本来想搜索+部分打表 后来想了想还是放弃吧

#include
     
      #define LL long longusing namespace std;const int N = 1000005;int n, a[N];int main(){    scanf("%d", &n);    if (n == 2){        puts("Impossible");        return 0;    }    for (int i = 1; i < n; ++i){        a[(LL)i * i % n] = 1;    }    for (int i = 1; i < n; ++i) {        printf("%d", 1 - a[i]);    }}