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() {
long long int n, i = 0, l = -2, k = INT_MAX, j, a = 0, f = 0, c = 0, b = 0,
x = 100000000007, d, y, m, zero = 0, one = 0, z;
string s = "qwertyuiopasdfghjkl;zxcvbnm,./", mar, maria;
char cha;
vector<long long int> vc, vct, vctt;
vec... | 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 = [0]*n
maxarr[n-1] = arr[-1]
curmax = maxarr[-1]
index = n-1
for val in arr[::-1]:
maxarr[index] = max(val, curmax)
curmax = maxarr[index]
index -= 1
for index, val in enumerate(arr):
out = 0
if maxarr[index] == val:
if index + 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;
long long h[100000], ans[100000];
cin >> n;
for (int i = 0; i < n; i++) cin >> h[i];
ans[n - 1] = 0;
for (int i = n - 2; i >= 0; i--) {
if (h[i] <= h[i + 1])
ans[i] = h[i + 1] - h[i] + 1;
else
ans[i] = 0;
h[i] = max(h[... | 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()))
l=[]
m=h.pop()
l.append(0)
while h:
if h[-1]<=m:
l.append(m-h[-1]+1)
m=max(m,h.pop())
else:
l.append(0)
m=max(m,h.pop())
for i in range(len(l)-1,-1,-1):
print(l[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 | import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class B {
public B() {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
List<Integer> list = new ArrayList<Integer>();
for(int i=0; i<n; i++) {
list.add(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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
long long int arr[n];
for (int(i) = 0; (i) < (n); ++(i)) {
cin >> arr[i];
}
long long int m = arr[n - 1];
long long int out[n];
out[n - 1] = 0;
for (int(i) = (n - 2); (i) >= (0); --(i)) {
if (arr[i] > m) {
out[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.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 | n = int(raw_input());
inp = raw_input().split();
inp.reverse();
mx = -1;
ans=[];
for el in inp:
x = int(el);
if mx == -1 or x > mx:
ans.append('0')
else:
ans.append(str(mx - x + 1))
mx = max(mx, x);
ans.reverse();
print " ".join(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 |
import java.util.Arrays;
import java.util.Scanner;
import java.util.Vector;
public class Luxurious_Houses {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
Vector<Object> v = new Vector<>();
long[] arr = new long[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 ar[100001];
int maxi[100001];
int main() {
int n, i;
scanf("%d", &n);
maxi[n] = 0;
for (int(i) = (0); (i) < (n); ++(i)) {
scanf("%d", &ar[i]);
}
for (int(i) = (n - 1); (i) >= (0); --(i)) {
maxi[i] = max(maxi[i + 1], ar[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.PrintWriter;
import java.util.Scanner;
public class luxuryhooses {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
int n = scan.nextInt();
int[] houses = new int[n];
for(int i = 0; i < n; i++) houses[i] = scan.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;
multiset<int> ss;
int a[100007];
int main() {
int n;
while (cin >> n) {
ss.clear();
for (int i = 1; i <= n; i++) {
cin >> a[i];
ss.insert(a[i]);
}
int t, v;
for (int i = 1; i <= n; i++) {
if (i == n) {
if (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 | n = int(input())
h = [int(i) for i in input().split()]
l = []
l_2 = [0] * n
l_2[n-1] = h[n-1]
for i in range (n-2, 0, -1):
l_2[i] = max( l_2[i+1], h[i])
l_2 = l_2 + [0]
l = [0] * n
for i in range (n):
m = l_2[i+1]
g = max ( m - h[i] +1, 0)
l[i] = g
for i in l:
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 | 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.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.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 | #include <bits/stdc++.h>
using namespace std;
const int MAX = 100005;
int main() {
int n, A[MAX], M[MAX];
while (scanf("%d", &n) == 1) {
for (int i = 0; i < n; i++) scanf("%d", &A[i]);
M[n - 1] = -1;
for (int i = n - 2; i > -1; i--) M[i] = max(M[i + 1], A[i + 1]);
for (int i = 0; i < n; i++)
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;
using namespace std;
template <class T1, class T2>
using p = pair<T1, T2>;
