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;
bool isPrime[1005];
void sieve(int N) {
for (int i = 0; i <= N; ++i) {
isPrime[i] = true;
}
isPrime[0] = false;
isPrime[1] = false;
for (int i = 2; i * i <= N; ++i) {
if (isPrime[i] == true) {
for (int j = i * i; j <= N; j += i) isPrime[j] = false;
... | 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.io.PrintStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Iterator;
public class Main_Round322Div2_B {
public static void main(String[] args) {
Scanner s... | 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())
i = [int(x) for x in raw_input().split()]
out = []
m = 0
for x in range(len(i), 0, -1):
if i[x-1] > m:
m = i[x-1]
out.append(0)
else:
out.append(m - i[x-1] + 1)
for i in range(len(out), 0, -1):
print out[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 | #include <bits/stdc++.h>
using namespace std;
int arr[100005], ans[100005];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
int maxx = arr[n - 1];
ans[n - 1] = 0;
for (int i = n - 2; i >= 0; i--) {
if (arr[i] > maxx) {
maxx = arr[i];
ans[i] = 0;
} else {
... | 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 | //package com.competitions.year2015.julytoseptember.round322div2;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.Writer;
import java.util.InputMismatchException;
public fi... | 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.*;
import java.io.*;
public final class Solution
{
public static void main(String[] args)
{
Reader input = new Reader();
PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
int n = input.nextInt();
int[] arr = new int[n];
int[] res... | 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())
l = map(int, raw_input().split())
M = 0
ans = [0]*n
for i in xrange(n-1, -1, -1):
ans[i] = max(0, M - l[i] + 1)
M = max(M, l[i])
print ' '.join(map(str, ans)) | 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 = 100001;
void process(int a[], int dp[][maxN], int N) {
int i, j;
for (i = 0; i < N; i++) dp[0][i] = a[i];
for (i = 1; i <= (int)(log2(N)); i++)
for (j = 0; j + (1 << (i - 1)) < N; j++)
dp[i][j] = max(dp[i - 1][j], dp[i - 1][j + (1 << (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 | def solve(n, h_arr):
h_arr = list(h_arr)
h_increment = [0] * n
largest = h_arr[n-1]
for i in range(n-2, -1, -1):
if h_arr[i] <= largest:
h_increment[i] = largest - h_arr[i] + 1
else:
largest = h_arr[i]
return " ".join(str(h_inc) for h_inc in h_increment)
... | 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.*;
import java.util.*;
public class B {
public static void main(String[] args) throws IOException {
int n = nextInt();
int[] floors = new int[n + 1];
for (int i = 0; i < n; ++i) {
floors[i] = nextInt();
}
int[] max = new int[n + 1];
floors[n] = -1;
for (int i = n - 1; 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 | #include <bits/stdc++.h>
using namespace std;
int n, h[200000], a[200000];
int main() {
cin >> n;
for (int i = 0; i < n; i++) cin >> h[i];
for (int i = n - 1; i >= 0; i--) a[i] = max(a[i + 1], h[i + 1]);
for (int i = 0; i < n; i++) cout << max(a[i] - h[i] + 1, 0) << ' ';
return 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;
const int N = 1e5;
vector<int> vc;
int a[N + 10];
int main() {
int n, mx = 0;
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
for (int i = n - 1; i >= 0; i--)
if (a[i] > mx)
vc.push_back(0), mx = a[i];
else
vc.push_back(mx - a[i] + 1);
for (... | 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[100007], b[100007];
cin >> n;
for (int i = 1; i <= n; ++i) cin >> a[i];
int maxi = 0;
for (int i = n; i >= 1; --i) {
b[i] = maxi;
maxi = max(maxi, a[i]);
}
for (int i = 1; i <= n; ++i)
if (a[i] <= b[i])
cout << abs(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())
arrh = map(int, raw_input().split())
ans = []
ans.append(0)
mx = arrh[n-1]
for i in xrange(n-2, -1, -1):
mx = max(arrh[i+1], mx)
ans.append(max(mx-arrh[i]+1, 0))
for i in xrange(n-1, -1, -1):
print ans[i],
print | 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 = input()
