Submission #385849

#	Time	Username	Problem	Language	Result	Execution time	Memory
385849		alireza_kaviani	Gondola (IOI14_gondola)	C++11	65 / 100	25 ms	3948 KiB

gondola

#include <bits/stdc++.h>
#include "gondola.h"
using namespace std;

typedef long long ll;

#define X			first
#define Y			second

const int MAXN = 3e5 + 10;
const int MOD = 1e9 + 9;

ll poww(ll x , ll y){
	ll ans = 1;
	for(; y ; y >>= 1 , x = x * x % MOD)	if(y & 1)	ans = ans * x % MOD;
	return ans;
}

int fact[MAXN] , inv[MAXN] , res[MAXN] , mark[MAXN];

int valid(int n, int inputSeq[]){
	int ind = -1;
	for(int i = 0 ; i < n ; i++){
		if(inputSeq[i] <= n)	ind = i;
	}
	if(ind == -1){
		for(int i = 0 ; i < n ; i++)	res[i] = i + 1;
		return 1;
	}
	for(int i = 0 ; i < n ; i++){
		res[(ind + i) % n] = inputSeq[ind] + i;
		if(res[(ind + i) % n] > n)	res[(ind + i) % n] -= n;
	}
	for(int i = 0 ; i < n ; i++){
		if(inputSeq[i] <= n && res[i] != inputSeq[i])	return 0;
	}
	return 1;
}

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

int replacement(int n, int gondolaSeq[], int replacementSeq[]){
	if(valid(n , gondolaSeq) == 0)	return -1;
	vector<pair<int , int>> vec; int len = 0;
	for(int i = 0 ; i < n ; i++){
		if(gondolaSeq[i] > n)	vec.push_back({gondolaSeq[i] , res[i]});
	}
	sort(vec.begin() , vec.end());
	int prv = n;
	for(auto &i : vec){
		replacementSeq[len++] = i.Y;
		for(int j = prv + 1 ; j < i.X ; j++)	replacementSeq[len++] = j;
		prv = i.X;
	}
	return len;
}

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

int countReplacement(int n, int inputSeq[]){
	if(valid(n , inputSeq) == 0)	return 0;
	fact[0] = 1;
	for(int i = 1 ; i < MAXN ; i++)	fact[i] = 1ll * fact[i - 1] * i % MOD;
	inv[MAXN - 1] = poww(fact[MAXN - 1] , MOD - 2);
	for(int i = MAXN - 2 ; i >= 0 ; i--)	inv[i] = 1ll * inv[i + 1] * (i + 1) % MOD;
	int mx = 0 , k = 0;
	for(int i = 0 ; i < n ; i++){
		mx = max(mx , inputSeq[i]);
		mark[inputSeq[i]] = 1; k += (inputSeq[i] > n);
	}
	ll ans = 1; int cnt = mx - n;
	for(int i = mx ; i >= 1 ; i--){
		if(mark[i])	continue;
		//cout << i << endl;
		ans = ans * (cnt - max(0 , i - n)) % MOD; cnt--;
	}
	return ans * inv[k] % MOD;
}

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