CodeForces 779E Bitwise Formula

位运算，枚举。

按按分开计算，枚举$？$是$0$还是$1$，分别计算出$sum$，然后就可以知道该位需要填$1$还是$0$了。

#include<map>
#include<set>
#include<ctime>
#include<cmath>
#include<queue>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
 
int n,m;
char s[][];
map<string,int>z;
char t[];
 
int ans1[],ans2[];
 
int q[];
 
struct X
{
    int op;
    int x1,x2;
}p[];
 
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=;i<=n;i++)
    {
        scanf("%s",t);
        z[t]=i;
        scanf("%s",t);
 
        scanf("%s",t);
        if(t[]==''||t[]=='')
        {
            p[i].op=;
            strcpy(s[i],t);
        }
        else
        {
            if(t[]=='?') p[i].x1=-;
            else p[i].x1 = z[t];
 
            scanf("%s",t);
            if(t[]=='O') p[i].op=;
            if(t[]=='X') p[i].op=;
            if(t[]=='A') p[i].op=;
 
            scanf("%s",t);
            if(t[]=='?') p[i].x2=-;
            else p[i].x2 = z[t];
        }
    }
 
    for(int j=;j<m;j++)
    {
        int sum[]; sum[]=sum[]=;
        for(int tx=;tx<=;tx++)
        {
            for(int i=;i<=n;i++)
            {
                if(p[i].op==) q[i] = s[i][j]-'';
                else
                {
                    int x1 = p[i].x1, x2 = p[i].x2;
                    if(x1==-) x1 = tx; else x1 = q[x1];
                    if(x2==-) x2 = tx; else x2 = q[x2];
 
                    if(p[i].op==) q[i] = (x1|x2);
                    else if(p[i].op==) q[i] = (x1^x2);
                    else if(p[i].op==) q[i] = (x1&x2);
                }
            }
            for(int i=;i<=n;i++) sum[tx]+=q[i];
        }
 
        if(sum[]==sum[]) ans1[j]=ans2[j]=;
        else if(sum[]<sum[]) ans1[j]=,ans2[j]=;
        else  ans1[j]=,ans2[j]=;
    }
 
    for(int i=;i<m;i++) printf("%d",ans1[i]); printf("\n");
    for(int i=;i<m;i++) printf("%d",ans2[i]); printf("\n");
 
    return ;
}