prompt string | response string |
|---|---|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.*;
import java.util.Stack;
import java.util.StringTokenizer;
public class Main {
static BufferedReader bufferedReader;
static PrintWriter printWriter;
static StringTokenizer stringTokenizer;
static void solve() throws Exception {
int houses = nextInt();
int max;
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, a[100009], b[100009], i, flag = 0;
cin >> n;
for (i = 0; i < n; i++) cin >> a[i];
for (i = n - 1; i >= 0; i--) {
if (a[i] > flag)
b[i] = -1;
else
b[i] = max(a[i], flag);
flag = max(flag, a[i]);
}
for (i = 0; i < n; i++... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
size_t n;
cin >> n;
vector<int> houses(n);
for (size_t i = 0; i < n; ++i) {
cin >> houses[i];
}
vector<int> res(n);
int max_elem = 0;
for (size_t i = 0; i < n; ++i) {
size_t j = n - 1 - i;
if (houses[j] > max_elem) {
max_elem =... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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()); li = list(map(int, input().split(" ")))
lu = []
k = 0
for i in range(n-1, -1, -1):
lu.append(max(0, k+1-li[i]))
k = max(k, li[i])
lu.reverse()
print(' '.join(map(str, lu)))
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
void solve() {
int n, m = 0;
cin >> n;
vector<pair<int, int>> a(n);
for (auto &u : a) cin >> u.first;
a[n - 1].second = 0;
for (int i = n - 2; i >= 0; i--) {
m = max(m, a[i + 1].first);
a[i].second = max(m, a[i + 1].first);
}
for (auto &u : a) {
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.util.Scanner;
public class codeforces581B
{
public static void main(String[] args)
{
Scanner s=new Scanner(System.in);
StringBuilder sb = new StringBuilder();
int n=s.nextInt(),i;
int[] a = new int[n];
int[] m = new int[n];
for(i=0;i<n;i++)
a[i]=s.nextInt();
m[n-1]=a[n-1];
for(i=n-... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #~ import io
#~ import sys
#~ import time
#~ start = time.clock()
#~ test = '''5
#~ 1 2 3 1 2'''
#~ test = '''4
#~ 3 2 1 4'''
#~ sys.stdin = io.StringIO(test)
n = int(input())
floors = list(map(int, input().split()))
#~ print(floors)
ma = floors[:]
max_so_far = float("-inf")
for i in range(n-1,-1,-1):
ma[i] = max(... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-9;
const double PI = acos(-1.0);
const int MOD = 1000 * 1000 * 1000 + 7;
const int INF = 2000 * 1000 * 1000;
const int MAXN = 100010;
template <typename T>
inline T sqr(T n) {
return n * n;
}
int n;
int a[MAXN], ans[MAXN], suff;
int main() {
scanf(... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
long long a[100100], n, b[100100], mx = -1e10;
set<pair<int, int> > ss;
int main() {
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
b[n - 1] = a[n - 1];
for (int i = n - 2; i >= 0; i--) {
b[i] = max(b[i + 1], a[i]);
}
for (int i = 0; i < n - 1; i++) {
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(raw_input())
l = map(int, raw_input().split())
m = [0]*n
max_value = 0
for i in xrange(n-1, -1,-1):
max_value = max(max_value, l[i])
m[i] = max_value
for i in xrange(n-1):
ans = m[i+1] - l[i]
if ans < 0: ans = 0
else: ans += 1
print ans,
print 0
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n=int(raw_input())
l=map(int,raw_input().split())
ans=[0]
cnt,chk=l[-1],1
for i in l[-2::-1]:
if i<=cnt:
ans.append(cnt-i+1)
else:
cnt=i
ans.append(0)
print ' '.join(map(str,ans[::-1])) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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];
int max_r;
int b[n];
fill(b, b + n, 0);
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
max_r = arr[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (max_r < arr[i]) {
max_r = arr[i];
} else
b... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(raw_input())
inp = raw_input()
inp =inp.split()
houses = [int(i) for i in inp]
added=[0 for i in inp]
max=0
for i in range(n-1,-1,-1):
if houses[i]<=max:
added[i]= max - houses[i]+1
if houses[i]>max:
max=houses[i]
for i in added:
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... | #include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int a[N], b[N];
int main() {
int n;