template <class T1, class T2>
using m = vector<pair<T1, T2>>;
template <class T>
using vv = vector<vector<T>>;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &t) {
os << "{";
... | 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()))
cur = arr[n - 1]
dp= [0] * (n)
for i in range(n - 2 , -1 , - 1):
if arr[i] > cur :
dp[i] = 0
cur = arr[i]
else:
dp[i] = cur - arr[i] + 1
print(*dp) | 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.math.*;
import java.util.*;
public class LuxioriousHouses {
public static void main(String[] args) {
Scanner sc = new Scanner();
int n=sc.nextInt();
int[] arr=new int[n];
long[] b=new long[n];
arr=sc.nextIntArray(n);
long max=arr[n-1];
b[n-1]=0;
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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, a[200000], maxs[200000];
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a[i];
}
maxs[n - 1] = a[n - 1];
for (int i = n - 2; i >= 0; i--) {
maxs[i] = max(maxs[i + 1], a[i]);
}
for (int i = 0; i < n - 1; i++) {
cout << max(maxs[... | 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.*;
import java.io.*;
public class Main {
public static void main(String [] args)throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
int array[] = new int[n];
StringTokenizer st = new StringTokenizer(br.readLine());
fo... | 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, max = 0;
cin >> n;
vector<long long> v(n), out(n);
for (int i = 0; i < n; i++) {
cin >> v[i];
}
max = v[n - 1];
out[n - 1] = 0;
for (int i = n - 2; i >= 0; i--) {
if (v[i] <= max) {
out[i] = (max - v[i] + 1);
} 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 | n = int(input())
arr = [int(x) for x in input().split()]
m = [0] * (n+1)
ans = [0] * (n+1)
for i in range(n-2,-1,-1):
m[i] = max(m[i+1], arr[i+1])
ans[i] = max(m[i] - arr[i] + 1, 0)
print(*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 | n = int(input())
input = list(map(int, input().split()))
output = [0 for x in range(len(input))]
m = 0
for i in range(n - 1, -1, -1):
output[i] = max(0, m - input[i] + 1)
m = max(m, input[i])
print(' '.join(str(output[i]) for i in range(len(output)))) | 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[100005];
int Tree[400005];
void build(int node, int left, int right) {
if (right < left) return;
if (left == right) {
Tree[node] = A[left];
return;
}
build(2 * node, left, (left + right) / 2);
build(2 * node + 1, (left + right) / 2 + 1, right);
Tre... | 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 mod = 1E9 + 7;
const int intmax = 1E9 + 7;
using namespace std;
int aa[100005];
int bb[100005];
int main() {
ios::sync_with_stdio(0);
int test, a, b, c;
int n;
cin >> n;
for (int i = 0; i < n; i++) cin >> aa[i];
bb[n - 1] = aa[n - 1];
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 | a = input()
b = map(int,raw_input().split())
c = []
maxH = 0
for i in b[::-1]:
c.append(max(0,maxH-i+1))
maxH = max(i,maxH)
u = ""
for i in c[::-1]:
u += str(i)+" "
print u
| 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())
a = list(map(int, input().split(' ')[:n]))
b = [0 for i in range(n)]
m = 0
for i in range(n-1, -1, -1):
b[i] = max(0, m - a[i] + 1)
m = max(m, a[i])
print(*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 | n=int(input())
l=[int(x) for x in input().split()]
a=[0]*n
m=l[n-1]
for i in range(n-2,-1,-1):
if l[i]>m:
a[i]=0
m=l[i]
else:
a[i]=m-l[i]+1
for x in a:
print(x,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 | #include <bits/stdc++.h>
using namespace std;
int ara[100004], ara2[100004];
int main() {
int n, i, j, k, l, ans = 0;
cin >> n;
for (i = 0; i < n; i++) cin >> ara[i];
for (i = n - 1; i >= 0; i--) {
ara2[i] = max(ara2[i + 1], ara[i + 1]);
}
for (i = 0; i < n; i++) {
ans = max(ara2[i] - ara[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.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt(),i,max = 0,z;
int[] houses = new int[n];
int[] levelup = new int[n];
for(i=0;i<n;i++){
houses[i] = 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 | n = int(input())
l = list(map(int,input().split()))
