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... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class LuxoriousHouses {
public static void main(String[] args) throws IOException {
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int []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>
int main() {
int n, i, s[200000], t[200000], j, x = 1, y, max = 0;
scanf("%d", &n);
for (i = 1; i <= n; i++) {
scanf("%d", &s[i]);
}
max = s[n];
t[n] = 0;
for (i = n - 1; i >= 1; i--) {
if (s[i] > max) {
t[i] = 0;
max = s[i];
} else {
t[i] = max + 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 a[100009];
long long max2[100009];
bool isnew[100009];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a[i];
}
max2[n - 1] = a[n - 1];
isnew[n - 1] = true;
for (int i = n - 2; i >= 0; i--) {
if (a[i] > max2[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... | n=int(raw_input(''))
s=raw_input('')
a=s.split(" ")
arr=[]
for i in range(n):
arr.append(int(a[i]))
max=-1
b=[]
for i in range(n):
k=arr[n-i-1]
if k>max:
b.append(0)
max=k
elif k==max:
b.append(1)
else:
b.append(max-k+1)
b=b[::-1]
print ' '.join(map(str,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... | import java.util.HashSet;
import java.util.Scanner;
public class codeforcesB {
public static void main(String[] args){
Scanner in = new Scanner(System.in);
int n = in.nextInt();
long[] h = new long[n];
for (int i = 0; i<n; i++) h[i] = in.nextLong();
long max = h[n-1];
HashSet<Integer> e = new HashSet<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())
a=list(map(int,input().split()))
ans=[0]*n
curr_max=-1
for i in range(n-1,-1,-1):
if a[i]>curr_max:
ans[i]=0
curr_max=a[i]
else:
ans[i]=curr_max-a[i]+1
for i in range(n):
print(ans[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... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n + 1];
int b[n + 1];
for (int i = 0; i < n; i++) cin >> a[i];
int mx = INT_MIN;
for (int i = n - 2; i >= 0; i--) {
mx = max(mx, a[i + 1]);
b[i] = mx;
}
for (int i = 0; i < n - 1; i++) {
cout << max(b[i] - 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 int n, max1 = 0;
cin >> n;
long long int a[n];
long long int b[n];
for (int i = 0; i < n; i++) {
cin >> a[i];
}
b[n - 1] = 0;
max1 = a[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (a[i] > max1) {
b[i] = 0;
max1 =... |
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, max = 0, a[100005], b[100005];
scanf("%d", &n);
for (i = 0; i < n; i++) {
scanf("%d", &a[i]);
}
max = a[n - 1];
b[n - 1] = 0;
for (i = n - 2; i >= 0; i--) {
if (a[i] <= max) {
b[i] = max - a[i] + 1;
} else {
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 n;
vector<int> h, s;
int main() {
ios_base::sync_with_stdio(false);
cin >> n;
h.resize(n);
for (int i = 0; i < n; ++i) cin >> h[i];
s = h;
for (int i = (int)s.size() - 2; i >= 0; --i) s[i] = max(s[i], s[i + 1]);
for (int i = 0; i < (int)s.size(); ++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() {
long long int n, a;
cin >> n;
vector<long long int> vec(n), copie;
for (long long int i(0); i < n; i++) {
cin >> a;
vec.push_back(a);
}
reverse(begin(vec), end(vec));
copie.push_back(0);
int maxi = vec[0];
for (long long int i(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 = input()
buildings = input().split(' ')
m = 0
lenght = len(buildings) - 1
result = []
for i in range(lenght, -1, -1):
b = int(buildings[i])
x = m - b
if x < 0:
result.append('0')
m = b
else:
result.append(str(x + 1))
print(' '.join([result[i] for i in range(lenght, -1, -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... | n = input()
buildings = list(map(int, input().split(' ')))
m = 0
result = []
for b in buildings[::-1]:
x = m - b
if x < 0:
result.append(0)
m = b
else:
result.append(x + 1)
