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() {
long long n;
cin >> n;
long long p[n], q[n];
for (long long i = 0; i < n; i++) {
cin >> p[i];
}
long long m = p[n - 1];
for (long long i = n - 2; i >= 0; i--) {
if (p[i] <= m) {
q[i] = (m - p[i]) + 1;
} else {
q[i] = 0;
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
/**
* @author Don Li
*/
public class LuxuriousHouses {
void solve() {
int n = in.nextInt();
int[] H = new int[n];
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... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, *a;
cin >> n;
a = new int[n];
for (int i = 0; i < n; i++) cin >> a[i];
int mx = a[n - 1];
vector<int> v;
v.push_back(0);
for (int i = n - 2; i >= 0; i--) {
mx = max(mx, a[i + 1]);
v.push_back(max(mx - a[i] + 1, 0));
}
for (int... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import math
def fact(n):
ans = 1
for i in range(2, n+1):
ans*= i
return ans
def comb(n, c):
return fact(n)//(fact(n-c)*c)
n = int(input())
nums= list(map(int, input().split()))
m= 0
ans = []
for i in range(n-1, -1, -1):
if(nums[i] <= m):
ans.append(str(m-nums[i]+1))
else:
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #!/bin/python
from sys import stdin
n=int(stdin.readline().strip())
arr=map(int,stdin.readline().strip().split())
max=arr[n-1]+1
ans=[0]*n
for i in range(n-2,-1,-1):
if arr[i]<max:
ans[i]=max-arr[i]
else:
ans[i]=0
max=arr[i]+1
for i in range(n):
print 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())
mas = list(map(int,input().split(" ")))
submas=[0]*(n)
ma=0
for i in range(n-1,-1,-1):
submas[i]=str(max(ma-mas[i]+1,0))
if (ma<mas[i]):
ma =mas[i]
print(" ".join(submas))
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 MAXN = (long long)((1e5) + 100);
long long cuberoot(long long x) {
long long lo = 1, hi = min(2000000ll, x);
while (hi - lo > 1) {
long long mid = (lo + hi) / 2;
if (mid * mid * mid < x) {
lo = mid;
} else
hi = mid;
}
if (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... | n = input()
a = map(int, raw_input().split())
out = [0]*n
m = 0
for i in range(n) :
e = a[n-1-i]
if e <= m :
out[n-1-i] = m - e + 1
else :
m = e
for o in out:
print o, |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | __author__ = 'yarsanich'
n = int(input())
a = list(map(int,input().split()))
b = [0 for i in range(n)]
m = 0
for i in range(n-1, -1, -1):
b[i] = max(0, m - a[i] + 1)
m = max(m, a[i])
print(*b) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class B {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
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... | import java.io.BufferedInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.util.InputMismatchException;
public class Main
{
class MyScanner
{
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
BufferedInputStream bis = new BufferedInpu... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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.ArrayList;
import java.util.StringTokenizer;
/**
* Created by tawsif on 9/29/15.
*
* @Time 12:07 PM
*/
public class LuxuriousHouses322B {
final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;
PrintWriter out;
long timeBegin, timeEnd;
public void runIO() th... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.*;
import java.util.*;
public class B {
public static void main(String[]args)throws IOException{
BufferedReader x=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(x.readLine());
StringTokenizer st=new StringTokenizer(x.readLine());
int[]arr... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | i = int(input())
l = list(reversed(list(map(int,input().split()))))
m = 0
t = []
for x in l:
if x > m:
t.append(0)
else:
t.append((m+1) - x)
if x > m:
m = x
print(' '.join([str(elem) for elem in list(reversed(t))])) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 isP(long int hj) {
long int op;
for (op = 2; op <= sqrt(hj); op++) {
if (hj % op == 0) return 0;
}
return 1;
}
void swap(long long int *p, long long int *q) {
long long int tmp = *p;
*p = *q;
*q = tmp;
}
int mind(long long int p) {
int mindd = 10;
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.util.Scanner;
public class B_581 {
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
Scanner 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... | n = int(input())
h = [int(i) for i in input().split()] + [0]
v = [0 for i in h]
for p in range(n,-1,-1):
if p == n: v[p] = h[p]
else: v[p] = max(v[p+1], h[p])
