Search is not available for this dataset
name stringlengths 2 112 | description stringlengths 29 13k | source int64 1 7 | difficulty int64 0 25 | solution stringlengths 7 983k | language stringclasses 4
values |
|---|---|---|---|---|---|
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m;
cin >> n;
int* a = new int[n];
int* b = new int[n];
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
if (n == 1)
cout << 0;
else {
m = a[n - 1];
b[n - 1] = 0;
for (int i = n - 2; i >= 0; --i) {
if (m < a[i]) {
... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
public class B {
int n;
long h[];
void readInput() throws NumberFormatException, IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
//Scanner sc = new Scann... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n], f[n];
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
f[n - 1] = a[n - 1];
f[n] = 0;
for (int i = n - 2; i >= 0; i--) f[i] = max(f[i + 1], a[i]);
for (int i = 0; i < n; i++)
if (f[i] == a[i] && f[i + 1] != f[i])
... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n = input()
m = map(int, raw_input().split())
max_so_far = 0
vals = [0] * n
for i in range(n - 1, -1, -1):
val = 0
if max_so_far >= m[i]:
val = max_so_far - m[i] + 1
vals[i] = val
max_so_far = max(max_so_far, m[i])
print ' '.join(str(v) for v in vals)
| PYTHON |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.util.*;
public class test {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] arr = new int[n];
int[] anss = new int[n];
int[] dh = new int[n];
for (int i = 0; i < n; i++)
arr[i] = sc.nextInt();
dh[n-1] = 0;
anss[n-1]=0;
String... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
template <class T1, class T2>
bool upmin(T1 &x, T2 v) {
if (x > v) {
x = v;
return true;
}
return false;
}
template <class T1, class T2>
bool upmax(T1 &x, T2 v) {
if (x < v) {
x = v;
return true;
}
return false;
}
void solve() {
int N;
cin >>... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.util.Scanner;
public class Main {
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int a[] = new int[n];
for(int i = 0; i < n; i++) {
a[i] = scanner.nextInt();
}
int b[] = new int[n+... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 |
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main{
public static void main(String[] args) throws Exception {
ModScanner ms=new ModScanner();
int n=ms.nextInt();
long[] a=new long[n];
for(int i=0;i<n;i++)
a[... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | '''
Created on 28-Sep-2015
@author: Venkatesh
'''
def read_int_list():
return [int(x) for x in raw_input().split()]
def read_int():
return int(raw_input())
def print_next_max_num(houses):
last_max = houses[-1]
result = [0]
houses.pop()
for ele in houses[::-1]:
if ele <= last_max:
... | PYTHON |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.util.*;
import java.lang.*;
import java.io.*;
/*
int n = scan.nextInt();
int[] arr = new int[n];
for(int i = 0; i < n; i++) arr[i] = scan.nextInt();
long n = scan.nextLong();
long[] arr = new long[n];
for(int = 0; i < n; i++) arr[i] = scan.nextLong();
*/
public class Main {
pu... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.util.Scanner ;
public class jkl {
public static void main (String [] args){
Scanner s = new Scanner (System.in);
int size = s.nextInt();
int [] a = new int [size];
int [] b = new int [size];
for (int j = 0 ; j<size ; j++){
int k = s.nex... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int n;
int a[100100];
pair<int, int> p[100100];
int main() {
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", a + i);
p[i] = make_pair(a[i], i);
}
sort(p, p + n);
reverse(p, p + n);
int len = 0;
for (int i = 0; i < n - 1; i++) {
while (p[... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
map<string, long long int> mapp;
long long int a[1000000], b[1000000];
int main() {
long long int n;
cin >> n;
for (long long int i = 0; i < n; i++) {
cin >> a[i];
}
long long int max = 0;
for (long long int i = n - 1; i >= 0; i--) {
if (a[i] > max)
... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
