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 | n=int(input())
s=[int(c) for c in input().split()]
s.sort()
s.reverse()
ans=1
def f(a):
i=0
b=True
while i<len(a) and b:
if a[i]<len(a)-i-1:
b=False
return b
i+=1
return b
while True:
p=[[] for i in range(ans)]
i=0
while i<n:
p[i%ans].append(... | 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())
arr = list(map(int,input().split()))
arr.sort()
unq, ans, req = set(arr), 1, 0
for ele in unq:
count = arr.count(ele)
req += count
if req//(ele+1) != req/(ele+1):
ans = max(1 + req//(ele+1), ans)
else:
ans = max(req//(ele+1), ans)
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 | n=input()
d=sorted(map(int,raw_input().split()))
for k in range(1,101):
b=1
for i in range(n):
if d[i] < i/k:
b=0
if b:
print k
exit(0) | 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.*;
import java.util.Map.Entry;
public class zz{
public static void main(String[] args) throws IOException{
MScanner sc = new MScanner(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 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
int n, arr[200];
while (cin >> n) {
int ans = 0, arr1[200];
for (int i = 0; i < n; i++) cin >> arr[i];
sort(arr, arr + n);
for (int i = 0; i < n; i++) {
bool flag = 0;
for (int j = 0; j < ans; 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 | R=lambda:map(int,raw_input().split())
n = input()
a = R()
a.sort()
res = 1
l = [[a[0]]]
for i in range(1, n):
ins = False
for j in range(len(l)):
if len(l[j]) <= a[i]:
l[j].append(a[i])
ins = True
break
if not ins:
l.append([a[i]])
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 data[100];
int freq[101];
int min(int a, int b) {
if (a < 0) return 0;
if (b < 0) return 0;
if (a < b) return a;
return b;
}
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) scanf("%d", data + i);
for (int i = 0; i < 101; i++) freq[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;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
const int INF = 1 << 29;
inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n >> b) & 1; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int 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 | #include <bits/stdc++.h>
using namespace std;
int a[110];
int vis[110];
int n;
int main() {
int i, j;
while (scanf("%d", &n) != EOF) {
memset(vis, 0, sizeof(vis));
for (i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
sort(a + 1, a + n + 1);
int ans = 0;
int cnt = 0;
for (i = 1; i <= 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 | #include <bits/stdc++.h>
using namespace std;
int main() {
vector<int> v1, v2;
int n, ctr, counter, flag;
cin >> n;
v1.resize(n);
v2.resize(n);
for (int i = 0; i < n; i++) {
cin >> v1[i];
v2[i] = 0;
}
sort(v1.begin(), v1.end());
counter = 0;
while (1) {
flag = 0;
for (int i = 0; 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 sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
# from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect... | 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
#input=sys.stdin.buffer.readline
t=1
mod=10**9+7
for __ in range(t):
#a=[]
n=int(input())
#n,m=map(int,input().split())
l=list(map(int,input().split()))
l.sort()
cnt=0
maxi=0
ans=0
# print(l)
while cnt<n:
cnt1=0
for i in range(n):
if l[i]!... | 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 CFR2283 {
public static void main (String[] args) {
Scanner sc = new Scanner (System.in);
PrintStream op = System.out;
int N = sc.nextInt();
int[] X = new int [N];
int[] pile = new int [N];
for (int i = 0; i < N; i++)
X[i] = sc.nextInt();
Arrays.sort... | 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 PI = 2.0 * acos(0.0);
const double EPS = 1e-9;
int cases = 1;
double getdist(pair<int, int> a, pair<int, int> b) {
return sqrt(pow(a.first - b.first, 2) + pow(a.second - b.second, 2));
}
void read(void) { return; }
int n, box[105];
int main() {
int test;
... | 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())
ar=list(map(int,input().strip().split(' ')))
ar.sort()
i=0
li=list()
cnt=[0]*(101)
for i in ar:
cnt[i]+=1
for i in range(101):
if cnt[i]==0:
continue
if len(li)==00:
x=(i+1)
li.extend([x]*(cnt[i]//x))
if cnt[i]%x!=0:
li.append(cnt[i]%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 | #!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
#from bisect import bisect_left as bl #c++ lowerbound bl(array,element)
#from bisect import bisect_right as br #c++ upperbound br(array,element)
def main():
n=int(input())
a=list(map(int,input().split(" ")... | 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.*;
/**
*
* @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;
... | 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 int MAXN = 100 + 10;
int n, a[MAXN], mark[MAXN], l;
void input() {
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a[i];
}
sort(a, a + n);
}
int main() {
input();
int l = 0;
bool chek;
for (int i = 0; i < n; i++) {
chek = true;
for (int 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.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args) {
InputStream input... | 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 x[101];
int towers = 0;
int n;
cin >> n;
for (int i = 0; i < n; i++) cin >> x[i];
sort(x, x + n);
int count = 0;
while (count < n) {
int cur = 0;
for (int i = 0; i < n; i++) {
if (x[i] >= cur) {
count++;
cur++;
... | 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.*;
/*
@author Mikhail Linkov
*/
public class TaskC {
private final static int MAX = 10000001;
public void solve() {
InputReader reader = new InputReader(System.in);
PrintWriter writer = new PrintWriter(System.out, true);
int n = reader.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 | /*
Keep solving problems.