print(' '.join(["%s" % max(0, v[i+1]+1-h[i]) for i in range(n)])) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, a[100002], m[100002], i, j, ans[100002];
scanf("%d", &n);
for (i = 0; i < n; i++) {
scanf("%d", &a[i]);
}
m[n - 1] = 0;
ans[n - 1] = 0;
for (i = n - 2; i >= 0; i--) {
m[i] = max(m[i + 1], a[i + 1]);
ans[i] = max(0, m[i] - 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 = input()
h = map(int, raw_input().split())
maxvec = [ 0 for i in xrange(n)]
for i in xrange(n-2,-1,-1):
maxvec[i]=max(maxvec[i+1],h[i+1])
newh=[max(maxvec[i]-h[i]+1,0) for i in xrange(n)]
for i in newh:
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.util.*;
import java.io.*;
public class temp {
public static void main(String[] args) throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
int[] arr = new int[n];
StringTokenizer st = new StringTokenizer(br.r... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6;
const int MAX = 1e7 + 1;
double pi = 3.1415926535897932384626433832795;
const long long inf = 1e18;
long long mod = 1e6;
long long powr(long long a, long long b) {
if (b == 0) return 1LL;
long long x = powr(a, b / 2);
x = (x * x) % mod;
if (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.*;
import java.io.*;
public class solution {
public static void merge(long arr[], int l, int m, int r) {
// Find sizes of two subarrays to be merged
int n1 = m - l + 1;
int n2 = r - m;
/* Create temp arrays */
long L[] = new long[n1];
long R[] = new... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
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... | import java.util.Scanner;
public class B {
public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
int n = cin.nextInt();
int[] a = new int[n];
int[] max = new int[n];
boolean[] visited = new boolean[n];
for (int i = 0; i < n; i++) {
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(input())
l = list(map(int , input().split()))
a = list(range(n+1)); y = 0
for i in range(n-1 , -1 , -1):
y = max(y , l[i])
a[i] = y
a[n] = 0
for i in range(n):
if (l[i] == a[i]) & (l[i] != a[i+1]):
print('0 ' , end = '')
else:
print(a[i]-l[i]+1 , end = '')
print(' ' , 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.Scanner;
public class LuxuriousHouse {
public static void main(String asd[])throws Exception
{
Scanner in=new Scanner(System.in);
int n=in.nextInt();
int a[]=new int[n];
int b[]=new int[n];
int max=Integer.MIN_VALUE;int k=0;
for(int i=0;i<n;i++)... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.*;
import java.util.*;
public class div_2_322_b {
public static void main(String args[]){
FScanner in = new FScanner();
PrintWriter out = new PrintWriter(System.out);
//int t = in.nextInt();
//while(t-->0) { }
int n=in.nextInt();
int a[]=in.readArray(n);
int arr[]=new int[n];
int max=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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class LuxuriousHouses {
public static void main(String[] args) {
out = new PrintWriter(new BufferedOutputStream(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... | input()
h = list(map(int,input().split()))
m = h[-1]
n = []
for x in h[::-1]:
n.append(str(max(0,m-x)))
m = max(m,x+1)
print (' '.join(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(void) {
int total;
cin >> total;
int *input = new int[total];
int *save = new int[total];
for (int i = 0; i < total; i++) cin >> input[i];
save[total - 1] = 0;
int highest = input[total - 1];
for (int i = total - 2; i > -1; i--) {
if (input[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=[int(i) for i in input().split()]
l.reverse()
n1=1
ans=[0]
mx=l[0]
if n>1:
if l[0]<l[1]:
ans.append(0)
else:
ans.append(l[0]-l[1]+1)
while n1<n-1:
num1=l[n1]
num2=l[n1+1]
mx=max(num1,l[0],mx)
if num2<=mx:
ans.append(mx-num2+1)
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author karanjobanputra
*/
public ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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, typename U>
inline void amin(T &x, U y) {
if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
if (x < y) x = y;
}
int main() {
int n;
cin >> n;
int a[n];
for (int i = 0; i < n; i++) {
cin >> a[i];
}
i... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.util.Scanner;
public class LuxuriousHouses {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int arr[] = new int[n];
int ans[] = new int[n];
int max = 0;
for(int i=0;i<n;i++)
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... |
n = int(input())
