Search is not available for this dataset
name stringlengths 2 112 | description stringlengths 29 13k | source int64 1 7 | difficulty int64 0 25 | solution stringlengths 7 983k | language stringclasses 4
values |
|---|---|---|---|---|---|
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
template <typename T>
T power(T a, int n, int mod) {
T res = 1;
while (n) {
if (n % 2 == 1) res = (res * a) % mod;
n /= 2;
a = (a * a) % mod;
}
return res;
}
int maxSeq(vector<int>& v) {
int lv = v[0];
int li = 1;
vector<int> v2;
for (int i = 1; ... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
#pragma comment(linker, "/STACK:2000000")
#pragma comment(linker, "/HEAP:2000000")
using namespace std;
void print_width(long long x) {
std::cout << std::fixed;
std::cout << std::setprecision(x);
}
long long power(long long x, long long y, long long p = 1000000007) {
long long res = 1;
... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
multiset<int> S;
multiset<int>::iterator it;
int main() {
int n, x, res = 0;
scanf("%d", &n);
for (int i = 0; i < n; ++i) scanf("%d", &x), S.insert(x);
while (!S.empty()) {
++res;
int cur = *S.begin(), cnt = 1;
S.erase(S.begin());
while (!S.empty()) ... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int MAX_N = 110;
int x[MAX_N], N, nr[MAX_N];
int main() {
cin >> N;
for (int i = 1; i <= N; ++i) cin >> x[i];
sort(x + 1, x + N + 1);
int l = 1, r = N, ans = 0;
while (l <= r) {
int mid = (l + r) / 2, aux = 0, curr = 1;
bool ev = true;
for (int i... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.Arrays;
import java.util.Scanner;
public class FoxBoxAcc {
static int x[];
static boolean taken[];
static int n;
static void init() {
Scanner sc = new Scanner(System.in);
n = sc.nextInt();
x = new int[n];
taken = new boolean[n];
for(int i = 0; i < n; i++)
x[i] = sc.nextInt();
... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int i, j, t, n, ans, count = 0, a[105], b[105] = {0}, k;
cin >> n;
for (i = 0; i < n; i++) {
cin >> a[i];
b[i]++;
}
sort(a, a + n);
for (i = 0; i < n; i++) {
for (j = i + 1; j < n; j++) {
if (b[i] <= a[j]) {
b[j] = b[i] + b... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import math
class CodeforcesTask388ASolution:
def __init__(self):
self.result = ''
self.n = 0
self.boxes = []
def read_input(self):
self.n = int(input())
self.boxes = [int(x) for x in input().split(" ")]
def process_task(self):
counts = [self.boxes.count(x... | PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import sys
import math
n = int(sys.stdin.readline())
xn = [int(x) for x in (sys.stdin.readline()).split()]
xn.sort()
k = [1]
res = 1
for i in xn[1:]:
flag = False
for t in range(res):
if(i >= k[t]):
k[t] += 1
flag = True
break
if(flag == False):
... | PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | __author__ = 'martslaaf'
a = raw_input()
table = map(int, raw_input().split(' '))
stops = 0
while table:
top = min(table)
table.remove(top)
weight = 1
while True:
sides = [x for x in table if x >= weight]
if sides:
elem = min(sides)
table.remove(elem)
... | PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n = input()
s = sorted(map(int,raw_input().split()))
a = []
for i in s:
if len(a)==0:
a.append(1)
else:
if i<min(a):
a.append(1)
else:
a[a.index(min(a))]+=1
print len(a) | PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class Soly
{
static final int INF = Integer.MAX_VALUE;
static void mergeSort(int[] a,int [] c, int b, int e)
{
if(b < e)... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #coding=utf-8
n = int(raw_input())
x = map(int, raw_input().split())
x.sort()
L = list()
for i in x:
flag = True
for j in range(len(L)):
if L[j] <= i:
L[j] += 1
flag = False
break
if flag:
L.append(1)
print len(L)
| PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int boxes[105];
bool used[105];
int main() {
int piles = 0, n;
cin >> n;
for (int i = 0; i < n; i++) cin >> boxes[i];
sort(boxes, boxes + n);
for (int i = 0; i < n; i++) {
