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;
int main() {
int n, m;
cin >> n;
int* a = new int[n];
int* b = new int[n];
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
if (n == 1)
cout << 0;
else {
m = a[n - 1];
b[n - 1] = 0;
for (int i = n - 2; i >= 0; --i) {
if (m < 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... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
public class B {
int n;
long h[];
void readInput() throws NumberFormatException, IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
//Scanner sc = new Scann... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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], f[n];
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
f[n - 1] = a[n - 1];
f[n] = 0;
for (int i = n - 2; i >= 0; i--) f[i] = max(f[i + 1], a[i]);
for (int i = 0; i < n; i++)
if (f[i] == a[i] && f[i + 1] != f[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()
m = map(int, raw_input().split())
max_so_far = 0
vals = [0] * n
for i in range(n - 1, -1, -1):
val = 0
if max_so_far >= m[i]:
val = max_so_far - m[i] + 1
vals[i] = val
max_so_far = max(max_so_far, m[i])
print ' '.join(str(v) for v in vals)
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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... | #include <bits/stdc++.h>
using namespace std;
template <class T1, class T2>
bool upmin(T1 &x, T2 v) {
if (x > v) {
x = v;
return true;
}
return false;
}
template <class T1, class T2>
bool upmax(T1 &x, T2 v) {
if (x < v) {
x = v;
return true;
}
return false;
}
void solve() {
int N;
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... | import java.util.Scanner;
public class Main {
public static void main(String[] args){
Scanner 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+... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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.StringTokenizer;
public class Main{
public static void main(String[] args) throws Exception {
ModScanner ms=new ModScanner();
int n=ms.nextInt();
long[] a=new long[n];
for(int i=0;i<n;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... | '''
Created on 28-Sep-2015
@author: Venkatesh
'''
def read_int_list():
return [int(x) for x in raw_input().split()]
def read_int():
return int(raw_input())
def print_next_max_num(houses):
last_max = houses[-1]
result = [0]
houses.pop()
for ele in houses[::-1]:
if ele <= last_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... | import java.util.*;
import java.lang.*;
import java.io.*;
/*
int n = scan.nextInt();
int[] arr = new int[n];
for(int i = 0; i < n; i++) arr[i] = scan.nextInt();
long n = scan.nextLong();
long[] arr = new long[n];
for(int = 0; i < n; i++) arr[i] = scan.nextLong();
*/
public class Main {
pu... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 jkl {
public static void main (String [] args){
Scanner s = new Scanner (System.in);
int size = s.nextInt();
int [] a = new int [size];
int [] b = new int [size];
for (int j = 0 ; j<size ; j++){
int k = s.nex... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 a[100100];
pair<int, int> p[100100];
int main() {
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", a + i);
p[i] = make_pair(a[i], i);
}
sort(p, p + n);
reverse(p, p + n);
int len = 0;
for (int i = 0; i < n - 1; i++) {
while (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... | #include <bits/stdc++.h>
using namespace std;
map<string, long long int> mapp;
long long int a[1000000], b[1000000];
int main() {
long long int n;
cin >> n;
for (long long int i = 0; i < n; i++) {
cin >> a[i];
}
long long int max = 0;
for (long long int i = n - 1; i >= 0; i--) {
if (a[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 a[100005], m[100005], n, t, i, temp;
scanf("%d", &n);
for (i = 0; i < n; i++) scanf("%d", &a[i]);
m[n - 1] = a[n - 1];
m[n] = 0;
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; 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())
arr = list(map(int, input().split()))
maxi=0
e=[]
for i in range(len(arr)):
if arr[len(arr)-1-i] > maxi:
maxi = arr[len(arr)-1-i]
flag=len(arr)-1-i
if len(arr)-1-i ==flag:
e.append((arr[flag]- arr[len(arr)-1-i]))
else:
e.append((arr[flag]- arr[len(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... | # In this template you are not required to write code in main
import sys
inf = float("inf")
#from collections import deque, Counter, OrderedDict,defaultdict
#from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
from math import ceil,floor,log,sqrt,factorial,pow,pi,gcd
#from bisect import 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... | /* / οΎοΎβ β β β β β β β β β β β γ
/ )\β β β β β β β β β β β β Y
(β β | ( Ν‘Β° ΝΚ Ν‘Β°οΌβ β(β γ
(β οΎβ Y βγ½-γ __οΌ
| _β ｑγ| γｑ |/
(β γΌ '_δΊΊ`γΌ οΎ
β |\ οΏ£ _δΊΊ'彑οΎ
β )\β β ｑβ β /
β β (\β #β /
β /β β β /α½£====================D-
/β β β /β \ \β β \
( (β )β β β β ) ).β )
(β β )β β β β β ( | /
|β /β β β β β β | /
[_] β β β β β [___] */
// Main Code at the Bottom
import ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 B {
/**
* @param args
*/
public static void main(String[] args) {
Scanner x=new Scanner(System.in);
int n=x.nextInt();
int arr[]=new int [n];
for(int i=0;i<n;i++){
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... | // CoDE deVELOPed uNDEr AcE proDUcTiONS (enJOY!!!!)
