Codeforces Round #553 (Div. 2）A

模拟，连续的4个 

#include <bits/stdc++.h>
using namespace std;
char s[100];
int ans = 100000000;
int solve(char a, char b)
{
	a = a - 'A', b = b - 'A';
	return min((a - b + 26) % 26, (b - a + 26) % 26);
}
int main()
{
	int n; cin>>n;
	scanf("%s",s+1);
	string c="ACTG";
	for (int i = 1; i <= n - 4 + 1; i++)
	{
		int dd=0;
		for (int j = 0; j < 4; j++)
			dd += solve(s[i + j], c[j]);
		ans = min(ans, dd);
	}
	cout << ans << endl;
	return 0;
}