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 | def trackIndexes(x):
global index
index += 1
return (x,index)
size = input(); index = 0
houses = map(lambda x: trackIndexes(int(x)),raw_input().split())
sortedHouses = sorted(houses,reverse = True); cache ,i= 0,0;
while i < size:
if houses[i][1] <= sortedHouses[cache][1]:
if houses[i][0] < sorte... | 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())
w = list(map(int,input().split()))
out = []
out.append(0)
ma = w[n-1]
for i in range(n-2,-1,-1):
if(w[i] < ma):
out.append((ma+1) - w[i])
elif(w[i] > ma):
out.append(0)
ma = w[i]
elif(w[i] == ma):
out.append(1)
for k in range(n-1,-1,-1):
print(out[k],... | 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>
int n;
int main() {
scanf("%d", &n);
int arr[n];
for (int i = 0; i < n; i++) {
scanf("%d", &arr[i]);
}
int brr[n], max = arr[n - 1];
brr[n - 1] = 0;
for (int i = n - 2; i >= 0; i--) {
if (max < arr[i + 1]) max = arr[i + 1];
if (arr[i] <= max) {
brr[i] = max - arr... | 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[222222];
scanf("%d\n", &n);
pair<int, int> p;
vector<pair<int, int> > v, v1;
for (int i = 1; i <= n; i++) {
int x;
scanf("%d", &x);
p.first = x;
p.second = 0;
v.push_back(p);
}
reverse(v.begin(), v.end());
int maxi =... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Scanner;
public class DevelopingSkill
{
public static void main(String args[]) throws IOException
{
@SuppressWarnings("resource")
Scanner sc=new Scan... | 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()))
if n==1:
print (0)
exit()
m=[-1]*n
maxx=l[-1]
m[-2]=maxx
for i in range(n-3,-1,-1):
maxx=max(maxx,m[i+1],l[i+1])
m[i]=maxx
ans=[]
# print (m)
for i in range(n):
if l[i]>m[i]:
ans.append(0)
else:
ans.append((m[i]-l[i]+1))
print (*ans) | PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T sqr(T x) {
T x_ = (x);
return x_ * x_;
}
template <typename T>
inline T qbr(T x) {
T x_ = (x);
return x_ * x_ * x_;
}
template <typename T>
inline int sign(T x) {
T x_ = (x);
return ((x_ > T(0)) - (x_ < T(0)));
}
template <typename... | 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 | input()
n, result = [int(x) for x in input().split()], []
max = -1
for i in range(len(n) - 1, -1, -1):
if n[i] <= max:
result.append(max - n[i] + 1)
else:
max = n[i]
result.append(0)
for i in range(len(result) - 1, -1, -1):
print(result[i], end = ' ')
print() | PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n = int(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 | #include <bits/stdc++.h>
int arr[100000] = {0}, arr2[100000] = {0};
int main() {
int i, n, max;
while (~scanf("%d", &n)) {
memset(arr, 0, sizeof(arr));
memset(arr2, 0, sizeof(arr2));
for (i = 0; i < n; i++) scanf("%d", &arr[i]);
max = arr[n - 1];
for (i = n - 1; i >= 0; i--) {
if (i == 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 n, i, mx = 0, mil, l, limit;
scanf("%d", &n);
int ara[n], ara2[n];
for (i = 0; i < n; i++) scanf("%d", &ara[i]);
for (l = n - 1; l >= 0; l--) {
if (ara[l] > mx) {
mx = ara[l];
ara2[l] = 0;
} else {
mil = mx + 1 - ara[l];
... | 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 MAX = 1e5 + 100;
int a[MAX], ma, ans[MAX];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
for (int i = n - 1; i >= 0; i--)
ans[i] = max(0, ma - a[i] + 1), ma = max(ma, a[i]);
for (int i = 0; i < n; i++) cout << ans[i] << " ";
... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n = int(input())
h = list(map(int, input().split()))
mx = h[-1]
mh = []
for i in range(n-1, -1, -1):
if mx < h[i]:
mx = h[i]
mh.append(mx)
mh = list(reversed(mh))
for i in range(n):
res = 0
if i+1 < n and mh[i] == mh[i+1]:
res = mh[i]-h[i]+1
print(res, 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 = list(map(int, input().split()))
max_height = 0
diff = n * [0]
for i in range(n).__reversed__():
if h[i] > max_height:
max_height = h[i]
diff[i] = 0