max1 = l[-1]
stack = [0]
for i in range(2,n+1):
if l[-i] == max1:
stack.append(1)
elif l[-i] > max1:
max1 = l[-i]
stack.append(0)
else:
stack.append(abs(l[-i]-max1)+1)
for i in range(len(stack)):
print(stack.pop(),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 = tuple(map(int, str.split(input())))
hm = [0] * (n)
ch = 0
for i, x in enumerate(reversed(h)):
hm[i] = max(0, ch - x + 1)
ch = max(ch, x)
print(str.join(" ", map(str, reversed(hm))))
| 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())
hs = list(map(int, input().split()))
def solve(n, hs):
dp = [0] * n
zeros = [0] * n
maxh = 0
for i in range(n-1, -1, -1):
if hs[i] > maxh:
zeros[i] = 1
maxh = hs[i]
dp[i] = maxh
result = []
for i in range(n):
if zeros[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 | __author__ = 'mowayao'
n = int(raw_input())
a = map(int,raw_input().split())
M = -1
ans = []
for i in xrange(n-1,-1,-1):
if M < a[i]:
ans.append(0)
M = a[i]
elif M==a[i]:
ans.append(1)
else:
ans.append(M-a[i]+1)
for i in xrange(n-1,-1,-1):
print ans[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 isP(long int hj) {
long int op;
for (op = 2; op <= sqrt(hj); op++) {
if (hj % op == 0) return 0;
}
return 1;
}
void swap(long long int *p, long long int *q) {
long long int tmp = *p;
*p = *q;
*q = tmp;
}
int mind(long long int p) {
int mindd = 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()
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(str(m_h - h + 1))
print(" ".join(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 | /**
* Created by lq on 15/9/30.
*/
import java.util.*;
import java.io.*;
public class b_Luxurious_Houses {
public static void main(String[] args){
// FileInputStream fis = null;
// try {
// fis = new FileInputStream("/Users/lq/leetcode/codeforces/virtual1/1.txt");
// }catch(Excep... | 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, arr[110000], b[110000];
int main() {
cin >> n;
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
for (int i = n - 2; i >= 0; i--) {
b[i] = arr[i + 1] - arr[i] + 1;
arr[i] = max(arr[i + 1], arr[i]);
if (b[i] < 0) b[i] = 0;
}
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 | #!/usr/bin/env python
I = raw_input
def ia(): return map(int, I().split())
def na(n): return [ia() for _ in range(n)]
I()
arr = ia()
res = [0]*len(arr)
most = 0
for i in range(len(arr)-1, -1, -1):
if arr[i] > most:
res[i] = 0
most = arr[i]
else:
res[i] = most-arr[i] +1
print " ".join... | 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>
int main() {
int a;
scanf("%d", &a);
long A[a], B[a];
for (int i = 0; i != a; ++i) {
scanf("%ld", &A[i]);
}
B[a - 1] = 0;
long mx = A[a - 1];
for (int i = a - 2; i >= 0; --i) {
if (mx < A[i])
B[i] = 0;
else
B[i] = mx + 1 - A[i];
mx = std::max(A[i], 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 | #include <bits/stdc++.h>
using namespace std;
int N, fl[111111], ans[111111], mfl;
int main() {
scanf("%d", &N);
for (int i = 0; i < N; i++) {
scanf("%d", &fl[i]);
}
for (int i = N - 1; i >= 0; i--) {
ans[i] = max(0, mfl + 1 - fl[i]);
mfl = max(mfl, fl[i]);
}
for (int i = 0; i < N; i++) {
pr... | 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.*;
public class fence
{
//static {System.out.println("hello"); }
public static void main(String args[])
{ Scanner scn=new Scanner(System.in);
int n=scn.nextInt();
if (n==1 ){ System.out.println(0); System.exit(0);}
int a[]=new int[n]; int max[]=new int[n]; int max2[]=new 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 | import java.io.BufferedReader;
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.lang.reflect.Array;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arra... | 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;
inline long long gcd(long long a, long long b) {
return b == 0 ? a : gcd(b, a % b);
}
inline long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
const long long mod = (long long)(1e9 + 7);
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL... | 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;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
