ZOJ 3818(substr函数的使用)

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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

题目大意：

　　这道题目是说，给你一个长度不超过55的文章，要求你把其中的类似与ABABA和ABABCAB的格式串找出来，如果能够找到这样的串，就输出Yes、

解题思路：

　　这道题是自己关于substr(i,j)函数的学习，substr(i,j)指的是，从str[i]开始的长度为j的字符串。刚刚好，我们枚举substr(0,i)作为A

　　substr(i,j)作为B，那么A的长度就是i，B的长度就是j，将这两个字符串进行拼接组合，看能否得到一个我们需要满足的字符串的形式。

　　关于对于C的枚举，我们可以用len-（i+j）*3得到C的长度然后通过substr((i+j)*2, len-(i+j)*3 ).

代码：

# include<cstdio>
# include<iostream>
# include<cstring>
# include<string>

using namespace std;

# define MAX 55

char str[MAX];
string s;

int main(void)
{
    int t; scanf("%d",&t);
    while ( t-- )
    {
        scanf("%s",str);
        int len = strlen(str);
        for ( int i = 0;i < len;i++ )
        {
            if ( islower(str[i])||isupper(str[i]) )
                s+=str[i];
        }
        //cout<<s<<endl;
        len = s.size();
        int flag = 0;
        for ( int i = 1;i < len&&flag==0;i++ )
        {
            for ( int j = 1;j < len&&flag==0;j++ )
            {
                string A = s.substr(0,i);
                string B = s.substr(i,j);
                if ( A==B )
                    continue;
                if ( A+B+A+B+A==s )
                {
                    flag = 1;
                    break;
                }
                if ( len-(i+j)*3>0 )
                {
                    string AB = A+B;
                    string C = 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");
        else
            puts("No");
        s.clear();
    }

    return 0;
}

ZOJ 3818(substr函数的使用)

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原文地址：http://www.cnblogs.com/wikioibai/p/4794950.html