else:
diff[i] = max_height - h[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())
a = list(map(int, input().split()))
add = [0]*n
m = a[n-1]
for i in range(n-2, -1, -1):
if a[i] > m:
add[i] = 0
m = a[i]
else:
add[i] = m - a[i] + 1
print(*add)
| 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 | number = raw_input()
houses = reversed(raw_input().split())
highest = 0
luxury = []
for house in houses:
if int(house) > highest:
highest = int(house)
luxury.append(0)
else:
luxury.append( highest - int(house) + 1 )
for aaa in reversed(luxury):
print aaa,
| 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.math.*;
public class Main1
{
static class Reader
{
private InputStream mIs;private byte[] buf = new byte[1024];private int curChar,numChars;public Reader() { this(System.in); }public Reader(InputStream is) { mIs = is;}
public int read() {if (n... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args) throws IOException {... | 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.awt.Point;
import java.io.*;
import java.lang.Integer;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;
import java.util.ArrayDeque;
import static java.lang.Math.*;
public class Main {
final boolean ONLINE_JUDGE = !new File("input.txt").exists();
BufferedReader 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 | #include <bits/stdc++.h>
using namespace std;
const int INF = 1000000000;
const int mod = 1000000007;
const double eps = 0.0000000001;
void solution() {
int n;
cin >> n;
vector<int> a(n + 1);
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
vector<int> dp(n + 1, 0);
dp[n] = a[n];
for (int i = n - 1; i ... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import itertools
n = int(input())
h = [int(x) for x in input().split()] + [0]
max_tail = list(itertools.accumulate(reversed(h), max))
max_tail.pop()
max_tail.reverse()
for i in range(n):
print(max(max_tail[i] - h[i] + 1, 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())
nums = [int(i) for i in input().split()]
floorstoadd = []
largesttoright = nums[-1]
for i in range(len(nums)-2, -1, -1):
if nums[i] <= largesttoright:
newone = str(largesttoright - nums[i] + 1)
floorstoadd.append(newone[::-1])
else:
largesttoright = nums[i]
floo... | 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 double EPS = 1e-9;
const double PI = acos(-1.0);
template <class T>
inline T _abs(T n) {
return ((n) < 0 ? -(n) : (n));
}
template <class T>
inline T _max(T a, T b) {
return (!((a) < (b)) ? (a) : (b));
}
template <class T>
inline T _min(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 | def luxhouse(n, h):
r = [0] * n
m = h[-1]
n -= 2
while n >= 0:
if m >= h[n]:
r[n] = m - h[n] + 1
else:
m = h[n]
n -= 1
return r
n = int(input())
h = raw_input().split(' ')
for i in range(len(h)):
h[i] = int(h[i])
for i in luxhouse(n, h):
p... | 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 | from collections import deque
n = input()
a = map(int, raw_input().split())
maxA = 0
res = deque()
for i in range(len(a)-1, -1, -1):
diff = maxA - a[i]
if diff < 0:
res.appendleft(str(0))
else:
res.appendleft(str(diff + 1))
if a[i] > maxA:
maxA = a[i]
print ' '.join(res)
| PYTHON |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import math
import itertools
import collections
def getdict(n):
d = {}
if type(n) is list:
for i in n:
if i in d:
d[i] += 1
else:
d[i] = 1
else:
for i in range(n):
t = ii()
if t in d:
d[t] += 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;
long long int mas[100001], ans[100001];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> mas[i];
}
long long int ma = 0;
for (int i = n - 1; i >= 0; i--) {
if (mas[i] > ma) {
ans[i] = 0;
ma = mas[i];
} else {
if (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.util.InputMismatchException;
import java.util.TreeSet;
//package PACKAGE_NAME;
/**
* Created by gaponec on 28.09.15.
