == AAA && B == BB && B == BBB && A != C && B != C)                    return true;            }        }    }    return false;}bool is_pretty(string s){    return is_pretty1(s) || is_pretty2(s);}int main(){    int T;    string tmp;    cin >> T;    while(T--) {        cin >> tmp;        string s = "";        for(int i = 0; i < tmp.length(); i++)            if(isalpha(tmp[i]))                s += tmp[i];        if(is_pretty(s)) cout << "Yes" << endl;        else cout << "No" << endl;    }    return 0;}