int main() {
int a[100005], m[100005], n, t, i, temp;
scanf("%d", &n);
for (i = 0; i < n; i++) scanf("%d", &a[i]);
m[n - 1] = a[n - 1];
m[n] = 0;
for (i = n - 2; i >= 0; i--) {
if (a[i] > m[i + 1])
m[i] = a[i];
else
m[i] = m[i + 1];
}
for (i = 0; i < n; i++) ... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n = int(input())
arr = list(map(int, input().split()))
maxi=0
e=[]
for i in range(len(arr)):
if arr[len(arr)-1-i] > maxi:
maxi = arr[len(arr)-1-i]
flag=len(arr)-1-i
if len(arr)-1-i ==flag:
e.append((arr[flag]- arr[len(arr)-1-i]))
else:
e.append((arr[flag]- arr[len(a... | PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | # In this template you are not required to write code in main
import sys
inf = float("inf")
#from collections import deque, Counter, OrderedDict,defaultdict
#from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
from math import ceil,floor,log,sqrt,factorial,pow,pi,gcd
#from bisect import b... | PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | /* / οΎοΎβ β β β β β β β β β β β γ
/ )\β β β β β β β β β β β β Y
(β β | ( Ν‘Β° ΝΚ Ν‘Β°οΌβ β(β γ
(β οΎβ Y βγ½-γ __οΌ
| _β ｑγ| γｑ |/
(β γΌ '_δΊΊ`γΌ οΎ
β |\ οΏ£ _δΊΊ'彑οΎ
β )\β β ｑβ β /
β β (\β #β /
β /β β β /α½£====================D-
/β β β /β \ \β β \
( (β )β β β β ) ).β )
(β β )β β β β β ( | /
|β /β β β β β β | /
[_] β β β β β [___] */
// Main Code at the Bottom
import ... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.util.ArrayList;
import java.util.Scanner;
public class B {
/**
* @param args
*/
public static void main(String[] args) {
Scanner x=new Scanner(System.in);
int n=x.nextInt();
int arr[]=new int [n];
for(int i=0;i<n;i++){
arr[i]... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | // CoDE deVELOPed uNDEr AcE proDUcTiONS (enJOY!!!!)
import java.util.*;
public class Ace
{
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int n=sc.nextInt(),i,j,k,max=0;
int[] ar=new int[n];
int[] br=new int[n];
for(i=0;i<n;i++) ar[i]=sc.nextInt();
max=ar[n-1];... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n, m, k, l, x, z, i;
cin >> n;
long long a[n + 5];
map<long long, long long> mp;
for (i = 1; i <= n; i++) {
cin >> a[i];
}
long long b[n + 5];
x = 0;
for (i = n; i >= 1; i--) {
x = max(a[i], x);
b[i] = x;
mp[b[i]]++;
... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n = int(input())
h = list(map(int, input().split())) + [0]
h_max = [0] * n
res = ['0'] * n
for i in range(n - 1, 0, -1):
h_max[i - 1] = max(h_max[i], h[i] + 1)
res[i] = str(max(h_max[i] - h[i], 0))
res[0] = str(max(h_max[0] - h[0], 0))
print(" ".join(res)) | PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n=int(input())
s=list(map(int,input().split()))
s.reverse()
d=s[0]
a=[]
for i in range(1,n):
f=int(s[i])
if f <= d:
k=str((d-f)+1)
a.append(k)
else:
a.append(0)
d=f
a.insert(0,"0")
a.reverse()
for j in a:
print(j,end=" ")
| PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n = int(input())
h = map(int, input().split()[::-1])
m_h = 0
res = []
for h in h:
if m_h < h:
res.append(0)
m_h = h
else:
res.append(m_h - h + 1)
print(" ".join(map(str, res[::-1]))) | PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int x[100009];
int a[100009];
int main() {
int n, mx;
cin >> n;
for (int i = 0; i < n; ++i) cin >> x[i];
a[n - 1] = 0;
mx = x[n - 1];
for (int i = n - 2; i >= 0; --i) {
if (mx == x[i]) {
a[i] = 1;
} else if (mx < x[i]) {
a[i] = 0;
mx = ... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
//Removed extra array
public class LuxuriousHouses {
public static void main(String[] args) throws IOException {
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(bf.readL... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n = input()
h = map(int, raw_input().split())
res = []
mx = 0
while h:
c = h.pop()
if c > mx:
res.append(0)
mx = c
else:
res.append((mx - c) + 1)
res.reverse()
print ' '.join(map(str, res))
| PYTHON |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
vector<int> One(1000007);
vector<int> Two(1000007);
int main() {
int n;
int maximun = -1000007;