import java.util.*;
public class Ace
{
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
int n=sc.nextInt(),i,j,k,max=0;
int[] ar=new int[n];
int[] br=new int[n];
for(i=0;i<n;i++) ar[i]=sc.nextInt();
max=ar[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... | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n, m, k, l, x, z, i;
cin >> n;
long long a[n + 5];
map<long long, long long> mp;
for (i = 1; i <= n; i++) {
cin >> a[i];
}
long long b[n + 5];
x = 0;
for (i = n; i >= 1; i--) {
x = max(a[i], x);
b[i] = x;
mp[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())
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] = str(max(h_max[i] - h[i], 0))
res[0] = str(max(h_max[0] - h[0], 0))
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... | n=int(input())
s=list(map(int,input().split()))
s.reverse()
d=s[0]
a=[]
for i in range(1,n):
f=int(s[i])
if f <= d:
k=str((d-f)+1)
a.append(k)
else:
a.append(0)
d=f
a.insert(0,"0")
a.reverse()
for j in a:
print(j,end=" ")
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(input())
h = map(int, input().split()[::-1])
m_h = 0
res = []
for h in h:
if m_h < h:
res.append(0)
m_h = h
else:
res.append(m_h - h + 1)
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... | #include <bits/stdc++.h>
using namespace std;
int x[100009];
int 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 = ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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;
//Removed extra array
public class LuxuriousHouses {
public static void main(String[] args) throws IOException {
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(bf.readL... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = input()
h = map(int, raw_input().split())
res = []
mx = 0
while h:
c = h.pop()
if c > mx:
res.append(0)
mx = c
else:
res.append((mx - c) + 1)
res.reverse()
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;
vector<int> One(1000007);
vector<int> Two(1000007);
int main() {
int n;
int maximun = -1000007;
cin >> n;
for (int i = 0; i < n; i++) cin >> One[i];
for (int i = n - 1; i >= 0; i--) {
if (One[i] > maximun)
Two[i] = One[i];
else
Two[i] = maximun... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 = h[n - 1]
maxes = [0] * n
maxes[n - 1] = 0
res = ''
for i in range(n - 1, 0, -1):
if h[i - 1] > maxi:
maxi = h[i - 1]
maxes[i - 1] = 0
else:
maxes[i - 1] = maxi - h[i - 1] + 1
for i in range(n):
res += str(maxes[i]) + ' '
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n=int(input())
h=list(map(int,input().split()))
ans=[0]*n
ans[n-1]=0
maxh=h[n-1]
for i in range(n-2,-1,-1):
if h[i]>maxh:
ans[i]=0
else:
ans[i]=maxh-h[i]+1
maxh=max(maxh,h[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... | #from time import time
n = int(input())
x = list(map(int, input().split()))
#start_time = time()
abc = x[-1]
a = [0]
for i in range(2, n + 1):
if x[-i] <= abc:
a.append(abc - x[-i] + 1)
elif x[-i] > abc:
abc = x[-i]
a.append(0)
print(*a[::-1])
#print(time() - start_time)
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | l=lambda:map(int,raw_input().split())
n=input()
h=l()
a=[0]*n
maxi=h[n-1]
for i in range(n-2,-1,-1):
if h[i]>maxi:
a[i]=0
maxi=h[i]
else:
a[i]=maxi+1-h[i]
for v in a:
print 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... | n=int(input())
l=[int(x) for x in input().split()]
g=[]
h=0
for i in range(n-1, -1, -1):
x=max(0, h+1-l[i])
h=max(h, l[i])
g.append(x)
g=g[::-1]
print(*g)
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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];
int main() {
int n, i, m, s;
cin >> n;
for (i = 1; i <= n; i++) cin >> a[i];
m = a[n];
a[n] = 0;
