Problem B – Buggy ICPC——思维，找规律

Problem B – Buggy ICPC
Author : Gabriel Poesia, Brasil
Alan Curing is a famous sports programmer. He is the creator of the theoretical model of computation
known as the Alan Curing Machine (ACM). He’s most famous for creating his own computer for programming competitions: the Integrated Computer for Programming Contests (ICPC). This computer
has a specialized operating system with commands for submitting code and testing executables on sample inputs, an input generator, a wide display for debugging, and a very soft keyboard. However, as it
happens even to the best, Alan’s creation has a nasty bug. Every time Alan types a vowel on the ICPC,
the content of the current line is reversed.
The bug has been extremely hard to track down, so Alan has decided to accept the challenge and
use the computer as it is. He is currently training touch typing on the ICPC. For now, he is only typing
strings using lowercase letters, and no spaces. When Alan types a consonant, it is appended to the end
of the current line, as one would expect. When he types a vowel, however, the typed character is first
added to the end of the line, but right after that the whole line is reversed. For example, if the current
line has “imc” and Alan types “a” (a vowel), for a brief moment the line will become “imca”, but then
the bug kicks in and turns the line into “acmi”. If after that he types the consonants “c”, “p” and “c”,
in that order, the line becomes “acmicpc”.
When practicing, Alan first thinks of the text he wants to type, and then tries to come up with a
sequence of characters he can type in order to obtain that text. He is having trouble, however, since he
realized that he cannot obtain some texts at all (such as “ca”), and there are multiple ways of obtaining
other texts (as “ac”, which is obtained whether he types “ac” or “ca”). Help Alan in his training by
telling him in how many ways he can type each text he wishes to type. A way of typing a text T can
be encoded by a string W with |T| characters such that if the characters are typed on the ICPC in the
order they appear in W (i.e. W1, W2, . . . , W|T|) the final result is equal to T, considering ICPC’s known
bug. Two ways are considered different if they are encoded by different strings. The letters that trigger
the bug in the ICPC when typed are “a”, “e”, “i”, “o” and “u”.
Input
The input consists of a single line that contains a non-empty string T of at most 105
lowercase
letters, representing the text Alan wants to type on the ICPC.
Output
Output a single line with an integer representing the number of distinct ways Alan can type the
desired text T considering ICPC’s known bug.
Sample input 1
ac
Sample output 1
2
Sample input 2
ca
Sample output 2
0
Sample input 3
acmicpc
Sample output 3
3

先从两个元音字母的情况入手：acmicpc(a，i)，因为只有两个元音字母（只会进行两次反转，那么在前面的元音字母一定后出现），所以要想符合原串，在加上a之前的字符串一定是：icm。此时问题就转换成了：有多少种添加i, c, m的方式，得到icm（cpc则不需处理，因为无法进行翻转）。当i的位置确定后，剩下的位置也就必定确定，所以i有几个摆放位置，就有几种方式。当多个元音字母时可以将问题分解成“1+（n -1）”的形式，仿照处理两个元音字母的方式计算即可。

按上述方法分析，可以找出如下规律：
1.全为辅音：1种
2.只有一个元音，且在开头：n种
3.只有一个元音或多个，开头为辅音：0种
4.多个元音：中间两个元音字符的距离

#include <iostream>
#include <string>
#include <set>
#include <algorithm>
#include <vector>
using namespace std;string s;
set<char> book{ 'a', 'e', 'i', 'o', 'u'};
vector<int> pos;int main()
{int cnt = 0;cin >> s;for (int i = 0; i < s.size(); i++)if (book.find(s[i]) != book.end()){cnt++;pos.push_back(i);}if (cnt == 0) cout << 1 << endl;else if (cnt == 1 && book.find(s[0]) != book.end()) cout << s.size();else if (book.find(s[0]) == book.end()) cout << 0 << endl;else{int p = (pos.size() + 1) / 2;cout <<  pos[p] - pos[p - 1];}return 0;
}

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