*/
public class B322 {
public static void main(String[] args) {
MyScanner sc = new MyScanner();
int n = sc.nextInt();
int[] arr =... | 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()
houses = map(int, raw_input().split())
result = [0] * n
last = len(houses) - 2
maxHeight = houses[-1]
while last >= 0:
if houses[last] <= maxHeight:
result[last] = maxHeight - houses[last] + 1
maxHeight = max(maxHeight, houses[last])
last -= 1
print ' '.join(map(str, result)) | 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 x, c = 0, w;
cin >> x;
int a[x], b[x];
for (int i = 0; i < x; i++) cin >> a[i];
w = a[x - 1];
for (int i = x - 1; i >= 0; i--) {
if (a[i] == w) b[i] = 1;
if (a[i] > w) {
w = a[i];
b[i] = 0;
}
if (a[i] < w) b[i] = abs(... | 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()))
ans=[0]*n
mx=a[n-1]
for i in range(n-2,-1,-1):
ans[i]=max(0,mx-a[i]+1)
if a[i]>mx:
mx=a[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 | import java.io.*;
import java.util.*;
public class LuxuriousHouses {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == nul... | 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 = map(int, numbers.split())
result = []
current = 0
for i in numbers[::-1]:
if i <= current:
result.append(current - i + 1)
else:
result.append(0)
current = i
retu... | 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 = 100005;
int n, h[maxn], ans[maxn];
int main() {
ios::sync_with_stdio(false);
cin >> n;
for (int i = 1; i <= n; i++) cin >> h[i];
int _max = h[n];
for (int i = n - 1; i >= 1; i--) {
if (h[i] > _max)
ans[i] = 0;
else
ans[i] = _ma... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
int n, i, mx, h[100005], t[100005];
int main() {
scanf("%d", &n);
for (i = 1; i <= n; i++) scanf("%d", &h[i]);
for (i = n; i >= 1; i--)
if (h[i] > mx)
mx = h[i];
else
t[i] = mx - h[i] + 1;
for (i = 1; i <= n; i++) printf("%d ", t[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.util.*;
import java.io.*;
import java.math.*;
import java.text.*;
public class Main
{
static FastScanner in = new FastScanner(System.in);
static StringBuilder sb = new StringBuilder();
static DecimalFormat df = new DecimalFormat();
public static void main(String[] args)
{
df.setMaximumFractio... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 |
import java.io.BufferedOutputStream;
import java.io.IOException;
import java.io.OutputStream;
import java.util.*;
public class test {
public static void main (String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
OutputStream out = new BufferedOutputStream ( System.out );
int n = sc.nextInt(... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
long long a[100010], b[100010];
int main() {
long long n, x, i = 0;
scanf("%lld", &n);
long long j = n - 1;
while (i < n) {
scanf("%lld", &a[i]);
i++;
}
b[j] = 0;
j--;
x = 0;
long long total = x + a[n - 1];
for (i = n - 2; i >= 0; i--) {
if (... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.util.*;
public class BF
{
public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
long a[]=new long[n];
long r[]=new long[n];
for(int i=0;i<n;i++)
a[i]=sc.nextLong();
long m=0;
for(int i=n-1;i>=0;i--)
{
if(a[i]<=m)
r[i]=m-a[i]+1;
m=Math.max(m,a[i]);
}
for(int i=0;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() {
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
int arr[n];
for (int i = 0; i < n; ++i) {
cin >> arr[i];
}
int mx = -100;
vector<int> ans;
for (int i = n - 1; i > -1; --i) {
if (arr[i] < mx)
ans.emplace_back(mx - arr[... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual soluti... | 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.IOException;
import java.io.InputStreamReader;
public class Main {
public static void main(String[] args) throws NumberFormatException, IOException {
// TODO Auto-generated method stub