t = list(map(int,input().split()))[::-1]
f=[0]
u=t[0]
for k in range(1,n):
if t[k]>u:
f.append(0)
if t[k]>u:
u=t[k]
else:
s = u-t[k]+1
f.append(s)
if t[k]>u:
u=t[k]
print(*f[::-1])
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n=int(input())
l=list(map(int,input().split()))
max1=0
arr1=[]
for i in range (n-1,-1, -1):
if l[i] >max1:
max1=l[i]
arr1.append(0)
elif l[i]==max1:
arr1.append(1)
else:
arr1.append(max1-l[i]+1)
arr1=arr1[::-1]
print(*arr1) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> v(n);
for (int i = 0; i < n; i++) {
cin >> v[i];
}
int max = v[v.size() - 1];
vector<int> res;
res.push_back(0);
for (int i = v.size() - 2; i >= 0; i--) {
if (max >= v[i]) {
res.push_back(max - v[i] +... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = input()
buildings = list(map(int, input().split(' ')))
m = 0
result = []
for i in range(len(buildings) - 1, -1, -1):
b = buildings[i]
x = m - b
if x < 0:
result.append(0)
m = b
else:
result.append(x + 1)
print(' '.join(map(str, result[::-1]))) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
long long r;
while (b != 0) {
r = a % b;
a = b;
b = r;
}
return a;
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const int maxn = 100010;
int n;
int h[maxn];
int ans[maxn];
void solve() ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import sys
input=sys.stdin.buffer.readline
n=int(input())
arr=list(map(int,input().split()))
right_max=arr[n-1]
ans=[0]
for i in range(n-2,-1,-1):
if arr[i]>right_max:
ans.append(0)
right_max=arr[i]
else:
ans.append(right_max-arr[i]+1)
ans=ans[::-1]
for x in ans:
print(x,end=' ')
#unknown_2433 |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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], maxi[100001];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a[i];
}
for (int i = n - 1; i >= 0; i--) {
maxi[i] = max(maxi[i + 1], a[i]);
}
for (int i = 0; i < n; i++) {
cout << max(maxi[i + 1] - a[i] + 1, 0) << '... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.util.Scanner;
public class _0909LuxuriousHouses {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n=sc.nextInt();
long[] arr = new long[n];
for(int i=0;i<n;i++) {
arr[i]=sc.nextLong();
}
long max=arr[n-1];
arr[n-1]=0;
for(int i=n-2;i>=0;i--) {
lo... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
import time
n = int(input())
A = [int(i) for i in input().split()]
start = time.time()
A = [ A[n-i-1] for i in range(n)]
ans = [ 0 for i in range(n) ]
m = 0
for i in range(n):
if A[i] > m:
m = A[i]
else:
ans[i] = 1+m-A[i]
for i in rang... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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[100000];
int b[100000];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a[i];
}
int max = a[n - 1];
b[0] = 0;
for (int i = n - 2; i >= 0; i--) {
if (a[i] <= max) {
b[i] = max - a[i] + 1;
} else {
b[i] = 0;
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
vector<int> v(100000), a(100000);
int max = -1, c;
cin >> c;
for (int i = 0; i < c; i++) {
cin >> v[i];
}
a[c - 1] = 0;
max = v[c - 1];
for (int i = c - 2; i >= 0; i--) {
if (v[i] > max) {
max = v[i];
a[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... | #include <bits/stdc++.h>
using namespace std;
int n, a[100005], maxx[100005];
int main(void) {
int i, j;
while (~scanf("%d", &n)) {
for (i = 0; i < n; i++) scanf("%d", &a[i]);
maxx[n - 1] = 0;
for (i = n - 2; i >= 0; i--) maxx[i] = max(maxx[i + 1], a[i + 1]);
for (i = 0; i < n; i++) {
if (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.BufferedWriter;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashMap;
public class Main {
public static void main... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int n;
int h[100000], max[100001];
cin >> n;
for (int i = 0; i < n; ++i) {
cin >> h[i];
}
max[n] = h[n - 1] - 1;
for (int i = n - 1; i >= 0; --i) {
if (h[i] > max[i + 1])
max[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... | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n, y = 0;
cin >> n;
long long a[n], b[n];
for (long long i = 0; i < n; i++) {
cin >> a[i];
}
b[n - 1] = 0;
y = a[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (y == a[i]) {
b[i] = 1;
} else {
y = max(y, 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... | #!/usr/bin/python
from collections import deque
def ir():
return int(raw_input())
def ia():
line = raw_input()
line = line.split()
return map(int, line)
# e + x = m + 1
# x = m + 1 - e
n = ir()
a = ia(); a.reverse()
m = 0; ans = []