print(' '.join(map(str, 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... | #include <bits/stdc++.h>
using namespace std;
const double eps = 1e-9;
const int MaxN = int(2e5) + 256;
const int MOD = int(1e9) + 7;
template <typename T>
inline T gcd(T a, T b) {
return b ? gcd(b, a % b) : a;
}
inline bool Palindrome(const string& s) {
return equal(s.begin(), s.end(), s.rbegin());
}
int s[MaxN], ... |
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 Counter
n=int(input())
a=[int(x) for x in input().split()]
mx=-1
ans=[0]*n
for i in reversed(range(n)):
ans[i]=max(0,mx+1-a[i])
mx=max(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... | #include <bits/stdc++.h>
using namespace std;
int n, a[100001];
int m[100002];
int main() {
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
m[n] = 0;
for (int i = n - 1; i >= 0; i--) m[i] = max(a[i], m[i + 1]);
for (int i = 0; i < n; i++)
cout << max(0, m[i + 1] - a[i] + 1) << (i == n - 1 ? '\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(raw_input())
a = [int(x) for x in raw_input().split()]
a.append(0)
c = 0
b = []
for i in xrange(n):
j = a[n - i - 1]
c = max(c , a[n - i])
if j > c:
k = 0
else:
k = c + 1 - j
b.append(k)
for i in xrange(n):
print b[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())
house = list(map(int,input().split()))
house.reverse()
final = []
largestVal = 0
for i in house:
if i > largestVal:
final.append("0")
else:
final.append(f"{largestVal - i + 1}")
largestVal = max([largestVal,i])
final.reverse()
print(' '.join(final)) |
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(i) for i in input().split()]
b = n*[0]
maxh = l[n-1]
for i in range(n-2,-1,-1):
b[i] = max(0,maxh+1-l[i])
maxh = max(maxh,l[i])
print(*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>
using namespace std;
set<long long> st;
long long b[200007], a[200007], t, r, x1, l, y2, x2, y4, n, i, k, ma, p, j, mi,
p1, p2, p3, p4, l1, l2, l3, l4, c;
string s, s1, s2, s3;
char ch;
int main() {
cin >> n;
for (i = 1; i <= n; i++) cin >> a[i];
ma = a[n];
b[n] = 0;
for (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 n, a[100005], maxi, b[100005];
int main() {
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &a[i]);
}
maxi = a[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (a[i] > maxi) {
maxi = a[i];
b[i] = 0;
} else
b[i] = maxi + 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.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... | import java.io.*;
import java.math.*;
import java.util.*;
/**
* Created by Sai on 2015/9/28.
*/
public class B {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
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 inf = 1e9 + 2;
const double pi = acos(-1.0);
const int N = 300010;
int n, a[N], suf[N], res[N];
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
suf[n] = a[n];
for (int i = n - 1; i >= 1; i--) suf[i] = max(suf[i + 1], 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;
void solve() {
long long int n;
cin >> n;
vector<long long int> v(n);
vector<long long int> v1(n);
for (long long int i = 0; i < n; i++) cin >> v[i];
long long int max = v[n - 1];
long long int cnt = 0;
for (long long int i = n - 1; i > -1; i--) {
if (v[... |
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 codeforces581B
{
public static void main(String[] args)
{
Scanner s=new Scanner(System.in);
StringBuilder sb = new StringBuilder();
int n=s.nextInt(),i;
int[] a = new int[n];
int[] m = new int[n];
for(i=0;i<n;i++)
a[i]=s.nextInt();
m[n-1]=a[n-1];
for(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, a[100000], i, m[100000];
cin >> n;
for (i = 0; i < n; i++) cin >> a[i];
m[n - 1] = a[n - 1];
for (i = n - 2; i > 0; i--) {
if (a[i] > m[i + 1])
m[i] = a[i];
else
m[i] = m[i + 1];
}
for (i = 0; i < n - 1; 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;