signed main() {
cin.tie(0);
cout.tie(0);
ios_base::sync_with_stdio(0);
cout.precision(3);
long long n;
cin >> n;
vector<long long> a(n);
for (long long i = 0; i < n; i++) cin >> 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 | # ===================================
# (c) MidAndFeed aka ASilentVoice
# ===================================
# import math
# import collections
# import string
# ===================================
n = int(input())
q = [int(x) for x in input().split()]
vmax = -1
for i in range(n-1, -1, -1):
t = q[i]
if (q[i] > vmax... | 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;
long long a[n], b[n];
for (int i = 0; i < n; i++) cin >> a[i];
long long mx = a[n - 1];
b[n - 1] = 0;
for (int i = n - 2; i >= 0; i--) {
if (a[i] <= mx)
b[i] = mx + 1 - a[i];
else
b[i] = 0;
mx = max(a[i], 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 | #include <bits/stdc++.h>
using namespace std;
int n, h[100001];
priority_queue<pair<int, int> > pq;
void input() {
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &h[i]);
pq.push(pair<int, int>(h[i], -i));
}
}
void solve() {
input();
for (int i = 0; i < n; i++) {
while (pq.size() && -pq... | 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 luxurious_houses():
houses = int(raw_input().strip())
height_of_houses = list(map(int, raw_input().strip().split()))
max_height_array = [0] * houses
max_height_array[houses - 1] = height_of_houses[houses - 1]
for i in range(houses - 2, -1, -1):
max_height_array[i] = max(max_height_array[... | 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 |
k = int(input())
houses = list((map(int, input().split())))
maxim = 0
answer = []
for i in range(len(houses)-1, -1, -1):
currentVal = houses[i]
if maxim < currentVal:
answer.append(0)
maxim = currentVal
else:
answer.append(maxim-currentVal+1)
print(" ".join(reversed([st... | 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 i, j, k, n, t = 1;
while (t--) {
cin >> n;
vector<long> v(n), b(n), ans(n, 0);
for (i = 0; i < n; i++) {
cin >> v[i];
b[i] = v[i];
}
for (i = n - 2; i >= 0; i--) {
if (b[i + 1] >= v[i]) ans[i] = 1 + b[i + 1] - v[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 |
#Solved by Fuad Ashraful BBabu
#soled date 12 july 2019
#verdict : AC
N=int(input())
tini=list(map(int,input().split()))
max_height=0
diff=N*[0]
for i in range(N).__reversed__():
if tini[i]>max_height:
max_height=tini[i]
diff[i]=0
else:
diff[i]=max_height-tini[i]+1
print(*diff)
| 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()))
m=max(h)
l=h.index(m)
i=n-2
a=[0]
maxh=h[n-1]
while(i>=0):
if(maxh==h[i]):
a.append(1)
else:
maxh = max(h[i], maxh)
if(maxh==h[i]):
a.append(0)
else:
a.append(maxh-h[i]+1)
i-=1
a.reverse()
for i 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;
void solve() {
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; i++) cin >> a[i];
vector<int> res(n, 0);
int mx = a[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (a[i] > mx) {
mx = a[i];
res[i] = 0;
} else {
int diff = mx - 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;
int main() {
int n;
cin >> n;
int h[n];
for (int i = 0; i < n; i++) cin >> h[i];
int max;
int a[n];
a[n - 1] = 0;
max = h[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (h[i] > max) {
a[i] = 0;
max = h[i];
} else {
a[i] = max - 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 | #include <bits/stdc++.h>
using namespace std;
const int E9 = 1e9;
const int E8 = 1e8;
const int E7 = 1e7;
const int E6 = 1e6;
const int E5 = 1e5;
const int E4 = 1e4;
const int E3 = 1e3;
const int N = 5e5 + 7;
const int INF = 1e9;
const long long inf = 1e18;
int a[N], mx[N];
bool us[N];
int main() {
int n, k = 0;
sc... | 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() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int n;
cin >> n;
vector<long long> v;
int num;
for (int i = 0; i < n; i++) {
cin >> num;
v.push_back(num);
}
int mx = 0;
vector<int> ans;
for (int i = v.size() - 1; 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 | import java.util.*;