cin >> n;
for (int i = 0; i < n; i++) cin >> One[i];
for (int i = n - 1; i >= 0; i--) {
if (One[i] > maximun)
Two[i] = One[i];
else
Two[i] = maximun... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n = int(input())
h = list(map(int, input().split()))
maxi = h[n - 1]
maxes = [0] * n
maxes[n - 1] = 0
res = ''
for i in range(n - 1, 0, -1):
if h[i - 1] > maxi:
maxi = h[i - 1]
maxes[i - 1] = 0
else:
maxes[i - 1] = maxi - h[i - 1] + 1
for i in range(n):
res += str(maxes[i]) + ' '
... | PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n=int(input())
h=list(map(int,input().split()))
ans=[0]*n
ans[n-1]=0
maxh=h[n-1]
for i in range(n-2,-1,-1):
if h[i]>maxh:
ans[i]=0
else:
ans[i]=maxh-h[i]+1
maxh=max(maxh,h[i])
print(' '.join(map(str,ans)))
| PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #from time import time
n = int(input())
x = list(map(int, input().split()))
#start_time = time()
abc = x[-1]
a = [0]
for i in range(2, n + 1):
if x[-i] <= abc:
a.append(abc - x[-i] + 1)
elif x[-i] > abc:
abc = x[-i]
a.append(0)
print(*a[::-1])
#print(time() - start_time)
| PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | l=lambda:map(int,raw_input().split())
n=input()
h=l()
a=[0]*n
maxi=h[n-1]
for i in range(n-2,-1,-1):
if h[i]>maxi:
a[i]=0
maxi=h[i]
else:
a[i]=maxi+1-h[i]
for v in a:
print v, | PYTHON |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n=int(input())
l=[int(x) for x in input().split()]
g=[]
h=0
for i in range(n-1, -1, -1):
x=max(0, h+1-l[i])
h=max(h, l[i])
g.append(x)
g=g[::-1]
print(*g)
| PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int a[100001];
int main() {
int n, i, m, s;
cin >> n;
for (i = 1; i <= n; i++) cin >> a[i];
m = a[n];
a[n] = 0;
for (i = n - 1; i > 0; i--) {
s = m + 1 - a[i];
m = max(m, a[i]);
a[i] = s < 0 ? 0 : s;
}
for (i = 1; i <= n; i++) printf("%d%c", a[i]... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class LuxoriousHouses {
public static void main(String[] args) throws IOException {
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int []max=... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 |
import java.io.BufferedOutputStream;
import java.io.IOException;
import java.io.OutputStream;
import java.util.*;
public class test {
public static void main (String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
OutputStream out = new BufferedOutputStream ( System.out );
int n = sc.nextInt(... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
int a[100005], b[100005];
int main() {
int n, max, l;
scanf("%d", &n);
int i;
for (i = 0; i < n; i++) {
scanf("%d", &a[i]);
}
max = 0;
for (i = n - 1; i >= 0; i--) {
l = 1;
if (a[i] > max) {
max = a[i];
l = 0;
}
b[i] = max - a[i] + l;
}
for (i =... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const double pi = acos(-1);
int a[100010], d[100010], flag[100010];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
d[n - 1] = a[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (d[i + 1] >= a[i]) {
d[i... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, max = 0;
cin >> n;
int a[n], h[n];
for (int i = 0; i < n; i++) cin >> a[i];
h[n - 1] = 0;
for (int i = n - 2; i >= 0; i--) {
if (a[i + 1] > max) {
max = a[i + 1];
h[i] = max;
} else {
h[i] = max;
}
}
for (int... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class P581B {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | # -*- coding:utf-8 -*-
import sys
import math
def some_func():
"""
"""
n = input()
max =0
cache = []
n_list = map(int,sys.stdin.readline().split())
for v in n_list[::-1]:
if v>max:
cache.append(0)
max=v
elif v ==max:
cache.append(1)
... | PYTHON |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 100000 + 10;
int a[MAXN];
int n, maxx;
int ans[MAXN];
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
reverse(a + 1, a + 1 + n);
maxx = 0;
for (int i = 1; i <= n; ++i) {
if (a[i] > maxx)
ans[n - i + 1] = 0;... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.io.BufferedReader;
import java.io.InputStreamReader;
public class Building {
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
