제출 #331531

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
331531		ignaciocanta	Timovi (COCI18_timovi)	C++14	80 / 80	19 ms	2156 KiB

timovi

#include <bits/stdc++.h>
 
using namespace std;
 
using tint = long long;
using ld = long double;
 
#define forsn(i, s, n) for(int i = s; i < int(n); i++)
#define forn(i, n) forsn(i, 0, n)
#define trav(a, x) for(auto& a : x)
 
using vi = vector<int>;
using vl = vector<tint>;
using vb = vector<bool>;
#define pb push_back
#define pf push_front
#define rsz resize
#define all(x) begin(x), end(x)
#define rall(x) x.rbegin(), x.rend() 
#define sz(x) (int)(x).size()
#define ins insert
#define esta(x,c) ((c).find(x) != (c).end())
 
using pi = pair<int,int>;
using pl = pair<tint,tint>;
#define f first
#define s second
#define mp make_pair
 
#define DBG(x) cerr << #x << " = " << x << endl;
 
const int MOD = 1e9+7;
const tint mod = 998244353;
const int MX = 2e5+5; 
const tint INF = 1e18; 
const int inf = 2e9;
const ld PI = acos(ld(-1)); 
const ld eps = 1e-8;
 
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, 1, -1};
 
template<class T> void remDup(vector<T> &v){ 
    sort(all(v)); v.erase(unique(all(v)),end(v));
}
 
 bool valid(int x, int y, int n, int m){
    return (0<=x && x<n && 0<=y && y<m);
}
 
int cdiv(int a, int b) { return a/b+((a^b)>0&&a%b); } //redondea p arriba
int fdiv(int a, int b) { return a/b-((a^b)<0&&a%b); } //redonde p abajo
 
void NACHO(string name = "cbarn"){
    ios_base::sync_with_stdio(0); cin.tie(0);
    //freopen((name+".in").c_str(), "r", stdin);
    //freopen((name+".out").c_str(), "w", stdout);
}

int main(){
	NACHO();
	int n, k, m; cin >> n >> k >> m;
	int a = fdiv(m, k);
	int vuelta = fdiv(a, n-1);
	vi ret (n, 0);
	ret[0] = fdiv(vuelta+1, 2)*k;
	ret[n-1] = fdiv(vuelta, 2)*k;
	forsn(i, 1, n-1) ret[i] = vuelta*k;
	m -= vuelta*k*(n-1);
	if(vuelta % 2 == 0){
		forn(i, n){
			ret[i] += min(k, m);
			m -= min(k, m);
			if(m == 0) break;
		}
	}else{
		for(int i = n-1; i >= 0; --i){
			ret[i] += min(k, m);
			m -= min(k, m);
			if(m == 0) break;
		}
	}
	forn(i, n) cout << ret[i] << " ";
}
//7 3 2000000000

• Subtask 1: 80 / 80