if (!used[i]) {
piles++;
int idx = i + 1;
int sum = 1;
while (id... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int n, arr[102], pile;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> n;
for (int i = 0; i < n; i++) {
cin >> arr[i];
arr[i]++;
}
sort(arr, arr + n);
for (int i = 0; i < n; i++)
if (pile * arr[i] <= i) pile++;
cout << pile << endl;... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, a[100], k = 1;
set<pair<int, int> > s;
cin >> n;
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
sort(a, a + n);
pair<int, int> p;
p = make_pair(1, 0);
s.insert(p);
for (int i = 1; i < n; i++) {
if (a[i] >= s.begin()->first) {
... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int s[110], v[110];
int main() {
int n, i, j, ans, sum, p;
while (cin >> n) {
for (i = 0; i < n; i++) {
cin >> s[i];
}
sort(s, s + n);
memset(v, 0, sizeof(v));
sum = 0;
ans = 0;
while (sum < n) {
ans++;
p = 0;
for (i =... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int arr[105];
int mark[105];
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; i++) cin >> arr[i];
sort(arr + 1, arr + n + 1);
int res = 0;
for (int i = 1; i <= n; i++) {
if (mark[i] == true) {
continue;
}
res++;
int cnt = 1;
for (i... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.*;
import java.util.*;
public final class code
// public class Main
// class code
// public class Solution
{
static int n,x[];
static void solve()throws IOException
{
n=nextInt();
x=new int[n+1];
for(int i=1;i<=n;i++)
x[i]=nextInt();
Arrays.sort(x);... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.*;
import java.math.*;
import java.io.*;
public class B{
static int[] dx={-1,1,0,0};
static int[] dy={0,0,1,-1};
static FastReader scan=new FastReader();
public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
static ArrayList<Pair>es;
static LinkedList<Inte... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n=int(input())
l=list(map(int,input().split()))
l.sort()
ans=0
while len(l)!=0:
A=[]
k=1
ans+=1
for j in range(1,len(l)):
if(l[j]!=0 and l[j]>=k):
k+=1
else:
A.append(l[j])
l=A
print(ans)
| PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int arr[1000] = {0};
int book[1000] = {0};
int n = 0;
int ct = 0;
int main() {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> arr[i];
}
sort(arr + 1, arr + n + 1);
int num = 1000;
while (1) {
num = 0;
for (int i = 1; i <= n; i++) {
if (!book[i... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | from itertools import *
n = int(raw_input())
x = map(int, raw_input().split())
x = sorted(x, reverse=True)
for p in xrange(1,n+1):
bins = [200 for _ in xrange(p)]
for e in x:
_,i = max(zip(bins,xrange(p)), key=lambda a: a[0])
bins[i] = min(e,bins[i]-1)
if len(filter(lambda b: b < 0, bins)) == 0:
pr... | PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
public class Coder{
sta... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;
import java.io.IOException;
public class Main {
public static void main(String[] args) {
OutputStream outputStream = System.out;
myScanner in = new myScanne... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.*;
public class A
{
public static void main(String ar[])
{
Scanner s=new Scanner(System.in);
int n=s.nextInt();
int a[]=new int[n];
for(int i=0;i<n;i++)
a[i]=s.nextInt();
Arrays.sort(a);
int piles=0; int l... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.List;
import java.util.AbstractCollection;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.util.LinkedList;
import java.io.InputStreamReader;
import java.io.Inpu... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int M = 1e9 + 7;
void solve() {
int n;
cin >> n;
int a[n];
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
for (int num = 1; num <= n; num++) {
int count[num], flag = 0;
for (int index = 0; index < num; index++) count[index] = 1e9;
for (int ... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.*;
import java.math.*;
import java.util.*;