public class L {
static Scanner sc = new Scanner(System.in);
public static void main(String[] args) {
int n = sc.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = sc.nextInt();
int tallest = -1;
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 | n = input()
h= map(int,raw_input().split())
maxh=0
ans=[]
for i in xrange(n-1,-1,-1):
if h[i]>maxh :
ans.append('0')
maxh = h[i]
else :
ans.append(str((maxh-h[i])+1))
print " ".join(ans[-1::-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>
int main() {
int n, i, j, k, m, l;
scanf("%d", &n);
int a[n], b[n];
for (i = 0; i < n; i++) {
scanf("%d", &a[i]);
}
m = a[n - 1];
b[n - 1] = 0;
for (i = n - 2; i >= 0; i--) {
if (a[i] > m) {
b[i] = 0;
m = a[i];
} else {
b[i] = m - 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 | if __name__ == '__main__':
n = str(input())
line = [int(it) for it in str(input()).split()]
line.reverse()
res = list()
mark = 0
for it in line:
res.append(str(max(1 + mark - it, 0)))
mark = max(mark, it)
res.reverse()
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.io.*;
import java.math.*;
import java.util.*;
import java.util.stream.*;
@SuppressWarnings("unchecked")
public class P581B {
public void run() throws Exception {
int n = nextInt();
int [] h = readInt(n);
int [] a = new int [n];
int mh = 0;
for (int i = n - 1; i >= 0; 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 | #include <bits/stdc++.h>
using namespace std;
long long a[100010], b[100010];
int main() {
long long n, m;
cin >> n;
m = -2147483647;
for (int i = 0; i < n; i++) cin >> a[i];
for (int i = n - 1; i >= 0; i--) {
if (a[i] <= m)
b[i] = m - a[i] + 1;
else
b[i] = 0;
m = max(m, a[i]);
}
f... | 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;
/**
* Created by yuu on 26/3/17.
*/
public class Problem581B {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] heights = new int[n];
int[] maxHeights = new int[n];
int[] results = 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 | n=int(input())
l=list(map(int,input().split()))
k=l[n-1]
y=[0]
for i in range(1,n):
if k<l[n-i-1]:
k=l[n-i-1]
y.append(0)
else:
y.append(k+1-l[n-i-1])
for i in range(n):
print(y[n-i-1],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 = [int(i) for i in input().split()]
l = []
l_2 = []
for i in range(n):
l_2 += [0]
l_2[n-1] = h[n-1]
for i in range (n-2, 0, -1):
l_2[i] = max( l_2[i+1], h[i])
l_2 = l_2 + [0]
for i in range (n):
m = l_2[i+1]
g = max ( m - h[i] +1, 0)
l += [g]
for i in l:
print(i, en... | 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.*;
import java.io.*;
public class B581
{
public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
long n = in.nextLong();
long[] h = new long[(int)n];
long temp=0;
for(int i=0;i<n;i++)
h[i]=in.nextLong();
for(int i=(int)n-1;i>=0;i--)
{
if(h[i]>te... | 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], b[n];
for (int i = 0; i < n; i++) cin >> a[i];
int mx = a[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (a[i] <= mx)
b[i] = mx + 1 - a[i];
else
mx = a[i], b[i] = 0;
}
b[n - 1] = 0;
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 | n = int(input())
a = [int(x) for x in input().split()][::-1]
ans = []
m = 0
for i in range(len(a)):
if i == 0:
ans.append(0)
m=a[i]
else:
if a[i] <= m:
ans.append(m-a[i]+1)
else:
ans.append(0)
m = a[i]
ans = ans[::-1]
for i in ans:
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 | import java.util.*;
public class lux{
public static void main(String[] args) {
Scanner s=new Scanner(System.in);
int n=s.nextInt();
int a[]=new int[n];
for(int i=0;i<n;i++){
a[i]=s.nextInt();
}
int max=0;
for(int i=n-1;i>=0;i--){
if(a[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;
long long int arr[100010];
long long int ans[100010];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
ans[n - 1] = 0;
long long int maxi = arr[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (arr[i] > maxi) {
ans[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.math.BigInteger;
import java.util.*;
import static java.lang.Math.*;
import static java.util.Arrays.fill;
