ARC094F Normalization

原题链接：https://beta.atcoder.jp/contests/arc094/tasks/arc094_d

Normalization

Problem Statement

You are given a string $S$ consisting of ‘a’,’b’ and ‘c’. Find the number of strings that can be possibly obtained by repeatedly performing the following operation zero or more times, modulo $998244353$:

Choose an integer $i$ such that $1≤i≤|S|−1$ and the $i$-th and $(i+1)$-th characters in $S$ are different. Replace each of the $i$-th and $(i+1)$-th characters in $S$ with the character that differs from both of them (among a, b and c).

Constraints

$2≤|S|≤2×10^5$
$S$ consists of ‘a’, ‘b’ and ‘c’.

Input

Input is given from Standard Input in the following format:

$S$

Output

Print the number of strings that can be possibly obtained by repeatedly performing the operation, modulo $998244353$.

Sample Input 1

abc

Sample Output 1

3
abc, aaa and ccc can be obtained.

Sample Input 2

abbac

Sample Output 2

65

Sample Input 3

babacabac

Sample Output 3

6310

Sample Input 4

ababacbcacbacacbcbbcbbacbaccacbacbacba

Sample Output 4

148010497

题目大意

给定一个由$a,b,c$组成的字符串$S（|S|<=2\times 10^5）$，每次可以选择两个相邻的不同字符，把它们修改成与两者都不同的字符，求能得到的不同的字符串的个数。

题解

$\%\%\%watson,watson\ is\ so\ strong!!!$

感觉这道题真神（虽然watson说很水），如果把$a,b,c$表示为$0,1,2$，可以发现上述替换操作在$mod\ 3$意义下是不改变整个序列的和的。

然后开始$dp$，$dp[i][j][k][p]$表示在第$i$个位置，前缀和为$j(mod\ 3)$，当前字符为$k$，有/无连续的相同字符时的字符个数，枚举一下$i,i+1$的字符分别是什么就可以转移了。

最后的答案为$dp[n][tot][0][1]+dp[n][tot][1][1]+dp[n][tot][2][1]+[原串没有连续的相同字母]$。

代码

#include<bits/stdc++.h>
using namespace std;
const int M=2e5+5,mod=998244353;
char ch[M];
int n,sum,i,j,k,a,b,dp[M][3][3][2];
bool flag=1;
void spj()
{
    for(i=2;i<=n;++i)if(ch[i]!=ch[i-1]){flag=0;break;}
    if(flag)puts("1"),exit(0);
    if(n<=3)(n==2?puts("2"):((ch[1]!=ch[2]&&ch[1]!=ch[3]&&ch[2]!=ch[3])?puts("3"):((ch[1]==ch[3])?puts("7"):puts("6")))),exit(0);
}
bool check(){for(int i=1;i<=n;++i)if(ch[i]==ch[i-1])return 0;return 1;}
void in(){scanf("%s",ch+1);}
void ac()
{
    n=strlen(ch+1);spj();
    for(i=1;i<=n;++i)sum=(sum+ch[i]-'a')%3;
    for(i=0;i<=2;++i)dp[1][i][i][0]=1;
    for(i=1;i<=n-1;++i)for(j=0;j<=2;++j)for(k=0;k<=2;++k)for(a=0;a<=1;++a)for(b=0;b<=2;++b)
    (dp[i+1][(j+b)%3][b][a|(k==b)]+=dp[i][j][k][a])%=mod;
    printf("%d",(0ll+dp[n][sum][0][1]+dp[n][sum][1][1]+dp[n][sum][2][1]+check())%mod);
}
int main(){in();ac();}