Codeforce219C-Color Stripe-蒲公英云

E. 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

Copy

6 3
ABBACC

output

Copy

2
ABCACA

input

Copy

3 2
BBB

output

Copy

1
BAB
题意：给你一行n个元素，k种颜色，要求相邻的元素颜色不能相同，输出最少改变颜色的数量和n个元素最后的颜色（若有多种可能可任意输出一种）
注意：k==2时必定是奇偶位不相同，要选一个最少改变方案，所以要知道奇位放A要改变的数多还是偶位放A要改变的数多，在训练时卡在这种情况了，
其实一开始想到了ABBA这种情况，但当时的思路是只改变相同颜色的元素，没想到不同颜色元素也能改变，所以当时就假设如果有偶数个相同颜色的元素，其两边元素颜色必不同，结果就一直wrong answer on the test 15...
1 #include<bits/stdc++.h>
  2 using namespace std;
  3 const int amn=5e5+5;
  4 char mp[amn];
  5 int a[amn],c[30];
  6 int main(){
  7     int n,k;
  8     ios::sync_with_stdio(0);
  9     cin>>n>>k;
 10     for(int i=1;i<=n;i++){
 11         cin>>mp[i];
 12         a[i]=mp[i]-'A'+1;
 13     }
 14     int ans=0,c1=0,c2=0;
 15     if(k==2){       ///k==2时必定是奇偶位不相同，要选一个最少改变方案，所以要知道奇位放A要改变的数多还是偶位放A要改变的数多
 16         for(int i=1;i<=n;i++){  /// 我们先假设奇位为A，偶位为B，因为有可能合法情况是偶位为A，奇数位为B，所以最后要比较哪个不合法的数量少，这样更改少的那个才能得到最少改变方案，故下面统计不合法数，
 17             if(i%2==0){     ///若偶位为A，则A的不合法数加1，否则B的不合法数加1
 18                 if(mp[i]=='A')c1++;
 19                 else c2++;
 20             }
 21             else{           ///若偶位为A，则B的不合法数加1，否则A的不合法数加1
 22                 if(mp[i]=='A')c2++;
 23                 else c1++;
 24             }
 25         }
 26         ans=min(c1,c2); ///选一个最小不合法数，得到最少改变方案
 27         for(int i=1;i<=n;i++){
 28             if(ans==c1){    ///若偶数位为A是不合法的，要将奇位变为A，偶位变为B
 29                 if(i%2) 
 30                     mp[i]='A';
 31                 else
 32                     mp[i]='B';
 33             }
 34             else{
 35                 if(i%2)     ///若奇数为位A是不合法的，要将偶数位变为A，奇数位变为B
 36                     mp[i]='B';
 37                 else
 38                     mp[i]='A';
 39             }
 40         }
 41     }
 42     else{
 43     a[n+1]=27;
 44     memset(c,0,sizeof c);
 45     bool f=0,d=0;
 46     int fr=1,ta=0,frc,mic,tac,it,cc;
 47     for(int i=2;i<=n;i++){
 48         if(a[i]==a[i-1]){
 49             d=1;
 50             if(fr>ta){
 51                 fr=i-1;
 52                 if(i-2>=1){
 53                     frc=a[fr-1];//cout<<frc<<'!'<<endl;
 54                     c[a[i]]=c[frc]=1;
 55                 }
 56                 else{
 57                     c[a[i]]=1;
 58                     f=1;
 59                 }
 60                 mic=a[i];
 61                 ta=i+1;
 62                 tac=a[ta];
 63                 c[tac]=1;
 64             }
 65             else{
 66 
 67                 ta=i+1;
 68                 tac=a[ta];
 69                 c[tac]=1;
 70             }
 71                 //printf("i:%d fr:%d ta:%d\n",i,fr,ta);
 72                 if(!f&&i==n){
 73                     ans+=(ta-fr)/2;
 74                     it=fr+1;
 75                     cc=frc;
 76                     while(it<ta){
 77                         a[it]=cc;
 78                         mp[it]=a[it]-1+'A';
 79                         it+=2;
 80                     }
 81                     break;
 82                 }
 83 
 84         }
 85         else if(d){
 86             //printf("i:%d fr:%d ta:%d tac:%d\n",i,fr,ta,tac);
 87             d=0;
 88             ans+=(ta-fr)/2;
 89             if(f){
 90                 cc=tac;
 91                 if(ta>n)
 92                     for(int i=1;i<=k;i++){
 93                         if(!c[i]){
 94                             cc=i;
 95                             break;
 96                         }
 97                     }
 98                 it=ta-2;
 99                 while(it>0){
100                     a[it]=cc;
101                     mp[it]=a[it]-1+'A';
102                     it-=2;
103                 }
104                 f=0;
105             }
106             else{
107                 cc=frc;
108                 //printf("---\ni:%d frc: %d tac: %d\n",i,frc,tac);
109                 //for(int i=1;i<=26;i++)cout<<c[i]<<' ';
110                 //cout<<"\n---\n";
111                 if(frc==tac){
112                     for(int i=1;i<=k;i++){
113                         if(!c[i]){
114                             cc=i;
115                             break;
116                         }
117                     }
118                 }
119                 //cout<<i<<'?'<<cc<<endl;
120                 memset(c,0,sizeof c);
121                     it=fr+1;
122                     while(it<ta){
123                         a[it]=cc;
124                         mp[it]=a[it]-1+'A';
125                         it+=2;
126                     }
127             }
128             fr=ta;
129             ta--;
130         }
131     }
132     if(f){
133         ans+=(ta-fr)/2;
134         for(int i=1;i<=k;i++){
135                         if(!c[i]){
136                             cc=i;
137                             break;
138             }
139         }
140         memset(c,0,sizeof c);
141                     it=fr+1;
142                     while(it<ta){
143                         a[it]=cc;
144                         mp[it]=a[it]-1+'A';
145                         it+=2;
146                     }
147             fr=ta;
148             ta--;
149     }
150     }
151     cout<<ans<<endl;
152     cout<<mp+1<<endl;
153 }