prompt string | response string |
|---|---|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int no_of_houses;
cin >> no_of_houses;
vector<int> floors;
vector<int> lux_floors;
int max_height = 0;
int temp;
for (int i = 0; i < no_of_houses; i++) {
cin >> temp;
floors.push_back(temp);
}
for (int i = no_of_houses - 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.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Bit... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 arr[(int)1e5 + 5];
int m[(int)1e5 + 5];
int main() {
ios_base::sync_with_stdio(false);
cin >> N;
for (int i = 0; i < N; i++) cin >> arr[i];
for (int i = N - 1; i >= 0; i--) m[i] = max(arr[i], m[i + 1]);
for (int i = 0; i < N; i++) {
if (arr[i] > 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... | #In the name of God
n = int(raw_input())
arr = map(int, raw_input().split())
maxx = [0 for i in range(n)]
build = [0 for i in range(n)]
maxx[-1] = arr[-1]
for i in range(n-2, -1, -1):
maxx[i] = max(arr[i+1], maxx[i+1])
build[i] = 0 if arr[i] > maxx[i] else maxx[i]-arr[i]+1
print ' '.join(map(str, build))
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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(x) for x in input().split()]
b = [0] * n
for i in range(n - 2, -1, -1):
b[i] = max(b[i + 1], a[i + 1])
print(' '.join(str(max(0, x - y + 1)) for x, y in zip(b, 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;
map<int, int> mm;
int main() {
int i, j, n;
cin >> n;
int a[n];
vector<int> res;
for (i = 0; i < n; i++) {
cin >> a[i];
}
int mx = -1;
for (i = n - 1; i >= 0; i--) {
mx = max(mx, a[i]);
mm[mx]++;
if (mx == a[i] && mm[mx] == 1)
res.push_... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 bb {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int num = sc.nextInt();
int[] l = new int[num];
int max = 0;
for (int i=0;i<num;i++) {
l[i] = sc.nextInt();
}
int[] ans = new int[num];
for (int i=num-1;i>-1;i--) {
if (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;
inline int rit() {
int x;
scanf("%d", &x);
return x;
}
int n;
int a[100005], ans[100005];
void read() {
int i, d;
n = rit();
for (i = 0; i < n; i++) a[i] = rit();
d = a[n - 1] + 1;
ans[n - 1] = 0;
for (i = n - 2; i >= 0; i--) {
ans[i] = max(0, d - 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.*;
import java.math.*;
import static java.lang.Math.*;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Double.parseDouble;
import static java.lang.String.*;
public class Main {
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... | #include <bits/stdc++.h>
using namespace std;
int n, m, i, j, k, t, s, x, y, h[100005], hh[100005], d, l;
int main() {
cin >> n;
for (i = 0; i < n; i++) scanf("%d", h + i);
m = h[n - 1];
for (i = n - 2; i >= 0; i--) {
if (h[i] <= m)
hh[i] = m - h[i] + 1;
else
m = h[i];
}
for (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.util.*;
public class LuxuriousHouses {
public static void main(String[] args) {
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 ans[]=new int[n];
int max=a[n-1];
for(int i=n-2;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.util.Scanner;
public class LuxuriousHouses {
static Scanner sc = new Scanner(System.in);
public static void main(String args[]) {
int n = sc.nextInt();
int[] t = new int[n];
int[] tab = new int[n];
for (int i = 0; i < n; i++)
t[i] = sc.nextInt();
tab[n - 1] = 0;
int max = t[n - 1];
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... | z=int(input())
x=list(map(int,input().split()))
x.reverse()
f=x[0]
d=[]
for i in range(z):
if f>=x[i] and i!=0:
d.append(str(f-x[i]+1))
elif f==x[i]:
d.append("0")
elif f<x[i]:
d.append("0")
f=x[i]
d.reverse()
print(" ".join(d).strip())
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import org.omg.Messaging.SYNC_WITH_TRANSPORT;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Houses {
public static void main(String... args) {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
try {
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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[100009];
pair<int, int> ma[100009];
cin >> n;
stack<int> s;
for (int i = 1; i <= n; i++) cin >> a[i];
ma[n].first = a[n], ma[n].second = n;
for (int i = n - 1; i > 0; i--) {
if (a[i] > ma[i + 1].first) {
ma[i].first = 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.math.BigInteger;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Map;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.TreeMap;
public class Main{
public stati... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | /**
* Created by InfoLight on 10/3/2015.