while (scanf("%d", &n) != EOF) {
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
int mmax = a[n];
b[n] = 0;
for (int i = n - 1; i > 0; i--) {
if (a[i] == mmax)
b[i] =... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(input())
h = list(map(int, input().split()))
cur = 0
l = [0] * n
for i in range(n - 1, -1, -1):
l[i] = max(0, cur + 1 - h[i])
cur = max(cur, h[i])
print (" ".join(map(str, 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;
int b, d, n, m, i, j, v[2100000], f[100001], maxi, a, ul, nrq = 1, nrr = 1,
newq, newr, ls, ld;
char s[100005], z;
vector<int> c;
double r, t, q, ff, pri;
bool cmm(int a, int b) {
if (a > b) return 1;
return 0;
}... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #!/usr/bin/python
if __name__ == '__main__':
size = int(raw_input().strip())
arr = [int(x) for x in raw_input().strip().split(" ")]
max_num = -1
op = []
for i in reversed(arr):
if i == max_num:
op.append(1)
elif i > max_num:
op.append(0)
max_num ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 Nmax = 1000000000;
int n;
vector<long long> a;
vector<long long> tree;
void add(int x, int i, int l, int r, int pos) {
if (i < l || i > r) return;
if (l == r) {
tree[pos] = x;
return;
}
int mid = (l + r) / 2;
add(x, i, l, mid, pos * 2);
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;
const int MXN = 1e5 + 3;
long long n, sum, mx = -MXN, id;
int main() {
long long a[MXN];
pair<int, int> b[MXN];
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
for (int i = n; i >= 1; i--) {
if (a[i] > mx) {
mx = a[i];
id = 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;
string Y = "YES";
string N = "NO";
string yy = "Yes";
string nn = "No";
int main() {
long long n;
while (cin >> n) {
long long ar[n];
for (long long i = 0; i < n; i++) {
cin >> ar[i];
}
long long mx = 0;
vector<long long> v(n);
for (long 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... | T = int(input())
st = raw_input().split()
st = list(map(int,st))
mx = 0
a = [0]*T
for i in range(T-1,-1,-1):
a[i] = max(0,mx + 1 - st[i])
mx = max(mx,st[i])
for i in range(0,T):
print a[i],
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
int n;
cin >> n;
int a[n + 1], ans[n + 1];
for (int i = 0; i < n; i++) cin >> a[i];
int max = a[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (a[i] > max)
ans[i] = 0, 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 = int(input())
a = list(map(int, input().split()))
_max = max(a)
b = [0] * n
b[n - 1] = 0
c = a[n - 1]
for i in range(n - 2, -1, -1):
if(a[i] > c):
b[i] = 0
c = a[i]
else:
c = max(c, a[i])
b[i] = c - a[i] + 1
for i in range(n):
print(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... | from sys import *
inp = lambda : stdin.readline()
def main():
n = int(inp())
l = [int(i) for i in inp().split()]
maxl,maxe = [0 for i in range(n)],0
for i in range(n-1,-1,-1):
maxe = max(maxe, l[i])
maxl[i] = maxe
for i in range(n-1):
print(max(0,maxl[i+1]-l[i]+1),end=" ")
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.BufferedReader;
import java.io.FileReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.TimeUnit;
public class Codeforces {
StringTokenizer tok;
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.Scanner;
public class B {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] hs = new int[n];
for (int i = 0; i < n; i++)
hs[i] = sc.nextInt();
sc.close();
long[] max = new long[n + 1];
for (int i = n - 1; i >= 0; i--)
max[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 Faruk
*/
import java.util.Arrays;
import java.util.Scanner;
import java.util.HashSet;
import java.util.HashMap;
import java.util.ArrayList;
public class tmp {
public static void main(String [] args){
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
long long i, mi, n, a[100005], b[100005];
cin >> n;
for (i = 0; i < n; i++) cin >> a[i];
mi = a[n - 1];
b[n - 1] = 0;
for (i = n - 2; i >= 0; i -= 1) {
if (a[i] > mi) {
b[i] = 0;
mi = a[i];
} else if (a[i] == mi) {
b[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 main() {
int n;
cin >> n;
int arr[n];
int ans[n];
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
int max = 0;
for (int i = n - 1; i >= 0; i--) {
