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[5100][5100];
map<string,int>z;
char t[5100];

int ans1[5100],ans2[5100];

int q[5100];

struct X
{
    int op;
    int x1,x2;
}p[5100];


int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%s",t);
        z[t]=i;
        scanf("%s",t);

        scanf("%s",t);
        if(t[0]=='0'||t[0]=='1')
        {
            p[i].op=1;
            strcpy(s[i],t);
        }
        else
        {
            if(t[0]=='?') p[i].x1=-1;
            else p[i].x1 = z[t];

            scanf("%s",t);
            if(t[0]=='O') p[i].op=2;
            if(t[0]=='X') p[i].op=3;
            if(t[0]=='A') p[i].op=4;

            scanf("%s",t);
            if(t[0]=='?') p[i].x2=-1;
            else p[i].x2 = z[t];
        }
    }

    for(int j=0;j<m;j++)
    {
        int sum[2]; sum[0]=sum[1]=0;
        for(int tx=0;tx<=1;tx++)
        {
            for(int i=1;i<=n;i++)
            {
                if(p[i].op==1) q[i] = s[i][j]-'0';
                else
                {
                    int x1 = p[i].x1, x2 = p[i].x2;
                    if(x1==-1) x1 = tx; else x1 = q[x1];
                    if(x2==-1) x2 = tx; else x2 = q[x2];

                    if(p[i].op==2) q[i] = (x1|x2);
                    else if(p[i].op==3) q[i] = (x1^x2);
                    else if(p[i].op==4) q[i] = (x1&x2);
                }
            }
            for(int i=1;i<=n;i++) sum[tx]+=q[i];
        }

        if(sum[0]==sum[1]) ans1[j]=ans2[j]=0;
        else if(sum[0]<sum[1]) ans1[j]=0,ans2[j]=1;
        else  ans1[j]=1,ans2[j]=0;
    }


    for(int i=0;i<m;i++) printf("%d",ans1[i]); printf("
");
    for(int i=0;i<m;i++) printf("%d",ans2[i]); printf("
");

    return 0;
}