prompt string | response string |
|---|---|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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],... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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 =... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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() |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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])) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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 -... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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];
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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] << " ";
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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=' ')
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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)
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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,
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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 {... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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;... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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 ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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 = ' ')
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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) ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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)
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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 =... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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)) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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(... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | # 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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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;
}
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... |
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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... |
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(... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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 (... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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;... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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[... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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]))
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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, ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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=" ") |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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] ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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))
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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();
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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);
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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;
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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[... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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 ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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=' ') |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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 {
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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])) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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 ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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,
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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)
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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=' ') |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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, |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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])
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | _ = 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)) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... |
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() |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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)])
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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;
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | """
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:
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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 ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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] = ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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;
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | //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++) ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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 =... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... |
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;
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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[... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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 -... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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=" ") |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... |
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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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(... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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;
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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];
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #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])... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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 - ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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... |
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