int flag = 0;
if (arr[i] > max) {
max = arr[i];
flag = 1;
}
if (max == arr[i]... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
/**
*
* @author Shreyas
*/
public class Height {
/**
* @param args the command line arguments
*/
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... | 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
*/
public class Main {
public static... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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() ] + [ 0 ]
s = [ 0 for x in range(n) ]
m = a[-1]
for i in range(n):
m = max(m , a[n - i])
s[n - i - 1] = m
res = ''
for i in range(n):
res += str(max(s[i] - a[i] + 1, 0)) + ' '
print(res[:-1])
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
public class P322_2B implements Runnable {
int nextInt(StreamTokenizer in) throws Exception {
in.nextToken();
return (int) in.nval;
}
StreamTokenizer in;
private int 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... | #!/usr/bin/env python
n = input()
a = map(int,raw_input().split())
b = [None] * n
b[n-1] = 0
i = n-2
tmp = 0
while i >= 0:
tmp = max(tmp, a[i+1])
b[i] = max(0, tmp+1-a[i])
i-=1
for x in range(0,n) : print b[x], |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1;
int a[N], n, b[N], maxn;
int main() {
cin >> n;
for (int i = 1; i <= n; i++) cin >> a[i];
b[n] = 0;
maxn = a[n];
for (int i = n - 1; i >= 1; i--) {
if (a[i] > maxn) {
b[i] = 0;
maxn = a[i];
} else {
b[i] = maxn - 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 sys
import string
from collections import Counter, defaultdict
from math import fsum, sqrt, gcd, ceil, factorial
from itertools import combinations, permutations
# input = sys.stdin.readline
flush = lambda: sys.stdout.flush
comb = lambda x, y: (factorial(x) // factorial(y)) // factorial(x - y)
# inputs
# ip ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 string YESNO[2] = {"NO", "YES"};
const string YesNo[2] = {"No", "Yes"};
const string yesno[2] = {"no", "yes"};
void YES(bool t = 1) { cout << YESNO[t] << '\n'; }
void Yes(bool t = 1) { cout << YesNo[t] << '\n'; }
void yes(bool t = 1) { cout << yesno[t] << '\n'; }
temp... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
int main() {
int n, i, j;
int max = INT_MIN;
scanf("%d", &n);
int a[n];
int b[n];
for (i = 0; i < n; ++i) {
scanf("%d", &a[i]);
}
for (j = n - 1; j >= 0; --j) {
if (a[j] <= max) {
b[j] = max + 1 - a[j];
} else {
b[j] = 0;
}
if (max < a[j]) {
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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], maxx[n + 1];
for (int i = 1; i < n + 1; ++i) cin >> a[i];
maxx[n] = a[n];
for (int i = n - 1; i > 0; --i) maxx[i] = max(maxx[i + 1], a[i]);
for (int i = 1; i < n; ++i)
if (maxx[i + 1] >= a[i])
cout << maxx[... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 = 100005;
int n;
int h[maxn], r[maxn];
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%d", &h[i]);
}
memset(r, 0, sizeof(r));
for (int i = n; i >= 1; --i) {
if (i == n) continue;
r[i] = max(h[i + 1], r[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;
void Rd(int &res) {
char c;
res = 0;
while (c = getchar(), !isdigit(c))
;
do {
res = (res << 3) + (res << 1) + (c ^ 48);
} while (c = getchar(), isdigit(c));
}
int A[100005], B[100005];
int main() {
int n, i, j, k;
scanf("%d", &n);
for (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... | #include <bits/stdc++.h>
using namespace std;
int mx, n, n1, k, a[100000], b[100000];
int main() {
cin >> n;
mx = 0;
for (int i = 0; i < n; i++) cin >> a[i];
for (int i = n - 1; i >= 0; i--) {
b[i] = max(0, (mx - a[i] + 1));
mx = max(mx, a[i]);
}
for (int i = 0; i < n; i++) cout << b[i] << " ";
re... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.*;
import java.util.*;
import java.math.*;
public class B {
public static void main(String args[]) {
try {
InputReader in = new InputReader(System.in);
BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
int n = in.readInt();
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 sun.font.FontRunIterator;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
/**
* Created by baba_beda on 9/22/15.