arr = map(int,raw_input().split())
new = [0]
arr = arr[::-1]
ma = arr[0]
for i in range(1,n):
if arr[i]<=ma:
ans = ma-arr[i] + 1
else:
ma = arr[i]
ans = 0
new.append(ans)
new = new[::-1]
print " ".join(map(str,new))
| 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.io.*;
import java.lang.Math;
public class C {
public static void main(String args[]) {
PrintWriter out = new PrintWriter(System.out);
FastScanner fs = new FastScanner();
int t=1;
for(int qq=1; qq <= t; qq++) {
int n=fs.nextInt();
int a[] = 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 | # http://codeforces.com/problemset/problem/581/B
def calculate(numbers):
numbers = [int(i) for i in numbers.split()]
result = []
current = 0
for i in numbers[::-1]:
if i <= current:
result.append(current - i + 1)
else:
result.append(0)
current = 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;
int main() {
int n;
cin >> n;
int arr[n + 1];
int ans[n + 1];
memset(ans, 0, sizeof(ans));
for (int i = 0; i < n; i++) cin >> arr[i];
int t = arr[n - 1];
for (int i = n - 2; i > -1; i--) {
t >= arr[i] ? ans[i] = t - arr[i] + 1 : t = arr[i];
}
for (in... | 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 s(a):
p=0
for i in a:
p+=i[1]
return p
n=input()
h=map(int,raw_input().split())
h.reverse()
diff=[0]
mh=h[0]
for i in range(1,n):
d=mh-h[i]
if d>=0:
diff.append(mh-h[i]+1)
else:
diff.append(0)
if h[i]>mh:
mh=h[i]
diff.reverse()
print ' '.join(ma... | 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.io.*;
public class Div2_322B {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] array = new int[n];
for(int i = 0; i < n; i++)
array[i] = sc.nextInt();
int[] ans = new int[n];
int max = array[n - 1];
for(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())
l=map(int,raw_input().split())
ans=[0]
cnt,chk=l[-1],1
for i in l[-2::-1]:
if i<=cnt and chk:
chk=0
ans.append(cnt-i+1)
elif i<=cnt:
ans.append(cnt-i+1)
else:
chk=1
cnt=i
ans.append(0)
print ' '.join(map(str,ans[::-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 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.StringTokenizer;
public class start {
static int x[];
public static void main(String[] args) throws IOException {
BufferedReader bf = new BufferedRead... | 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())
h = list(map(int, input().split()))
maxi = 0
c = []
for i in reversed(h):
if i > maxi:
maxi = i
c.append(0)
else:
c.append(maxi + 1 - i)
for i in reversed(c):
print(i, 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 | NumberOfHouses = int(input())
Floors = list(map(int, input().split()))
AddMeArr = [0]*NumberOfHouses
MaxNum = Floors[-1]
for i in range(NumberOfHouses-2, -1, -1):
if Floors[i] > MaxNum:
MaxNum = Floors[i]
else:
AddMeArr[i] = MaxNum - Floors[i] + 1
print(*AddMeArr, sep=' ') | 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 root(long long n) {
long long ans = 1;
while (1) {
if (ans * ans > n) {
return ans - 1;
break;
} else if (ans * ans == n) {
return ans;
break;
} else {
ans += 1;
}
}
}
int main() {
long long n, i, max;
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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
long long hi[n];
long long dp[n];
for (int i = 0; i < n; i++) {
cin >> hi[i];
}
dp[n - 1] = 0;
for (int i = n - 2; i >= 0; i--) {
dp[i] = max(dp[i + 1], hi[i + 1]);
}
if (hi[0] > dp[0]) {
cout << "0";
} else {
... | 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.BufferedWriter;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class B {
static StringTokenizer... | 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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.Random;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
im... | 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 = list(map(int, input().split()))
m = -1
ans = []
for i in range(n):
if a[n-i-1] > m:
ans.append("0")
else:
ans.append(str(m+1-a[n-i-1]))
m = max(a[n-i-1], m)
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 | /**
* Created by heat_wave on 12.01.15.