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int n... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import sys
n = int(input())
a = list(map(int, sys.stdin.readline().split()))
b = [0]*n
for i in range(n-2,-1,-1):
b[i] = max(a[i+1],b[i+1])
c = [max(b[i]+1-a[i],0) for i in range(n)]
c[-1] = 0
print(' '.join([str(i) for i in c]))
| 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];
stack<int> s;
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; ++i) scanf("%d", a + i);
int maxx = -1;
for (int i = n - 1; i >= 0; --i) {
if (maxx >= a[i])
s.push(maxx + 1 - a[i]);
else
s.push(0);
maxx = max(maxx, ... | 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()))
f = [0] * (n + 1)
r = [0] * (n + 1)
for i in range(n - 1, -1, -1):
f[i] = f[i + 1]
if f[i] >= a[i]:
r[i] = f[i] - a[i] + 1
f[i] = max(f[i], a[i])
for i in range(n):
print(r[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.*;
import java.util.*;
public class Main {
private final static InputReader ir = new InputReader(System.in);
private final static OutputWriter ow = new OutputWriter(System.out);
private final static int INF = Integer.MAX_VALUE;
private final static int NINF = Integer.MIN_VALUE;
priva... | 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 collections import deque
input()
A, max_h = deque(), 0
for h in reversed(list(map(int, input().split()))):
if h > max_h:
max_h = h
A.appendleft(0)
else:
A.appendleft(max_h - h + 1)
print(*A) | PYTHON3 |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int ara[100];
int main() {
int n;
cin >> n;
int ara[n];
int i;
for (i = 0; i < n; i++) {
cin >> ara[i];
}
int max = -1;
for (i = n - 1; i >= 0; i--) {
if (ara[i] > max) {
max = ara[i];
ara[i] = 0;
} else {
ara[i] = max - ara[i] ... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n=int(input())
h=list(map(int,input().split()))
h.reverse()
s=[]
m=0
for i in h:
if i>m:
m=i
a=0
elif m >=i:
a=(m-i)+1
s+=[str(a)]
s.reverse()
print(' '.join(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 | import java.util.*;
public class test {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] arr = new int[n];
int[] anss = new int[n];
int[] dh = new int[n];
for (int i = 0; i < n; i++)
arr[i] = sc.nextInt();
dh[n-1] = 0;
anss[n-1]=0;
String... | JAVA |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | raw_input()
h = map(int, raw_input().split(' '))
maxi = []
temp = h[len(h) - 1]
c_temp = 0
for i in xrange(len(h) - 1, -1, -1):
if temp != max(temp, h[i]):
c_temp = 0
temp = max(temp, h[i])
if temp == h[i]:
c_temp += 1
maxi.append((temp, c_temp))
# print temp, c_temp
# print maxi
for (i, j) in zip(xrange(len... | 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;
long long gcd(long long a, long long b) {
long long r;
while (b != 0) {
r = a % b;
a = b;
b = r;
}
return a;
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long n, A[100005], tmp, dem, Ma[100005];
int main() {
ios_base::sy... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.io.Writer;
impo... | 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 Ma {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
int[] ArrIn= new int[N];
int[] ArrOut= new int[N];
int i=0;
for (int k=0; k<N; k++) {
ArrIn[k] = 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 | import java.io.*;
import java.util.StringTokenizer;
/**
* Created by Alvin on 5/20/2016.
*/
public class Codeforces_round_322_div_2_LuxuriousHouses {
public static void main(String[] args) {
FScanner input = new FScanner();
out = new PrintWriter(new BufferedOutputStream(System.out), true);
... | 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];
for (int i = 0; i < n; i++) {
cin >> a[i];
}
int max;
max = a[n - 1];
int h[n];
h[n - 1] = 0;
for (int i = n - 2; i >= 0; i--) {
if (max < a[i]) {
h[i] = 0;
} else {
h[i] = max - a[i] + 1;
... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Stack;
