题面关键信息：横坐标&(2^1-1)&(2^2-1)&…&(2^n-1)，实际上在&(2^1-1)时后面位数的值就不需要考虑了，我们只需要算出答案的奇偶性即可

#include<algorithm>
#include<iostream>
#include<iomanip>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
using namespace std;
const int INF=0x3f3f3f3f;
int read()
{
   int s=0,bj=0;
   char ch=getchar();
   while(ch<'0'||ch>'9')bj|=(ch=='-'),ch=getchar();
   while(ch>='0'&&ch<='9')s=(s<<1)+(s<<3)+(ch^48),ch=getchar();
   return bj?-s:s;
}
void printnum(int x)
{
    if(x>9)printnum(x/10);
    putchar(x%10^48);
}
void print(int x,char ch)
{
    if(x<0){putchar('-');x=-x;}
    printnum(x);putchar(ch);
}
int n;
int len;
char ch[10005]; 
int num;
int main()
{
    n=read();
    for(int i=1;i<=n;++i)
    {
        scanf("%s",ch+1);len=strlen(ch+1);
        num=(num+ch[len]-'0')%2;//加上这一个数的最后一位
    }
    print(num,'\n');
    return 0;
}