제출 #389364

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
389364		Pichon5	곤돌라 (IOI14_gondola)	C++17	60 / 100	54 ms	4612 KiB

gondola

#include "gondola.h"
#include<bits/stdc++.h>
#define vi vector<int>
#define pb push_back
#define F first
#define S second
#define ll long long int
using namespace std;
const ll M=1e9+7;
int valid(int n, int inputSeq[]){
	set<int> seen;
	for(int i = 0; i < n; i++){
		if(seen.count(inputSeq[i])) return 0;
		seen.insert(inputSeq[i]);
	}
	int pos = -1, st;
	for(int i = 0; i < n; i++) if(inputSeq[i] <= n) pos = i, st = inputSeq[i];
	if(pos == -1) return 1;
	for(int j = 0; j < n; j++, pos++, pos %= n){
		if(inputSeq[pos] > n) continue;
		if(inputSeq[pos] != (st+j-1)%n+1) return 0;
	}
	return 1;
}
//----------------------

int replacement(int n, int v[], int R[])
{
    int pos1=0;
    for(int i=0;i<n;i++){
        if(v[i]<=n){
            int B=v[i]-1;
            if(i-B>=0){
                pos1=i-B;
            }else{
                B-=i;
                pos1=n-B;
            }
            break;
        }
    }
    int pos=pos1;
    vector<pair<int,int> >E;//gondola final y gondola inicial
    for(int i=1;i<=n;i++){
        E.pb({v[pos],i});
        pos++;
        if(pos==n)pos=0;
    }
    sort(E.begin(),E.end());
    int next=n+1;
    int res=0;
    for(int i=0;i<n;i++){
        if(E[i].F==E[i].S)continue;
        R[res]=E[i].S;next++;res++;
        //cout<<"pongo  "<<E[i].S<<endl;
        int curr=next-1;
        while(curr!=E[i].F){
            //cout<<"pongo   "<<curr<<endl;
            R[res]=curr;
            curr++;next++;res++;
        }
    }
    //cout<<"respuesta   "<<res<<endl;
    //cout<<endl;

    return res;
}

//----------------------

int countReplacement(int n, int inputSeq[]){
	if(!valid(n, inputSeq)) return 0;
	vector<int> a(250001);
	ll s = 1, m = n;
	for(int i = 0; i < n; i++) a[inputSeq[i]]++;
	for(int i = 249999; i > n; i--) a[i] += a[i+1];
	for(int i = n+1; i < 250000; i++){
		if(!a[i]) break;
		if(a[i] == a[i+1]) s *= a[i], s %= M;
	}
	for(int i = 0; i < n; i++) if(inputSeq[i] <= n) m = 1;
	s *= m, s %= M;
	return s;
}
//number of subtask
//n
//3 1 4
//5 1 2 3 4
//2 3 4 9 6 7 1

• Subtask 1: 5 / 5
• Subtask 2: 5 / 5
• Subtask 3: 10 / 10
• Subtask 4: 5 / 5
• Subtask 5: 10 / 10
• Subtask 6: 20 / 20
• Subtask 7: 5 / 5
• Subtask 8: 0 / 15
• Subtask 9: 0 / 15
• Subtask 10: 0 / 10