*/
import java.util.*;
import java.io.*;
public class CFA {
BufferedReader br;
PrintWriter out;
StringTokenizer st;
boolean eof;
final long MOD = 1000L * 1000L * 1000L + 7;
int[] dx = {0, -1, 0, 1};
int[] dy = {1, 0, -1, 0};
void solve() throws 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 |
n=int(input())
a=list(map(int,input().split()))
a.sort()
visited=[0]*n
i=0
res=0
while i<n:
if visited[i]==0:
visited[i]=0
j=i
c=0
while j<n:
if a[j]>=c and visited[j]==0:
visited[j]=1
c += 1
j+=1
res+=1
i += 1
pr... | 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 piles[105];
int main() {
int n;
cin >> n;
vector<int> A(n);
for (int i = 0; i < n; ++i) cin >> A[i];
sort(A.begin(), A.end());
for (int i = 0; i < n; ++i) {
bool flag = false;
for (int j = 0; j <= A[i]; ++j)
if (piles[j]) {
--piles[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;
int mm[105] = {0};
bool vis[105] = {0};
int main() {
int n, ans = 0, m;
cin >> n;
m = n;
for (int i = 0; i < n; i++) {
cin >> mm[i];
}
sort(mm, mm + n);
int we = 0;
while (m) {
for (int i = 0; i < n; i++) {
if (!vis[i] && mm[i] >= we) {
... | 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 sys
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
# t=int(input())
t=1
while t:
t-=1
n=int(input())
l=list(map(int,input().split()))
l.sort()
ans=1
for i in range(n):
if(l[i]<i//ans):
ans+=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.lang.reflect.Array;
import java.math.*;
import java.text.DecimalFormat;
import java.util.*;
public class Main {
private static int n;
private static int [] A;
private static boolean can(int m) {
int [] R = new int[m];
for (int i = 0;i < m;i++)
R... | 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 | def main():
mode="filee"
if mode=="file":f=open("test.txt","r")
#f.readline()
#input()
get = lambda :[int(x) for x in (f.readline() if mode=="file" else input()).split()]
[n]=get()
a=get()
a.sort()
p=[[]]
p[0].append(a[0])
for i in a[1:]:
p=sorted(p,key = lambda x:len... | 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.ArrayList;
import java.util.Comparator;
import java.util.Iterator;
import java.util.Scanner;
/**
* Created by PalmZE on 13.11.2015.
