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.util.*;
import java.io.*;
/**
* @author Alvex - GoldenReam1503@gmail.com
* Time: 1:41:47 PM
* Date: Nov 18, 2015
*/
public class _581B_Luxurious_Houses
{
private static class Output
{
private final PrintWriter printer;
public void printLine(Object...objects)
{
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 JavaApplication154 {
public static void main(String[] args) {
Scanner sc =new Scanner(System.in);
int a= sc.nextInt();
int[] b= new int[a];
for(int i=0;i<a;i++){
b[i]=sc.nextInt();
}
int max=b[a-1];
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | a = int(input())
b = list(map(int, input().split()))
b = b[::-1]
h = 0
g = []
for i in b:
if h < i:
h = i
g.append(0)
elif h == i:
g.append(1)
else:
g.append(h - i + 1)
g = g[::-1]
for i in g:
print(i, end = ' ') |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.util.ArrayList;
import java.util.Scanner;
import java.util.Stack;
public class LuxurioursHouses {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
ArrayList<Integer> houses = new ArrayList<Integer>();
for(int i = 0; i < n; i++){
houses.add(sc.... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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>
inline void remax(T& A, T B) {
if (A < B) A = B;
}
template <class T>
inline void remin(T& A, T B) {
if (A > B) A = B;
}
string ToString(long long num) {
string ret;
do {
ret += ((num % 10) + '0');
num /= 10;
} while (num);
reverse(ret... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 itertools import imap
I = lambda: map(int, raw_input().split())
n, = I()
houses = I()[:n]
_max = 0
lux = [0] * n
for i in range(1, n+1):
index = n - i
if houses[index] <= _max:
lux[index] = _max + 1
else:
_max = houses[index]
lux[index] = _max
print ' '.join(imap(lambda x, y: s... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... |
/* */
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new InputStreamReader(System.in));
}
String 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... | import java.io.*;
import java.util.StringTokenizer;
/**
* 581B
* ΞΈ(n) time
* ΞΈ(n) space
*
* @author artyom
*/
public class _581B implements Runnable {
private BufferedReader in;
private StringTokenizer tok;
private Object solve() throws IOException {
int n = nextInt();
int[] a = read... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 -*-
"""
Created on Thu Aug 13 23:52:16 2020
@author: DELL
"""
n=int(input())
h=list(map(int,input().split()))
h.reverse()
s=[]
m=0
for i in h:
if i>m:
m=i
a=0
elif m >=i:
a=(m-i)+1
s+=[str(a)]
s.reverse()
print(' '.join(s)) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int N;
cin >> N;
vector<int> h(N);
for (int i = 0; i < h.size(); ++i) {
cin >> h[i];
}
vector<int> res(N);
int tallest = -1;
for (int i = h.size() - 1; i >= 0; --i) {
if (h[i] <= tallest) {
res[i] = tallest + 1 - h[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... | #include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 10;
int a[maxn], maxh[maxn];
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", a + i);
for (int i = n; i > 0; i--) maxh[i] = max(a[i], maxh[i + 1]);
for (int i = 1; i < n; i++) {
if (a[i] > maxh[i + 1])
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int n;
int main() {
cin >> n;
vector<int> ar(n), supp(n);
for (int i = 0; i < n; i++) cin >> ar[i];
int t = ar[n - 1];
for (int i = n - 2; i >= 0; i--) {
supp[i] = t;
t = max(t, ar[i]);
}
for (int i = 0; i < n - 1; i++) {
if (ar[i] > supp[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.*;
import java.lang.*;
public class codeforces {
static class FastIO {
InputStream dis;
byte[] buffer = new byte[1 << 17];
int pointer = 0;
public FastIO(String fileName) throws Exception {
dis = new FileInputStream(fileName);
}
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 LuxuriousHouse1
{
public static void main(String args[])
{
int n,j,max=0,flag=0;
Scanner s=new Scanner(System.in);
n=s.nextInt();
//int[] a=new int[n+1];
