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