Codeforces Round #602 (Div. 2, based on Technocup 2020 Elimination Round 3) F2. Wrong Answer on test 233 (Hard Version) dp 数学

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].

题意

现在有n道题，每道题有k个答案，但是你现在犯傻了，把第一题的答案交到了第二题，第二题交到了第3题，第k题交到了第(k%n)+1题的位置上去。

现在想知道，有多少种填答案的方案，可以使得交换后的正确数量多于交换前的正确数量。

题解

数据范围小的话，dp[i][j]表示现在考虑到了第i题，交换后比交换前多得j分。

那么如果h[i]==h[i+1]的话，dp[i][j]=dp[i-1][j]，因为无论如何填什么正确得个数都不会变。

其他情况 dp[i][j] = dp[i-1][j+1]+dp[i-1][j-1]+(k-2)dp[i-1][j]，有一种情况是之前对了，转换后错了；之前错了，转换后对了；其他k-2种答案都保持不变。

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hard version我们要反着做，假设我们知道最后转换后和转换前分数一样得方案数为ans的话，那么k^n-ans表示的是转换后得分发生改变的方案数。

又因为转换前分数高和转换后分数高的方案数是一样的，因为对称，所以最后答案一定是 (k^n-ans)/2

那么这个ans怎么做呢，假设现在h[i]!=h[i+1]的个数为num个，因为相同的话没有意义，因为填什么都无所谓

我们枚举+1的位置有多少个，C(num,i);同样的-1也得i个C(num-i,i)，其他num-2i个位置有k-2种选择(k-2)^{(num-2i)，剩下n-num个位置都有k个选择k}(n-num)。

那么i个+1位置的方案数其实就是C(num,i)C(num-i,i)(k-2)^(num-2i)k(n-num)，最后用所有的方案数减去他再除以2就完事。

代码：

#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;
}