*/
public class B {
public static void main(String[] args) {
new B().run();
}
Scanner in;
void run() {
try {
in = 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... | #include <bits/stdc++.h>
using namespace std;
int main() {
long long int a[100009], i, j, k, l, m, n, b[100009];
while (cin >> n) {
for (i = 0; i < n; i++) {
cin >> a[i];
}
m = a[n - 1];
b[n - 1] = 0;
k = 0;
for (i = n - 2; i >= 0; i--) {
if (m - a[i] == 0) {
b[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())
v = list(map(int, input().split()))
v1 = [0]*n
m = 0
for i in range(n):
if v[-i-1]>m:
m=v[-i-1]
else:
v1[-i-1]=m-v[-i-1]+1
print(' '.join(map(str,v1))) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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;
/**
* Created by Microsoft on 18/10/2015.
*/
public class algo_contest1_q5 {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.util.Arrays;
import java.util.Scanner;
public class BigRecursion {
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[] b = new int[n];
b[n-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... | n=int(input())
lst=list(map(int,input().split()))
res,mx=[0],lst[-1]
for i in range(n-2,-1,-1):
x=lst[i]
if x<=mx:res.append(mx-x+1)
else:res.append(0);mx=x
res.reverse()
print(*res) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(input())
h = list(map(int, input().split())) + [0]
h_max = 0
res = [0] * n
for i in range(n, 0, -1):
h_max = max(h_max, h[i] + 1)
res[i - 1] = max(h_max - h[i - 1], 0)
print(" ".join(map(str, res))) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int p, n;
cin >> n;
int a[100001] = {}, dp[100001], h[100001] = {};
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
dp[n - 1] = a[n - 1];
for (int i = n - 2; i >= 0; --i) {
if (a[i] == dp[i + 1]) h[i] = 1;
dp[i] = max(dp[i + 1], a[i]);
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n=int(input())
l=list(map(int,input().split()))
s=[0]*n
a=[0]*n
s[-1]=l[-1]
for i in range(1,n):
if l[n-i-1]<=s[n-i]: a[n-i-1]=s[n-i]-l[n-i-1]+1
s[n-i-1]=max(l[n-i-1],s[n-i])
print(' '.join(map(str,a))) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
long double PI = acosl(-1);
bool compare_int(int a, int b) { return (a > b); }
bool compare_string(string a, string b) { return a.size() < b.size(); }
bool compare_pair(const pair<int, int> &a, const pair<int, int> &b) {
return (a.second < b.second);
}
long long int fact(... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 B322 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int arr[] = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = in.nextInt();
}
int max=0;
int ans[]= new int [n];
for(int i=arr.length-1;i>=0;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... | n = int(input())
h = [int(x) for x in input().split()]
tmp = []
for i in range (n):
tmp.append(0)
tmp[n - 1] = h[n - 1]
for i in range (n - 2, -1, -1):
if h[i] > tmp[i + 1]:
tmp[i] = h[i]
else:
tmp[i] = tmp[i + 1]
for i in range (n - 1):
if h[i] == tmp[i + 1]:
print(1, end = " ")... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... |
import java.io.*;
import java.util.*;
import java.math.*;
public class Main {
public static void main(String[] args) throws java.lang.Exception {
Reader pm =new Reader();
int n = pm.nextInt();
int[] a = new int[n];
for(int i=0;i<n;i++){
a[i] = pm.nextInt();
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(raw_input())
houses = [int(x) for x in raw_input().split()]
m = 0
ans = [0] * n
for i in xrange(n-1, -1, -1):
if houses[i] <= m:
ans[i] = m - houses[i] + 1
else:
ans[i] = 0
m = max(houses[i], m)
print " ".join([str(x) for x in ans])
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n=int(input())
a=list(map(int,input().split()))
a.reverse()
c=[]
c.append(0)
maxi=a[0]
for i in range(len(a)-1):
if(a[i+1]>maxi):
c.append(0)
maxi=a[i+1]
elif(a[i+1]<=maxi):
c.append(abs(1+maxi-a[i+1]))