int x[100009], a[100009];
int main() {
int n, mx;
cin >> n;
for (int i = 0; i < n; ++i) cin >> x[i];
a[n - 1] = 0;
mx = x[n - 1];
for (int i = n - 2; i >= 0; --i) {
if (mx == x[i]) {
a[i] = 1;
} else if (mx < x[i]) {
a[i] = 0;
mx = x[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() {
long long int n, ar[1000001], i, m;
vector<long long int> vec;
cin >> n;
for (i = 0; i < n; i++) {
cin >> ar[i];
}
m = ar[n - 1];
for (i = n - 2; i >= 0; i--) {
if (ar[i] == m + 1)
vec.push_back(0);
else if (ar[i] > m)
vec.... |
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> vec;
vector<int> dp;
vector<int> rein;
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) {
int x;
cin >> x;
vec.push_back(x);
}
int big = -2;
dp.resize(n);
rein.resize(n);
fill(rein.begin(), rein.end(), 0);
for (int j = 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(''))
arr=list(map(int,input().split()))
max=-1
b=[]
for i in range(0,n):
p=arr[n-1-i]
if p>max:
b.append(0)
max=p
else:
b.append(max-p+1)
for i in range(0,n):
print(b[n-i-1],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... | num=int(input())
a=list(map(int,input().split()))
b=[0]*num
maxx=a[-1]
for i in range(num - 2, -1, -1):
if a[i] <= maxx:
b[i] = maxx - a[i] + 1
else:
maxx = a[i]
print(*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>
using namespace std;
int a[101000], g[101000];
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
for (int i = n; i >= 1; i--) g[i] = max(g[i + 1], a[i]);
for (int i = 1; i <= n; i++) {
if (a[i] > g[i + 1])
printf("0 ");
else
prin... |
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()))
max_num=0
res=[0]*n
for i in range(n-1,-1,-1):
res[i]=max(0,max_num-a[i]+1)
max_num=max(max_num,a[i])
print(' '.join(map(str,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(input())
h = list(map(int, input().split())) + [0]
h_max = 0
res = []
for i in range(n, 0, -1):
h_max = max(h_max, h[i] + 1)
res.append(max(h_max - h[i - 1], 0))
print(" ".join(map(str, 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... | import java.util.*;
import java.io.*;
public class B581 {
public static void main(String[] args) throws IOException
{
input.init(System.in);
PrintWriter out = new PrintWriter(System.out);
int n = input.nextInt();
int[] a = new int[n];
for(int i = 0; i<n; i++) a[i] = input.nextInt();
int max = 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... | input()
a=list(map(int, input().split()))[::-1]
x=-1
b=[]
for v in a:
b+=[max(0,x-v+1)]
x=max(x,v)
print(*b[::-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... | from __future__ import print_function
t = int(input())
a = list(map(int , input().split()))
b=[]
max = a[t-1]
i=t-2
while i >= 0:
if a[i] <= max:
b.append(max-a[i]+1)
else:
max = a[i]
b.append(0)
i = i-1
b.insert( 0, "0")
i = t-1
while(i>=0):
print(b[i] , end = " ")
i = 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.util.ArrayList;
import java.util.Scanner;
public class cf581b {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n=in.nextInt();
ArrayList<Integer> houses = new ArrayList<>();
ArrayList<Integer> floors = new ArrayList<>();
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())
L = list(map(int,input().split()))
maxlist = [L[-1] for i in range(n)]
for i in range(n-2,-1,-1):
maxlist[i] = max(L[i],maxlist[i+1])
M = [None for i in range(n)]
for i in range(n):
if L[i] == maxlist[i]:
if i+1 < n and maxlist[i] == maxlist[i+1]:
M[i] = str(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... | #include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
long long n, i;
cin >> n;
long long a[n], revMax[n];
for (i = 0; i < n; i++) cin >> a[i];
revMax[n - 1] = a[n - 1];
for (i = n - 2; i >= 0; i--) {
revMax[i] = max(revMax[i + 1], a[i]);
}