int[] b=new int[n+1];
int[] c=new int[n+1];
for(int i=1;i<=n;i++)
{
b[i]=s.nextInt();
//b[i]=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... | a = int(input())
b = list(map(int, input().split()))
m = [0]*a
an =b[a-1]
for i in range(a-2, -1, -1):
m[i] = max(0, an - b[i] +1)
if b[i] > an:
an = b[i]
print(*m) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
struct debugger {
static void call(string::iterator it, string::iterator ed) {}
template <typename T, typename... aT>
static void call(string::iterator it, string::iterator ed, T a, aT... rest) {
string b;
for (; *it != ','; ++it)
if (*it != ' ') 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(raw_input())
h = map(int, raw_input().split())
ans = []
arr = [0]*n
for i in xrange(n-2, -1, -1):
arr[i] = max(arr[i+1], h[i+1])
for i in xrange(n):
if h[i] > arr[i]:
ans.append(0)
else:
ans.append(arr[i]-h[i]+1)
#print arr
#print ans
print ' '.join(map(str, ans)) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int a[100001], b[100001];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
int t = 0;
for (int i = n - 1; i >= 0; i--) {
b[i] = max(0, t - a[i] + 1);
t = max(a[i], t);
}
for (int i = 0; i < n; i++) cout << b[i] << " ";
return 0;... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | z = int(input())
l = list(map(int, input().rstrip().split(" ")))
a = [0]*z
m = l[-1]
if z ==1:
print(0)
else:
for i in range(z-2,0,-1):
if l[i] <= m:
a[i]=m-l[i]+1
if l[i]>m:
m = l[i]
if l[0] <= m:
a[0]=m-l[0]+1
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())
l=list(map(int,input().split()))
m=0
d=[]
for i in range(n-1,-1,-1):
s=max(0,m+1-l[i])
d.append(s)
if(l[i]>m):
m=l[i]
print(*d[::-1]) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.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... |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class LuxuriousHouses {
public static void main(String[] args) throws IOException {
MyScanner sc = new MyScanner(System.in);
int n = sc.nextInt();
l... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
long long powermod(long long _a, long long _b, long long _m) {
long long _r = 1;
while (_b) {
if (_b % 2 == 1) _r = (_r * _a) % _m;
_b /= 2;
_a = (_a * _a) % _m;
}
return _r;
}
long long string_to_number(string s) {
long long x = 0;
stringstream conv... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 b {
public static void main(String[] args) {
try(Scanner scan = new Scanner(System.in)) {
int n = scan.nextInt();
int[] list = new int[n];
for (int i = 0; i < n; i++) list[i]=scan.nextInt();
int[] cumfreq = new int[n];
cumfreq[n-1]=list[n-1];
for (int i... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
# from fractions import *
from heapq import*
from bisect import *
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
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.*;
public class B581codeforces {
public static void main(String args[]){
Scanner s = new Scanner(System.in);
int n;
n = s.nextInt();
int[] a = new int[n];
for(int i=0;i<n;i++){
a[i] = s.nextInt();
}
int max=0;
for(int i = a.length-1;i>=0;i--){
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... | import java.io.*;
import java.util.StringTokenizer;
public class B {
private static class Solution implements Runnable {
private static final long modulo = 1000000007;
private void solve() {
int n = in.nextInt();
int a[] = new int[n];
int segmentMax[] = new 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()))
arr.reverse()
count = 0
arr2 = []
for k in arr:
arr2.append(max(count+1-k, 0))
count = max(count, k)
arr2.reverse()
for kk in arr2:
print(kk, 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;
template <typename T>
T prod(const T &a, const T &b) {
return ((a % ((long long int)1e9 + 7)) * (b % ((long long int)1e9 + 7))) %
((long long int)1e9 + 7);
}
template <typename T>
T pow_(const T &a, const T &b) {
if (!b) return 1;
long long int p = pow_(a, b ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n, h, m[100000], k, mx = -1, m1[100000];