for (i = n - 1; i > 0; i--) {
s = m + 1 - a[i];
m = max(m, a[i]);
a[i] = s < 0 ? 0 : s;
}
for (i = 1; i <= n; i++) printf("%d%c", 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... | 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... |
import java.io.BufferedOutputStream;
import java.io.IOException;
import java.io.OutputStream;
import java.util.*;
public class test {
public static void main (String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
OutputStream out = new BufferedOutputStream ( System.out );
int n = sc.nextInt(... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
int a[100005], b[100005];
int main() {
int n, max, l;
scanf("%d", &n);
int i;
for (i = 0; i < n; i++) {
scanf("%d", &a[i]);
}
max = 0;
for (i = n - 1; i >= 0; i--) {
l = 1;
if (a[i] > max) {
max = a[i];
l = 0;
}
b[i] = max - a[i] + l;
}
for (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 mod = 1e9 + 7;
const double pi = acos(-1);
int a[100010], d[100010], flag[100010];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
d[n - 1] = a[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (d[i + 1] >= a[i]) {
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... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, max = 0;
cin >> n;
int a[n], h[n];
for (int i = 0; i < n; i++) cin >> a[i];
h[n - 1] = 0;
for (int i = n - 2; i >= 0; i--) {
if (a[i + 1] > max) {
max = a[i + 1];
h[i] = max;
} else {
h[i] = max;
}
}
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... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class P581B {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | # -*- coding:utf-8 -*-
import sys
import math
def some_func():
"""
"""
n = input()
max =0
cache = []
n_list = map(int,sys.stdin.readline().split())
for v in n_list[::-1]:
if v>max:
cache.append(0)
max=v
elif v ==max:
cache.append(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 int MAXN = 100000 + 10;
int a[MAXN];
int n, maxx;
int ans[MAXN];
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
reverse(a + 1, a + 1 + n);
maxx = 0;
for (int i = 1; i <= n; ++i) {
if (a[i] > maxx)
ans[n - 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... | import java.io.BufferedReader;
import java.io.InputStreamReader;
public class Building {
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
String S[] = br.readL... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 = 100005;
int h[N], ans[N];
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%d", h + i);
}
for (int mx = 0, i = n; i >= 1; --i) {
if (mx >= h[i]) ans[i] = mx - h[i] + 1;
mx = max(mx, h[i]);
}
for (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... | #include <bits/stdc++.h>
using namespace std;
int n;
int h[100100], b[100100];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
if (0 & 1) freopen("input", "r", stdin);
if (0 & 2) freopen("output", "w", stdout);
cin >> n;
for (int i = 0; i < n; i++) cin >> h[i];
for (int i = n - 1; i ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int h[n];
for (int i = 0; i < n; i++) cin >> h[i];
int z = 0;
vector<int> ans;
for (int i = n - 1; i >= 0; i--) {
if (z == h[i]) {
ans.push_back(1);
continue;
}
z = max(z, h[i]);
if (z - h[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 n, arr[100000], r, arr1[100000], z;
bool boo[100000];
int main() {
cin >> n;
for (int i = 0; i < n; i++) {
cin >> r;
arr[i] = r;
}
for (int i = 0; i < n; i++) arr1[i] = arr[i];
reverse(arr1, arr1 + n);
z = arr1[0];
boo[0] = 1;
for (int 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... | n = int(input())
arr = list(map(int, input().split()))
maxArr = arr.copy()
maxArr.reverse()
maxArr[0] = maxArr[0] * -1
for i in range(1, len(arr)):