*/
public class Task388A {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
Arr... | 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,ans,tot=input(),0,0
a=sorted(map(int, raw_input().split()))
taken = [False] * n
while tot < n:
curr=0
for i in range(n):
if (not taken[i]) and a[i] >= curr:
taken[i] = True
curr += 1
tot += curr
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;
const int N = 200;
int a[N];
bool used[N];
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
sort(a, a + n);
int cnt = 0;
int ans = 0;
while (cnt != n) {
int st = 0;
for (int i = 0; i < n; i++) {
if (!used[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 main() {
ios_base::sync_with_stdio(0);
int H[101] = {0};
int n, x;
cin >> n;
for (int i = 0; i < n; i++) cin >> x, H[x]++;
int ats = 0;
int box = 0;
while (box <= 100) {
if (H[box] == 0) {
box++;
continue;
}
ats++;
H[box]--;
... | 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.BufferedWriter;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.Arrays;
import java.util.Comparator;
import java.util.StringTokenizer;
public class Main
{
private static Buffered... | 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;
int x[110];
multiset<int, greater<int> > s;
int main() {
scanf("%d", &n);
for (int i = 0; i < n; i++) scanf("%d", &x[i]);
sort(x, x + n);
for (int i = 0; i < n; i++) {
auto y = s.lower_bound(x[i]);
if (y == s.end()) {
s.insert(1);
} else {
... | 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())
val = [[]]
for x in sorted(int(x) for x in input().split()):
for l in val:
if x >= len(l):
l.append(x)
break
if len(val[-1]) > 0:
val.append([])
print(len(val) - 1)
| 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.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;
public class T389C {
public void solve(int n, int[] x) {
Arrays.sort(x);
int cur = 0;
int ans... | 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 | //https://codeforces.com/problemset/problem/388/A
//A. Fox and Box Accumulation
import java.util.*;
import java.io.*;
public class CF_388_A{
public static void main(String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
PrintWriter pw = new PrintWriter(new Out... | 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 int maxn = 111;
int n, ans;
int a[maxn];
bool v[maxn];
void Init() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
}
void Work() {
sort(a + 1, a + 1 + n);
while (1) {
int pos = 1;
while (v[pos] && pos <= n) pos++;
if (pos == 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 | #include <bits/stdc++.h>
using namespace std;
const int N = 109;
int a[N], c[N];
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
sort(a, a + n);
reverse(a, a + n);
int ans = -1;
for (int i = 1; i <= n; ++i) {
memset(c, 0, sizeof(c));
for (int j = 0; j < i; ++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.util.*;
public class c228
{
public static void main(String ar[])
{
Scanner obj=new Scanner(System.in);
int n=obj.nextInt();
int a[]=new int[n];
boolean b[]=new boolean[n];
Arrays.fill(b,false);
for(int i=0;i<n;i++)
a[i]=obj.nextInt();
Arrays.sort(a);
int cnt=n;
int ret=0;
int cur=0;
while(cnt>0)
{
ret+=1;
... | 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, x[105], used[105], ans = 0;
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 (!used[i]) {
used[i] = 1;
int l = 1;
for (int j = i + 1; j < n; j++) {
if (!used[j] && l... | 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 = sorted(map(int, input().split()), reverse=True)
def ok(cnt):
stacks = [[] for _ in range(cnt)]
for i, val in enumerate(a):
stacks[i % cnt].append(val)
for stack in stacks:
for i, val in enumerate(stack):
if val < len(stack) - i - 1:
return Fa... | 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[101];
int main() {
int i, j, n, ans = 0, size = 0;
scanf("%d", &n);
int b[n];
for (i = 0; i < n; i++) {
scanf("%d", &b[i]);
arr[b[i]]++;
}
for (i = 0; i <= 100; i++) {
size = 0;
if (arr[i] > 0) {
ans++;
arr[i]--;
size++;... | 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.util.Arrays;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
* @author Mahmoud Aladdin <aladdin3>
*/
public class Main {
public static void main(String[] args) {
InputSt... | 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;
import java.io.OutputStream;
import java.io.IOException;
import java.util.Arrays;
import java.io.PrintWriter;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
* @author Himalay(himalayjoriwal@gmail.com)
*/