*/
import java.util.Scanner;
public class Main {
public static long max(long a, long b){
if (a > b) return a;
else return b;
}
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
int totalCount = 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... | n=int(input())
a=list(map(int,input().split()))
flag=0
b=[0]
k=a[-1]
for i in range(1,n):
if k>a[n-i-1]:
b.append(-a[n-i-1]+k+1)
else:
if k-a[n-i-1]+1>0:
b.append(k-a[n-i-1]+1)
else:
b.append(0)
k=a[n-i-1]
for i in b[::-1]:
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;
const int MOD = 1000000007;
void solve() {
int n;
cin >> n;
vector<int> vec(n);
for (int& i : vec) cin >> i;
vector<int> pref(n);
pref[n - 1] = vec[n - 1];
for (int i = (n - 2); i >= 0; i--) {
pref[i] = max(pref[i + 1], vec[i]);
}
vector<int> ans(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() {
long long n;
cin >> n;
vector<long long> v(n);
for (long long i = 0; i < n; i++) cin >> v[i];
vector<long long> g(n);
g[n - 1] = v[n - 1];
for (long long i = n - 2; i >= 0; i--) g[i] = max(v[i + 1], g[i + 1]);
for (long long i = 0; i < n - 1; i+... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int n;
long long int num[100005], maxn;
int main() {
while (~scanf("%d", &n)) {
for (int i = 0; i < n; i++) scanf("%lld", &num[i]);
maxn = num[n - 1];
num[n - 1] = 0;
for (int i = n - 2; i >= 0; i--) {
if (num[i] <= maxn) {
num[i] = maxn - nu... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.*;
import java.util.*;
public class Main{
public static void main(String[] args) {
MyScanner sc = new MyScanner();
out = new PrintWriter(new BufferedOutputStream(System.out));
int n=sc.nextInt();
long arr[]=new long[n];
for(int i=0;i<n;i++)
arr[i]=sc.nextLong();
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 = 100100;
pair<int, int> b[maxn];
int ans[maxn], a[maxn];
int main() {
ios::sync_with_stdio(false);
int n, x;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> x;
a[i] = x;
b[i] = make_pair(x, i);
}
sort(b, b + n, greater<pair<int, int> >... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n;
cin >> n;
long long a[n];
for (int i = 1; i <= n; i++) cin >> a[i];
int Max = a[n] - 1;
for (int i = n; i >= 1; i--) {
if (a[i] > Max) {
Max = a[i];
a[i] = 0;
} else
a[i] = Max - a[i] + 1;
}
for (int 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 A[100005];
int ans[100005];
int main() {
int N;
cin >> N;
for (int i = 0; i < N; i++) cin >> A[i];
int ma = 0;
for (int i = N - 1; i >= 0; i--) {
ans[i] = max(0, ma - A[i] + 1);
ma = max(ma, A[i]);
}
for (int i = 0; i < N; i++) cout << 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... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class ProblemB {
public static void main(String[] args) {
InputReader in = new InputReader();
PrintWriter out = new PrintWriter(System.ou... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 MAX = 100005;
long long a[MAX], b[MAX];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
long long max = a[n - 1];
b[n - 1] = 0;
for (int i = n - 2; i >= 0; i--) {
if (a[i] > max) {
max = a[i];
b[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... | l=int(input())
c=list(map(int,input().split()))
m=[0]
for i in range(l-1,-1,-1):
m.append(max(c[i],m[l-1-i]))
for i in range(l):
if (m[l-1-i]<c[i])or(i==l-1):
print(0,end=' ')
else:
print(m[l-1-i]+1-c[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.Scanner;
public class PB {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Communication S=new Communication();
int n=S.i();
int[] a=S.is();
int M=0;
for(int i=n-1;i>-1;i--){
if(a[i]>M){
M=a[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;
long long mod = 1000000007;
long double pi = 3.1415926535;
long long num_generator() { return rand() % 1000; }
void array_generator(long long a[], long long n) {
for (long long i = 0; i < n; i++) a[i] = rand() % 100;
}
void solve() {
