- Quiz
- AC
- 解説
- 余事象を包除原理で求める(下記画像参照)
int main(){ cin.tie(0); ios::sync_with_stdio(false); // input ll N; cin>>N; ll all = N; ll a=0; // 余事象 a += N/2; a += N/3; a += N/5; a += N/7; a -= N/(2*3); a -= N/(2*5); a -= N/(2*7); a -= N/(3*5); a -= N/(3*7); a -= N/(5*7); a += N/(2*3*5); a += N/(2*3*7); a += N/(2*5*7); a += N/(3*5*7); a -= N/(2*3*5*7); ll ans = N-a; p(ans); return 0; }