if maxArr[i] > abs(maxArr[i-1]):
maxArr[i] = maxArr[i] * -1
else:
maxArr[i] = abs(maxArr[i-1])
maxArr.reverse()
resultArr = []
for i in range(len... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, m[100000], mn[100000], mx;
cin >> n;
for (int i = 0; i < n; ++i) cin >> m[i];
reverse(m, m + n);
mn[0] = 0, mx = m[0];
for (int i = 1; i < n; i++)
if (m[i] > mx)
mn[i] = 0, mx = m[i];
else
mn[i] = mx - m[i] + 1;
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 n, i, m;
int b[100001];
int a[100001];
int main() {
cin >> n;
for (i = 0; i < n; i++) cin >> a[i];
m = a[n - 1];
for (i = n - 2; i >= 0; i--) {
if (m >= a[i]) b[i] = m - a[i] + 1;
if (m < a[i]) m = a[i];
}
for (i = 0; i < n; i++) cout << 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())
arr=list(map(int,input().split()))
dp=[0]*(n+1)
dp[-1]=-1;
for i in range(n-1,-1,-1):
maxx=max(dp[i+1],arr[i])
dp[i]=maxx
#print(*dp)
ans=[]
for i in range(n):
if arr[i]>dp[i+1]:ans.append(0)
else:ans.append(dp[i+1]-arr[i]+1)
print(*ans)
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> a(n), m(n);
for (int& x : a) cin >> x;
m[n - 1] = 0;
for (int i = n - 2; i >= 0; --i) m[i] = max(a[i + 1], m[i + 1]);
for (int i = 0; i < n; ++i)
cout << (a[i] > m[i] ? 0 : m[i] - a[i] + 1) << ' ';
}
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.util.Scanner;
public class LuxuriousHouses {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
int[] a = new int[n];
for(int i = 0; i < n; i++) a[i] = scan.nextInt();
int max = a[n-1];
int[] h = new int[n];
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... | n = int(input())
a = input().split()
for i in range(n):
a[i] = int(a[i])
m = -1
res = [0]*n
for i in range(n-1, -1, -1):
if a[i] > m:
res[i] = 0
else:
res[i] = m + 1 - a[i]
m = max(m, a[i])
for b in res:
print(b, end=" ") |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class LuxuriousHouses {
public static void main(String[] args) throws NumberFormatException, IOException {
BufferedReader scan = new BufferedReader(new InputStreamReader(System.in));
int houses = Integer.parseInt(... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 h[maxn], mx[maxn];
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; ++i) scanf("%d", h + i);
int cur = 0;
for (int i = n - 1; i >= 0; --i) {
mx[i] = cur;
cur = max(cur, h[i]);
}
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... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
while (scanf("%d", &n) == 1) {
vector<int> x(n);
for (int i = 0; i < n; i++) scanf("%d", &x[i]);
int maxi = INT_MIN;
vector<int> ans(n);
for (int i = n - 1; i >= 0; --i) {
if (maxi >= x[i]) {
ans[i] = maxi + 1 - 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 a[100005], n, i, max;
cin >> n;
for (i = 0; i < n; i++) {
cin >> a[i];
}
max = a[n - 1];
a[n - 1] = 0;
for (i = n - 2; i >= 0; i--) {
if (max > a[i]) {
a[i] = max - a[i] + 1;
} else if (a[i] > max) {
max = 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())
a = map(int, raw_input().split())
b = []
height = 0
for i in reversed(xrange(n)):
if a[i] <= height:
b.append(height - a[i] + 1)
else:
height = a[i]
b.append(0)
for x in reversed(b):
print x,
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
const int N = 100005;
long long a[N];
long long dp[N];
int main() {
long long n;
scanf("%lld", &n);
int max_key = 0;
int fg = 0;
for (int i = 0; i < n; ++i) {
scanf("%lld", &a[i]);
}
dp[n - 1] = 0;
for (int i = n - 2; i >= 0; --i) {
dp[i] = max(dp[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()))
m=n-1
b=[]
b.append(0)
for i in range(n-2,-1,-1):
if a[i]<=a[m]:
b.append(a[m]+1-a[i])
else:
m=i
b.append(0)