for (i = 0; i < n - 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.*;
public class LuxuriousHouses {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int a[] = new int[n];
for(int j = 0; j < n; ++j){
a[j] = scanner.nextInt();
}
int maxi = a[n-1];
int b[] = new int[n];
b[n-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... | n = int(input())
a = list([int(x) for x in input().split()])
maxs = a[-1]
needed = [0] * n
for i in range(n - 2, -1, -1):
need = max(0, maxs - a[i] + 1)
needed[i] = need
maxs = max(maxs, a[i])
print(*needed)
|
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[100100], h[100100], mx[100100];
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
scanf("%d", a + i);
}
for (int i = n; i >= 1; i--) {
mx[i] = max(mx[i + 1], a[i]);
}
for (int i = 1; i <= n; i++) {
if (a[i] > mx[i + 1]) continue;
... |
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, s;
cin >> n;
vector<long long> vc;
for (long long i = 0; i < n; i++) {
cin >> s;
vc.push_back(s);
}
long long mx = -9999;
vector<long long> build;
mx = vc[n - 1];
build.push_back(0);
for (long long 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... | import sys
n = int(sys.stdin.next())
floors = map(int, sys.stdin.next().split(' '))
m = 0
s = []
for h in floors[::-1]:
s.append('%i' % ((m - h + 1) if (m - h + 1) > 0 else 0))
m = max(h, m)
print(' '.join(s[::-1]).strip())
|
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 luxurioushouses
{
public static void main(String args[])
{
Scanner in=new Scanner(System.in);
int n=in.nextInt();
int a[]=new int[n];
int b[]=new int[n];
int i,pos=0,s=0,max=0;
for(i=0;i<n;i++)
a[i]=in.nextInt();
/*d... |
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, j;
long long int a[100003], b[100003];
cin >> n;
for (i = 0; i < n; i++) {
cin >> a[i];
}
b[n - 1] = 0;
long long int max = a[n - 1];
for (j = n - 2; j >= 0; j--) {
if (a[j] > max) {
max = a[j];
b[j] = 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.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Stack;
import java.util.StringTokenizer;
public class LuxHouse {
public static void main(String[] args) {
Input... |
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.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;
public class cf {
public static void main(String[] args) throws Exception {
Scanner in=new Scanner(System.in);
int n=in.nextInt();
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.*;
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... | n = int(input())
#n, m = map(int, input().split())
#s = input()
c = list(map(int, input().split()))
a = [0] * n
l = c[-1]
for i in range(n - 2, -1, -1):
if c[i] > l:
l = c[i]
else:
a[i] = l - c[i] + 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... | def GetLuxHts(hts):
heights = []
for i in hts:
heights.append(int(i))
lux_hts_revrsd = ''
lux_hts = ''
max_height = heights[len(hts)-1]
for i in range(len(hts)-1,-1,-1):
if heights[i] <= max_height and i!=len(hts)-1:
diff = max_height - heights[i] + 1
#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... | import java.io.*;
import java.util.*;
public class b1 {
public static void main(String[] args) throws FileNotFoundException {
Scanner in = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
int n = in.nextInt();
int h[] = new int[n];
for (int i = 0; i < n; i++) {
h[i] = in.ne... |
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())
array = list(map(int, input().split()))
maxH = -1
luxury = list()
for i in range(n - 1, -1, -1):
luxury.append(max(maxH + 1 - array[i], 0))
maxH = max(maxH, array[i])
print(*luxury[::-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... | n = int(input())
a = list(map(int, input().split()))
b = [0] * n
ma = a[-1]
for i in range(n - 2, -1, -1):
if a[i] <= ma:
b[i] = ma - a[i] + 1
else:
ma = a[i]
