题目描述
There is a string s of length 3 or greater. No two neighboring characters in s are equal.
Takahashi and Aoki will play a game against each other. The two players alternately performs the following operation, Takahashi going first:
Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.
The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.
Constraints
3≤|s|≤105
s consists of lowercase English letters.
No two neighboring characters in s are equal.
Takahashi and Aoki will play a game against each other. The two players alternately performs the following operation, Takahashi going first:
Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.
The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.
Constraints
3≤|s|≤105
s consists of lowercase English letters.
No two neighboring characters in s are equal.
输入
The input is given from Standard Input in the following format:
s
s
输出
If Takahashi will win, print First. If Aoki will win, print Second.
样例输入
aba
样例输出
Second
提示
Takahashi, who goes first, cannot perform the operation, since removal of the b, which is the only character not at either ends of s, would result in s becoming aa, with two as neighboring.
很很简单的一个找规律题，但是卡了好久。
头尾相同的时候，最后肯定是剩下奇数个字母的，比如ababa。那么如果位数是奇数那么中间肯定进行了偶数次操作，反正就是奇数次。
头尾不同的话，最后剩下偶数个，如果位数为奇数就操作了奇数次，否则偶数次。
偶数次操作那么第一个人就无法操作就输，奇数次操作第二个人无法操作即输。
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 int main() 5 { 6 string s; 7 cin>>s; 8 int len = s.length(); 9 if(s[0] == s[len-1]) 10 if(len&1)printf("Second "); 11 else printf("First "); 12 else 13 if(len&1)printf("First "); 14 else printf("Second "); 15 }