2021 ICPC Shaanxi warm up match b.code (bit operation)

B . CODE 

Title Description

A length of （2^n）-1 Of 01 character string , The probability that every byte may be wrong is 1/10^20.
character string s, character s[i], For each of these i（i=2^k）, s[i] Satisfy ： s[i]=⨁(s[j]),j&i=i .

【s[i] = the xor sum of all position j which k−th digit is 1 in binary representation.】 
eg:

s[3]=1,s[5]=1,s[6]=0,s[7]=1

then
s[1]=s[3]⨁s[5]⨁s[7]=1
s[2]=s[3]⨁s[6]⨁s[7]=0
s[4]=s[5]⨁s[6]⨁s[7]=0
The final message to be sent is 1010101.
Find the original string , You can assume that the given string has at most one error .

Input

3 3
1010101
1010111
1110101

Output

1010101
1010101
1010101

//#include<bits/stdc++.h>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<string>
#include<cstdio>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<deque>
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N=200200;
const int M=5002;
const int mod=998244353;
LL a[N],f[M][M];
LL mp[N];
int b[N];
bool vis[N];
int dx[]={-1,0,0,1},dy[]={0,1,-1,0};
int main()
{
    cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
    int T=1;
    cin>>T;
    int n;
    cin>>n;
    while(T--)
    {
        string s;
        cin>>s;
        int len=s.size();
        s=" "+s;
        int flag=0;
        for(int i=1;i<=len;i<<=1)
        {
            int sum=0;
            for(int j=1;j<=len;j++)
            {
                if((j&i)==i&&(i!=j))
                {
                    sum^=(s[j]-'0');
                }
            }
            if(sum!=s[i]-'0') flag|=i;
        }
        if(flag!=0)
        {
            if(s[flag]=='0') s[flag]='1';
            else s[flag]='0';
        }
        for(int i=1;i<=len;i++) cout<<s[i]; 
        cout<<endl;
    }
    return 0;
}