public class Main {
public static void main(String[] args) throws IOException {
run();
end();
}
static void run() throws IOException {
int n = 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 | n = int(input())
l = list(map(int, input().split()))
m = l[-1]
l[-1] = 0
for i in range(n-2, -1, -1):
if l[i] == m:
l[i] = 1
elif l[i] > m:
m = l[i]
l[i] = 0
else:
l[i] = m - l[i] + 1
print(" ".join(map(str, l))) | 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 n;
vector<long long> a, b;
long long q, max_et;
int main() {
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%I64d", &q);
a.push_back(q);
}
b.push_back(0);
max_et = a[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (a[i] > max_et) {
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;
const int N = 2e5 + 10;
long long a[N], b[N];
int main() {
int n, d;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a[i];
b[i] = 0;
}
long long maxx = a[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (a[i] <= maxx) {
b[i] = maxx + 1 - 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;
int main() {
int n;
cin >> n;
int arr[n], suf[n];
for (int i = 0; i < n; i++) cin >> arr[i];
suf[n - 1] = arr[n - 1];
for (int i = n - 2; i >= 0; i--) {
suf[i] = max(arr[i], suf[i + 1]);
}
for (int i = 0; i < n - 1; i++) {
if (suf[i + 1] >= 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 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
;
long long int n, t;
cin >> n;
long long int a[n], b[n];
b[n - 1] = 0;
for (long long int i = 0; i < n; i++) cin >> a[i];
t = a[n - 1];
for (long long int i = n - 2; 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 | buildingsNum = int(raw_input())
firstLine = raw_input().split(" ")
buildings = []
for i in range(0, buildingsNum):
buildings.append(firstLine[i])
addedFloors = []
addedFloors.append(0)
for i in range (len(buildings) - 2, -1, -1):
if addedFloors[len(addedFloors) - 1] == 0:
a = int(buildings[i + 1]) + 1 - int(build... | 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())
a=map(int,raw_input().split())
i, mh = n-1, a[n-1]
while i>=0:
if mh<a[i]:
mh=a[i]
a[i]=0
else:
a[i]=mh-a[i] + 1
i-=1
a[n-1]=0
print(" ".join(map(str,a)))
| 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 | # Description of the problem can be found at http://codeforces.com/problemset/problem/581/B
n = int(input())
l_n = list(map(int, input().split()))
m = 0
l_a = [0 for i in range(n)]
for i in range(n):
l_a[n - i - 1] = max(0, m + 1 - l_n[n - i - 1])
m = max(m, l_n[n - i - 1])
print(" ".join(str(x) for x in l_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 the name of Allah the Most Merciful.
* Author
* Md. Toufiqul Islam
* Dept. Of CSE
* Ahsanullah University Of Science And Technology
*/
//package CodeForces;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;
public cla... | 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 a[N];
int i, j, n;
void input() {
cin >> n;
for (i = 0; i < n; ++i) {
cin >> a[i];
}
}
int ans;
void output() { cout << ans << '\n'; }
int main() {
std::ios_base::sync_with_stdio(false);
std::cin.tie(nullptr);
input();
int b[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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int a, b[100005], i, j;
cin >> a;
for (i = 1; i <= a; i++) {
cin >> b[i];
}
vector<int> v;
int maxn = 0;
for (i = a; i >= 1; i--) {
if (maxn < b[i]) {
v.push_back(0);
} else if (maxn > b[i]) {
v.push_back(maxn - b[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 | n=input()
a=map(int,raw_input().split())
maxi=0
l=[]
for i in xrange(n-1,0,-1):
if a[i]>maxi:
maxi=a[i]
l.append(maxi)
copy=l[::-1]
for i in xrange(n-1):
if a[i]<=copy[i]:
print copy[i]-a[i]+1,
else:
print 0,
print 0 | 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(raw_input())
a = map(int, raw_input().split())
b = [0]*(n)
cur_max = a[n - 1]
b[n - 1] = 0
for i in xrange(n - 2, -1, -1):
if a[i] <= cur_max:
b[i] = cur_max - a[i] + 1
cur_max = max(a[i], cur_max)
for i in xrange(len(b)):
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.ArrayList;
import java.util.Scanner;