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... | import java.util.Scanner;
public class Codeforce
{
public static void main(String[] args)
{
int n;
Scanner s =new Scanner (System.in);
n=s.nextInt();
int arr[]=new int [n];
for(int i=0;i<n;i++)
arr[i]=s.nextInt();
s.close();
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;
template <class T>
T _abs(T n) {
return (n < 0 ? -n : n);
}
template <class T>
T _max(T a, T b) {
return (!(a < b) ? a : b);
}
template <class T>
T _min(T a, T b) {
return (a < b ? a : b);
}
template <class T>
T sq(T x) {
return x * x;
}
template <class T>
T gcd(T 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... |
import java.util.Scanner;
public class Luxurious_Houses {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int[] arr = new int[n];
int[] ans = new int[n];
for(int i = 0; i<n; i++)
{
arr[i] = in.nextInt();
}
int max = -1;
for(int i = n-1; 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, index = -1, max;
cin >> n;
int *a = new int[n];
for (int i = 0; i < n; i++) cin >> a[i];
for (int i = 0; i < n - 1; i++) {
max = 0;
for (int j = i + 1; j < n; j++)
if (max < a[j]) max = a[j];
if (max < a[i])
cout << "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():
num_of_house = int(raw_input())
house_heights = map(int, raw_input().split())
cur_max = 0
answer = [0 for i in range(num_of_house)]
for i in range(num_of_house-1, -1, -1):
if i < num_of_house - 1:
if cur_max + 1 > house_heights[i]:
answer[i] = cu... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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, hs[100001], ans[100001];
int main() {
cin >> n;
for (int i = 1; i <= n; ++i) cin >> hs[i];
int maxx = 0;
for (int i = n; i; --i) {
ans[i] = max(maxx - hs[i] + 1, 0);
maxx = max(maxx, hs[i]);
}
for (int i = 1; i < n; ++i) cout << ans[i] << ' ';
c... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, a, Max = 0;
vector<int> v;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a;
v.push_back(a);
}
Max = v.back();
v[v.size() - 1] = 0;
for (int i = v.size() - 2; i >= 0; i--) {
if (v[i] < Max)
v[i] = (Max + 1) - v[i];
e... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
const int N = 1e5;
int h[N], max[N];
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; ++i) {
scanf("%d", h + i);
}
max[n - 1] = h[n - 1];
for (int i = n - 2; i >= 0; --i) {
max[i] = h[i] > max[i + 1] ? h[i] : max[i + 1];
}
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... | #include <bits/stdc++.h>
using namespace std;
int a[400000];
void Up(int l, int r, int k, int u, int v) {
if (l == r) {
a[u] = v;
return;
}
int m = (l + r) / 2;
if (m >= k)
Up(l, m, k, u * 2, v);
else
Up(m + 1, r, k, u * 2 + 1, v);
a[u] = max(a[u * 2], a[u * 2 + 1]);
}
int main() {
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... | n=int(input())
l=list(map(int,input().split()))[::-1]
ma=0
ans=[]
for i in range(n):
if l[i]>ma: ans+=['0']; ma=l[i]
else: ans+=[str(ma-l[i]+1)]
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... | def solve(arr):
maxRight = arr[-1]
arr[-1] = 0
for i in range(len(arr) - 2, - 1, - 1):
if maxRight >= arr[i]:
arr[i] = maxRight + 1
else:
maxRight = arr[i]
return arr
n = input()
arr = map(int,raw_input().split())
ans = arr
ans = list(ans)
arr = solve(arr)
for 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=int(input())
l=list(map(int,input().split()))
d,c=[0]*a,l[-1]
for x in range(2,a+1):
d[-x]=max(0,c-l[-x]+1)
c=max(c,l[-x])
print(*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... | n=int(raw_input())
N=map(int,raw_input().split())
a=0
b=[]
for i in range(n):
b.append(max(0,a+1-N[n-i-1]))
a=max(a,N[n-i-1])
for i in range(n):