long long n;
cin >> n;
long long... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import itertools
n = int(input())
a = list(map(int, input().split()))
b = [0] * n
m = a[-1]
for i in range(n - 2, -1, -1):
b[i] = max(0, m - a[i] + 1)
m = max(m, a[i])
print(' '.join(map(str, b)))
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int A[n], B[n];
for (int i = 0; i < n; i++) {
cin >> A[i];
}
int max = A[n - 1];
B[n - 1] = 0;
for (int i = n - 2; i >= 0; i--) {
if (max < A[i]) {
max = A[i];
B[i] = 0;
} else {
B[i] = max - A[i] +... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(input())
arr = [int(i) for i in input().split()]
stack = []
res = [0 for _ in range(n)]
for i in range(n-1, -1, -1):
if (stack and stack[-1] >= arr[i]):
res[i] = stack[-1] - arr[i] + 1
if (not stack):
stack.append(arr[i])
elif (stack[-1] < arr[i]):
stack.append(arr[i])
pri... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | if __name__ == '__main__':
n = int(raw_input())
heights = [int(num) for num in raw_input().split()]
max_height, adds = 0, [0]*n
for i in reversed(range(n)):
if heights[i] <= max_height:
adds[i] = max_height - heights[i] + 1
max_height = max(heights[i], max_height)
prin... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;
public class j_luxurious_houses
{
static class Reader
{
final private int BUFFER_SIZE = 1 << 16;
private ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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())
mass1 = list(map(int,input().split()))
maxx = 0
mass2 = []
for i in range(n - 1, -1, -1):
if mass1[i] > maxx:
mass2.append(0)
maxx = mass1[i]
else:
mass2.append(maxx - mass1[i] + 1)
for i in range(n - 1, -1, -1):
print(mass2[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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class LuxuriousHouses {
public static void main(String[] args) throws IOException {
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(bf.readLine());
long[] house... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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()
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... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Houses {
public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(in.readLine());
String[] temp = in.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... | #include <bits/stdc++.h>
using namespace std;
const long long INF = 1e11 + 5;
const int MAX = 1e5 + 10;
int arr[MAX], answer[MAX];
int main() {
int t;
scanf("%d", &t);
for (int i = 0; i < t; ++i) scanf("%d", &arr[i]);
int ttl = -1;
for (int i = t - 1; i >= 0; --i) {
if (arr[i] > ttl)
answer[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... | n = int(input())
h = [int(i) for i in input().split()]
maxHeight = h[-1]
heightRequired = [0]
for i in range(len(h)-2, -1, -1):
heightRequired.append(max(0, maxHeight - h[i] + 1))
maxHeight = max(maxHeight, h[i])
heightRequired.reverse()
for i in range(len(heightRequired)):
print(heightRequired[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>
const long long INF = 2e9;
const long long N = 1e5 + 1;
const long long mod = 1e9 + 7;
const long double eps = 1E-7;
using namespace std;
bool used[10001];
void solve() {
int n, a[N];
vector<int> v;
cin >> n;
for (int i = 0; i < n; ++i) cin >> a[i];
int mx = 0;
for (int i = n - 1; i... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(input())
h = list(map(int, input().split()))
m_h = 0
res = []
for h in h[::-1]:
if m_h < h:
res.append(0)
m_h = h
else:
res.append(m_h - h + 1)
print(" ".join(map(str, res[::-1]))) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
void swapll(long long *a, long long *b) {
long long tmp = *a;
*a = *b;
*b = tmp;
}
void swapc(char *a, char *b) {
char tmp = *a;
*a = *b;
*b = tmp;
}
void solve() {
long long n;
cin >> n;
vector<long long> v(n);
