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
3 6
ABBACC
output
Copy
2
ABCACA
input
Copy
2 3 a
BBB
output
Copy
. 1
BAB-
question is intended: to give you a row of n elements, k colors, requires not the same color adjacent element, changing the final color output least n elements and the number of colors (if any output may be more likely one)
Note: k == 2 parity bits are not necessarily the same, to choose the least change a program, so you know you want to change the odd bit put a number of more or even bit a place to change a few more, this card during training case, and
in fact, the beginning thought ABBA this case, but it was the only idea is to change the elements of the same color, I did not expect a different color elements also change, so it was assumed that if there are an even number of elements of the same color, its sides element color will be different, the results have been 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 is } 14 int ANS = 0 , C1 = 0 , C2 = 0 ; 15 IF (== K 2 ) { /// time must k == 2 is a parity bit is not the same, to choose a minimum change program, so to know the odd bits discharge are more or even bit a to change the discharge for multiple a to change 16 for ( int I = . 1 ; I <= n-; I ++) { /// let's assume that the odd bits of a, even bit is B, because there may be a legitimate case even bit as a, odd bits B, so in the end you want to compare what is not legitimate small number, so that changes can be small to give the least change scheme, it is not statistically valid number below, . 17 IF (I% 2 == 0 ) { /// If the a bit is even, the number of illegal a plus 1, or invalid number B plus 1 18 is IF (MP [I] == ' A ' ) C1 ++ ; . 19 the else C2 ++ ; 20 is } 21 is the else { /// If the A bit is even, the number of illegal B plus 1, or invalid number of A plus 1 22 is IF (MP [I] == ' a ' ) C2 ++ ; 23 is the else C1 ++ ; 24 } 25 } 26 is ANS = min (C1, C2); /// selected from a number of illegal minimum, the least change scheme to obtain 27 for ( int I = . 1 ; I <= n-; I ++ ) { 28 IF (C1 == ANS) { /// If A is even bit is not legal, to the odd bit is set to A, becomes even bit B 29 IF (% I 2 ) 30 MP [I] = ' A ' ; 31 is the else 32 MP [I] = ' B ' ; 33 is } 34 is the else { 35 IF (% I 2 ) /// If the a bit is an odd number not legal, to a becomes even bit, odd bit variant is B 36 MP [I] = ' B ' ; 37 [ the else 38 is 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 }