String S[] = br.readL... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int N = 100005;
int h[N], ans[N];
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%d", h + i);
}
for (int mx = 0, i = n; i >= 1; --i) {
if (mx >= h[i]) ans[i] = mx - h[i] + 1;
mx = max(mx, h[i]);
}
for (int i = 1; i... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int n;
int h[100100], b[100100];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
if (0 & 1) freopen("input", "r", stdin);
if (0 & 2) freopen("output", "w", stdout);
cin >> n;
for (int i = 0; i < n; i++) cin >> h[i];
for (int i = n - 1; i ... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int h[n];
for (int i = 0; i < n; i++) cin >> h[i];
int z = 0;
vector<int> ans;
for (int i = n - 1; i >= 0; i--) {
if (z == h[i]) {
ans.push_back(1);
continue;
}
z = max(z, h[i]);
if (z - h[i] != 0)
... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int n, arr[100000], r, arr1[100000], z;
bool boo[100000];
int main() {
cin >> n;
for (int i = 0; i < n; i++) {
cin >> r;
arr[i] = r;
}
for (int i = 0; i < n; i++) arr1[i] = arr[i];
reverse(arr1, arr1 + n);
z = arr1[0];
boo[0] = 1;
for (int i = 0; i <... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n = int(input())
arr = list(map(int, input().split()))
maxArr = arr.copy()
maxArr.reverse()
maxArr[0] = maxArr[0] * -1
for i in range(1, len(arr)):
if maxArr[i] > abs(maxArr[i-1]):
maxArr[i] = maxArr[i] * -1
else:
maxArr[i] = abs(maxArr[i-1])
maxArr.reverse()
resultArr = []
for i in range(len... | PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m[100000], mn[100000], mx;
cin >> n;
for (int i = 0; i < n; ++i) cin >> m[i];
reverse(m, m + n);
mn[0] = 0, mx = m[0];
for (int i = 1; i < n; i++)
if (m[i] > mx)
mn[i] = 0, mx = m[i];
else
mn[i] = mx - m[i] + 1;
for (int... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int n, i, m;
int b[100001];
int a[100001];
int main() {
cin >> n;
for (i = 0; i < n; i++) cin >> a[i];
m = a[n - 1];
for (i = n - 2; i >= 0; i--) {
if (m >= a[i]) b[i] = m - a[i] + 1;
if (m < a[i]) m = a[i];
}
for (i = 0; i < n; i++) cout << b[i] << ' ';... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n=int(input())
arr=list(map(int,input().split()))
dp=[0]*(n+1)
dp[-1]=-1;
for i in range(n-1,-1,-1):
maxx=max(dp[i+1],arr[i])
dp[i]=maxx
#print(*dp)
ans=[]
for i in range(n):
if arr[i]>dp[i+1]:ans.append(0)
else:ans.append(dp[i+1]-arr[i]+1)
print(*ans)
| PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> a(n), m(n);
for (int& x : a) cin >> x;
m[n - 1] = 0;
for (int i = n - 2; i >= 0; --i) m[i] = max(a[i + 1], m[i + 1]);
for (int i = 0; i < n; ++i)
cout << (a[i] > m[i] ? 0 : m[i] - a[i] + 1) << ' ';
}
| CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.util.Scanner;
public class LuxuriousHouses {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
int[] a = new int[n];
for(int i = 0; i < n; i++) a[i] = scan.nextInt();
int max = a[n-1];
int[] h = new int[n];
for(int i = n-2; i >= 0; i--){... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n = int(input())
a = input().split()
for i in range(n):
a[i] = int(a[i])
m = -1
res = [0]*n
for i in range(n-1, -1, -1):
if a[i] > m:
res[i] = 0
else:
res[i] = m + 1 - a[i]
m = max(m, a[i])
for b in res:
print(b, end=" ") | PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class LuxuriousHouses {
public static void main(String[] args) throws NumberFormatException, IOException {
BufferedReader scan = new BufferedReader(new InputStreamReader(System.in));
int houses = Integer.parseInt(... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int h[maxn], mx[maxn];
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; ++i) scanf("%d", h + i);
int cur = 0;
for (int i = n - 1; i >= 0; --i) {
mx[i] = cur;
cur = max(cur, h[i]);
}
for (int i = 0; i < n - 1; ++i) ... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
while (scanf("%d", &n) == 1) {
vector<int> x(n);
for (int i = 0; i < n; i++) scanf("%d", &x[i]);
int maxi = INT_MIN;
vector<int> ans(n);
for (int i = n - 1; i >= 0; --i) {
if (maxi >= x[i]) {
