D - Say No to Palindromes

列舉

可觀察到只有類似 abcabcabcabc..., bacbacbac... 等 abc 三個字母都迴圈出現才滿足要求

可記錄 \(cnt[i][j][k]\)，前 \(i\) 箇中 \(a, b, c\) 分別在 模 \(3\) 餘 \(0,1,2\) 的個數

因此列舉 abc 的 6 種排列，用字首和求最小改動次數即可

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>

using namespace std;
typedef long long ll;
const int N = 2e5 + 10;
string s;
int cnt[N][3][3];//cnt[i][j][k]為前i個字母中(a, b, c)分別在(0, 1, 2)三個位置的個數
int n, m;
int l, r;

int solve(int a, int b, int c)
{
	int cnt_a = cnt[r][a][0] - cnt[l-1][a][0];
	int cnt_b = cnt[r][b][1] - cnt[l-1][b][1];	
	int cnt_c = cnt[r][c][2] - cnt[l-1][c][2];	
	return r - l + 1 - cnt_a - cnt_b - cnt_c;
}

int main()
{
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n >> m;
	cin >> s;
	s = " " + s;
	for (int i = 1; i < s.size(); i++)
	{
		for (int j = 0; j < 3; j++)
			for (int k = 0; k < 3; k++)
				cnt[i][j][k] = cnt[i-1][j][k];
		char ch = s[i];
		int t = i % 3;
		if (ch == 'a')
			cnt[i][0][t]++;
		else if (ch == 'b')
			cnt[i][1][t]++;
		else
			cnt[i][2][t]++;
	}
	
	while(m--)
	{
		cin >> l >> r;
		int ans = 1e9;
		ans = min(ans, solve(0, 1, 2));
		ans = min(ans, solve(0, 2, 1));
		ans = min(ans, solve(1, 0, 2));
		ans = min(ans, solve(1, 2, 0));
		ans = min(ans, solve(2, 0, 1));
		ans = min(ans, solve(2, 1, 0));
		cout << ans << endl;
	}
	return 0;
}