import java.lang.*;
public class Main{
public static InputStream inputStream = System.in;
public static OutputStream outputStream = System.out;
public static FastReader in = new FastReader(inputStream);
public static PrintWriter out = new Prin... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n;
cin >> n;
int str[n];
for (int i = 0; i < n; i++) cin >> str[i];
sort(str, str + n);
int piles = 1;
for (int boxseen = 1; boxseen < n; boxseen++) {
if (str[boxseen] < (boxseen / piles... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int n, y, cnt = 0;
vector<pair<int, int> > x;
int main() {
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &y);
x.push_back(make_pair(y, 0));
}
sort((x.begin()), (x.end()));
for (int j = 1; j < n; j++) {
int i = j - 1;
while (i >= 0) {
... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
while (cin >> n) {
int v[n];
for (int i = 0; i < n; i++) cin >> v[i];
sort(v, v + n);
int c = 0;
for (int i = 0; i < n; i++) {
if (v[i] == -1) continue;
int p = 1;
c++;
for (int j = i + 1; j < n; j++) {
... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.PriorityQueue;
import java.util.StringT... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | from collections import Counter as cs
n=int(input())
ls=[int(a) for a in input().split()]
ls.sort()
ls1=dict(cs(ls))
ctr1,ctr2=1,0
for i in range(n):
if ls1[ls[i]]:
ls1[ls[i]]-=1
for j in range(i,n):
if i!=j and ls[j]>=ctr1 and ls1[ls[j]]:
ctr1+=1
ls1[ls[j... | PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.*;
import java.util.*;
public class cf388a {
static FastIO in = new FastIO(), out = in;
public static void main(String[] args) {
int n = in.nextInt();
int[] v = new int[n];
for(int i=0; i<n; i++) v[i] = in.nextInt();
Arrays.sort(v);
int lo = 1, hi = n;
while(hi - lo > 2) {
... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int x[102], n, k;
int main() {
cin >> n;
for (int i = 0; i < n; i++) cin >> x[i];
sort(x, x + n);
for (int i = 0; i < n; i++) {
if ((x[i] + 1) * k <= i) k++;
}
cout << k;
return 0;
}
| CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;
import java.util.StringTokenizer;
/**
* Created with IntelliJ IDEA.
* User: AUtemuratov
* Date: 07.04.14
* Time: ... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | # -*- coding: utf-8 -*-
n=int(raw_input())
s=map(int,raw_input().split())
s.sort()
piles=0
while s.count(-1)<n:
h=0
found=0
for i,t in enumerate(s):
if t>=h:
s[i]=-1
h+=1
piles+=1
print piles
| PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long dx[4] = {-1, 1, 0, 0};
const long long dy[4] = {0, 0, -1, 1};
long long XX[] = {-1, -1, -1, 0, 0, 1, 1, 1};
long long YY[] = {-1, 0, 1, -1, 1, -1, 0, 1};
const long long N = (long long)(6 * 1e5 + 10);
const long long M = 1e9 + 7;
long long fact[N], invfact[N... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | # python2
import sys, threading, os.path
import collections, heapq, math,bisect
import string
sys.setrecursionlimit(10**6) # max depth of recursion
threading.stack_size(2**27)
def is_valid(i,j,n):
if i >=0 and j >=0 and i<n and j<n:
return True
else:
return False
def main():
if os.path.exi... | PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n = input()
arr = map(int,raw_input().split())
arr.sort()
# arr=arr[::-1]
piles=[0]*101
s=0
c=0
pile=[1]
for i in arr[1:]:
# print i
f=False
for j in range(len(pile)):
if pile[j]<=i:
pile[j]+=1
f=True
break
if f==False:
pile.append(1)
# pile.sort()... | PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.*;
import java.io.*;
public class Fox_and_Box_Accumulation
{
public static void main(String args[]) throws Exception
{
BufferedReader f=new BufferedReader(new InputStreamReader(System.in));
// BufferedReader f=new BufferedReader(new FileReader("Fox_and_Box_Accumulation.in"));
int runs=Integer.p... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n=int(input())
a=list(map(int,input().split()))
st=0
a.sort()
while len(a)!=0:
wt=1;a[0]='a'