*/
import java.io.*;
import java.util.Scanner;
import java.util.StringTokenizer;
public class A {
Scanner in = new Scanner(System.in);
public void solve() {
int n = in.nextInt();
long [] h = new long[n];
for (int i = 0; i < n; 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())
h=[int(i) for i in input().split()]
p=n*[0]
p[-1]=0
gg=n*[0]
gg[-1]=0
for i in range(n-2,-1,-1):
gg[i]=max(gg[i+1],h[i+1])
for i in range(n):
d=gg[i]-h[i]
if d<0:
d=0
p[i]=d
else:
p[i]=d+1
for i in range(n):
print(p[i],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())
ans = []
ma = 0
for h in list(reversed(list(map(int, input().split())))):
ans += [max(0, ma - h + 1)]
ma = max(ma, h)
print(' '.join(map(str, reversed(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;
vector<pair<long long int, long long int> > v;
pair<long long int, long long int> p;
int main() {
long long int x, y, z, n, a[100002], b[100002], c, i, j, need;
while (cin >> n) {
v.clear();
for (i = 0; i < n; i++) {
cin >> x;
b[i] = x;
}
for... | 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()
a = map(int, raw_input().split())
b = [0]*n
b[0] = 0
a.reverse()
max = a[0]
for i in range(1, n):
if a[i] > max:
b[i] = 0
max = a[i]
else:
b[i] = max + 1 - a[i]
b.reverse()
for i in range(n):
print b[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 | n = int(input())
houses = [int(c) for c in input().split()]
ans = [0] * n
max_ = -1
for i in range(n - 1, -1, -1):
ans[i] = max_ - houses[i] + 1 if houses[i] <= max_ else 0
max_ = max(max_, houses[i])
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;
long long int n, height[100005], maxR[100005];
int main() {
cin >> n;
for (int i = 0; i < n; i++) cin >> height[i];
maxR[n] = height[n - 1] - 1;
maxR[n - 1] = height[n - 1];
for (int i = n - 2; i >= 0; i--) maxR[i] = max(maxR[i + 1], height[i]);
for (int 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 | import java.io.*;
import java.util.*;
public class LuxuriousHouses {
public static void main(String args[]){
FScanner in = new FScanner();
PrintWriter out = new PrintWriter(System.out);
int n = in.nextInt();
int arr[]=in.readArray(n);
int copy[]=new int[n+1];
int max=arr[n-1],a=0;long sum=0;
for(int 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 | __author__ = 'Admin'
n = int(input())
mas1 = list(map(int, input().split()))
masans = [0]
max = mas1[n - 1]
for i in range(n - 2, -1, -1):
if mas1[i] > max:
max = mas1[i]
masans.append(0)
else:
masans.append(max - mas1[i] + 1)
masans.reverse()
print(*masans) | 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())
seq = [int(x) for x in raw_input().split()]
seq.reverse()
cur_max = seq[0]
result = [0]
for elem in seq[1:]:
if elem > cur_max:
result.append(0)
cur_max = elem
else:
result.append(cur_max - elem + 1)
for elem in result[::-1]:
print elem,
| 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;
void solve() {
long long n;
cin >> n;
;
long long a[n];
for (long long i = 0; i < n; i++) cin >> a[i];
long long mx[n];
mx[n - 1] = 0;
for (long long i = n - 2; i >= 0; i--) {
mx[i] = max(a[i + 1], mx[i + 1]);
}
for (long long 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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int data[n];
map<int, int> cnt;
for (int i = 0; i < n; ++i) {
cin >> data[i];
++cnt[data[i]];
}
for (int i = 0; i < n; ++i) {
if (cnt.crbegin()->first == data[i])
cout << (cnt.crbegin()->second == 1 ? 0 : 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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.PrintStream;
import java.util.Scanner;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author Mehul Sharma
*/
public class Main {
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 | n = int(input())
string = input()
numbers = string.split(" ")
for x in range(n):
numbers[x] = int(numbers[x])
floors = []
maximum = numbers[n - 1]
for x in range(n - 2, -1, -1):
a = numbers[x]
if a > maximum:
maximum = a
floors.append("0")
else:
floors.append(str(maximum - a + 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.io.IOException;
import java.io.PrintWriter;
import java.util.Scanner;
public class luxuriousHouses {
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
// BufferedReader b=new BufferedReader(new InputStreamReader(System.in));
Scanner scan=new Scanner(Syste... | 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())