for (long long i = 0; i < n; i++) cin >> 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... | /******/ (function(modules) { // webpackBootstrap
/******/ // The module cache
/******/ var installedModules = {};
/******/ // The require function
/******/ function __webpack_require__(moduleId) {
/******/ // Check if module is in cache
/******/ if(installedModules[moduleId])
/******/ return installedModu... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | //package Codeforces;
import java.io.BufferedReader;
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.StringTokenizer;
/**
* Created by mudi... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | /* package whatever; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Ideone
{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
Scanner s = 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... | #include <bits/stdc++.h>
using namespace std;
template <class T>
T power(T N, T P) {
return (P == 0) ? 1 : N * power(N, P - 1);
}
int main() {
long long int n;
long long int ara[100100];
cin >> n;
long long int boro = -1;
long long int maxi[100100];
memset(maxi, -1, sizeof maxi);
for (long long 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... |
//package solution;
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class Solution implements Runnable {
BufferedReader in;
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;
long long int n, a[100005], m, s, x, y, z;
string k;
int main() {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
for (int i = n; i > 0; i--) {
if (a[i] > m)
z = 0;
else
z = 1;
m = max(a[i], m);
a[i] = m - a[i] + z;
}
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>
int main() {
long long n, max, i;
scanf("%lld", &n);
long long a[n], b[n];
for (i = 0; i < n; i++) scanf("%lld", &a[i]);
b[n - 1] = 0;
max = a[n - 1];
for (i = n - 2; i >= 0; i--) {
if (max >= a[i]) {
b[i] = max - a[i] + 1;
} else if (max < a[i]) {
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 x;
int main() {
scanf("%i", &x);
int *A1 = new int[x];
for (int i = 0; i < x; i++) scanf("%i", &A1[i]);
int mx = A1[x - 1], sum = 0;
vector<int> A2;
A2.push_back(0);
for (int i = x - 2; i > -1; i--) {
A2.push_back(max(0, mx - A1[i] + 1));
mx = max(... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
long long int n, h, m, mx;
vector<int> vp;
cin >> n;
m = n;
while (n--) {
cin >> h;
vp.push_back(h);
}
mx = vp[m - 1];
vp[m - 1] = 0;
for (int i = m - 2; i >= 0; i--) {
if (vp[i] <= mx)
vp[i] = mx - vp[i] + 1;
else if (vp... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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.OutputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
/**
* Built using CHelper plug-in
* Actual solution is at the top
* @author Tifuera
*/
public class Main {
public static void main(Strin... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n=int(input())
l=list(map(int,input().split()))
v=0
S=[]
for i in range(n-1,-1,-1) :
if l[i]<=v :
S.append(abs(l[i]-v)+1)
else :
S.append(0)
if l[i]>v :
v=l[i]
S1=[]
for i in range(n-1,-1,-1) :
S1.append(str(S[i]))
print(' '.join(S1))
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken())... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | if __name__ == '__main__':
n = int(input())
l = list(map(int, input().split()))
a = []
max1 = 0
for i in range (n-1,-1,-1):
if l[i] > max1 :
max1 = l[i]
a.append(0)
else:
a.append((max1-l[i])+1)
a.reverse()
print(*a) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.InputStreamReader;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
* @author Erasyl Abenov
*
*
*/
pub... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 bucky{
public static void main (String args[]) {
Scanner input=new Scanner(System.in);
HashMap<Integer, Integer> map= new HashMap<Integer, Integer>();
int n=input.nextInt();