print b[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... | var n = parseInt(readline());
var input = readline().split(' ').map(function (item) {
return parseInt(item);
})
var result = [];
input.reduceRight(function (prev, curr) {
if(curr > prev){
result.push(0);
} else if (curr === prev) {
result.push(1);
} else {
result.push(prev - curr + 1);
}
return Math.max(pre... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 = 100000 + 10;
int a[maxn], o[maxn];
int main() {
int n;
while (scanf("%d", &n) != EOF) {
for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
int _max = 0;
for (int i = n - 1; i >= 0; --i)
if (a[i] > _max)
o[i] = 0, _max = 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 N = 100005;
int n;
int v[N], seg[4 * N + 1];
int a, b;
void init(int r, int i, int j) {
if (i == j)
seg[r] = v[i];
else {
init(2 * r, i, (i + j) / 2);
init(2 * r + 1, (i + j) / 2 + 1, j);
seg[r] = max(seg[2 * r], seg[2 * r + 1]);
}
}
int quer... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 maxsize = 30;
using namespace std;
int main() {
int n;
int A[100005];
cin >> n;
for (int i = 1; i <= n; i++) cin >> A[i];
int m = 0;
int ans[100005];
ans[n] = 0;
for (int i = n - 1; i >= 1; i--) {
m = max(m, A[i + 1]);
ans[i] = m + 1 - A[i];
if (ans[i] < 0) ans[i... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n=int(input())
a=map(int,input().split(" "))
rmax=0
res=[]
for x in reversed(list(a)):
t = rmax - x
res.append(0 if t<0 else t+1)
rmax=max(x, rmax)
for x in reversed(res):
print(x,end=" ")
print()
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(raw_input())
numbers = map(int, raw_input().split())
maxx = numbers[n - 1]
numbers[n - 1] = 0
for it in xrange(n - 2, -1, -1):
if numbers[it] > maxx:
maxx = numbers[it]
numbers[it] = 0
else:
numbers[it] = maxx + 1 - numbers[it]
print ' '.join(map(str, numbers)) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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[100009];
int main() {
int i, j, m, n, t, k;
cin >> n;
int ma = 0;
for (i = 0; i < n; i++) {
cin >> a[i];
}
int sum = 0;
int b[100009];
for (i = n - 1, j = 0; i >= 0; i--) {
if (a[i] > ma) {
ma = a[i];
b[j++] = 0;
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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int maxn = 100005;
const int maxm = 10005;
int h[maxn], ans[maxn];
void solve() {
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> h[i];
}
int maxh = 0;
for (int i = n; i >= 1; 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... | #include <bits/stdc++.h>
using namespace std;
int mx, n, ara[1000000];
int main() {
scanf("%d", &(n));
for (int i = (0); i < (n); i++) scanf("%d", &(ara[i]));
mx = ara[n - 1];
for (int i = (n - 2); i >= (0); i--) {
if (mx >= ara[i])
ara[i] = mx - ara[i] + 1;
else {
mx = ara[i];
ara[i] ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.Reader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.Writer;
impor... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 double pi = acos(-1.0);
const double eps = 1e-9;
int n, a[200100], m[200100];
int main() {
cin >> n;
for (int i = 1; i <= n; i++) cin >> a[i];
for (int i = n; i >= 1; i--) m[i] = max(m[i + 1], a[i]);
for (int i = 1; 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 maxi[100000 + 100];
bool mark[100000 + 100];
int main() {
int n;
cin >> n;
int h[n];
for (int i = 0; i < n; i++) cin >> h[i];
maxi[n - 1] = h[n - 1];
mark[n - 1] = true;
for (int i = n - 2; i >= 0; i--) {
if (h[i] > maxi[i + 1]) {
maxi[i] = 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... | n=int(raw_input())
a=map(int,raw_input().split(' '))
m=0
b=[0]*n
m=a[-1]
for i in xrange(len(a)-2,-1,-1):
if(a[i]<=m):
b[i]=m-a[i]+1
else:
m=a[i]