for i in range(len(a)):
if a[i]!='a' and a[i]>=wt:
wt+=1;a[i]='a'
for i in range(a.count('a')):
a.remove('a')
st+=1
print(st)
| PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | /*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
//package fox.and.box.accumulation;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
impo... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | INF = 1000000
if __name__ == '__main__':
n = int(raw_input())
a = sorted([int(x) for x in raw_input().split()], key=lambda x: -x)
for num_piles in range(1, n + 1):
piles = [INF] * num_piles
i = 0
j = 0
while i < n:
for k in range(num_piles):
if p... | PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | t = []
n = int(input())
a = sorted(list(map(int , input().split())))
for i in range(n):
t = sorted(t)[::-1]
flg = True;
for j in range(len(t)):
if(a[i] >= t[j]):
t[j] += 1;flg = False;break
if(flg):t.append(1)
print(len(t))
| PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
bool used[200];
int x[200];
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) cin >> x[i];
sort(x, x + n);
int ans = 0;
for (int i = 0; i < n; i++) {
if (used[i]) continue;
ans++;
used[i] = true;
int remain = 1;
for (int j = i + 1; j ... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long gcd(long long x, long long y) {
if (y == 0)
return x;
else
return gcd(y, x % y);
}
long long expo(long long n, long long m) {
long long r = 1;
while (m > 0) {
if (m % 2) r = (r * n) % 1000000007;
n = (n * n) % 1000000007;
m = m / 2;
}... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | /* 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
{
Scanner sc=new Scanner(System.in);
int n... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int a[105];
bool kt[105];
int n, kq;
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
sort(a + 1, a + 1 + n);
memset(kt, 1, sizeof(kt));
for (int i = 1; i <= n; i++)
if (kt[i]) {
kq++;
int d = 1;
int t = a[i];
... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;
import java.util.TreeSet;
public class C {
static int N;
static int[] a;
public static void main(String[] a... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.Arrays;
import java.util.Scanner;
public class FoxAndBox {
public static void main(String[] args) throws Throwable {
Scanner in = new Scanner(System.in);
int N = in.nextInt();
int[] B = new int[N];
boolean[] used = new boolean[N];
... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.Scanner;
public class FoxandBoxAccumulation228 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int a[] = new int[101];
for (int i = 0; i < n; i++) {
a[sc.nextInt()]++;
}
boolean end ... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.*;
import java.util.*;
import java.lang.*;
import java.math.*;
public
class Main
{
public static int GCD(int a, int b)
{
if (b==0) return a;
return GCD(b,a%b);
}
public static void main(String args[]) throws IOException
{
try{
BufferedReader br = new BufferedReader(new InputStreamRead... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import sys
input = sys.stdin.readline
'''
'''
#from heapq import heapify, heappop, heappush
li = lambda: list(map(int, input().split()))
n = int(input())
x = li()
x.sort()
def solve(n, x, num_piles):
piles = [x[-i] for i in range(1, num_piles+1)]
for i in reversed(range(n-num_piles)):
xi = x[... | PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.*;
import java.util.Arrays;
import java.util.Vector;
public class Main {
public static Reader in = new Reader();
public static Writer out = new Writer();
public static void main(String[] args) {
int n = in.readInt();
int[] x = new int[n];
for(int i=0; i<n; i++) {
x[i] = in.readInt(... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class P388A {
public static void main(String[] args) {
InputStream inputStream = System.... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.*;import java.io.*;import java.math.*;
public class CF389C