Long nums[]=new Long[n];
Long valori[]=new Long[n];
Long max=Long.valueOf(-1);
for (int ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
long a[100005];
int main() {
int N;
scanf("%d", &N);
for (int i = 1; i <= N; ++i) scanf("%ld", &a[i]);
long maxi = INT_MIN;
for (int i = N; i >= 1; --i) {
if (a[i] > maxi) {
maxi = a[i];
a[i] = 0;
} else if (a[i] < maxi) {
a[i] = maxi - a... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n, mx = 0;
cin >> n;
vector<long long> v(n), v2(n);
if (n == 1) {
cout << 0;
return 0;
}
for (int i = 0; i < n; i++) {
cin >> v[i];
}
mx = v[n - 1];
v2.push_back(0);
for (int i = n - 2; i >= 0; i--) {
if (v[i] - mx ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.*;
import java.util.*;
import java.math.*;
public class Div2_322_B {
public static void main(String[] args) throws IOException {
BufferedInputStream bis = new BufferedInputStream(System.in);
BufferedReader br = new BufferedReader(new InputStreamReader(bis));
int n = Integer.parseInt(br.readLine()... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int i, j, a[100005], b[100005], n, s, sm;
cin >> n;
for (i = 0; i < n; i++) cin >> a[i];
sm = 0;
for (i = n - 1; i >= 0; i--) {
b[i] = sm;
if (a[i] > sm) sm = a[i];
}
for (i = 0; i < n; i++) {
if (a[i] <= b[i])
cout << b[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 = int(raw_input())
r = raw_input()
l = r.split(' ')
l.reverse()
L = []
highest = 0
for y in l:
x = int(y)
if x > highest:
highest= x
L.append('0')
else:
L.append(str(highest-x+1))
L.reverse()
print ' '.join(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... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
public class CF_322_B {
public static void main(String[] args) throws IOException{
PrintWriter pw = new PrintWriter(System.out, true);
BufferedReader input = new BufferedReader(... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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()))
R = ["" for _ in range(n)]
R[n-1] = "0"
m = L[n-1]
for k in range(n-2,-1,-1):
if m >= L[k]:
R[k] = str(m-L[k]+1)
else:
R[k] = "0"
m = max(L[k],m)
print(" ".join(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;
int main() {
int n;
cin >> n;
int a[n + 1], i, g;
for (i = 1; i <= n; ++i) cin >> a[i];
g = a[n];
for (i = n - 1; i > 0; --i) {
if (a[i] > g) {
g = a[i];
a[i] = 0;
} else if (a[i] <= g) {
a[i] = (g + 1 - a[i]);
}
}
a[n] = 0;
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;
inline double min(double a, double b) {
if (a < b) return a;
return b;
}
int n;
int arr[100010];
int dp[100010];
int main() {
cin >> n;
for (int(i) = (0); (i) < (n); i++) cin >> arr[i];
dp[n - 1] = 0;
for (int i = n - 2; i >= 0; i--) {
dp[i] = max(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... | n=int(input())
a=list(map(int,input().split()))
b=(n-1)
c=[]
c.append(0)
for i in range(n-2,-1,-1):
if a[i]<=a[b]:
c.append(a[b]-a[i]+1)
else:
b=i
c.append(0)
print(*c[::-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 long long MAX = 1e5 + 10;
long long n, a[MAX];
long long Max[MAX];
int main() {
cin >> n;
for (long long i = 1; i <= n; i++) cin >> a[i];
Max[n] = a[n];
for (long long i = n - 1; i >= 1; i--) {
if (a[i] > Max[i + 1])
Max[i] = a[i];
else
Max... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main(int argc, const char* argv[]) {
int n, maxf;
cin >> n;
int h[n];
for (int i = 0; i < n; i++) cin >> h[i];
maxf = h[n - 1];
h[n - 1] = 0;
for (int i = n - 2; i >= 0; i--) {
if (h[i] > maxf) {
maxf = h[i];
h[i] = 0;
} else
h[i]... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, i, c = 0;
scanf("%i", &n);
vector<int> b(n), a(n);
for (i = 0; i < n; i++) scanf("%i", &b[i]);
for (i = b.size() - 1; i >= 0; c = max(c, b[i--]))
a[i] = max(0, c + 1 - b[i]);
for (int i = 0; i < n; i++) printf("%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... | #include <bits/stdc++.h>
