qscqesze
Published on 2017-09-02 11:31 in 暂未分类 with qscqesze

# 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.

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].

## 题解

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。

## 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[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=inv=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;
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)%mod<<endl;
}
posted @ 2019-11-24 23:26  qscqesze  阅读(262)  评论(0编辑  收藏