ans[i] = maxi + 1 - x[i]... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long int a[100005], n, i, max;
cin >> n;
for (i = 0; i < n; i++) {
cin >> a[i];
}
max = a[n - 1];
a[n - 1] = 0;
for (i = n - 2; i >= 0; i--) {
if (max > a[i]) {
a[i] = max - a[i] + 1;
} else if (a[i] > max) {
max = a[i... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n = int(raw_input())
a = map(int, raw_input().split())
b = []
height = 0
for i in reversed(xrange(n)):
if a[i] <= height:
b.append(height - a[i] + 1)
else:
height = a[i]
b.append(0)
for x in reversed(b):
print x,
| PYTHON |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int N = 100005;
long long a[N];
long long dp[N];
int main() {
long long n;
scanf("%lld", &n);
int max_key = 0;
int fg = 0;
for (int i = 0; i < n; ++i) {
scanf("%lld", &a[i]);
}
dp[n - 1] = 0;
for (int i = n - 2; i >= 0; --i) {
dp[i] = max(dp[i ... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n=int(input())
a=list(map(int,input().split()))
m=n-1
b=[]
b.append(0)
for i in range(n-2,-1,-1):
if a[i]<=a[m]:
b.append(a[m]+1-a[i])
else:
m=i
b.append(0)
print(*b[::-1])
| PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.util.Scanner;
public class Codeforce
{
public static void main(String[] args)
{
int n;
Scanner s =new Scanner (System.in);
n=s.nextInt();
int arr[]=new int [n];
for(int i=0;i<n;i++)
arr[i]=s.nextInt();
s.close();
int a... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
template <class T>
T _abs(T n) {
return (n < 0 ? -n : n);
}
template <class T>
T _max(T a, T b) {
return (!(a < b) ? a : b);
}
template <class T>
T _min(T a, T b) {
return (a < b ? a : b);
}
template <class T>
T sq(T x) {
return x * x;
}
template <class T>
T gcd(T a... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 |
import java.util.Scanner;
public class Luxurious_Houses {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int[] arr = new int[n];
int[] ans = new int[n];
for(int i = 0; i<n; i++)
{
arr[i] = in.nextInt();
}
int max = -1;
for(int i = n-1; i >= 0... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, index = -1, max;
cin >> n;
int *a = new int[n];
for (int i = 0; i < n; i++) cin >> a[i];
for (int i = 0; i < n - 1; i++) {
max = 0;
for (int j = i + 1; j < n; j++)
if (max < a[j]) max = a[j];
if (max < a[i])
cout << "0 "... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 |
def main():
num_of_house = int(raw_input())
house_heights = map(int, raw_input().split())
cur_max = 0
answer = [0 for i in range(num_of_house)]
for i in range(num_of_house-1, -1, -1):
if i < num_of_house - 1:
if cur_max + 1 > house_heights[i]:
answer[i] = cu... | PYTHON |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int n, hs[100001], ans[100001];
int main() {
cin >> n;
for (int i = 1; i <= n; ++i) cin >> hs[i];
int maxx = 0;
for (int i = n; i; --i) {
ans[i] = max(maxx - hs[i] + 1, 0);
maxx = max(maxx, hs[i]);
}
for (int i = 1; i < n; ++i) cout << ans[i] << ' ';
c... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, a, Max = 0;
vector<int> v;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a;
v.push_back(a);
}
Max = v.back();
v[v.size() - 1] = 0;
for (int i = v.size() - 2; i >= 0; i--) {
if (v[i] < Max)
v[i] = (Max + 1) - v[i];
e... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
const int N = 1e5;
int h[N], max[N];
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; ++i) {
scanf("%d", h + i);
}
max[n - 1] = h[n - 1];
for (int i = n - 2; i >= 0; --i) {
max[i] = h[i] > max[i + 1] ? h[i] : max[i + 1];
}
for (int i = 0; i < n - 1; ++i) {
... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int a[400000];
void Up(int l, int r, int k, int u, int v) {
if (l == r) {
a[u] = v;
return;
}
int m = (l + r) / 2;
if (m >= k)
Up(l, m, k, u * 2, v);
else
Up(m + 1, r, k, u * 2 + 1, v);
a[u] = max(a[u * 2], a[u * 2 + 1]);
}
int main() {
int n, ... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n=int(input())
l=list(map(int,input().split()))[::-1]
ma=0
ans=[]
for i in range(n):
if l[i]>ma: ans+=['0']; ma=l[i]
else: ans+=[str(ma-l[i]+1)]