print(*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>
using namespace std;
int n, i, a[100001], b[100001], k;
int main() {
cin >> n;
for (; i < n; i++) cin >> a[i];
while (i--) {
k = max(k, a[i]);
b[i] = k;
}
for (i = 0; i < n; i++)
cout << (b[i] == b[i + 1] ? b[i] - a[i] + 1 : 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... | 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
else:
hh[i] = h0 + 1 - h
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;
const int MAXN = 1e5 + 10;
int n, h[MAXN], mx[MAXN];
int main() {
cin >> n;
for (int i = 0; i < n; i++) cin >> h[i];
for (int i = n; i >= 0; i--) mx[i] = max(h[i + 1], mx[i + 1]);
for (int i = 0; i < n; i++) cout << max(h[i], mx[i] + 1) - h[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;
vector<long long> a(n), res(n);
for (int i = 0; i < n; i++) cin >> a[i];
long long mx = INT_MIN;
for (int i = n - 1; i >= 0; i--) {
res[i] = (mx >= a[i] ? mx - a[i] + 1 : 0);
mx = max(mx, 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... | import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args)throws IOException{
PrintWriter pw = new PrintWriter(System.out);
InputReader in = new InputReader(System.in);
int n = in.nextInt();
int[] a = 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;
int main() {
int n;
scanf("%d", &n);
int h[n], freq[n], mflr = 0;
for (int i = 0; i < n; i++) scanf("%d", &h[i]);
for (int i = n - 1; i >= 0; i--) {
if (h[i] > mflr) {
freq[i] = 0;
mflr = h[i];
} else
freq[i] = mflr + 1 - h[i];
}
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;
const int N = 1007, K = 1007, INF = LONG_MAX;
int n, a[150000], i, j, m[150000];
int main() {
cin >> n;
for (i = 0; i < n; i++) {
cin >> a[i];
}
int mx = 0;
for (i = n - 1; i >= 0; i--) {
mx = max(mx, a[i]);
m[i] = mx;
}
for (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>
int a[1000005];
int b[100005];
int main() {
int n;
scanf("%d", &n);
int i;
for (i = 1; i <= n; i++) scanf("%d", &a[i]);
int ma = 0;
for (i = n; i >= 1; i--) {
if (a[i] > ma) {
ma = a[i];
b[i] = 0;
} else {
b[i] = ma + 1 - a[i];
}
}
for (i = 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 java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math... |
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- Kishan_25
* BTech 2nd Year DAIICT
*/
import java.io.*;
import java.math.*;
import java.util.*;
import javax.print.attribute.SetOfIntegerSyntax;
public class code {
private static InputStream stream;
private static 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... | n = int(input())
h = list(map(int, input().split())) + [0]
h_max = [0] * n
res = [0] * n
for i in range(n - 1, 0, -1):
h_max[i - 1] = max(h_max[i], h[i] + 1)
res[i] = max(h_max[i] - h[i], 0)
res[0] = max(h_max[0] - h[0], 0)
print(" ".join(map(str, 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... | #include <bits/stdc++.h>
using namespace std;
long long a[100008], a2[100008], n;
int main() {
cin >> n;
for (long int i = 1; i <= n; i++) cin >> a[i];
long long maxi = a[n];
for (long i = n - 1; i >= 1; i--) {
if (a[i] > maxi) {
a2[i] = a[i] - 1;
maxi = a[i];
} else if (a[i] == 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.Stack;
import java.util.StringTokenizer;
public class B581Alt {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedR... |
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()))
res = [0]
k = 0
for i in range(-2, -len(lst)-1, -1):
if k < lst[i+1]:
k = lst[i+1]
if k >= lst[i]:
res.append(k-lst[i]+1)
else:
res.append(0)
print(*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... | N = int(input())
H = [int(_) for _ in input().split()]
suffix_max = [0] * (N + 1)
for i in range(N - 1, -1, -1):
suffix_max[i] = max(H[i], suffix_max[i + 1])
ans = [0] * N
for i in range(N):