c.reverse()
print(*c)
|
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | 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.pop();
} else {
if (last_floor === prev_floor) {
result.unshift(... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = int(input())
a = list(map(int, input().split()))
b = [0] * n
ma = a[-1]
for i in range(n - 2, -1, -1):
if a[i] <= ma:
b[i] = ma - a[i] + 1
else:
ma = a[i]
string_ints = [str(int) for int in b]
print(" ".join(string_ints)) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 arr[100005], arr2[100005];
map<long long, int> cnt;
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> arr[i];
cnt[arr[i]]++;
arr2[i] = arr[i];
}
for (int i = n - 2; i >= 0; i--) {
arr2[i] = max(arr2[i], arr2[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 = [int(x) for x in input().split()]
m = [0] * (n + 1)
m[n] = - 10 ** 9 - 7
for i in range(n-1,0,-1):
m[i] = max(m[i+1], a[i])
for i in range(n):
print(max(a[i], m[i+1] + 1) - a[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... | n = int(input())
a = list(map(int, input().split()))
ans = []
m = 0
for i in range(n - 1, -1, -1):
ans.append(max(0, (m + 1) - a[i]))
m = max(m, a[i])
ans.reverse()
for i in ans:
print(i, end = ' ') |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = input("")
hString = raw_input("")
h = hString.split(" ")
h = map(int, h)
result = ["0"]
prevMax = n - 1
for i in range(n-2, -1, -1):
if h[prevMax] < h[i]:
result.append("0")
prevMax = i
else:
result.append(str(h[prevMax] - h[i] + 1))
print " ".join(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;
int main() {
int n;
cin >> n;
int arr[n];
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
int tmp = 0, index;
for (int i = n - 1; i >= 0; i--) {
if (tmp < arr[i]) {
tmp = arr[i];
index = i;
}
if (i == index) {
arr[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.util.*;
public class _581B_Luxurious_Houses {
public static void main(String[] args){
Scanner leer = new Scanner(System.in);
int n = leer.nextInt();
long h[]= new long[n];
for (int i = 0; i < n; i++){
h[i]= leer.nextLong();
}
long T[]= new long[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() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.precision(20);
int n;
cin >> n;
int a[n];
vector<int> v;
for (int i = 0; i < n; i++) cin >> a[i];
v.push_back(0);
int m = a[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (a[i] > 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... | #!/usr/bin/env python3
# 20150929
# 11:10PM-
def main():
n = int(input())
a = list(map(int, input().split()))
b = [0]*len(a)
m = a[-1]
for i in range(len(a)-2,-1,-1):
if a[i] > m:
m = a[i]
else:
b[i] = m - a[i] + 1
print(' '.join(map(str,b)))
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 array
def main():
n = int(input())
hs = input().split()
maxi = 0
p = array.array('Q', n * [0])
for i in range(n - 1, -1, -1):
hi = int(hs[i])
p[i] = max(0, maxi - hi + 1)
maxi = max(maxi, hi)
for i in range(n - 1):
print(p[i], '', end='')
print(p[n... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.util.*;
import java.io.*;
public class Luxurious_Houses
{
public static void main(String args[]) throws Exception
{
BufferedReader f=new BufferedReader(new InputStreamReader(System.in));
int runs=Integer.parseInt(f.readLine());
StringTokenizer st=new StringTokenizer(f.readLine());
int[] arr=new 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... | # -*- coding: utf-8 -*-
import sys,math,heapq,itertools as it,fractions,re,bisect,collections as coll
n = int(raw_input())
h = map(int, raw_input().split())
mx = [0] * n
for i in xrange(n - 2, -1, -1):
mx[i] = max(mx[i + 1], h[i + 1])
ans = [max(0, mx[i] - h[i] + 1) for i in xrange(n)]
print " ".join(map(str, an... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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 testP02 {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int[] h=new int[n];
