https://codeforces.com/problemset/problem/1066/E
思路:下串往右移的过程中可以算出下串的每一位的1和上串的1&产生的贡献,可以On跑出上串的一个前缀和,累加过来就是答案了。
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<deque>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=2e5+1000;
typedef long long LL;
const LL mod=998244353;
inline LL read(){LL x=0,f=1;char ch=getchar(); while (!isdigit(ch)){if (ch=='-') f=-1;ch=getchar();}while (isdigit(ch)){x=x*10+ch-48;ch=getchar();}
return x*f;}
LL suf[maxn];
deque<char>s,p;
LL ksm(LL a,LL k){
LL res=1;
while(k>0){
if(k&1) res=res*a%mod;
k>>=1;
a=a*a%mod;
}
return res%mod;
}
int main(void)
{
cin.tie(0);std::ios::sync_with_stdio(false);
LL n,m;cin>>n>>m;
for(LL i=1;i<=n;i++) {char op;cin>>op;s.push_back(op);}
for(LL i=1;i<=m;i++) {char op;cin>>op;p.push_back(op);}
LL cnt=abs(n-m);
if(n<m){
for(LL i=1;i<=cnt;i++) s.push_front('0');
}
else if(n>m){
for(LL i=1;i<=cnt;i++) p.push_front('0');
}
LL ans=0;cnt=0;
for(LL i=max(n,m);i>=1;i--){
char now=s.back();s.pop_back();
suf[i]=suf[i+1];
if(now=='1') suf[i]=(suf[i]%mod+ksm(2,cnt%mod)%mod)%mod;
char down=p.back();p.pop_back();
if(down=='1'){
ans=(ans%mod+suf[i]%mod)%mod;
}
cnt++;
}
cout<<ans%mod<<"\n";
return 0;
}