狀壓板子。
\(dp_{i,j}\) 表示考慮到第 \(i\) 為狀態為 \(j\) 的方案數。
時間複雜度 \(O(n \times 2^k)\)。
程式碼:
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/*
Tips:
你陣列開小了嗎?
你MLE了嗎?
你覺得是貪心,是不是該想想dp?
一個小時沒調出來,是不是該考慮換題?
打 cf 不要用 umap!!!
記住,rating 是身外之物。
該衝正解時衝正解!
Problem:
演算法:
思路:
*/
#include<bits/stdc++.h>
using namespace std;
//#define map unordered_map
#define forl(i,a,b) for(register long long i=a;i<=b;i++)
#define forr(i,a,b) for(register long long i=a;i>=b;i--)
#define forll(i,a,b,c) for(register long long i=a;i<=b;i+=c)
#define forrr(i,a,b,c) for(register long long i=a;i>=b;i-=c)
#define lc(x) x<<1
#define rc(x) x<<1|1
#define mid ((l+r)>>1)
#define cin(x) scanf("%lld",&x)
#define cout(x) printf("%lld",x)
#define lowbit(x) (x&-x)
#define pb push_back
#define pf push_front
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define endl '\n'
#define QwQ return 0;
#define ll long long
#define ull unsigned long long
#define lcm(x,y) x/__gcd(x,y)*y
#define Sum(x,y) 1ll*(x+y)*(y-x+1)/2
#define aty cout<<"Yes\n";
#define atn cout<<"No\n";
#define cfy cout<<"YES\n";
#define cfn cout<<"NO\n";
#define xxy cout<<"yes\n";
#define xxn cout<<"no\n";
#define printcf(x) x?cout<<"YES\n":cout<<"NO\n";
#define printat(x) x?cout<<"Yes\n":cout<<"No\n";
#define printxx(x) x?cout<<"yes\n":cout<<"no\n";
ll t;
string s;
ll n,m,ans;
ll dp[1010][1<<11],mod=998244353;
bool pd(ll num,ll wz)
{
if(wz<m)
return 1;
string s;
ll len=0;
while(num)
{
len++;
if(num%2)
s+='1';
else
s+='0';
num/=2;
}
while(s.size()!=m)
s=s+'0';
string ss=s;
reverse(ss.begin(),ss.end());
return s!=ss;
}
void solve()
{
cin>>n>>m>>s;
s=' '+s;
forl(i,1,n)
{
if(s[i]=='?')
s[i]='2';
else
s[i]=(s[i]=='A')+'0';
}
dp[0][0]=1;
// cout<<s;
forl(i,1,n)
{
forl(time,0,(1ll<<m)-1)
{
if(s[i]=='1')
{
ll now=time*2;
now%=1<<m;
now|=1;
if(pd(now,i))
{
dp[i][now]+=dp[i-1][time];
dp[i][now]%=mod;
}
}
else if(s[i]=='0')
{
ll now=time*2;
now%=1<<m;
now|=0;
now%=1<<m;
if(pd(now,i))
{
dp[i][now]+=dp[i-1][time];
dp[i][now]%=mod;
}
}
else
{
ll now=time*2;
now%=1<<m;
now|=0;
now%=1<<m;
if(pd(now,i))
{
dp[i][now]+=dp[i-1][time];
dp[i][now]%=mod;
}
now=time*2;
now%=1<<m;
now|=1;
now%=1<<m;
if(pd(now,i))
{
dp[i][now]+=dp[i-1][time];
dp[i][now]%=mod;
}
}
}
}
forl(i,0,(1<<m)-1)
if(pd(i,n))
ans+=dp[n][i],ans%=mod;//,cout<<i<<":"<<dp[n][i]<<endl;
cout<<ans<<endl;
}
int main()
{
IOS;
t=1;
// cin>>t;
while(t--)
solve();
/******************/
/*while(L<q[i].l) */
/* del(a[L++]);*/
/*while(L>q[i].l) */
/* add(a[--L]);*/
/*while(R<q[i].r) */
/* add(a[++R]);*/
/*while(R>q[i].r) */
/* del(a[R--]);*/
/******************/
QwQ;
}