题面在这里！
比较显而易见的暴力，O(2^(2n) + 2^n * 100) 就可以直接做了
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=4105;
#define pb push_backinline int getint(){int x=0; char ch=getchar();for(;!isdigit(ch);ch=getchar());for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';return x;
}inline int getbit(){int x=0; char ch=getchar();for(;ch!='0'&&ch!='1';ch=getchar());for(;ch=='0'||ch=='1';ch=getchar()) x=(x<<1)|(ch=='1');return x;
}void W(int x){ if(x>=10) W(x/10); putchar(x%10+'0');}vector<int> g[N][105];
int w[N],n,m,q,ans[500005],cnt[N],all,qz[105];inline void solve(){for(int i=0;i<=all;i++){memset(qz,0,sizeof(qz));for(int j=0,x;j<=all;j++){x=w[i^j^all];if(x<=100) qz[x]+=cnt[j];}for(int j=0;j<=100;j++){if(j) qz[j]+=qz[j-1];for(int k:g[i][j]) ans[k]=qz[j];}}
}int main(){n=getint(),m=getint(),q=getint(),all=(1<<n)-1;for(int i=n-1;i>=0;i--) w[1<<i]=getint();for(int i=1;i<=all;i++) w[i]=w[i&-i]+w[i^(i&-i)];for(int i=0;i<m;i++) cnt[getbit()]++;for(int i=0,X,Y;i<q;i++) X=getbit(),Y=getint(),g[X][Y].pb(i);solve();for(int i=0;i<q;i++) W(ans[i]),puts("");return 0;
}
)
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