{
static final Random random=new Random();
static boolean memory = true;
static void ruffleSort(int[] a) {
int n = a.length;
for (int i=0; i<n; i++) {
int oi=random.nextInt(n), temp=a[oi];
... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int n, a[101];
bool can(int mid) {
for (int i = 0; i < n; ++i)
if (i / mid > a[i]) return false;
return true;
}
int main() {
scanf("%d", &n);
for (int i = 0; i < n; ++i) scanf("%d", a + i);
sort(a, a + n);
for (int i = 1; i < 101; ++i)
if (can(i)) {
... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
public class hals {
public static void main(String[] args)throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Main
{
public static void main(String[] args) throws NumberFormatException, IOException {
... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int a[105], n;
bool vis[105];
int main() {
scanf("%d", &n);
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
sort(a, a + n);
int sum = n;
int ans = 0;
int now = 0;
while (sum) {
for (int i = 0; i < n; i++) {
if (!vis[i]) {
if (a[i] >= now) {
... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 |
def main():
n = int(raw_input())
B = sorted( int(x) for x in raw_input().split() )
res = solve(B,n)
print res
def solve(B,n):
count = 0
piles = []
for x in B:
placed = False
for p in xrange(count):
if x>=piles[p]:
piles[p] += 1
placed = True
break
if not placed:
piles.append(1)
coun... | PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long GCDFast(long long a, long long b) {
while (b) b ^= a ^= b ^= a %= b;
return a;
}
int dx[8] = {-1, 0, 1, 0, -1, 1, 1, -1};
int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
long long a[100005], b[105];
int main() {
long long n, i, j, k, temp, m;
long long x, y, z, s... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | from math import inf as inf
from math import *
from collections import *
import sys
input=sys.stdin.readline
t=1
while(t):
t-=1
n=int(input())
a=list(map(int,input().split()))
a.sort()
re=a[0]
s=0
for i in range(n):
if(s*(a[i]+1)<=i):
s+=1
print(s)
| PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.*;
public class GoodWork {
static Scanner in =new Scanner(System.in);
static int max=1,l=1;
public static void main(String[] args) {
int n=in.nextInt();
int x[]=new int[101],p[]=new int [101];
for(int i=0;i<n;i++){int o=in.nextInt();x[o]++;}
for(int i=0;i<=100;i++){i... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const double EPS = -1e8;
const double Pi = acos(-1);
bool inline equ(double a, double b) { return fabs(a - b) < EPS; }
const int MAXN = 110;
int n;
vector<int> in;
int main() {
ios_base::sync_with_stdio(0);
cin >> n;
in.resize(n);
for (int i = (0); i <= (n - 1); i++... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
std::multiset<int> x;
std::multiset<int>::iterator it;
int n;
int kutije[102];
bool Moze(int broj) {
for (int i = 0; i < n; ++i) x.insert(kutije[i]);
int brojac = 0;
while (!x.empty()) {
for (int i = 0; i < broj; ++i) {
if (x.lower_bound(brojac) != x.end())
x.erase(x.low... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import os
import sys
from io import BytesIO, IOBase
from collections import Counter
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write... | PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n=int(input())
boxes=[int(x) for x in input().split()]
boxes.sort(reverse=True)
piles=[]
while boxes:
box=boxes.pop()
for pile in piles:
if box>=len(pile):
pile.append(box)
break
else:
piles.append([box])
print(len(piles))
| PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n = input()
x = sorted(map(int, raw_input().split()))
ans = 1
for i in range(n):
if x[i] < i / ans: ans += 1
print ans
| PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> v;
for (int i = 0; i < n; ++i) {
int curr;
cin >> curr;
v.push_back(curr);
}
sort(v.begin(), v.end());
vector<vector<int>> piles;
for (int curr : v) {
bool placed = false;