print(' '.join(ans[::-1])) | PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | def solve(arr):
maxRight = arr[-1]
arr[-1] = 0
for i in range(len(arr) - 2, - 1, - 1):
if maxRight >= arr[i]:
arr[i] = maxRight + 1
else:
maxRight = arr[i]
return arr
n = input()
arr = map(int,raw_input().split())
ans = arr
ans = list(ans)
arr = solve(arr)
for i ... | PYTHON |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | a=int(input())
l=list(map(int,input().split()))
d,c=[0]*a,l[-1]
for x in range(2,a+1):
d[-x]=max(0,c-l[-x]+1)
c=max(c,l[-x])
print(*d)
| PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n=int(raw_input())
N=map(int,raw_input().split())
a=0
b=[]
for i in range(n):
b.append(max(0,a+1-N[n-i-1]))
a=max(a,N[n-i-1])
for i in range(n):
print b[n-i-1], | PYTHON |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | var n = parseInt(readline());
var input = readline().split(' ').map(function (item) {
return parseInt(item);
})
var result = [];
input.reduceRight(function (prev, curr) {
if(curr > prev){
result.push(0);
} else if (curr === prev) {
result.push(1);
} else {
result.push(prev - curr + 1);
}
return Math.max(pre... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 10;
int a[maxn], o[maxn];
int main() {
int n;
while (scanf("%d", &n) != EOF) {
for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
int _max = 0;
for (int i = n - 1; i >= 0; --i)
if (a[i] > _max)
o[i] = 0, _max = a[i];
... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int N = 100005;
int n;
int v[N], seg[4 * N + 1];
int a, b;
void init(int r, int i, int j) {
if (i == j)
seg[r] = v[i];
else {
init(2 * r, i, (i + j) / 2);
init(2 * r + 1, (i + j) / 2 + 1, j);
seg[r] = max(seg[2 * r], seg[2 * r + 1]);
}
}
int quer... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
int maxsize = 30;
using namespace std;
int main() {
int n;
int A[100005];
cin >> n;
for (int i = 1; i <= n; i++) cin >> A[i];
int m = 0;
int ans[100005];
ans[n] = 0;
for (int i = n - 1; i >= 1; i--) {
m = max(m, A[i + 1]);
ans[i] = m + 1 - A[i];
if (ans[i] < 0) ans[i... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n=int(input())
a=map(int,input().split(" "))
rmax=0
res=[]
for x in reversed(list(a)):
t = rmax - x
res.append(0 if t<0 else t+1)
rmax=max(x, rmax)
for x in reversed(res):
print(x,end=" ")
print()
| PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n = int(raw_input())
numbers = map(int, raw_input().split())
maxx = numbers[n - 1]
numbers[n - 1] = 0
for it in xrange(n - 2, -1, -1):
if numbers[it] > maxx:
maxx = numbers[it]
numbers[it] = 0
else:
numbers[it] = maxx + 1 - numbers[it]
print ' '.join(map(str, numbers)) | PYTHON |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int a[100009];
int main() {
int i, j, m, n, t, k;
cin >> n;
int ma = 0;
for (i = 0; i < n; i++) {
cin >> a[i];
}
int sum = 0;
int b[100009];
for (i = n - 1, j = 0; i >= 0; i--) {
if (a[i] > ma) {
ma = a[i];
b[j++] = 0;
continue;
... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int maxn = 100005;
const int maxm = 10005;
int h[maxn], ans[maxn];
void solve() {
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> h[i];
}
int maxh = 0;
for (int i = n; i >= 1; i--) {
if (... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int mx, n, ara[1000000];
int main() {
scanf("%d", &(n));
for (int i = (0); i < (n); i++) scanf("%d", &(ara[i]));
mx = ara[n - 1];
for (int i = (n - 2); i >= (0); i--) {
if (mx >= ara[i])
ara[i] = mx - ara[i] + 1;
else {
mx = ara[i];
ara[i] ... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.Reader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.Writer;
impor... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int inf = (int)1e9;
const double pi = acos(-1.0);
const double eps = 1e-9;
int n, a[200100], m[200100];
int main() {
cin >> n;
for (int i = 1; i <= n; i++) cin >> a[i];
for (int i = n; i >= 1; i--) m[i] = max(m[i + 1], a[i]);
for (int i = 1; i <= n; i++) {
... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int maxi[100000 + 100];
bool mark[100000 + 100];
int main() {
int n;
cin >> n;
int h[n];
