F2. Wrong Answer on test 233 (Hard Version)
Your program fails again. This time it gets "Wrong answer on test 233"
.
This is the harder version of the problem. In this version, 1≤n≤2⋅105. You can hack this problem if you locked it. But you can hack the previous problem only if you locked both problems.
The problem is to finish n one-choice-questions. Each of the questions contains k options, and only one of them is correct. The answer to the i-th question is hi, and if your answer of the question i is hi, you earn 1 point, otherwise, you earn 0 points for this question. The values h1,h2,…,hn are known to you in this problem.
However, you have a mistake in your program. It moves the answer clockwise! Consider all the n answers are written in a circle. Due to the mistake in your program, they are shifted by one cyclically.
Formally, the mistake moves the answer for the question i to the question imodn+1. So it moves the answer for the question 1 to question 2, the answer for the question 2 to the question 3, ..., the answer for the question n to the question 1.
We call all the n answers together an answer suit. There are kn possible answer suits in total.
You're wondering, how many answer suits satisfy the following condition: after moving clockwise by 1, the total number of points of the new answer suit is strictly larger than the number of points of the old one. You need to find the answer modulo 998244353.
For example, if n=5, and your answer suit is a=[1,2,3,4,5], it will submitted as a′=[5,1,2,3,4] because of a mistake. If the correct answer suit is h=[5,2,2,3,4], the answer suit a earns 1 point and the answer suite a′ earns 4 points. Since 4>1, the answer suit a=[1,2,3,4,5] should be counted.
Input
The first line contains two integers n, k (1≤n≤2⋅105, 1≤k≤109) — the number of questions and the number of possible answers to each question.
The following line contains n integers h1,h2,…,hn, (1≤hi≤k) — answers to the questions.
Output
Output one integer: the number of answers suits satisfying the given condition, modulo 998244353.
Examples
input
3 3
1 3 1
output
9
input
5 5
1 1 4 2 2
output
1000
input
6 2
1 1 2 2 1 1
output
16
Note
For the first example, valid answer suits are [2,1,1],[2,1,2],[2,1,3],[3,1,1],[3,1,2],[3,1,3],[3,2,1],[3,2,2],[3,2,3].
The meaning of problems
There are n questions, each question has k answers, but you silly now, the answer to the first question referred to the second question, the second question was referred to the third question, the second question was referred to the first k (k% n) the title of the position +1 up.
The correct number would like to know, how many there are programs to fill the answer, you can make the exchange more than the correct amount before the exchange.
answer
A small range of data words, DP [i] [j] denotes the i-th now given to the question, after an exchange of little more than j exchange minutes ago.
Then if h [i] == h [i + 1], then, dp [i] [j] = dp [i-1] [j], as have the correct number of what in any case do not become filled.
Other cases dp [i] [j] = dp [i-1] [j + 1] + dp [i-1] [j-1] + (k-2) dp [i-1] [j], a the case is before, after conversion was wrong; wrong before, after conversion; the other k-2 kinds of answers remain the same.
hard version we want to do the opposite of, say we know and have the same number of points before the conversion program to convert the last ans, then k ^ n-ans represents the converted score of the number of program changes.
Because the front and high conversion fraction and converting high score for the program number is the same, because of symmetry, the final answer must be (k ^ n-ans) / 2
So this ans how to do it, assuming now h [i]! = Number of h [i + 1] is a num because the same words does not make sense, because it does not matter what fill
We enumerate the number of positions +1, C (num, i); they have the same i-th -1 (num-i, i) C, num-2i other positions have choices k-2 (- K- 2) ^ (num-2i), the remaining n-num selection positions have k k ^ (n-num).
Then the number of the i + 1-position program is actually C (num, i) C (num-i, i) (k-2) ^ (num-2i) k ^ (n-num), and finally the number of programs in all he subtracted and divided by 2 to get away.
Code:
#include<bits/stdc++.h>
using namespace std;
const int maxn = 2005;
const int mod = 998244353;
int h[maxn];
long long dp[maxn][maxn*2],base=2003,k,n;
int main(){
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++)
scanf("%d",&h[i]);
if(k==1){
cout<<"0"<<endl;
return 0;
}
dp[0][base]=1;
for(int i=1;i<=n;i++){
for(int j=base-2000;j<=base+2000;j++){
if(h[i]==h[i%n+1]){
dp[i][j]=dp[i-1][j]*k%mod;
}else{
dp[i][j]=(dp[i-1][j+1]+dp[i-1][j-1]+dp[i-1][j]*(k-2))%mod;
}
}
}
long long ans = 0;
for(int i=1;i<=n;i++){
ans=(ans+dp[n][base+i])%mod;
}
cout<<ans<<endl;
}
#include<bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const int maxn = 2e5+7;
int n,k,h[maxn];
long long powmod(long long a,long long b){
if(b==0)return 1;
return b%2==0?powmod(a*a%mod,b/2):powmod(a*a%mod,b/2)*a%mod;
}
long long fac[maxn],inv[maxn];
long long C(int a,int b){
if(b<0||b>n)return 0;
return (fac[a]*inv[b]%mod)*inv[a-b]%mod;
}
int main(){
fac[0]=inv[0]=1;
for(int i=1;i<maxn;i++){
fac[i]=i*fac[i-1]%mod;
inv[i]=powmod(i,mod-2)*inv[i-1]%mod;
}
cin>>n>>k;
if(k==1){
cout<<"0"<<endl;
return 0;
}
for(int i=0;i<n;i++)
cin>>h[i];
int num = 0;
h[n]=h[0];
for(int i=0;i<n;i++){
if(h[i]!=h[i+1])num++;
}
long long ans = 0;
for(int i=0;i*2<=num;i++){
long long tmp = C(num,i)*C(num-i,i)%mod*powmod(k-2,num-2*i)%mod*powmod(k,n-num);
ans=(ans+tmp)%mod;
}
cout<<((powmod(k,n)-ans+mod)*inv[2])%mod<<endl;
}