cin >> n;
for (int i = 0; i < n; i++) {
cin >> m[i];
}
reverse(m, m + n);
for (int i = 0; i < n; i++) {
if (m[i] > mx) {
mx = m[i];
m1[i] = 0;
} else
m1[i] = mx - m[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.lang.*;
import java.io.*;
import java.math.BigDecimal;
public class R322B {
public static void main (String[] args) throws java.lang.Exception {
InputReader in = new InputReader(System.in);
PrintWriter w = new PrintWriter(System.out);
int n = 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... | n=int(raw_input())
h=map(int,raw_input().split())
h2=h[:]
for i in xrange(n-2,-1,-1):h[i]=max(h[i],h[i+1])
for i in xrange(n-1):print max(0,h[i+1]-h2[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... | import java.util.*;
import java.io.*;
public class Main {
// File file = new File("input.txt");
// Scanner in = new Scanner(file);
// PrintWriter out = new PrintWriter(new FileWriter("output.txt"));
public static void main(String[] args) {
// Scanner in = new Scanner(System.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())
a=map(int,raw_input().split())
b=[0]*n
b[-1]=a[-1]
for i in range(n-1):
b[n-i-2]=max(a[n-i-2],b[n-i-1])
for i in range(n-1):
if a[i]>b[i+1]:
print 0,
else:
print b[i+1]-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... | n = int(raw_input())
x = raw_input().split(" ")
maxsofar = 0
ans = []
for i in range(n-1, -1, -1):
x[i] = int(x[i])
if x[i] > maxsofar:
maxsofar = x[i]
ans.append("0")
else :
c = abs(x[i] - maxsofar) + 1
ans.append(str(c))
ans = ans[::-1]
print " ".join(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;
int a[n], i, z = 0;
for (i = 0; i < n; i++) cin >> a[i];
for (i = n - 1; i >= 0; i--) {
if (a[i] <= z)
a[i] = z - a[i] + 1;
else {
z = a[i];
a[i] = 0;
}
}
for (i = 0; i < n; i++) cout << 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.StringTokenizer;
public class B {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.rea... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 get(data):
return data[1]
def get_answer(houses):
maxv = []
local_max = 0
for i in reversed(range(0 , len(houses))):
if houses[i] > local_max:
local_max = houses[i]
maxv.append(-1)
else:
maxv.append(local_max)
maxv = list(reversed(maxv))
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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.awt.Point;
import java.io.*;
import java.math.BigInteger;
public class CodeForces
{
FastScanner in;
PrintWriter out;
public void solve() throws IOException
{
int n = in.nextInt();
int mas[] = in.nextIntArray(n);
int res[] = ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(input())
a = list(map(int,input().split()))
z,s=0,[]
for i in range(n-1,-1,-1):
if a[i]>z:s.append(0);z = a[i]
else:s.append(max(0,z+1-a[i]))
s.reverse()
print(*s) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int n, ma;
int a[100010];
int b[100010];
int main() {
cin >> n;
for (int i = 1; i <= n; i++) cin >> a[i];
b[n] = 0;
ma = a[n];
if (n > 1)
for (int i = n - 1; i; i--) {
if (ma < a[i])
b[i] = 0;
else
b[i] = ma - a[i] + 1;
ma = m... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... |
n = int(input())
arr = list(map(int,input().split()))
k = arr[n - 1]
dp = [0] * n
for i in range(n - 2 , -1 , - 1):
if arr[i] > k :
dp[i] = 0
k = arr[i]
else:
dp[i] = k - arr[i] + 1
print(*dp)
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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, q[100005], maxi = 0, ans[100005];
cin >> n;
for (long long i = 0; i < n; i++) cin >> q[i];
for (long long i = n - 1; i >= 0; i--) {
if (q[i] > maxi) {
ans[i] = 0;
maxi = q[i];
} else {
ans[i] = maxi - q[i] + 1;
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 100 * 1000;
int n;
int h[MAXN];
int main() {
cin >> n;
for (int a = 0; a < n; ++a) {
cin >> h[a];
}
int i = 0;
for (int a = n - 1; a >= 0; --a) {
int h1 = h[a];
h[a] = max(0, i - h1 + 1);
i = max(i, h1);
}
for (int a = 0; a < 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())
a = list(map(int, input().split()))