for(int i=0;i<n;i++)
h[i]=sc.nextInt();
int[] a=new int[n];
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int n, h[100008], mayor, resp[100008], mayores[100008];
int main() {
scanf("%d", &n);
for (int i = 0; i < n; ++i) {
scanf("%d", &h[i]);
}
mayor = -1;
for (int j = n - 1; j >= 0; j--) {
mayores[j] = mayor;
if (h[j] > mayor) {
mayor = h[j];
}
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from 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.Locale;
import java.util.StringTokenizer;
public class Main implements Runnable {
final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;
BufferedReader in;
PrintWriter out;
StringTokenizer tok = new StringTokenizer("");
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... | n=input()
a=map(int,raw_input().split())
rs=[0]*n
mx=0
for i in xrange(n-1,-1,-1):
rs[i]=max(mx+1,a[i])
mx=max(mx,a[i])
for i in xrange(n):
print rs[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;
long long n, arr[100005], Max, ans[100005];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cin >> n;
for (long long i = (long long)(1); i <= (long long)(n); i++) cin >> arr[i];
Max = arr[n];
ans[n] = 0;
for (long long i = (long long)(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;
long long h[100010], a[100010];
int main() {
int n;
while (cin >> n) {
memset(a, 0, sizeof(a));
for (int i = 0; i < n; ++i) cin >> h[i];
a[n - 1] = 0;
if (n > 1) a[n - 2] = h[n - 1];
for (int i = n - 3; i >= 0; --i) {
a[i] = max(a[i + 1], h[i +... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
bool compare(int a, int b) { return a > b; }
class Compare {
public:
bool operator()(int a, int b) { return a > b; }
};
int mod(int a, int b) { return ((a % b) + b) % b; }
int a[100009], n, suffix[100009];
int main() {
scanf("%d", &n);
for (int i = 0; i < n; i++) sca... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n, a[N];
cin >> 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--) {
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.*;
import java.util.*;
public class test {
int INF = (int)1e9;
long MOD = 1000000007;
void solve(InputReader in, PrintWriter out) throws IOException {
int n = in.nextInt();
int[] a = new int[n];
for(int i=0; i<n; i++) {
a[i] = in.nextInt();
... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
Input... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
long long int pwr(long long int a, long long int b, long long int mod) {
a %= mod;
if (a < 0) a += mod;
long long int ans = 1;
while (b) {
if (b & 1) ans = (ans * a) % mod;
a = (a * a) % mod;
b /= 2;
}
return ans;
}
long long int gcd(long long int a,... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | import java.util.*;
public class HelloWorld{
public static void main(String []args){
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int[] arr=new int[n];
for(int i=0;i<n;i++){
arr[i]=sc.nextInt();
}
int[] ans=new int[n];
ans[n-1]... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
long long i, n, a, max = -1;
cin >> n;
vector<int> v, m;
while (n--) {
cin >> a;
v.push_back(a);
}
for (i = v.size() - 1; i >= 0; i--) {
if (v[i] > max) max = v[i];
m.push_back(max);
}
for (i = 0; i < v.size() - 1; i++) {
if ... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | # your code goes here
n = int(raw_input())
h = map(int, raw_input().split())
l=[0]*n
max = 0
arr = h[::-1]
# print h, arr
for i in range(n):
if arr[i] <= max:
l[i] = max - arr[i] + 1
else:
max = arr[i]
# print max
l = l[::-1]
print ' '.join(str(x) for x in 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... | n = int(input())
h = [i for i in map(int, input().split())]
h = h[::-1]
ans = []
maxH = h[0]
for i in range(1, len(h)):
if maxH - h[i] + 1 < 0:
ans.append(0)
else:
ans.append(maxH - h[i] + 1)
maxH = max(maxH, h[i])
