Codeforce219C——贪心——Color Stripe

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.

Sample test(s)

input

6 3
ABBACC

output

2
ABCACA

input

3 2
BBB

output

1
BAB

/*
对m == 2 的情况特殊处理  假顶A出现在奇数项，B出现在奇数项，比较出现的次数，如果A出现次数少于B，那么该A，反之改B
对于 m > 2 的情况 正常处理
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAX = 500000 + 10;
int main()
{
    int n, m;
    char s[MAX];
    while(~scanf("%d%d", &n, &m)){
        scanf("%s", s);
        int len = strlen(s);
        char ch;
        int ans = 0;
        if(m == 2){
           int c1 = 0, c2 = 0;
           for(int i = 0 ; i < len; i++){
                   if(i % 2 == 1) {
                           if(s[i] == 'A') c1++;
                           else c2++;
                   }
                   else {
                           if(s[i] == 'A') c2++;
                           else c1++;
                   }
           }
                   int ans = min(c1, c2);
                   for(int i = 0; i < len; i++){
                           if(ans == c1){
                                if(i % 2 == 1){
                                        s[i] = 'B';
                                }
                                else s[i] = 'A';
                           }
                           else {
                                if(i % 2 == 1){
                                        s[i] = 'A';
                                }
                                else s[i] = 'B';
                           }
                   }
                   printf("%d\n", ans);
                   printf("%s\n",s);
        }
        else {
        for(int i = 1 ; i < len; i++){
                if(s[i-1] == s[i]){
                        ans++;
                        for(int j = 1; j <= m ;j++){
                                ch = 'A' + j - 1;
                                if(s[i] == ch || s[i+1] == ch);
                                else {
                                                s[i] = ch;
                                                break;
                                        }
                                }
                        }
                }
        printf("%d\n%s\n", ans, s);
        }
    }
    return 0;
}