maxH = 0
add = []
add.append(0)
for i in range(n - 1):
maxH = max(a[n - 1 - i], maxH)
add.append(max(0, maxH + 1 - a[n - 2 - i]))
for i in range(n):
print(add[n - 1 - 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;
map<int, int> m;
int a[1000001], ans[100000];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
for (int i = 1; i <= n; i++) cin >> a[i];
int mx = a[n];
for (int i = n - 1; i > 0; i--) {
ans[i] = max(0, mx + 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... | import java.util.Scanner;
/**
*
* @author Rishabh-PC
*/
public class Main {
public static void main (String args[])
{
Scanner stdIn = new Scanner(System.in);
int n,i,max=0,temp=0;
n = stdIn.nextInt();
int arr[] = new int[n];
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 = list(map(int, input().split()))
a = list(map(int, input().split()))
a.reverse()
mx = a[0]
ans = list()
ans.append(0)
for i in a[1:]:
ans.append(max(0, mx - i + 1))
mx = max(mx, i)
ans.reverse()
p = ''
for i in ans:
p += i.__str__() + ' '
print(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;
int GCD(int x, int y) {
if (x % y == 0)
return y;
else
return (GCD(y, x % y));
}
int main() {
int n, x, i;
while (scanf("%d", &n) == 1) {
int ara[100005];
int ar[100005] = {0};
for (i = 0; i < n; i++) {
scanf("%d", &x);
ara[i] = 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;
struct bui {
int fl;
int id;
} b[100100];
int ans[100100];
int main() {
int n;
while (~scanf("%d", &n)) {
for (int i = 1; i <= n; i++) {
scanf("%d", &b[i].fl);
;
b[i].id = i;
}
int index = 1;
int maxn = -1;
for (int 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 ans[100001];
int a[100001];
int main() {
int n, i, max1;
cin >> n;
max1 = 0;
for (i = 1; i <= n; i++) cin >> a[i];
for (i = n; i >= 1; i--) {
if (max1 > a[i]) {
ans[i - 1] = max1;
} else {
ans[i - 1] = a[i];
max1 = a[i];
}
}
f... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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;
using ll = long long;
int main() {
int n;
cin >> n;
vector<int> a(n);
for (int &i : a) cin >> i;
vector<int> r(n);
int mx = 0;
for (int i = n - 1; i >= 0; --i)
r[i] = max(mx - a[i] + 1, 0), mx = max(mx, a[i]);
for (int i : r) cout << i << ' ';
return 0... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
int n;
cin >> n;
int ara[n];
for (int i = 0; i < n; i++) cin >> ara[i];
int max = ara[n - 1];
ara[n - 1] = 0;
for (int i = n - 2; i >= 0; i--) {
if (ara[i] <= max)
ara[i] = max - ara[i] + 1;
else {
ma... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
const int N = 2e6;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n;
cin >> n;
int h[n];
for (int i = 0; i < n; ++i) {
cin >> h[i];
}
vector<int> v(n, 0);
int maxx = 0;
for (int i = n - 1; i >= 0; i--) {
v[i] = max(0, maxx - h[... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int arr[100005], ans[100005];
int main(void) {
int n;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
ans[n - 1] = 0;
int large = arr[n - 1];
for (int i = n - 2; i > -1; i--) {
if (arr[i] > large) {
ans[i] = 0;
large = 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... |
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class B {
static StringTokenizer st;
static BufferedReader br;
static PrintW... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 long long MOD = 1e9 + 7;
const double PI = 2 * acos(0.0);
pair<long long, long long> arrR[100005];
long long arr[100005];
int main() {
long long n, fi = 0;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
arrR[n - 1].first = arr[n - 1];
for (in... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
int main() {
int n, max = 0;
scanf("%d", &n);
int ara[n], ans[n], i, j;
for (i = 0; i < n; i++) {
scanf("%d", &ara[i]);
}
for (i = n - 1; i >= 0; i--) {
if (ara[i] > max) {
max = ara[i];
j = i;
}
if (max - ara[i] == 0 && i == j)
ans[i] = 0;
else... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.*;
public class Main {