houses = list(map(int, input().split(' ')))
answers = []
mh = 0
for h in houses[::-1]:
if mh >= h:
answers.append(mh - h + 1)
else:
answers.append(0)
mh = max(mh, h)
print(' '.join(map(str, answers[::-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 main() {
int n, x;
ios_base::sync_with_stdio(0);
cin >> n;
vector<int> v(n);
for (int i = 0; i < n; ++i) {
cin >> x;
v[i] = x;
}
int max = -1, tmp;
vector<int> m(n);
for (int i = v.size() - 1; i >= 0; --i) {
tmp = max;
if (v[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 | n = int(input())
lst = list(map(int, input().split()))
dp = [-1000] * n
dp[n - 1] = 0
for i in range(len(lst) - 2, -1, -1):
dp[i] = max(dp[i + 1], lst[i + 1])
for i in range(n):
if dp[i] >= lst[i]:
print(dp[i] - lst[i] + 1, end=' ')
else:
print(0, 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())
a=list(map(int,input().split()))
x=1
r=[]
for i in range(n-1,-1,-1):r+=[max(x-a[i],0)];x=max(x,a[i]+1)
print(' '.join(map(str,r[::-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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution ... | 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, arr[100005], arrAns[100005];
cin >> n;
for (int i = 0; i < n; i++) cin >> arr[i];
int max = arr[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (arr[i] <= max)
arrAns[i] = (max - arr[i]) + 1;
else {
arrAns[i] = 0;
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>
using namespace std;
int N = (int)1e5 + 1;
int main() {
std::ios::sync_with_stdio(0);
long long int n;
cin >> n;
long long int a[n], max[n];
for (int i = 0; i < n; i++) cin >> a[i];
long long int ma = a[n - 1];
max[n - 1] = a[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (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 | #include <bits/stdc++.h>
using namespace std;
vector<int> v;
int main() {
int n, i, val, sub = 0;
cin >> n;
int arr[n + 5];
for (i = 0; i < n; i++) {
cin >> arr[i];
}
val = arr[n - 1];
v.push_back(0);
for (i = n - 2; i >= 0; i--) {
if (arr[i] <= val) {
sub = (val - arr[i]) + 1;
v.pus... | 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 | #codeforces
if __name__=="__main__":
n=int(input())
a=list(map(int,input().split()))
ma=a[n-1]
ans=[0]
i=n-2
while i>=0:
if a[i]>ma:
ans.append(0)
ma=a[i]
else:
d=(ma-a[i])+1
ans.append(d)
i=i-1
ans.reverse()
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 | __author__ = 'hariprahal'
def main():
n = int(raw_input())
array = raw_input().split()
array = [int(i) for i in array]
max1 = array[n - 1]
array[n - 1] = 0
str1 = ''
for i in range(n - 2, -1, -1):
if array[i] > max1:
max1 = array[i]
array[i] = 0
el... | 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().strip())
arr = list(map(int, input().strip().split()))
currMax = 0
ans = [0]*n
for i in range(n-1, -1, -1):
ans[i] = max(0, currMax-arr[i]+1)
currMax = max(currMax, arr[i])
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 | x = int(input())
inp = input().split()
out = []
maxh = 0
for i in reversed(range(0,x)):
j = int(inp[i])
if j <= maxh:
out.append(maxh - j + 1)
else:
out.append(0)
maxh = j
s = ""
for i in reversed(out):
s += str(i) + " "
print(s) | 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 h[100000];
int main() {
int n;
cin >> n;
for (int j = 0; j < n; ++j) {
cin >> h[j];
}
long long max = h[n - 1];
long long answer[100000];
answer[n - 1] = 0;
for (int i = n - 2; i >= 0; --i) {
if (h[i] > max) {
max = h[i];
answer... | 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()))
out = [0]
best = h[-1]
for i in range(n-2, -1, -1):
if h[i] < best:
out.append(best - h[i] + 1)
elif h[i] > best:
best = h[i]
out.append(0)
else:
out.append(1)
print(" ".join(map(str, reversed(out)))) | 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())
l = [int(x) for x in input().split()]
assert len(l) == n
l.reverse()
pre = [0]
for i in range(1, n):
pre.append(max(pre[-1], l[i-1]))
out = [max(0, 1+pre[i]-l[i]) for i in range(n)]
out.reverse()
print(' '.join(str(x) for x in out))
| 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())
a=map(int,raw_input().split())
b=[]
mx=0
for i in xrange(n-1,-1,-1):
if mx>=a[i]: b.append(mx-a[i]+1)
else : b.append(0)
mx=max(mx,a[i])
#print a[i]
for i in xrange(len(b)-1,-1,-1):
print b[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.util.Scanner;
/**
* Created on 28/09/2015.