using namespace std;
int n, Max, a[100005], b[100005];
int main() {
cin >> n;
for (int i = 0; i < n; ++i) cin >> a[i];
for (int i = n - 1; i >= 0; i--) {
b[i] = max(0, Max - a[i] + 1);
Max = max(Max, a[i]);
}
for (int i = 0; i < n; ++i) cout << b[i] << " ";
}
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.util.*;
public class solution {
public static void main(String[] args) {
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 max=Integer.MIN_VALUE;
int ans[]=new int[n];
for(int i=n-1;i>=0;i--) {
if(a[i]>max) {
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | N = int(input())
H = list(map(int, input().split()))
maxh = H[-1]
maxlst = []
for i in range(N):
maxh = max(maxh, H[N - i - 1])
maxlst.append(maxh)
maxlst.reverse()
ans = []
for i in range(N):
if i == N - 1:
a = 0
else:
maxv = maxlst[i + 1]
if H[i] == maxv:
a = 1
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
public class CFLuxHouse {
public static void main(String[] args)throws IOException {
// TODO Auto-generated method stub
BufferedReader buff = new BufferedReader(new InputStreamReader(System.in));
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... |
import java.util.Scanner;
public class B321 {
public static void main(String[] args) {
int n;
Scanner in=new Scanner(System.in);
n=in.nextInt();
int[] a=new int[n];
int[] b=new int[n];
for(int i=0;i<n;i++){
a[i]=in.nextInt();
}
int m=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.Arrays;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] h = new int[n];
for (int i = 0; i < h.length; i++) {
h[i] = sc.nextInt();
}
System.out.println(String.join(" ", Arrays.stream... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int ar[n], ans[n];
for (int i = 0; i < n; i++) cin >> ar[i];
int maxi = 0;
for (int i = n - 1; i >= 0; i--) {
ans[i] = max(maxi - ar[i] + 1, 0);
maxi = max(maxi, ar[i]);
}
for (int i = 0; i < n; i++) cout << 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... | import java.util.*;
import java.io.*;
import java.util.Scanner;
public class main {
public static void main (String [] args) throws Exception {
new main();
}
BufferedReader in;
PrintWriter out;
StringTokenizer st;
public String next() throws Exception { //takes next word from input
if (st == null || !st.... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 nn[100010];
bool bb[100010];
long long maxx[100010];
int main() {
int n;
while (scanf("%d", &n) != EOF) {
memset(bb, 0, sizeof(bb));
memset(nn, 0, sizeof(nn));
memset(maxx, -1, sizeof(maxx));
for (int i = 0; i < n; i++) {
scanf("%I64d", &... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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
raw_input()
houses = [int(i) for i in raw_input().strip().split(" ")]
N = len(houses)
maxh = 0
rs = [0] * N
for i in xrange(N-1, -1, -1):
# print i
h = houses[i]
if maxh < h:
maxh = h
r = 0
else:
r = maxh + 1 - h
rs[i] = str(r)
print " ".join(rs)
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int a[n + 1];
a[n] = 0;
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
vector<int> res;
for (int i = n - 1; i >= 0; --i) {
res.push_back(max(0, a[i + 1] - a[i] + 1));
a[i] = max(a[i], a[i + 1]);
}
for (int i = n - ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int a[100020];
long long b[100020];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
long long max = a[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (a[i] <= max)
b[i] = max - a[i] + 1;
else
max = a[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... | #!/usr/bin/env python
#-*- coding: utf-8 -*-
def main_fun():
n = int(raw_input().strip())
line = raw_input().strip().split(' ')
d = {n-1:0}
for x in range(n-2,-1,-1):
if int(line[x+1]) > d[x+1]:
d[x] = int(line[x+1])
else:
d[x] = d[x+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=[int(i) for i in raw_input().split()]
m=-1000000007