public class domik {
public static void main(String[] args) throws IOException {
Stack stek=new Stack();
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
String s = 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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> nums;
for (int i = 0; i < n; i++) {
int nxt;
cin >> nxt;
nums.push_back(nxt);
}
int suffmax[100000];
suffmax[n - 1] = -999999;
for (int i = n - 2; i >= 0; i--) {
suffmax[i] = max(suffmax[i + 1], nums[... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class LuxuriousHouses {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int house;
cin >> house;
int *floor = (int *)malloc(sizeof(int) * house);
int *added = (int *)malloc(sizeof(int) * house);
fill(added, added + house, 0);
for (int i = 0; i < house; i++) cin >> floor[i];
int max = floor[house - 1];
for (int i = 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())
arr=list(map(int,input().split()))
temp=[None]*n
m=0
last=0
for i in range(n):
if last>m:
m=last
last=arr[n-1-i]
temp[n-1-i]=m
for i in range(len(arr)):
if arr[i]<temp[i]+1:
print(temp[i]+1-arr[i],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 | import java.io.*;
import java.lang.reflect.Array;
import java.net.Inet4Address;
import java.nio.ByteBuffer;
import java.util.*;
public class test {
int max = 0;
long t[];
boolean visited[];
ArrayList<Integer> queue = new ArrayList<>();
public static void main(String args[]) throws IOException {
... | 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.*;
public class B {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();long max=0;
int[] a = new int[n];
long[] b = new long[n];
for(int i=0;i<n;i++)
{
a[i]=sc.nextInt();
}
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 | input()
nums = [int(t) for t in input().split()]
max_ = nums[-1] - 1
for i in range(len(nums) - 1, -1, -1):
newVal = nums[i] if nums[i] > max_ else max_ + 1
max_ = max(max_, nums[i])
nums[i] = newVal - nums[i]
# print(nums)
for i, k in enumerate(nums):
print(k, end=' ' if i != (len(nums) - 1) else '\n... | 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()
buildings = input().split(' ')
m = 0
result = []
for building in buildings[::-1]:
b = int(building)
x = m - b
if x < 0:
result.append('0')
m = b
else:
result.append(str(x + 1))
print(' '.join(result[::-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 Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] a = new int[n];
for(int i=0; i<n; i++)
a[i] = sc.nextInt();
int max = a[n-1];
int[] ans = new int[n];
ans[n-1] = 0;
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 a[200000], b[200000], n;
int main() {
cin >> n;
for (int i = 1; i <= n; i++) cin >> a[i];
for (int i = n; i >= 1; i--) b[i] = max(a[i], b[i + 1]);
for (int i = 1; i <= n; i++)
if (b[i + 1] >= a[i])
cout << b[i + 1] - a[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=input()
a=map(int,raw_input().split())
h=0
for i in range(n-1,-1,-1):
x=max(0,h+1-a[i])
h=max(h,a[i])
a[i]=x
for x in a:
print x,
| PYTHON |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int n;
int h[100001];
int r[100001];
int main() {
scanf("%d", &n);
for (int i = 0; i < n; i++) scanf("%d", h + i);
int maxis = 0;
for (int i = n - 1; i >= 0; i--) {
int k = maxis - h[i] + 1;
if (k >= 0)
r[i] = k;
else
r[i] = 0;
maxis = ma... | 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;
bool flag;
void fast() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
string con(long long od) {
stringstream st;
st << od;
string pr;
st >> pr;
return pr;
}
int main() {
fast();
int n, mn = (INT_MIN);
cin >> n;
vector<int> v(n, 0);
vector<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()))
L = [0] * n
mx = 0
for i in range(n-1, -1, -1):
if a[i] > mx: L[i] = 0
else: L[i] = mx + 1 - a[i]
mx = max(mx, a[i])
