Codeforces Round #135 (Div. 2)-C. Color Stripe

原题链接

C. Color Stripe

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 3ABBACC

output

2ABCACA

input

3 2BBB

output

1BAB

1.如果n == 1则不需要改动。

2.如果k == 2 && n > 1则比较将原字符串改为ABABABA..和BABABAB...的代价哪个小

3.如果k > 2 && n > 1从下标1开始遍历整个字符串str, 若str[i] == str[i-1]则改动str[i]这个字符使之不等于str[i-1]和str[i+1]

#include <bits/stdc++.h>#define maxn 500005using namespace std;typedef long long ll;char str[maxn];char p[26];int main(){//freopen("in.txt", "r", stdin);for(int i = 0; i < 26; i++) p[i] = 'A' + i;int n, k;scanf("%d%d%s", &n, &k, str);int len = strlen(str);if(n == 1){puts("0");puts(str);}else if(k == 2){int k1 = 0, k2 = 0, d1 = 0, d2 = 1;for(int i = 0; i < len; i++){if(str[i] != p[d1]){k1++;}if(str[i] != p[d2]){k2++;}d1 ^= 1;d2 ^= 1;}if(k1 < k2){d1 = 0;for(int i = 0; str[i]; i++, d1 ^= 1) str[i] = p[d1];printf("%d\n", k1);puts(str);}else{d2 = 1;for(int i = 0; str[i]; i++, d2 ^= 1)  str[i] = p[d2];printf("%d\n", k2);puts(str);}}else{int ans = 0;for(int i = 1; i < len; i++){if(str[i] == str[i-1]){for(int j = 0; j < k; j++){if(p[j] != str[i-1]){if(i < len-1){if(p[j] != str[i+1]){str[i] = p[j];ans++;break;} }else{ str[i] = p[j]; ans++; break;    }}}}}printf("%d\n", ans);puts(str);}return 0;}