for i in b:
print i, |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
public class Gots {
public static long mod = 1000000000 + 7;
public static long gcd(long a, long b) {
return (b == 0) ? a : gcd(b, a % b);
}
public static int counter = 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... | n=int(input())
z=list(map(int,input().split()))
lst=[0]*n;m=z[-1];lst[-1]=0
for i in range(n-2,-1,-1):
if z[i] > m:
lst[i]=0
m = z[i]
else:
lst[i]= m-z[i]+1
print(*lst)
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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()))
maxi = -1
for i in range(n - 1, -1, -1):
t = a[i]
if a[i] > maxi:
a[i] = 0
else:
a[i] = maxi - a[i] + 1
maxi = max(maxi, t)
print(*a)
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n;
cin >> n;
vector<long long> v(n), s(n);
for (int i = 0; i < n; i++) cin >> v[i], s[i] = v[i];
reverse(s.begin(), s.end());
long long mx = 0;
for (int i = 0; i < n; i++) s[i] = max(s[i], mx), mx = s[i];
reverse(s.begin(), s.end());
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(input())
h = list(map(int, input().split()))
a = [0]*n
m = -1
for i in range(n-1,-1,-1):
a[i] = max(m+1-h[i],0)
m = max(m, h[i])
print(*a)
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
long long int a[1000000], b[1000000], n, k, l, x, m, s, p, i, j, t;
int main() {
map<long long, long long> v;
cin >> n;
for (i = 0; i < n; i++) {
cin >> a[i];
v[a[i]]++;
}
for (i = 0; i < n; i++) b[i] = a[i];
sort(b, b + n);
reverse(b, b + n);
m = 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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
public class a {
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... | #include <bits/stdc++.h>
using namespace std;
int main() {
vector<int> h, ans;
int n, t = 0;
ios::sync_with_stdio(false);
cin.tie(0);
cin >> n;
h.resize(n + 1);
for (int i = 1; i <= n; i++) cin >> h[i];
ans.push_back(0);
t = h[n];
for (int i = n - 1; i > 0; i--) {
if (h[i] > t)
ans.push_ba... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int a[100005];
int suff[100005];
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--) {
suff[i] = max(suff[i + 1], a[i]);
}
for (int i = 1; i <= n; i++) {
printf("%d ", max(0, suff[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... | num = int(raw_input())
houses = [int(x) for x in raw_input().split()]
maxAt = [0] * num
highest = 0
for i in range(num-1, -1, -1):
if houses[i] > highest:
highest = houses[i]
maxAt[i] = highest
for i in range(0, num-1):
highest = maxAt[i+1]
if houses[i] <= highest:
print (hi... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 TestOnly.Div2B_322.Code1;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class Main
{
FastScanner in;
PrintWriter out;
public void solve() throws IOException
{
int n = in.nextInt();
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())
li=map(int, raw_input().split())
ans=[li[n-1]]
temp=li[n-1]
for i in xrange(n-2, -1, -1):
if li[i]<=temp:
ans.append(temp+1)
else:
ans.append(li[i])
temp=li[i]
ans=ans[::-1]
for i in xrange(len(ans)):
print ans[i]-li[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;
int n;
bool flag;
int a[MaxN + 5], b[MaxN + 5];
int Max = 1 >> 30;
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
for (int i = n; i >= 1; i--) {
flag = 0;
if (a[i] > Max) {
Max = a[i];
flag = 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;
cin >> n;
vector<int> q(n);
for (int i = 0; i < n; i++) {
cin >> q[i];
}
int mx = 0;
vector<int> ans;
for (int i = n - 1; i >= 0; i--) {
ans.push_back(max(0, mx + 1 - q[i]));
mx = max(mx, q[i]);
}
reverse(ans.begin(), ans.... |
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