print(*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 | n = int(input())
h = list(map(int, input().split()))
h.reverse()
ans = [0]
m = h[0]
for i in range(1, n):
if h[i] > m:
ans.append(0)
m = h[i]
else:
ans.append(m-h[i] + 1)
ans.reverse()
print(*ans, 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 | N = int(raw_input())
A = map(int , raw_input().split())
l = len(A)
B = [0]
max_to_right = A[-1]
for i in range(2,len(A)+1):
max_to_right = max(max_to_right, A[-i+1])
B.append(max(0, max_to_right-A[-i]+1))
B.reverse()
for i in B:
print i, | PYTHON |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n = int(input())
mat = list(map(int,input().split()))[::-1]
rmnd = 0
sat = []
for i in mat:
if rmnd >= i:
sat.append(rmnd - i + 1)
else:
sat.append(0)
rmnd = i
print(*sat[::-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 | _ = input()
a = [int(x) for x in input().split()]
mx = 0
ans = []
for x in reversed(a):
if x > mx:
ans.append(0)
mx = x
else:
ans.append(mx - x + 1)
ans = reversed(ans)
print(' '.join(str(x) for x in 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 |
def main():
raw_input()
L = [ int(i) for i in raw_input().split(" ") ]
res = []
maxi = 0
for i in range(len(L)-1,-1,-1):
res.append( str( max(0,maxi+1-L[i])) )
maxi = max(L[i],maxi)
res.reverse()
print " ".join(res)
main() | 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()
h = map(int, raw_input().split()) + [0]
goal = [0 for i in xrange(n+1)]
for i in xrange(n-1, -1, -1):
goal[i] = max(goal[i+1], h[i+1]+1)
print " ".join([str(max(0, goal[i] - h[i])) for i in xrange(n)])
| 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 | mod = 10 ** 9 + 7
ii = lambda : int(input())
si = lambda : input()
dgl = lambda : list(map(int, input()))
f = lambda : map(int, input().split())
il = lambda : list(map(int, input().split()))
ls = lambda : list(input())
n=ii()
l=il()
l1=[0]*n
mx=0
for i in range(n-1, -1, -1):
if l[i]<=mx:
l1[i]=mx-l[i]+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>
int main(int argc, char const *argv[]) {
int n, i, max;
scanf("%d", &n);
int a[n];
int r[n];
for (i = 0; i < n; i++) scanf("%d", &a[i]);
r[n - 1] = 0;
max = a[n - 1];
for (i = n - 2; i >= 0; i--) {
if (a[i] <= max) {
r[i] = max - a[i] + 1;
} else {
r[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 | """
Codeforces Round #322 (Div. 2)
Problem 581 B. Luxurious Houses
@author yamaton
@date 2015-10-06
"""
import itertools as it
import functools
import operator
import collections
import math
import sys
def solve(xs):
max_height = 0
result = []
for x in reversed(xs):
if x > max_height:
... | 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 mod = 1e9 + 7;
int main() {
int arr[123456];
int maxx[123456];
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &arr[i]);
}
maxx[n] = 0;
for (int i = n - 1; i >= 0; i--) {
maxx[i] = max(maxx[i + 1], arr[i]);
}
for (int ... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
int n;
cin >> n;
vector<long long int> vec(n), ans(n);
for (int i = 0; n > i; i++) {
cin >> vec[i];
}
long long int maxi = vec[n - 1], sifir = 0;
ans[n - 1] = 0;
for (int i = n - 2; i >= 0; i--) {
ans[i] = ... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, b, mx = 0;
cin >> n;
int a[n + 1];
vector<int> t;
for (int k = 1; k <= n; k++) cin >> a[k];
for (int k = n; k >= 1; k--) {
t.push_back(max(0, mx - a[k] + 1));
mx = max(a[k], mx);
}
for (int k = t.size() - 1; k >= 0; k--) {
cou... | 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, maxFloor = 0;
cin >> n;
int h[n];
int ans[n];
for (int i = 0; i < n; i++) {
cin >> h[i];
}
maxFloor = h[n - 1];
ans[n - 1] = 0;
for (int i = n - 2; i >= 0; i--) {
if (maxFloor < h[i]) {
maxFloor = h[i];
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 | //package CodeForces.Round322;
import java.util.Scanner;
/**
* Created by Ilya Sergeev on 28.09.2015.
