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
Question: Given a string consisting of n characters, which only contains k kinds of characters, ask how many characters can be modified at least to get an adjacent character string that does not have the same string.
Idea: There are two situations, if k>2 is normal greedy (change of i can make i-1 and i+1 satisfy the conditions at the same time), if k=2 points (ABAB or BABA)
AC code:
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=500010;
char s[N];
int a[N];
int main()
{
int n,k;
cin>>n>>k;
scanf("%s",s+1);
int ans=0;
for(int i=1;i<=n;i++)
a[i]=s[i]-'A';
int l=0,r=0;
if(k==2)
{
for(int i=1;i<=n;i++)
{
if(i%2&&s[i]=='A'||(i%2==0&&s[i]=='B'))
r++;
if(i%2&&s[i]=='B'||(i%2==0&&s[i]=='A'))
l++;
}
if(l<r)
{
cout<<l<<endl;
for(int i=1;i<=n;i++)
{
if(i%2)
cout<<'A';
else
cout<<'B';
}
}
else
{
cout<<r<<endl;
for(int i=1;i<=n;i++)
{
if(i%2)
cout<<'B';
else
cout<<'A';
}
}
cout<<endl;
}
else
{
a[0]=-1;
for(int i=1;i<=n;i++)
{
if(a[i]==a[i-1])
{
ans++;
a[i]=(a[i]+1)%k;
if(a[i]==a[i+1])
a[i]=(a[i]+1)%k;
}
}
cout<<ans<<endl;
for(int i=1;i<=n;i++)
cout<<char(a[i]+'A');
cout<<endl;
}
}