static StreamTokenizer in=new StreamTokenizer(new BufferedReader(new InputStreamReader(Syst... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 = 1000 * 100;
int h[MAXn + 5], Max[MAXn + 5];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) cin >> h[i];
Max[n - 1] = h[n - 1];
for (int i = n - 2; i >= 0; i--) Max[i] = max(Max[i + 1], h[i]);
for (int i = 0; i < n - 1; i++) cout << ma... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n=int(input())
l=[int(q) for q in input().split()]
c=0
d=l[::-1]
m=d[0]
ans=[0]
for i in range(1,len(l)):
ans.append(max(0,m-d[i]+1))
if d[i]>m:
m=d[i]
print(*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... | //package CF;
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class A {
public static void main(String[] args) throws Exception
{
Scanner 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()))
stat = [0] * n
for i in range(n - 2, -1, -1):
stat[i] = max(stat[i + 1], h[i + 1])
for j in range(n):
if h[j] > stat[j]:
print('0', end = ' ')
else:
print(stat[j] - h[j] + 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... | n = int(raw_input())
a = map(int, raw_input().split())
lol = []
ma = a[-1]
lol.append(0)
for i in range(n-2,-1,-1):
if a[i]>ma:
ma = a[i]
lol.append(0)
else:
lol.append(ma-a[i]+1)
for i in range(n-1,-1,-1):
print lol[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 i, n, a[100001], x[100001], max;
cin >> n;
for (i = 1; i <= n; i++) cin >> a[i];
x[n] = 0;
max = a[n];
for (i = n - 1; i > 0; i--) {
if (a[i] <= max)
x[i] = max - a[i] + 1;
else {
x[i] = 0;
max = a[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;
long long a[100000] = {0};
long long r[100000] = {0};
int main() {
ios_base::sync_with_stdio(false);
long long n;
cin >> n;
for (long long i = 0; i < n; ++i) cin >> a[i];
long long m = a[n - 1];
for (long long i = n - 2; i >= 0; --i) {
if (a[i] <= m) r[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())
ls = list(map(int, input().split()))
ans = [0] * n
curr_max = ls[-1]
for i in range(n-2, -1, -1):
if ls[i] > curr_max:
curr_max = ls[i]
else:
ans[i] = (curr_max - ls[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... |
import java.util.Scanner;
/**
* Created by MenonS on 02-10-2015.
*/
public class B322 {
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
int h = scanner.nextInt();
int[] floors = new int[h];
int[] output = new int[h];
for(int i=0;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... | import java.util.*;
public class CodeForces581{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int n = input.nextInt();
int[] a = new int[n];
for(int i = 0;i<n;i++){
a[i] = input.nextInt();
}
int max = a[n-1];
a[n-1] = 0;
for(int i = n-2;i>=0;i--){
if(a[i] <= ma... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int r[100001];
int main() {
int n;
scanf("%d", &n);
int ar[n + 1];
for (int i = 0; i < n; i++) {
scanf("%d", &ar[i]);
}
int ma = ar[n - 1];
for (int i = n - 2; i > -1; i--) {
if (ma >= ar[i])
r[i] = ma - ar[i] + 1;
else
ma = ar[i];
}
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long int n, arr[100005] = {}, ans[100005] = {};
cin >> n;
long long int mx = -1e9;
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
for (int i = n - 1; i >= 0; i--) {
if (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() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n = 0, max = 0;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; i++) cin >> a[i];
vector<int> ans(n);
for (int i = n - 1; i >= 0; i--) {
if (a[i] > max) {
ans[i] = 0;
max... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
bool is_digit(char c) {
if (c == '1' || c == '2' || c == '3' || c == '4' || c == '5' || c == '6' ||
c == '7' || c == '8' || c == '9') {
return true;
}
return false;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n;
ci... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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[111111], n, B[111111];
int main() {
int i, ma = -1;
cin >> n;
for (i = 0; i < n; i++) cin >> A[i];
for (i = n - 1; i >= 0; i--) {
if (A[i] > ma) {
ma = A[i];
B[i] = 0;
