二次联通门 : LibreOJ #525. 「LibreOJ β Round #4」多项式
官方题解 :
/*
LibreOJ #525. 「LibreOJ β Round #4」多项式 由于会有多种解
所以只需要找出一组特殊解即可 */
#include <cstdio>
#include <iostream> void read (int &now)
{
register char c = getchar ();
for (now = ; !isdigit (c); c = getchar ());
for (; isdigit (c); now = now * + c - '', c = getchar ());
} int Main ()
{
int N; read (N); register int i, j; int L = N;
int phi = ; if (N == ) return printf ("-1"), ;
for (i = ; i * i <= L; ++ i)
if (L % i == )
{
L /= i, phi *= (i - );
for (; L % i == ; L /= i, phi *= i);
}
if (L != ) phi *= L - ;
printf ("%d\n", L = phi << | ); ++ phi;
for (i = ; i < phi; ++ i) printf ("0 ");
printf ("%d ", N - );
for (++ i; i < L; ++ i) printf ("0 ");
printf ("");
return ;
} int ZlycerQan = Main ();
int main (int argc, char *argv[]) {;}