public class CodeForces {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i]=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>
using namespace std;
const int N = 100001;
const double eps = 1e-11;
const int inf = (int)2e9;
const int mod = (int)1e9 + 7;
int a[N];
int s[N];
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", a + i);
}
for (int i = n - 1; i >= 0; i--) {
s[i] = 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 | #include <bits/stdc++.h>
using namespace std;
long long n, m = -1000, k, l, i, j, a[110000], b[110000];
int main() {
cin >> n;
for (i = 1; i <= n; i++) cin >> a[i];
for (i = n; i >= 1; i--) {
b[i] = max(m - a[i] + 1, 0ll);
m = max(m, a[i]);
}
for (i = 1; i <= n; i++) cout << b[i] << " ";
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 | import java.io.InputStreamReader;
import java.io.BufferedReader;
public class BCF518B {
public static void main(String[]args)throws Exception
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
int arr[]=new int[n];
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())
bF = list(map(int, input().split()))
max_ahead = [0]
for i in range(n - 2, -1, -1):
max_ahead.append(max(bF[i + 1], max_ahead[-1]))
max_ahead = max_ahead[::-1]
for i in range(0, n):
if bF[i] > max_ahead[i]:
print(0, end=" ")
else:
print(max_ahead[i] - bF[i] + 1, 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 | #include <bits/stdc++.h>
using namespace std;
long long mx, n, i, j, a, x, t[100000], v[100000];
int main() {
cin >> n;
for (i = 0; i < n; i++) cin >> t[i];
v[n - 1] = 0;
mx = t[n - 1];
for (i = n - 2; i >= 0; i--) {
if (t[i] > mx) {
v[i] = 0;
mx = t[i];
} else
v[i] = mx - t[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 | n = int(input())
a = list(map(int, input().split()))
b = []
mx = -(10**10)
for i in range(n-1, -1, -1):
b.append(max(0, mx+1-a[i]))
mx = max(mx, a[i])
b = b[::-1]
for i in b:
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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[100000], b[100000];
for (int i = 0; i < n; i++) cin >> a[i];
b[n - 1] = 0;
int mx = a[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (a[i] > mx) {
mx = a[i];
b[i] = 0;
} else
b[i] = 1 + mx - 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;
public class Main {
public static void main(String[] args) throws NumberFormatException, IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.r... | 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 | from itertools import imap
I = lambda: map(int, raw_input().split())
n, = I()
houses = I()[:n]
_max = 0
lux = [0] * n
for index in range(n-1, -1, -1):
if houses[index] <= _max:
lux[index] = _max + 1
else:
_max = houses[index]
lux[index] = _max
print ' '.join(imap(lambda x, y: str(x - y)... | 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 inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int maxn = 1e6 + 10;
const int maxv = 1e3 + 10;
const double eps = 1e-9;
inline int read() {
char c = getchar();
int f = 1;
while (!isdigit(c)) {
if (c == '-') f = -1;
c = getchar();
}
int x = 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 | # https://vjudge.net/contest/379796#problem/M
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 = [int(s) for s in input().split()]
b = [0]*n
i = n-1
sum = 0
while i > 0:
sum = max(sum, a[i])
b[i-1] = max(0, sum + 1 - a[i-1])
i = i - 1
for elem in b:
print(elem, 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 sys
import math
#import random
#sys.setrecursionlimit(1000000)
input = sys.stdin.readline
############ ---- USER DEFINED INPUT FUNCTIONS ---- ############
def inp():
return(int(input()))
def inara():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def... | 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 const N = 1e6 + 5;
int n, m, arr[N], b[N], cnt = 0, ma;
int main() {
cin >> n;
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
ma = arr[n];
for (int i = n - 1; i >= 0; i--) {
if (ma < arr[i]) {
ma = arr[i];
continue;
} else {
b[i] ... | CPP |
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