Codeforces Round #462 (Div. 2) B. A Prosperous Lot
問題ページ
Problem - B - Codeforces
考えたこと
- loopが2個の8を18個並べた36個が上限に見える
- ループが2個の8とループが1個の4,6,9あたりを使って適当に並べればよさそう
- leading zeroにだけ気をつけて出すと通る
AよりBの方が簡単ですが…
#include <bits/stdc++.h> using namespace std; using ll = long long; #define int ll using VI = vector<int>; using VVI = vector<VI>; using PII = pair<int, int>; #define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i) #define REP(i, n) FOR(i, 0, n) #define ALL(x) x.begin(), x.end() #define PB push_back const ll LLINF = (1LL<<60); const int INF = (1LL<<30); const int MOD = 1000000007; template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); } template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); } template <typename T> bool IN(T a, T b, T x) { return a<=x&&x<b; } template<typename T> T ceil(T a, T b) { return a/b + !!(a%b); } template<class S,class T> ostream &operator <<(ostream& out,const pair<S,T>& a){ out<<'('<<a.first<<','<<a.second<<')'; return out; } template<class T> ostream &operator <<(ostream& out,const vector<T>& a){ out<<'['; REP(i, a.size()) {out<<a[i];if(i!=a.size()-1)out<<',';} out<<']'; return out; } int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0}; signed main(void) { int k; cin >> k; if(k > 36) { cout << -1 << endl; } else if(k%2 == 0) { cout << string(k/2, '8') << endl; } else { cout << string(k/2, '8') << '4' << endl; } return 0; }