codefores 1066 E. Binary Numbers AND Sum

这个题吧

你画一下就知道了

就拿这个例子来讲

4 5
1001
10101

对于b串的话第5位只会经过a串的第4位，b串的第4位会经过a串的第3位和第4位。。。。。b串的第1和第2位会经过a串的每一位

由于是&操作，计算1经过a串每一位所能带来的权值，

对b串进行处理，相加即可

#include <bits/stdc++.h>
#define ll long long
#define mp make_pair
#define x first
#define y second
using namespace std;
const int maxn = 2e5+;
const ll mod = ;
ll fac[maxn];
ll rea[maxn];
ll ans1,ans2;
char a[maxn];
char b[maxn];
int main()
{
    //freopen("in.txt","r",stdin);
    ll n,m;
    scanf("%lld%lld",&n,&m);
    scanf("%s %s",a,b);
    fac[] = ;
    for(int i = ; i < n; ++i){
        fac[i] = *fac[i-]%mod;
    }

    for(int i = ; i < n/; ++i)
    {
        swap(a[i],a[n--i]);
    }
    for(int i = ; i < m/; ++i)
    {
        swap(b[i],b[m--i]);
    }

    rea[] = ;
    if(a[] == '')
        rea[] = ;
    for(int i = ; i < n; ++i)
    {
        rea[i] = rea[i-];
        if(a[i] == '')
            rea[i] = (rea[i]+fac[i]);
    }

    for(int i = ; i < n; ++i)
    {
        if(a[] == '')
            ans1 = (ans1 + rea[i])%mod;
    }

    ll ans = ;
    for(int i = ; i < n && i < m; ++i)
    {
        if(b[i] == '')
        {
            ans = (ans+rea[i])%mod;
        }
    }

    for(int i = n; i < m; ++i)
    {
        if(b[i] == '')
            ans = (ans+rea[n-])%mod;
    }

    cout << ans << endl;
}