for e in a:
ans.append(max(m + 1 - e, 0))
if e > m: 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;
long long int maxt(long long int x, long long int y) { return (x > y) ? x : y; }
int main() {
long long int n;
cin >> n;
long long int arr[n], i;
for (i = 0; i < n; ++i) cin >> arr[i];
long long int ans[n];
ans[n - 1] = arr[n - 1];
for (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())
l=input().split()
s=[]
for i in range(n):
l[i]=int(l[i])
s.append(0)
i=0
m=l[n-1]
while(i<n):
k=n-i-1
if(m>=l[k] and i!=0):
s[k]=m-l[k]+1
else:
m=l[k]
i+=1
for i in range(n):
print(s[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;
long long int h[n];
for (int i = 0; i < n; i++) {
cin >> h[i];
}
long long int b[n];
long long int max = 0;
for (int j = 0; j < n; j++) {
if (j == 0) {
max = h[n - j - 1];
b[n - j - 1] = max - 1;
} else {... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
int[] array = new int[n];
for (int i = 0; i < n; i++) {
array[i] = scan.nextInt();
}
int[] temp = new int[n];
temp[n-1]=0;
int max = array[n-1];
fo... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.util.*;
import java.io.*;
import java.lang.*;
public class Luxury
{
static class InputReader {
private final InputStream stream;
private final byte[] buf = new byte[8192];
private int curChar, snumChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.strea... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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()))
l1=[0]*n; mx=0; ans=""
for i in range(n-1,-1,-1):
mx=max(mx,h[i])
l1[i]=mx
for i in range(n-1):
if h[i]>l1[i+1]: ans+="0 "
else: ans+=f"{l1[i]-h[i]+1} "
ans+='0'
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.*;
public class Test {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int[] arr = new int[sc.nextInt()];
for(int i = 0; i < arr.length; i++){
arr[i] = sc.nextInt();
}
sc.close();
int[] max = new int[arr.length];
for(int i = max.length - 2; i >= 0; i--){... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int n, i, j, a[100000], k, x, b[100000];
int main() {
cin >> n;
j = n - 2;
b[n - 1] = 0;
for (i = 0; i < n; i++) cin >> a[i];
x = a[n - 1];
for (i = n - 2; i >= 0; i--) {
if (a[i] <= x) {
b[j] = x - a[i] + 1;
j--;
} else {
b[j] = 0;
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, i;
cin >> n;
long long int a[n], suf[n];
for (i = 0; i < n; i++) {
cin >> a[i];
}
suf[n - 1] = 0;
for (i = n - 2; i >= 0; i--) {
suf[i] = max(a[i + 1], suf[i + 1]);
}
for (i = 0; i < n; i++) {
cout << max(suf[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... | n=input()
l=map(int,raw_input().split())
x=[0]*n
maxi=l[-1]
for i in xrange(n-2,-1,-1):
if l[i]>maxi:
maxi=l[i]
else:
x[i]=maxi-l[i]+1
for i in x:
print i,
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(input())
a = list(map(int, input().split()))
a.reverse()
mx = 0
out = [0] * n
for i in range(n):
out[i] = max(mx + 1, a[i]) - a[i]
mx = max(a[i], mx)
out.reverse()
print(' '.join(map(str, out)))
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | N = int(input())
A = list(map(int, input().split()))
B = A[:]
for i in range(N-2, -1, -1):
if(B[i] < B[i+1]):
B[i] = B[i+1]
for i in range(0, N - 1):
if(A[i] > B[i+1]):
print("0",end=" ")
else:
# print(A[i], end = "----")
print(B[i+1] + 1 - A[i], end=" ")
print("0") |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
double pi = acos(-1);
const int MXN = 1e5 + 5;
const long long mod = 1e9 + 7;
void printv(vector<long long> v, long long n) {
for (long long i = 0; i < n; i++) {
cout << v[i] << " ";
}
}
vector<long long> inputv(vector<long long> v, long long n) {
for (long long 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()))
ans = [0] * n
x = a[-1]
for i in range(n-2, -1, -1):
ans[i] = max(0, x-a[i]+1)
x = max(a[i], x)
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 num[100010];
int cop[100010];
int ans[100010];
map<int, int> pic;
int cmp(int a, int b) { return a > b; }
int main() {
int n, flagc = 0;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &num[i]);
pic[num[i]]++;
cop[i] = num[i];
}
sort(cop... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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, m, v = int(input()), 0, []
for x in map(int, reversed(input().split())):
v.append(max(0, m - x + 1))
m = max(m, x)
print(' '.join(map(str, reversed(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... | /*
* To change this license header, choose License Headers in Project Properties.
* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
import java.util.*;
/**
*
* @author MyPC
*/
public class _581B_LuxuriousHouses {
public static void main(String args[]){
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 f(l):
n = len(l)
rl = [0]*n
rm = l[-1]
for i in range(n-2,-1,-1):
rl[i] = max(rm+1-l[i],0)
if l[i]>rm:
rm = l[i]
return rl
_ = input() #1e5
l = list(map(int,input().split()))
print(*f(l))
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
int a[maxn], b[maxn], c[maxn];
int main() {
int n;
scanf("%d", &n);
for (int i = (1); i <= (n); i++) scanf("%d", &a[i]);
b[n + 1] = 0;
for (int i = (n); i >= (1); i--) {
if (a[i] > b[i + 1]) {
b[i] = a[i];
} else {
b[i]... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(input())
house = list(map(int,input().split(' ')))
dos = [0]*n
maximum = house[n-1]
for i in range(n-2,-1,-1):
dos[i]=max(0,maximum-house[i]+1)
if house[i]>maximum:
maximum=house[i]
print(*dos)
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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()))
maxr = [0 for i in range(n)]
result = [0 for i in range(n)]
for i in range(n-2, -1, -1):
maxr[i] = max(maxr[i+1], a[i+1])
for i in range(n):
result[i] = maxr[i] - a[i] + 1
if result[i] < 0:
result[i] = 0
print(' '.join(str(r) for r in result))
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, num[100002], i, j, ans[100002], max;
cin >> n;
for (i = 1; i <= n; i++) scanf("%d", &num[i]);
ans[n] = 0;
max = 0;
for (i = n; i >= 1; i--) {
if (max <= num[i]) {
max = num[i];
ans[i] = 0;
++max;
} else
ans[i] ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 1;
long long n, i, j, k, m;
long long dem = 0, tong = 0, x, y, z, l, r;
char k1, k2;
long long a[maxn], b[maxn], c[maxn], h[maxn];
long long d[maxn];
string s, s1, s2;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cin >> 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;
static const int MAXN = 1e5 + 10;
int n;
int data[MAXN];
int tail[MAXN];
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%d", data + i);
}
for (int i = n; i > 0; --i) {
tail[i] = max(tail[i + 1], data[i + 1]);
}
for (int i = 1; i <= n... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | N, X, Answer = int(input()), list(map(int, input().split())), [0]
Max = X[-1]
for i in X[:N - 1:][::-1]:
Answer.append(max(0, Max - i + 1))
Max = max(Max, i)
print(*Answer[::-1])
# Come together for getting better !!!!
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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;
int main() {
int n;
cin >> n;
int Arr[n];
for (int i = 0; i < n; ++i) cin >> Arr[i];
int mFl = -1;
for (int i = n - 1; i >= 0; --i) {
if (Arr[i] > mFl) {
mFl = Arr[i];
Arr[i] = 0;
} else
Arr[i] = (mFl + 1) - Arr[i];
}
for (int i = 0... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | r=lambda:map(int,raw_input().split())
n=input()
h=r()
M=0
a=[]
for i in reversed(h):
a.append(max(0,(M+1)-i))
M=max(M,i)
print ' '.join(map(str,reversed(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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Bui... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual soluti... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
using ll = long long int;
int arr_min(int arr[], int a, int b) {
int min = arr[0];
for (int i = a; i != b; i++) {
if (min > arr[i]) min = arr[i];
}
return min;
}
int arr_max(int arr[], int a, int b) {
int max = arr[0];
for (int i = a; i != b; i++) {
if (... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... |
import java.util.Scanner;
public class B {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int h[] = new int[n+2];
for(int i = 0 ; i < n; i++) {
h[i] = in.nextInt();
}
int max[] = new int[n+2];
max[n] = 0;
for(int i = n-1; i >= 0; i--) ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... |
import java.awt.Point;
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.Arr... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> v(n), x(n);
for (int i = 0; i < n; i++) cin >> v[i];
int mx = 0;
for (int i = n - 1; i >= 0; i--) {
if (v[i] > mx) {
x[i] = 0;
mx = v[i];
} else if (v[i] < mx)
x[i] = mx + 1 - v[i];
else
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(raw_input())
hs = map(int, raw_input().split(' '))
max_h = 0
a = []
for h in hs[::-1]:
a.append(0 if h > max_h else max_h + 1 - h)