for (auto &pile : piles)... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int n, x[200], ans, cur;
bool vis[200];
int main() {
cin >> n;
for (int i = 0; i < n; i++) cin >> x[i];
sort(x, x + n);
for (int i = 0; i < n; i++) {
if (vis[i]) continue;
ans++;
for (int j = i + 1; j < n; j++) {
if (x[j] > cur && !vis[j]) vis[j] =... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n=int(input())
a=list(map(int,input().split()))
a.sort()
dp = [0]
for i in a:
add = False
for j in range(len(dp)):
if(dp[j]<=i):
dp[j]+=1
add = True
break
if not add :
dp.append(1)
print(len(dp))
| PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n;
cin >> n;
int arr[n];
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
int ans = 0;
sort(arr, arr + n);
vector<int> v;
for (int i = 0; i < n; i++) {
if (v.empty())
v.push_... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, tmp;
cin >> n;
multiset<int> a;
for (int i = 0; i < n; i++) {
cin >> tmp;
a.insert(tmp);
}
int ans = 0;
while (!a.empty()) {
int h = 0;
while (a.lower_bound(h) != a.end()) {
a.erase(a.lower_bound(h));
h++;
}
... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n = int(input())
arr = list(map(int,input().split()))
arr.sort(reverse = True)
vis = [False]*n
counter = 0
ans = 0
while counter<n:
ans+=1
init = 0
for i in range(n-1,-1,-1):
if not vis[i] and arr[i]>=init:
vis[i] = True
counter+=1
init+=1
print(ans)
| PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.*;
import java.math.*;
import java.util.*;
import java.lang.*;
public class Main{
public static InputStream inputStream = System.in;
public static OutputStream outputStream = System.out;
public static FastReader in = new FastReader(inputStream);
public static PrintWriter out = new PrintWriter(outp... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Stack;
import java.util.StringTokenizer;
public class Main {
/**
* @param args
*/
p... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
int n, a[102] = {0}, i, k;
cin >> n;
for (i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
for (k = 1; k <= n; k++) {
bool ans = true;
for (i = 0; i < n; i++) {
ans = ans && (a[i] >= (i / k));
}
... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
import math
n=int(input())
box=[int(x) for x in input().split()]
box.sort()
k=0
for i in range(n):
if k*(box[i]+1)<=i:
k+=1
print(k)
| PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int fx[] = {0, 1, 1, 1, 0, -1, -1, -1};
int fy[] = {1, 1, 0, -1, -1, -1, 0, 1};
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n, i, j, k, l, f = 0;
cin >> n;
int ara[n];
for (i = 0; i < n; i++) {
cin >> ara[i];
}
sort(ara, ara... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | def arr():
return map(int,raw_input().split())
n=int(raw_input())
a=arr()
a=sorted(a)
count=0
piles=0
flag=True
while flag:
flag=False
count=0
for i in range(n):
if a[i]==-1:
continue
if a[i]>=count:
count+=1
a[i]=-1
else:
flag=True... | PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int a[101], vis[101] = {0};
int main() {
ios_base::sync_with_stdio(0);
;
int i, n;
cin >> n;
for (i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
int ans = 0, boxes = 0;
while (boxes < n) {
ans++;
int c = 0;
for (i = 0; i < n; i++) {
if (a[... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import heapq
#br = open('a.in')
f = lambda: map(int, raw_input().strip().split())
n = f()[0]
a = sorted(f())
b = []
for i in a:
h = 0
for j, k in enumerate(b):
if k <= i:
b[j] += 1
h = 1
break
if h == 0:
heapq.heappush(b, 1)
print len(b)
| PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.*;
public class TaskC
{
Scanner in;
PrintWriter out;
int n;
ArrayList<Integer> elements = new ArrayList<Integer>();
int ans;
public static void main(String[] args)
{
TaskC mainTest = new TaskC();
... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
public class Main2 {
public static void main(String args[]) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 |
import java.awt.*;