ans[i] = max(suffix_max[i + 1] - H[i] + 1, 0)
print(' '.join(str(_) for _ 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... | n=int(input())
l=list(map(int,input().split()))
m=l[-1]
z=[0]*n
for i in range(n-2,-1,-1):
if l[i]>m:
m=l[i]
else:
z[i]=m-l[i]+1
print(*z) |
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().strip().split(' ')))
#n,m = map(int, input().strip().split(' '))
i=n-1
l2=[]
m=0
while(i>=0):
if lst[i]>m:
l2.append(0)
m=lst[i]
elif m==lst[i]:
l2.append(1)
else:
l2.append(m+1-lst[i])
i-=1
for j in range(n):
print(l2[n-j-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.*;
import java.util.*;
import java.util.stream.Collectors;
public class Solution {
public static void main(String[] args) throws IOException {
IO io = new IO();
int n = io.getInt() ;
List<Integer> ls = io.getIntegerArray(n) ;
List<Integer> res = new ArrayList<>() ;
... |
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
try:
while True:
n = int(input())
val = list(map(int, input().split()))
maxval = 0
for i in range(n-1, -1, -1):
if maxval >= val[i]:
val[i] = maxval - val[i] + 1
else:
maxval = val[i]
val[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;
long long flag, cnt, j, i, k, ans, maxy;
long long n;
int main() {
cin >> n;
vector<long long> v(n), temp(n), ans(n);
ans[n - 1] = 0;
maxy = v[n - 1];
for (i = 0; i < n; i++) {
cin >> v[i];
temp[i] = v[i];
}
for (i = n - 2; i >= 0; i--) {
if (maxy ... |
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[100010];
int m[100010];
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
long long k;
scanf("%lld", &k);
a[i] = k;
}
m[n - 1] = a[n - 1];
for (int i = n - 2; i >= 0; i--) {
m[i] = max(m[i + 1], a[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... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, i, k = 0;
cin >> n;
int h[n], a[n];
for (i = 0; i < n; i++) {
cin >> h[i];
}
h[n] = 0;
for (i = n - 1; i >= 0; i--) {
k = max(k, h[i + 1]);
a[i] = k;
}
for (i = 0; i < n; i++) {
cout << max(a[i] - h[i] + 1, 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() {
long int n;
cin >> n;
long int large = 0;
vector<int> v(n);
vector<int> answer(n);
for (int i = 0; i < n; i++) {
cin >> v[i];
}
for (int i = n - 1; i >= 0; i--) {
if (v[i] > large) {
answer[i] = 0;
large = v[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... | import java.util.*;
public class Main{
public static void main(String [] args)
{
Scanner scan=new Scanner(System.in);
int n=scan.nextInt();
int a[]=new int[n];
int i;
for(i=0;i<n;i++)
{
a[i]=scan.nextInt();
}
int b[]=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... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, i, a[100001], max[100001], maxx = 0;
cin >> n;
for (i = 0; i < n; i++) {
cin >> a[i];
max[i] = 0;
}
for (i = n - 1; i >= 0; i--) {
if (a[i] > maxx) {
maxx = a[i];
max[i] = maxx - 1;
continue;
}
max[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... | #include <bits/stdc++.h>
using namespace std;
int conversion(string p) {
int o;
o = atoi(p.c_str());
return o;
}
string toString(int h) {
stringstream ss;
ss << h;
return ss.str();
}
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long lcm(long long a, long long b) { re... |
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 + 10;
int a[maxn], maxx[maxn];
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &a[i]);
}
maxx[n - 1] = a[n - 1];
for (int i = n - 2; i >= 0; i--) {
maxx[i] = max(maxx[i + 1], a[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.BufferedReader;
import java.io.Closeable;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class LuxuriousHouses implements Closeable {
private InputRead... |
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... | #!/usr/bin/env python
from sys import stdin as cin
def main():
n = int(next(cin))