*/
public class B {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n = s.nextInt();
int[] houses = new int[n];
for (int i = 0; i < n; i++) {
houses[i] = s.nextInt();
}
int[] maxHeights = new int[n];
for (int... | 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 i, maxi, n;
cin >> n;
long long A[n];
for (i = 0; i < n; i++) cin >> A[i];
maxi = A[n - 1];
long long maxii[n];
maxii[n - 1] = 0;
for (i = n - 2; i >= 0; i--) {
if (A[i] <= maxi)
maxii[i] = maxi + 1 - A[i];
else {
m... | 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.IOException;
import java.io.InputStream;
import java.util.Arrays;
import java.util.InputMismatchException;
public class B581 {
static InputStream is;
public static void main(String[] args) {
is = System.in;
solve();
}
static void solve() {
int n = ni();
int[] h = new int[n];
for (int ... | 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())
ai = list(map(int,input().split()))
maxm = ai[n-1]
bi = [0]*n
for i in range(len(ai)-2,-1,-1):
bi[i] = max(0,maxm+1-ai[i])
if ai[i] > maxm:
maxm = ai[i]
print(*bi) | 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 main():
input()
hh = list(map(int, input().split()))[::-1]
h0 = 0
for i, h in enumerate(hh):
if h0 < h:
h0 = h
hh[i] = 0
elif h0 > h:
hh[i] = h0 + 1 - h
else:
hh[i] = 1
print(*hh[::-1])
if __name__ == '__main__':
main(... | 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> v(n);
for (int i = 0; i < n; i++) cin >> v.at(i);
vector<int> w(n);
w.at(n - 1) = 0;
int k = v.at(n - 1);
for (int i = n - 2; i >= 0; i--) {
if (v.at(i) > k) {
w.at(i) = 0;
k = v.at(i);
} else {
... | 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 | //package codeForces;
import java.util.Scanner;
public class LuxuriousHouse {
static int findMax(int r,int l,int[]arr){
int max=arr[r];
for(int i=r;i<l;i++){
if(arr[i]>max)max=arr[i];
}
return max;
}
public static void main(String[]args){
Scanner scanner... | 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 Codeforces581BLuxeryHouse {
public static void main(String[] args) {
int a, b, sum = 0;
Scanner scanner = new Scanner(System.in);
a = scanner.nextInt();
int[] arr = new int[a];
for (int i = 0; i < a; i++) {
arr[i] = scanner.nextInt();
}
int[] answer = new 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;
int main() {
long long n, i, j, arr[100001], b[100001] = {0}, temp, ans, maxright = 0;
cin >> n;
for (i = 0; i < n; i++) {
cin >> arr[i];
}
b[n - 1] = arr[n - 1];
maxright = arr[n - 1];
for (i = n - 2; i >= 0; i--) {
temp = arr[i];
b[i] = maxright;... | 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()
house = map(int,raw_input().split())
ans = [0] * n
maxi = house[-1]
for i in range(n-2,-1,-1):
ans[i] = max(maxi+1-house[i],0)
maxi = max(maxi,house[i])
print ' '.join(str(i) for i in ans)
| 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;
void next(long long int a[], long long int n) {
long long int i;
long long int max = a[n - 1];
a[n - 1] = -1;
for (i = n - 2; i >= 0; i--) {
long long int temp = a[i];
a[i] = max + 1;
if (max < temp) max = temp;
}
}
int main() {
long long int n, a[10... | 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()
l = map(int, raw_input().split())
ma = 0
a = [0]*n
for i in range(n - 1, -1, -1):
a[i] = max(0, ma + 1 - l[i])
ma = max(ma, l[i])
for i in range(n):
print a[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 + 100;
long long int a[maxn], dp[maxn];
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; i++) cin >> a[i];
for (int i = n; i >= 1; i--) {
dp[i] = max(a[i], dp[i + 1]);
if (a[i] > dp[i + 1]) {
a[i] = 0;
} else {
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 | a = input()
b = map(int, raw_input().split())
def luxhouse(a, b):
c = []
lux = 0
for x in reversed(xrange(a)):
if b[x] > lux:
lux = b[x]
c.append(0)
else:
c.append(lux+1-b[x])
print " ".join(str(x) for x in reversed(c))
luxhouse(a, b)
| 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>
const double EPS = 1e-24;
const long long int MOD = 1000000007ll;
const long long int MOD1 = 1000000009ll;
const long long int MOD2 = 1100000009ll;
const double PI = 3.14159265359;
int INF = 2147483645;
long long int INFINF = 9223372036854775807;
template <class T>
T Max2(T a, T b) {
return 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 | /* package whatever; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Ideone
{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
int n,b;
Scanne... | 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())
h = [int(s) for s in input().split()]
# list of maxima: element i is max(h[i:])
m = [0] * n
m[n-1] = h[n-1]
for i in reversed(range(n-1)):
m[i] = max(h[i], m[i+1])
a = [max(0, m[i+1]+1 - h) for i, h in enumerate(h[:-1])] + [0]
print(" ".join(str(x) for x in 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;
const int maxn = 1e5 + 500;
int n, p[maxn], maxv[maxn];
int main(int argc, char *argv[]) {
std::ios::sync_with_stdio(false);
std::cin.tie(0);
cin >> n;
for (int i = 0; i < n; ++i) cin >> p[i];
int ptr = 0;
for (int i = n - 1; i >= 0; --i) {
maxv[i] = ptr;
... | 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.Arrays;
import java.util.Scanner;
public class Luxurioushouse {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner input = new Scanner(System.in);
int n = input.nextInt();
int[] arr = new int[n];
int[] brr = new int[n];
for(int... | 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;
long long n, i, a[100005], r[100005], m;
int main() {
cin >> n;
for (i = 1; i <= n; i++) {
cin >> a[i];
}
r[n] = 0;
for (i = n - 1; i >= 1; i--) {
m = max(m, a[i + 1]);
r[i] = m;
}
for (i = 1; i <= n; i++) {
if (r[i] >= a[i]) {
cout << r[... | 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 INF = (int)1e9;
const int SIZE = 100005;
int main() {
ios_base::sync_with_stdio(false);
int n;
cin >> n;
vector<int> v(n), res(n);
vector<pair<int, int> > vp;
for (int i = 0; i < n; i++) cin >> v[i];
int mx = 0;
vp.push_back(make_pair(n - 1, v[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 | __author__ = 'aste'
def main():
n = int(raw_input())
h = [int(x) for x in raw_input().split()]
m = 0
i = n - 1
a = []
while i >= 0:
a.append(max(0, m + 1 - h[i]))
m = max(m, h[i])
i -= 1
i = n - 1
while i >= 0:
print a[i],
i -= 1
if __name__ == "... | 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 main() {
int n, m = 0;
cin >> n;
int a[100005], b[100005];
for (int i = 0; i < n; i++) {
cin >> a[i];
}
for (int i = n - 1; i >= 0; i--) {
if (a[i] > m) {
m = a[i];
b[i] = 0;
} else {
b[i] = m - a[i] + 1;
}
}
for (int 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>
#pragma optimize("O3")
using namespace std;
const long long MOD = 1e9 + 7;
const long long INF = 1e18 + 7;
const int base = 2e5 + 1;
const long long MAX = 1e6;
const double EPS = 1e-9;
const double PI = acos(-1.);
const int MAXN = 2 * 1e5 + 147;
mt19937 rng(chrono::steady_clock::now().time_sinc... | 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()))
a = [0]
max1 = l[-1]
for i in range(n - 2, -1, -1):
if l[i] > max1:
max1 = l[i]
a.append(0)
else:
a.append(max1 - l[i] + 1)
print(*a[::-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 Driver {
public static Scanner scanner;
public static void main(String[] args) {
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];
b[n-1]=0;
for(int i=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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> A(n), M(n);
for (int i = 0; i < n; i++) {
cin >> A[i];
}
int maxi = -1;
for (int i = n - 1; i >= 1; i--) {
maxi = max(maxi, A[i]);
M[i] = maxi;
}
vector<int> res;