for (int i = 0; i < n; i++) cin >> h[i];
maxi[n - 1] = h[n - 1];
mark[n - 1] = true;
for (int i = n - 2; i >= 0; i--) {
if (h[i] > maxi[i + 1]) {
maxi[i] = h[i];... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n=int(raw_input())
a=map(int,raw_input().split(' '))
m=0
b=[0]*n
m=a[-1]
for i in xrange(len(a)-2,-1,-1):
if(a[i]<=m):
b[i]=m-a[i]+1
else:
m=a[i]
for i in b:
print i, | PYTHON |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
public class Gots {
public static long mod = 1000000000 + 7;
public static long gcd(long a, long b) {
return (b == 0) ? a : gcd(b, a % b);
}
public static int counter = 0... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n=int(input())
z=list(map(int,input().split()))
lst=[0]*n;m=z[-1];lst[-1]=0
for i in range(n-2,-1,-1):
if z[i] > m:
lst[i]=0
m = z[i]
else:
lst[i]= m-z[i]+1
print(*lst)
| PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n = int(input())
a = list(map(int, input().split()))
maxi = -1
for i in range(n - 1, -1, -1):
t = a[i]
if a[i] > maxi:
a[i] = 0
else:
a[i] = maxi - a[i] + 1
maxi = max(maxi, t)
print(*a)
| PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n;
cin >> n;
vector<long long> v(n), s(n);
for (int i = 0; i < n; i++) cin >> v[i], s[i] = v[i];
reverse(s.begin(), s.end());
long long mx = 0;
for (int i = 0; i < n; i++) s[i] = max(s[i], mx), mx = s[i];
reverse(s.begin(), s.end());
... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n = int(input())
h = list(map(int, input().split()))
a = [0]*n
m = -1
for i in range(n-1,-1,-1):
a[i] = max(m+1-h[i],0)
m = max(m, h[i])
print(*a)
| PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
long long int a[1000000], b[1000000], n, k, l, x, m, s, p, i, j, t;
int main() {
map<long long, long long> v;
cin >> n;
for (i = 0; i < n; i++) {
cin >> a[i];
v[a[i]]++;
}
for (i = 0; i < n; i++) b[i] = a[i];
sort(b, b + n);
reverse(b, b + n);
m = b[... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
public class a {
public static void main(Strin... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
vector<int> h, ans;
int n, t = 0;
ios::sync_with_stdio(false);
cin.tie(0);
cin >> n;
h.resize(n + 1);
for (int i = 1; i <= n; i++) cin >> h[i];
ans.push_back(0);
t = h[n];
for (int i = n - 1; i > 0; i--) {
if (h[i] > t)
ans.push_ba... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int a[100005];
int suff[100005];
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
for (int i = n; i >= 1; i--) {
suff[i] = max(suff[i + 1], a[i]);
}
for (int i = 1; i <= n; i++) {
printf("%d ", max(0, suff[i... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | num = int(raw_input())
houses = [int(x) for x in raw_input().split()]
maxAt = [0] * num
highest = 0
for i in range(num-1, -1, -1):
if houses[i] > highest:
highest = houses[i]
maxAt[i] = highest
for i in range(0, num-1):
highest = maxAt[i+1]
if houses[i] <= highest:
print (hi... | PYTHON |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | //package TestOnly.Div2B_322.Code1;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class Main
{
FastScanner in;
PrintWriter out;
public void solve() throws IOException
{
int n = in.nextInt();
in... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n=int(raw_input())
li=map(int, raw_input().split())
ans=[li[n-1]]
temp=li[n-1]
for i in xrange(n-2, -1, -1):
if li[i]<=temp:
ans.append(temp+1)
else:
ans.append(li[i])
temp=li[i]
ans=ans[::-1]
for i in xrange(len(ans)):
print ans[i]-li[i], | PYTHON |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int MaxN = 1e5;
int n;
bool flag;
int a[MaxN + 5], b[MaxN + 5];
int Max = 1 >> 30;
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
for (int i = n; i >= 1; i--) {
flag = 0;
if (a[i] > Max) {
Max = a[i];
flag = 1;... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> q(n);
for (int i = 0; i < n; i++) {
cin >> q[i];
}
int mx = 0;
vector<int> ans;
for (int i = n - 1; i >= 0; i--) {
ans.push_back(max(0, mx + 1 - q[i]));
mx = max(mx, q[i]);
}
reverse(ans.begin(), ans.... | CPP |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.