} else {
B[i] = ma - A[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())
l = list(map(int,input().split()))
ans = [0]
tmp = l[-1]
for i in range(n-2,-1,-1):
ans.append(max(0 , tmp-l[i]+1))
if l[i] > tmp:
tmp = l[i]
#print(tmp)
print(*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... | #include <bits/stdc++.h>
using namespace std;
const int M = 1e9 + 7;
long long fastpower(long long x, long long n, long long M) {
if (n == 0)
return 1;
else if (n % 2 == 0)
return fastpower((x * x) % M, n / 2, M);
else
return (x * fastpower((x * x) % M, (n - 1) / 2, M)) % M;
}
long long GCD(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... | #include <bits/stdc++.h>
using namespace std;
int main() {
long long int n;
cin >> n;
long long int a1, i;
vector<long long int> a;
for (i = 0; i < n; i++) {
cin >> a1;
a.push_back(a1);
}
long long int maxa = 0;
long long int maxa1 = 0;
for (i = n - 1; i >= 0; i--) {
maxa1 = maxa;
maxa... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 i, n;
long long a[100005], high = 0, temp;
std::cin >> n;
for (i = 0; i < n; i++) {
std::cin >> a[i];
}
for (i = n - 1; i >= 0; i--) {
temp = a[i];
a[i] = max(0LL, high + 1 - a[i]);
high = max(temp, high);
}
for (i = 0; i < n... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
while (cin >> n) {
vector<int> a(n);
for (int i = 0; i < n; i++) cin >> a[i];
vector<int> mx(n);
for (int i = n - 2; i >= 0; i--) mx[i] = max(a[i + 1], mx[i + 1]);
for (int i = 0; i < n; i++) cout << max(0, mx[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;
int main() {
int n, max = 0;
scanf("%d", &n);
int arr[n], result[n];
for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
result[n - 1] = 0;
max = arr[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (arr[i] > max) {
max = arr[i];
result[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;
const int maxn = 100000 + 5;
int ar[maxn], br[maxn];
int n, ans, cns, MAX, MIN;
int main() {
while (cin >> n) {
memset(br, 0, sizeof(br));
for (int i = 0; i < n; i++) scanf("%d", &ar[i]);
MAX = 0;
for (int i = n - 1; i >= 0; i--) {
if (MAX < ar[i]) {... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.util.*;
public class First {
public static void main(String [] argv) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int []a = new int[n];
for(int i = 0; i<n; i++)
a[i] = sc.nextInt();
int localMax = a[n - 1];
ArrayList<Integer> ar = new ArrayList<Integer>();
ar.add(0);
for... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n=int(input())
arr=[int(x) for x in input().split()]
ans=[0 for i in range(n)]
maxi=arr[-1]
for i in range(n-2,-1,-1):
if arr[i]>maxi:
maxi=arr[i]
ans[i]=0
else:
ans[i]=maxi-arr[i]+1
for i in ans:
print(i,end=' ') |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int n, a[100005], mx[100005];
int main() {
ios_base::sync_with_stdio(false);
cin >> n;
for (int i = 1; i <= n; i++) cin >> a[i];
mx[n] = a[n];
mx[n + 1] = 0;
for (int i = n - 1; i >= 1; i--) mx[i] = max(mx[i + 1], a[i]);
for (int i = 1; i <= n; 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 const N = 1e6 + 5;
int n, m, arr[N], b[N], cnt = 0, ma;
int main() {
cin >> n;
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
ma = -100;
for (int i = n - 1; i >= 0; i--) {
if (ma < arr[i]) {
ma = arr[i];
} else {
b[i] = ma - arr[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 com.sun.corba.se.impl.orbutil.ORBConstants;
import java.io.PrintWriter;
import java.util.*;
import java.util.Arrays ;
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.lang.reflect.Array;
public class Test{
static PrintWriter pw = new PrintWr... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class B322 {
static int[]array;
public static void main(String [] args) throws IOException{
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
int n = Integ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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.Collections;
import java.util.Scanner;
/**
* Created by sanjayarvind on 08/02/2017 AD.