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 | n=int(input())
a=list(map(int,input().split()))
b=[0]*n
mx=a[n-1]
for i in range(n-2,-1,-1):
b[i]=max(0,mx-a[i]+1)
if a[i]>mx:mx=a[i]
print(*b)
# Made By Mostafa_Khaled | 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.*;
import java.util.*;
public class main {
static BufferedReader in;
static PrintWriter out;
static StringTokenizer tok;
static String next() throws IOException{
while ( tok == null || !tok.hasMoreTokens()){
tok = new StringTokenizer(in.readLine());
}
return tok.nextT... | 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() {
ios::sync_with_stdio(0);
cin.tie(NULL);
int n, mx;
cin >> n;
int a[n], b[n];
for (int i = 0; i < n; ++i) cin >> a[i];
b[n - 1] = 0;
mx = a[n - 1];
for (int i = n - 2; i >= 0; --i) {
mx = max(mx, a[i + 1]);
if (mx >= a[i])
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 | #include <bits/stdc++.h>
using namespace std;
int A[100002];
int Max[100002];
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int N;
cin >> N;
for (int i = 1; i <= N; i++) {
cin >> A[i];
}
Max[N] = 0;
for (int i = N - 1; i > 0; i--) {
Max[i] = max(A[i + 1], Max[i + 1]);
}
for (int 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;
long long n, a[500005], cmax, d[500005], ans;
int main() {
ios::sync_with_stdio(0);
;
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
for (int i = n - 1; i >= 0; i--) {
d[i] = cmax;
cmax = max(cmax, a[i]);
}
for (int i = 0; i < n; i++) {
if (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 | from sys import *
input = lambda:stdin.readline()
int_arr = lambda : list(map(int,stdin.readline().strip().split()))
str_arr = lambda :list(map(str,stdin.readline().split()))
get_str = lambda : map(str,stdin.readline().strip().split())
get_int = lambda: map(int,stdin.readline().strip().split())
get_float = lambda : m... | 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.BufferedInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.InputMismatchException;
public class Main
{
class MyScanner
{
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
BufferedInputS... | 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 int n;
cin >> n;
long long int a[n];
for (int i = 0; i < n; i++) cin >> a[i];
vector<long long int> b(n, 0);
for (int i = n - 2; i >= 0; i--) {
b[i] = a[i + 1] - a[i];
}
long long int s = 0;
for (int i = n - 2; 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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int x;
scanf("%d", &x);
int arr[x];
for (int i = 0; i < x; i++) {
scanf("%d", &arr[i]);
}
int maxi = arr[x - 1];
vector<int> v;
v.push_back(0);
for (int i = x - 2; i >= 0; i--) {
if (arr[i] <= maxi) {
v.push_back(maxi - arr[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 | from sys import stdin
n = int(stdin.readline())
hi = list(map(int, stdin.readline().split()))
res = [''] * n
vmax = 0
for i in range(n - 1, -1, -1):
if vmax < hi[i]:
vmax = hi[i]
res[i] = '0'
else:
res[i] = str(vmax - hi[i] + 1)
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 | import java.util.Scanner;
import java.util.Vector;
public final class e{
final int N=205123;
public void solve(){
Scanner in = new Scanner(System.in);
int a[]=new int[N];
int b[]=new int[N];
int n,mx;
n=in.nextInt();
for(int i=0;i<n;++i)
a[i]=in.nextI... | 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 | #------------------------------what is this I don't know....just makes my mess faster--------------------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer =... | 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=input()
t=map(int,raw_input().split())
t.reverse()
m=t[0]
li=[0]
for i in xrange(1,n):
if t[i]>m:
m=t[i]
li+=[0]
else:
li+=[m+1-t[i]]
li.reverse()
for i in li:
print i,
| PYTHON |
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