J1B What beats RiOI

思路

RiOI 这个字符串较短，打表出每一种可能的大小写情况，遍历时用 find 函数判断即可。

复杂度

令 $|s|$ 为字符串 $s$ 的长度。

时间复杂度:

$O(|x|+|y|)$

空间复杂度:

$O(|x|+|y|)$

Code

#include <bits/stdc++.h>
using namespace std;
int pd(string a)
{
	string b[105] = {"rioi","Rioi","rIoi","riOi","rioI","RIoi","RiOi","RioI","rIOi","rIoI","riOI","RIOi","rIOI","RIoI","RiOI","RIOI"};
	for (int i = 0;i < 15;i++)
	{
		if (a.find(b[i]) != -1) return 1;
	}
	return 0;
}
int main()
{
	string a,b;
	cin >> a >> b;
	int x = pd(a),y = pd(b);
	if (x && y) cout << "Either is ok!";
	else if (x) cout << a << " for sure!";
	else if (y) cout << b << " for sure!";
	else cout << "Try again!";
	return 0;
}