k=[]
for i in range(n-1,-1,-1):
x=l[i]
if x>m:
m=x
k.append(0)
else:
k.append(m+1-x)
for i in range(n):
print k.pop(), |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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.util.*;
import java.io.*;
import java.util.stream.*;
public class b {
private void solve() throws IOException {
int n = nextInt();
Integer[] a = new Integer[n];
for(int i = 0; i < n; ++i) {
a[i] = nextInt();
}
int curMax = 0;
for(int i = n -... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long power(long long a, long long b) {
long long ret = 1;
while (b) {
if (b % 2 == 1) ret = (ret * a) % ((int)(1e9) + 7);
b /= 2;
a = (a * a) % ((int)(1e9) + 7);
}
return... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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];
fill(sol, sol + n, 0);
int mx = a[n - 1];
sol[n - 1] = 0;
for (int i = n - 2; i >= 0; i--) {
if (a[i] <= mx) {
sol[i] = 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... | //Author: Mo Abjal, MJP Rohilkhand University Bareilly, UP, India 2018.
///////////////////////////////////////////////////////////////////////////////////////////////////////////
/////////////////////////////////////////********************//////////////////////////////////////////////
////////////////////////////////... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | var numeric = readline(),
floor = readline().split(' '),
result = [0];
for (; floor.length; ) {
var last_floor = +floor[floor.length - 1],
prev_floor = +floor[floor.length - 2];
if (prev_floor > last_floor) {
result.unshift(0);
floor.splice(floor.length - 1, 1);
} else {
if (last_floor === prev_floor... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
long long int n;
cin >> n;
long long int a[n], b[n], max = 0;
b[n - 1] = 0;
for (long long int i = 0; i < n; i++) cin >> a[i];
for (long long int i = n - 1; i >= 0; i--) {
if (i == n - 1)
max = a[i];
else {
if (a[i] <= max) {
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n, houses = int(raw_input()), map(int, raw_input().split())
m, l = houses[n-1], [0]
for i in xrange(n-2, -1, -1):
l.append(0 if houses[i] > m else m + 1 - houses[i])
m = max(m, houses[i])
print ' '.join(map(str, reversed(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... | import java.util.Scanner;
public class B581 {
public static void main(String[] args) {
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 max = a[n-1];
int[] add = new int[n];
for (int i = n-2; i >= 0; i--) {
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | input()
a=map(int,raw_input().split())
r,m=[],0
for d in a[::-1]:
r+=max(m+1-d,0),
m=max(m,d)
for d in r[::-1]:
print d,
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(raw_input().strip())
houses = map(int, raw_input().strip().split(' '))
answers = [-1] * n
maxfloors = -1
for i in xrange(len(houses) - 1, -1, -1):
answers[i] = max(maxfloors + 1 - houses[i], 0)
if houses[i] > maxfloors:
maxfloors = houses[i]
print ' '.join(map(str, answers)) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 + 20;
int a[MAXN], ans[MAXN];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n, maximum = 0;
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
ans[n - 1] = 0, maximum = max(maximum, a[n - 1]);
for (int i = n... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> v;
for (int i = 0; i < n; i++) {
int a;
cin >> a;
v.push_back(a);
}
int Max = v.back();
v.back() = 0;
for (int i = v.size() - 2; i >= 0; i--) {
if (v[i] < Max)
v[i] = ((Max - v[i]) + 1);
els... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 arr[100000 + 7], mx, d, ans[100000 + 7], n, i, j, k;
while (scanf("%lld", &n) == 1) {
for (i = 1; i <= n; i++) {
scanf("%lld", &arr[i]);
}
mx = 0;
for (i = n; i > 0; i--) {
if (arr[i] > mx) {
mx = arr[i];
... |
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