[组合数学]Neither AB nor BA

题目描述

Find the number of strings s of length N consisting of 'A', 'B', and 'C' that satisfy the following condition:
·s can be converted to the empty string by repeating the following operation:

·Choose two consecutive characters in s and erase them. However, choosing 'AB' or 'BA' is not allowed.

For example, 'ABBC' satisfies the condition for N=4, because we can convert it as follows: 'ABBC' → (erase 'B'B) → 'AC' → (erase 'AC') → '(empty)'.

The answer can be enormous, so compute the count modulo 998244353.

Constraints
·2≤N≤107
·N is an even number.

输入

Input is given from Standard Input in the following format:
N

输出

Print the number of strings that satisfy the conditions, modulo 998244353.

样例输入 Copy

2

样例输出 Copy

7

提示

Except 'AB' and 'BA', all possible strings satisfy the conditions.

思路：首先不考虑'C'。把奇数位'A' 'B'互换，问题等价转换为只能消除相异的相邻两个字符，最终能消完的串有多少个。

这个问题的答案是'A' 'B'字符数相同的（即都等于n/2）串的个数。

那么考虑'C'的话，'C'可以被看做是'A' 'B'里任意一个，所以只要满足'A' 'B'字符个数的差不超过'C'的个数，那么这个串就能被消完。

所以不能被消完的串  就是num['A']>n/2 或 num['B']>n/2 的串。枚举统计有多少。

最后答案是总个数-不能被消完的个数。

#include<bits/stdc++.h>
#pragma GCC optimize(2)
typedef long long ll;
using namespace std;
const ll mod=998244353;

ll f[10000005],invf[10000005];

inline ll qpow(ll a,ll b){
  ll ret=1;
  while(b){
    if(b&1) ret=ret*a%mod;
    a=a*a%mod;
    b>>=1;
  }
  return ret;
}

inline void init(ll n){
  f[0]=1;
  for(ll i=1;i<=n;i++) f[i]=f[i-1]*i%mod;
  invf[n]=qpow(f[n],mod-2);
  for(ll i=n-1;i>=0;i--) invf[i]=invf[i+1]*(i+1)%mod;
}

inline ll C(ll n,ll m){
  if(m<0||m>n) return 0;
  if(m==0||m==n) return 1;
  return f[n]*invf[m]%mod*invf[n-m]%mod;
}

int main()
{
    ll n;scanf("%lld",&n);
    init(n);
    ll ans=0;
    for(ll i=n/2+1;i<=n;i++){
        ans=(ans+C(n,i)*qpow(2,n-i))%mod;
    }
    ans=(qpow(3,n)-2*ans%mod+mod)%mod;
    printf("%lld
",ans);
    return 0;
}