ans = ans[::-1]
ans.append(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.Scanner;
public class probACon322 {
/**
* @param args
*/
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = 0;
n = sc.nextInt();
int[] houses = new int[n];
int[] diff = new int[n];
for (int i =... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | n = input()
s = map(int,raw_input().split())
a = s[-1]
b = ['0']
for i in xrange(n-2,-1,-1):
if s[i]==a:
b.append('1')
elif s[i]>a:
b.append('0')
else:
b.append(str(a+1-s[i]))
a = max(a,s[i])
print ' '.join(b[::-1]) |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | from math import *
from Queue import *
from sys import *
from datetime import *
n = int(raw_input())
h = map(int, raw_input().split())
m = [0 for i in range(n)]
m[n-1] = h[n-1]
for i in range(n-2, -1, -1):
m[i] = max(m[i+1], h[i])
res = []
for i in range(n-1):
res.append(max(m[i+1]-h[i]+1,0))
res.append(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.File;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main {
public static void main(String args[]) throws IOException {
BufferedReader in;
File f = new File("entrada.txt");
Str... |
Problem: The capital of Berland has n multifloor buildings. The architect who built up the capital was very creative, so all the houses were built in one row.
Let's enumerate all the houses from left to right, starting with one. A house is considered to be luxurious if the number of floors in it is strictly greater th... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
scanf("%d", &n);
int a[n], r[n];
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
int mx = a[n - 1];
r[n - 1] = 0;
for (int i = n - 2; i > -1; i--) {
if (mx >= a[i])
r[i] = mx - a[i] + 1;
else
mx = a[i], r[i] = 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;
int main() {
int n;
cin >> n;
vector<int> v(n), res(n);
for (int i = 0; i < n; i++) cin >> v[i];
reverse(v.begin(), v.end());
int mx = v[0];
res[0] = 0;
for (int i = 1; i < n; i++) {
if (v[i] > mx) {
mx = v[i];
res[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 main() {
long long int n, i, max;
cin >> n;
long long int a[n], b[n + 1];
for (i = 0; i < n; i++) cin >> a[i];
max = a[n - 1];
for (i = n - 1; i >= 0; i--) {
if (a[i] >= max) {
b[i] = a[i];
max = a[i];
} else
b[i] = b[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()))
x=a[n-1]
b=[0]*n
for i in range(n-2,-1,-1):
b[i]=max(0,x-a[i]+1)
if a[i]>x:x=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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.Input... |
Problem: Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, a... | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.util.StringTokenizer;
import java.io.Writer;
... |
Problem: Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, a... | import java.io.*;
import java.util.*;
public class A603 {
public static void main(String[] args) throws IOException {
input.init(System.in);
PrintWriter out = new PrintWriter(System.out);
int n = input.nextInt();
int res = 1;
char[] s = input.next().toCharArray();
for(int i= 1; i<n; i++) if(s[i... |
Problem: Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, a... | input()
a = input()
s = []
for c in a:
if not s or s[-1][0] != c:
s.append([c, 1])
else:
s[-1][1] += 1
s2 = sorted(s, key=lambda x: x[1])
delta = 0
if s2[-1][1] >= 3 or len(s2) > 1 and s2[-2][1] >= 2:
delta = 2
elif s2[-1][1] >= 2:
delta = 1
print(len(s) + delta) |
Problem: Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, a... | import java.io.*;
import java.math.*;
import java.util.*;
/**
*
* @author Saju
*
*/
public class Main {
private static int dx[] = { 1, 0, -1, 0 };
private static int dy[] = { 0, -1, 0, 1 };
private static final long INF = Long.MAX_VALUE;
private static final int INT_INF = Integer.MAX_VALUE;
... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.