CF--思维练习--CodeForces - 219C Color Stripe （思维）

ACM思维题训练集合 A colored stripe is represented by a horizontal row of n square cells, each cell is pained one of k colors. Your task is to repaint the minimum number of cells so that no two neighbouring cells are of the same color. You can use any color from 1 to k to repaint the cells.

Input The first input line contains two integers n and k (1 ≤ n ≤ 5·105; 2 ≤ k ≤ 26). The second line contains n uppercase English letters. Letter "A" stands for the first color, letter "B" stands for the second color and so on. The first k English letters may be used. Each letter represents the color of the corresponding cell of the stripe.

Output

Print a single integer — the required minimum number of repaintings. In the second line print any possible variant of the repainted stripe.

Examples Input 6 3 ABBACC

Output 2 ABCACA Input

3 2 BBB Output 1 BAB

题目很简单，类比平面图的四色定理，线性上有3色定理，即如果k>=3那么无论遇到什么情况AABB的时候，我们都可以将相同的变成第三种颜色，所以这里要特判K=2的时候，我一开始想了很久，就是不知道怎么处理，但是k=2的串只有ABABA……和BABABA……这两种情况，判断当前穿变成这两种那种花费少，就是答案了。

#include <bits/stdc++.h>
using namespace std;
int n, m, cnt = 0;
char check(char a, char b)
{
    for (char i = 'A'; i <= 'A' + m - 1; i++)
    {
        if (i == a || i == b)
            continue;
        else
            return i;
    }
}
int main()
{

    cin >> n >> m;
    string s;
    cin >> s;
    if (m == 2)
    {
        int cnt1 = 0, cnt2 = 0;
        for (int i = 0; i < n; i++)
        {
            if (s[i] != ('A' + (i % 2)))
                cnt1++;
            else
                cnt2++;
        }
        if (cnt2 > cnt1)
        {
            cout << cnt1 << endl;
            for (int i = 0; i < n; i++)

                if (i % 2 == 0)
                    putchar('A');
                else
                    putchar('B');
        }
        else
        {
            cout << cnt2 << endl;
            for (int i = 1; i <= n; i++)

                if (i % 2 == 0)
                    putchar('A');
                else
                    putchar('B');
        }
        puts("");
        return 0;
    }
    for (int i = 1; i < s.length();)
    {
        if (s[i] == s[i - 1])
        {
            s[i] = check(s[i - 1], s[i + 1]);
            i = i + 2;
            cnt++;
        }
        else
            i++;
    }
    cout << cnt << endl
         << s << endl;
}