import java.io.*;
import java.util.*;
public class Main {
static Main.MyScanner sc = new Main.MyScanner();
static PrintWriter out = new PrintWriter(System.out);
// static PrintStream out = System.out;
public static void main(String[] args) {
... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long int pile[1000001], a[1000001], n;
int ok(long long int x) {
memset(pile, 0, sizeof pile);
if (n <= x) return 1;
for (long long int i = 1; i <= x; i++) pile[i] = a[i];
for (long long int i = x + 1; i <= n; i++) {
long long int no;
no = i % x;
if... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.InputMismatchException;
public class B {
public static void main(String[] args) throws Exception {
// TODO Auto-generated method stub
InputReader s = new I... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import math
n=int(input())
arr=[int(x) for x in input().split()]
arr.sort()
k=0
for i in range(n):
if k*(arr[i]+1)<=i:
k+=1
print(k) | PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
bool cmp(int a, int b) { return a > b; }
int m[200], n;
int mx[200];
bool isok(int r) {
for (int i = 0; i < r; ++i) mx[i] = m[i];
int j = 0;
for (int i = r; i < n; ++i) {
if (mx[j]) {
--mx[j];
mx[j] = min(mx[j], m[i]);
++j;
if (j == r) j = ... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n=int(input())
a=[int(x) for x in input().split()]
a.sort()
x=[]
for i in range(n):
c=0
for j in range(len(x)):
if x[j]<=a[i]:
c=1
x[j]+=1
break
if not c:
x.append(1)
#print(x)
print(len(x)) | PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
int main(void) {
int n;
scanf("%d", &n);
int x[101] = {0};
for (int i = 0, temp; i < n; i++) {
scanf("%d", &temp);
x[temp]++;
}
int ans = x[0], remain = 0;
for (int i = 1; i <= 100; i++) {
remain += ans;
if (x[i] <= remain)
remain -= x[i];
else {
in... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | # coding: utf-8
n = int(raw_input())
values = [int(s) for s in raw_input().split()]
values.sort()
used = [False for _ in range(n)]
res = 0
for i in range(n):
if used[i]:
continue
total = 1
used[i] = True
res += 1
for k in range(n-i):
j = k+i
if used[j]:
continue
... | PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.*;
public class C{
public static void main(String... args){
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
ArrayList<Integer> list = new ArrayList<Integer>();
while(n-->0) list.add(sc.nextInt());
int pile = 0;
while(list.size()>0){
pile++;
Collections.sort(list);
list.... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long BIGER = 1000000000000000;
const int BIG = 1000000000;
int n;
int a[101];
int f[101];
int main() {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
sort(a + 1, a + n + 1);
int s = 0;
int g = 0;
int t = 0;
while (s < n) {
t = 0;
... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 |
import java.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Solution {
static PrintWriter fop = new PrintWriter(System.out);
public static void main(String[] a... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.*;
import java.io.*;
public class C
{
Reader in;
PrintWriter out;
int i = 0, j = 0;
void solve()
{
//START//
int n = in.nextInt();
int freq[] = new int[101];
int cur = 0;
for (i = 0; i < n; i++)
{
cur = in.nextI... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.*;
import static java.util.Arrays.*;
import static java.lang.Math.*;
public class C {
Scanner sc = new Scanner(System.in);
void doIt()
{
int n = sc.nextInt();
int [] x = new int[n];
boolean [] used = new boolean[n]; // false
for(int i = 0; i < n; i++) x[i] = sc.nextInt();
sort(x);
int cn... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n = int(raw_input())
f = raw_input().split(' ')
for i in range(0, n):
f[i] = int(f[i])
f.sort()
ans = 0;
while(len(f)):
c = 0
i = 0
while(True):
if i >= len(f):
break;
if f[i] >= c:
f.pop(i)
c = c + 1
else:
i = i + 1
ans = ans ... | PYTHON |
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