*/
public class CR294B {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] buildings=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() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
int a;
cin >> a;
long* x = new long[a];
long* z = new long[a];
long max = 0;
for (long i = 0; i < a; i++) {
cin >> z[i];
}
bool in = 0;
for (long i = a - 1; i >= 0; i--) {
if (max < z[i]... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
long long v[100005];
long long st[400005];
void init(int idx, int a, int b) {
if (a == b) {
st[idx] = v[a];
} else {
int m = (a + b) / 2;
init(idx * 2 + 1, a, m);
init(idx * 2 + 2, m + 1, b);
st[idx] = max(st[idx * 2 + 1], st[idx * 2 + 2]);
}
}
lon... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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, maxi = 0, temp;
cin >> n;
int a[n];
for (int i = 0; i < n; i++) {
cin >> a[i];
}
for (int i = n - 1; i >= 0; i--) {
if (maxi >= a[i])
a[i] = maxi - a[i] + 1;
else {
maxi = a[i];
a[i] = 0;
}
}
for (int i =... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(input())
arr = list(map(int, input().split()))
res = [0] * n
mx = -1
for i in range(n-1, -1, -1):
if arr[i] > mx:
mx = arr[i]
res[i] = 0
else:
add = max(0, mx + 1 - arr[i])
res[i] = add
print(*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;
const double EPS = 1e-9;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
int n;
while (~scanf("%d", &n)) {
vector<long long int> vec;
long long int a[n];
for (long long int i = 0; i < n; i++) cin >> a[i];
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... | #include <bits/stdc++.h>
using namespace std;
int h[100002], p[100002];
int main() {
int n, m = 0;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> h[i];
}
for (int i = n - 1; i >= 0; i--) {
if (h[i] > m) {
p[i] = 0;
m = h[i];
} else {
p[i] = (m + 1) - h[i];
}
}
for (int i ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n];
for (int i = 0; i < n; i++) cin >> a[i];
int sol[n];
int maxh = 0;
for (int i = n - 1; i >= 0; i--) {
sol[i] = max(0, maxh + 1 - a[i]);
if (a[i] > maxh) maxh = a[i];
}
for (int i = 0; i < n; i++) cout << sol[... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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()))
mx=0
ans=[]
for i in range (n-1,-1, -1):
if l[i] >mx:
mx=l[i]
ans.append(0)
elif l[i]==mx:
ans.append(1)
else:
ans.append(mx-l[i]+1)
ans.reverse()
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 a[100005], ans[100005];
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
int tmp = 0;
for (int i = n; i >= 1; --i) {
ans[i] = max(tmp + 1 - a[i], 0);
tmp = max(tmp, a[i]);
}
for (int i = 1; i <= n; ++i) printf("... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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()))
etaj = [0] * n
ma = A[-1]
for i in range(-2, -n-1, -1):
if A[i] > ma:
ma = A[i]
elif A[i] == ma:
etaj[i] = 1
else:
etaj[i] = ma - A[i] + 1
for elem in etaj: print(elem, 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())
home = list(map(int, input().split()))
suff = [0] * n
for i in range(n - 2, -1, -1):
suff[i] = max(suff[i + 1], home[i + 1])
for i in range(n):
if home[i] > suff[i]:
print(0, end=" ")
else:
print(suff[i] - home[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... | n = int(input())
a = [int(i) for i in input().split()]
ans = []
m = 0
for i in range(n - 1, -1, -1):
if a[i] <= m:
ans.append(m - a[i] + 1)
elif a[i] > m:
ans.append(0)
m = a[i]
print(*ans[::-1]) |
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