if max_h < h:
max_h = h
for aa in a[::-1]:
print aa, |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(input())
houses = list(reversed(list(map(int, input().split()))))
ans = [0]*(n)
highestToTheRight = houses[0]
for i in range(1,n):
if houses[i] > highestToTheRight:
highestToTheRight = houses[i]
ans[i] = 0
else:
ans[i] = highestToTheRight - houses[i] + 1
print(*reversed(ans)) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(input())
A = list(map(int, input().split()))
etaj = [0] * n
l = len(A)
ma = A[-1]
for i in range(-2, -l-1, -1):
if A[i] > ma:
ma = A[i]
etaj[i] = 0
"""
ma = A[i]
if A[i] <= A[i+1]:
etaj[i] = etaj[i+1] + A[i+1] - A[i]
elif (A[i] > A[i+1]) and (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... | #!/usr/bin/env python
#
# http://codeforces.com/problemset/problem/581/B
try:
n = int(raw_input())
h = raw_input().split()
for i in range(0, n):
h[i] = int(h[i])
t = 0
for i in reversed(range(0, n)):
if h[i] > t:
t = h[i]
h[i] = 0
else:
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... | import java.util.Scanner;
public class contest_1_2 {
public static void main(String[] args) {
int n;
Scanner cin=new Scanner(System.in);
n=cin.nextInt();
int[] array=new int[n];
int[] sol=new int[n];
for(int i=0;i<n;i++)
{
array[i]=cin.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(input())
houses = [int(x) for x in input().split(' ')]
max_sizes = [0] * n
current_max = 0
for i in reversed(range(n)):
max_sizes[i] = current_max
if houses[i] > current_max:
current_max = houses[i]
for i in range(n):
floors_added = max_sizes[i] - houses[i] + 1
if floors_added < 0:
floors_added = 0
i... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
template <typename T1, typename T2>
void mapped(map<T1, T2> m) {
for (auto ite = m.begin(); ite != m.end(); ++ite) {
cout << "'" << ite->first << "'"
<< " : " << ite->second << endl;
}
}
template <typename T1, typename T2>
void ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 MX = (int)1e6 + 17;
const int MOD = (int)1e9 + 7;
const long long oo = (long long)1e18 + 7;
const int INF = (int)999999999;
const int N = (int)1e5 + 17;
const int di[4] = {-1, 0, 1, 0};
const int dj[4] = {0, 1, 0, -1};
inline long long IN() {
long long x = 0, ch... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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.*;
/**
* Created by bobyk on 22/08/15.
*/
public class Main {
public static void main(String[] args) throws IOException {
new Main().Run();
}
InputReader reader;
PrintWriter pw;
StringTokenizer stok;
private void Run() throws IOException {
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
int a[100010];
int b[100010];
int main() {
int n;
scanf("%d", &n);
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
int max1 = a[n - 1];
for (int i = n - 2; i >= 0; i--) {
b[i] = max1 - a[i] + 1;
if (b[i] < 0) b[i] = 0;
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) throws FileNotFoundException {
/*Read and Write*/
//Scanner in=new Scanner(new FileReader("testA.txt"));
//FileWriter fw=new FileWriter("output... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 + 50;
int data[maxn], answer[maxn] = {0};
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; ++i) scanf("%d", &data[i]);
int high = data[n - 1];
for (int i = n - 2; i >= 0; --i) {
if (data[i] > high)
high = data[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;
int main() {
int n;
long long int a[100006], max;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%lld", &a[i]);
}
max = a[n - 1];
a[n - 1] = 0;
for (int i = n - 2; i >= 0; i--) {
if (a[i] <= max) {
a[i] = max - a[i] + 1;
} else {
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.util.Scanner;
public class Program {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner (System.in);
int n = Integer.parseInt(sc.nextLine());
int [] kq = new int [n];
String [] tmp = sc.nextLine().split(" ");
int max = Integer.parseInt(tmp[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;
void pl(long long a) { printf("%lld\n", a); }
void pll(long long a, long long b) { printf("%lld %lld\n", a, b); }
void plll(long long a, long long b, long long c) {
printf("%lld %lld %lld ", a, b, c);
}
void sss(string s) { cout << s, printf("\n"); }
long long string_to_l... |
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