*/
public class Belitn {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] mass = new int[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 | #include <bits/stdc++.h>
using namespace std;
long long a[200000], b[200000];
int main() {
ios_base::sync_with_stdio(0);
long long n, mx = 0, i, kol = 1;
cin >> n;
for (i = 1; i <= n; i++) cin >> a[i];
for (i = n; i >= 1; i--) {
b[kol] = max(0ll, mx - a[i] + 1);
mx = max(mx, a[i]);
kol++;
}
fo... | 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 t, a[100005], c[100005];
cin >> t;
for (int x = 0; x < t; x++) cin >> a[x];
c[t - 1] = 0;
int flag = a[t - 1];
for (int x = t - 2; x >= 0; x--) {
a[x] > flag ? c[x] = 0 : c[x] = flag - a[x] + 1;
flag = max(flag, a[x]);
}
for (int x =... | 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.StringTokenizer;
public class Luxhouses
{
public static void main(String arg[]) throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] stra;
StringBuilder sb=new StringBuilder();
int[] arr;
... | 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;
void maxe(long long int *array, long long int begin, long long int end,
long long int *max, long long int *index) {
long long int i, maxi = -9;
for (i = begin; i < end; i++) {
if (array[i] >= maxi) {
maxi = array[i];
*index = i;
}
}
*ma... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n, _max;
vector<long long> h, lux;
cin >> n;
h.resize(n);
lux.resize(n);
for (long long i = 0; i < n; i++) cin >> h[i];
lux[n - 1] = 0;
_max = n - 1;
for (long long i = n - 2; i >= 0; i--) {
if (h[i] <= h[_max])
lux[i] = 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 | #include <bits/stdc++.h>
using namespace std;
const double pi = 2 * acos(0.0);
const int OO = 0x3f3f3f3f;
const int N = 1e5 + 5;
int arr[N], n;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n;
for (int i = 0; i < n; i++) cin >> arr[i];
int mx = -1;
vector<int> v;
for (int i = n -... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | n=int(input())
l=list(map(int,input().split()))
maxh=0
a=[]
for i in range(n-1,-1,-1):
a.append((max(0,maxh+1-l[i])))
if l[i]>maxh:
maxh=l[i]
for i in range(n-1,-1,-1):
print(a[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())
a= [int(i) for i in input().split()]
pref= [0]*n
from collections import defaultdict
d = defaultdict(int)
pref[-1] = a[-1]
d[pref[-1]]=n-1
for i in range(n-2,-1,-1):
pref[i] = max(pref[i+1],a[i])
if d[pref[i]]==0:
d[pref[i]]=i
out = [0]*n
for i in range(n):
if a[i]==pref[i] and... | 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, i, arr[1 << 18];
cin >> n;
for (i = 0; i < n; i++) cin >> arr[i];
vector<int> v1;
int max1 = -1;
for (i = n - 1; i >= 0; i--) {
if (max1 == -1 || max1 < arr[i])
v1.push_back(0);
else
v1.push_back(max1 - arr[i] + 1);
ma... | 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, c = 0;
int zero = 0;
vector<int> v;
cin >> n;
int arr[n];
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
for (int i = n - 1; i >= 0; i--) {
if (arr[i] > c) {
v.push_back(zero);
c = arr[i];
} else
v.push_back(... | 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 t, n, i, j, ma;
cin >> n;
int arr[n], b[n];
for (i = 0; i < n; i++) {
cin >> arr[i];
}
ma = arr[n - 1];
b[n - 1] = 0;
for (i = n - 2; i >= 0; i--) {
if (arr[i] <= ma) {
b[i] = (ma - arr[i]) + 1;
} else {
b[i] = 0;
... | CPP |
581_B. Luxurious Houses | The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater than in all... | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n, i, j;
int max = INT_MIN;
cin >> n;
int a[n];
int b[n];
for (i = 0; i < n; i++) {
cin >> a[i];
}
for (j = n - 1; j >= 0; j--) {
if (a[j] <= max) {
b[j] = max + 1 - a[j];
... | 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 MX = 100005;
using namespace std;
int main() {
int n;
cin >> n;
long long int ar[n + 5];
long long int maxi[n + 5];
for (int i = 0; i <= n - 1; i++) cin >> ar[i];
maxi[n - 1] = ar[n - 1];
for (int i = n - 2; i >= 0; i--) {
maxi[i] = ((ar[i]) > (maxi[i + 1]) ? (ar[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.Scanner;
public class LuxuriousHouses {
static int K;
static int[] intarray;
static Scanner scan;
static int[] results;
static int max;
public static void main(String[] args) {
scan = new Scanner(System.in);
K = scan.nextInt();
intarray = new int[K];
results = new int[K];
results[K - ... | 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.*;
import java.lang.reflect.Array;
import java.util.*;
public class CodeForces {
public static void main(String[] args)throws IOException {
//MyScanner sc = new MyScanner(new FileReader("input.txt"));
//PrintWriter out = new PrintWriter("output.txt");
PrintWriter out = new Pr... | JAVA |
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