a = map(int, next(cin).split())
b = [0] * len(a)
for i in range(len(a)-1, 0, -1):
b[i-1] = max(a[i], b[i])
return [max(b[i]-a[i]+1, 0) for i in range(len(a))]
print ' '.join(map(str, 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... | #BeEf_Killer_______
# / _ _ \
# / (.) (.) \
# ( _________ )
# \`-V-|-V-'/
# \ | /
# \ ^ /
# \ \
# \ `-_
# `-_ -_
# -_ -_
# _- _-
# _- _-
# _- _-
#... |
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 = int(input())
b = list(map(int, input().split()))
mark = 0
answer = list()
for x in range(a):
if b[a - x - 1] > mark:
answer.append(0)
mark = b[a - x - 1]
else:
answer.append(mark - b[a - x - 1] + 1)
for x in range(a):
print(answer[a - x - 1], "", end="", flush=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 a[100001] = {0}, b[100001] = {0};
int i, k, l, m, n, o, p;
int main() {
scanf("%d\n", &n);
for (int i = 0; i < n; i++) {
scanf("%d ", &a[i]);
}
b[n] = a[n - 1] - 1;
for (int i = n - 1; i > -1; i--) {
b[i] = max(b[i + 1], a[i + 1]);
}
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 P581B{
static int[] run(int[] ar){
int[] max_ar = new int[ar.length];
for (int i=ar.length-1; i>-1; i--) {
if(i==ar.length-1)
max_ar[i] = 0;
else{
if(max_ar[i+1]>ar[i+1])
max_ar[i] = max_ar[i+1];
else
max_ar[i] = ar[i+1];
}
}
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 = input()
a = map(int, raw_input().split())
m = []
mxm = 0
for i in range(n-1, -1, -1):
mxm = max(a[i], mxm)
m.append(mxm)
for i in range(n-1):
if a[i]>m[n-2-i]:
print 0,
else:
print m[n-2-i]-a[i]+1,
print 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() {
int n;
cin >> n;
int ar[n], br[n], cr[n];
for (int i = 0; i < n; i++) {
cin >> ar[i];
br[i] = 0;
cr[i] = -1;
}
int max = ar[n - 1];
br[n - 1] = max;
cr[n - 1] = 0;
for (int i = n - 1; i >= 0; i--) {
if (max < ar[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... | #include <bits/stdc++.h>
int main() {
int s, i, y;
int f = 0;
scanf("%d", &s);
int n[s];
int r[s];
for (i = 0; i < s; i++) {
scanf("%d", &n[i]);
}
int g = n[s - 1];
for (i = s - 2; i >= 0; i--) {
if (n[i] > g) {
g = n[i];
r[i] = 0;
} else {
r[i] = (g - n[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.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... | import java.util.*;
public class Main
{
public static void main(String[] args)
{
Scanner s=new Scanner (System.in);
int n=s.nextInt();
int arr[] = new int[n];
int ans[] =new int[n];
for (int i=0;i<n;i++)
{
arr[i]=s.nextInt();
}
int mx = arr[n-1];
ans[n-1]=0;
for (int 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.Scanner;
import java.util.Stack;
public class q2
{
public static void main(String aegs[])
{
Scanner s=new Scanner(System.in);
int n=s.nextInt();
int a[]=new int[n];
int ng[]=new int[n];
Stack<Integer> st=new Stack<Integer>();
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... |
import java.util.*;
import java.io.*;
public class B {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int[] height = new int[n];
int[] max = new int[n];
for (int i = 0; i < n; i++) {
height[i] = in.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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.Input... |
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;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
bool isPrime(long long int n) {
for (long long int i = 2; i * i <= n; i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
bool isPowerOfTwo(long long... |
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... | import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;
public class CF581B {
public static void main(String[] args)throws Exception {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
for(String ln;(ln=in.readLine())!=n... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.