public class Main {
public static ... | 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())
x = sorted(list(map(int, input().split())))
a = [0] * n
ans = 0
for i in range(n):
ok = False
for j in range(ans):
if x[i] >= a[j]:
a[j] += 1
ok = True
break
if ok == False:
ans+=1
a[ans - 1] = 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 | n,ans = input(),0
a = map(int,raw_input().split())
h = [(len(filter(lambda x:x <= i,a))+i)/(i+1) for i in range(101)]
print max(h)
| 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 n;
int a[110], d[110];
void solve() {
int i, j, ans, con, dem;
memset(d, 0, sizeof(d));
sort(a, a + n);
ans = 0;
con = n;
while (con) {
dem = 0;
for (i = 0; i < n; i++)
if (d[i] == 0 && a[i] >= dem) {
d[i] = 1;
dem++;
co... | 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() {
ios_base::sync_with_stdio(false);
int n, ans = 0;
cin >> n;
vector<int> a(n), c(n, 1), p(n, -1);
for (int i = 0; i < n; i++) cin >> a[i];
sort(a.begin(), a.end());
while (a.size()) {
fill(c.begin(), c.begin() + a.size(), 1);
fill(p.begin()... | 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 long double eps = 1e-9;
const int inf = (1 << 30) - 1;
const long long inf64 = ((long long)1 << 62) - 1;
const long double pi = 3.1415926535897932384626433832795;
template <class T>
T sqr(T x) {
return x * x;
}
template <class T>
T abs(T x) {
return x < 0 ? -x : x... | 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.Arrays;
import java.util.StringTokenizer;
public class A {
public static void main(String[] args) throws Exception{
int n = readInt();
int[] t = new int[n];
for(int i = 0; i < n; i++){
t[i] = readInt();
}
Arrays.sort(t);
... | 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.ArrayList;
import java.util.Collections;
import java.util.Iterator;
import java.util.Scanner;
public class round389problemc {
static Scanner in = new Scanner(System.in);
public static void main(String[] args) {
int n=in.nextInt();
ArrayList<Integer> a=new ArrayList<Integer>();
for(int i=0;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 java.io.*;
import java.util.*;
public class CodeForces {
public void solve() throws IOException {
int n = nextInt();
List<Integer> arr = new ArrayList<Integer>();
int res = 0;
for (int i = 0; i < n; i++) {
arr.add(nextInt());
}
Collections.sort(arr);
while (arr.size() > 0) {
int h = 0;
... | 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 i, j, k, n, ans, used;
int main() {
cin >> n;
int a[n];
for (i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
while (used < n) {
ans++;
k = 0;
for (i = 0; i < n; i++)
if (a[i] >= k) {
a[i] = -1;
k++;
used++;
}
}... | 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(raw_input())
l=map(int,raw_input().split())
ans=chk=cnt=0
l.sort()
while n:
if n==chk:
chk=0
cnt=0
if cnt==0:
ans+=1
l.pop(0)
cnt+=1
n-=1
elif cnt<=l[chk]:
l.pop(chk)
cnt+=1
n-=1
else:
chk+=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 | def correct(a):
n = len(a)
for i in range(n):
for j in range(len(a[i])):
if (a[i][j] < len(a[i]) - j - 1):
return False
return True
n = int(input())
a = list(map(int, input().split()))
a.sort(reverse=True)
ans = n
for i in range(n, 0, -1):
cur = []
for j in rang... | 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.Scanner;
import java.util.TreeMap;
public class C228 {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n = s.nextInt();
int[] w = new int[101];
for(int i = 0; i < n; i++)
w[s.nextInt()]++;
int ans = 0; int cov = 0;
int r = 100; int l = 0; int t = 0;... | 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);
long n, i, j, x = 0, ans = 0, term = 0;
long a[101];
bool vis[101] = {};
cin >> n;
for (i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
while (x < n) {
ans++, term = 0;
for (i = 0; i < n; i++)
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 | N = input()
a = map(int, raw_input().split())
a.sort()
p = [1]
for x in a[1:]:
mini = min(p)
if x >= mini:
p[p.index(mini)] += 1
else:
p.append(1)
print len(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 | n = int(input())
ip = input().split(' ')
ip2 = sorted(list(map(int, ip)))
count = 0
keys = set(ip2)
while len(ip2) != 0:
stack = []
for x in keys:
while (len(stack)<= x and ip2.count(x) > 0):
stack.append(x)
del[ip2[ip2.index(x)]]
count = count + 1
print(count)
... | 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.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
import java.util.TreeMap;
public class C {
static BufferedReader stdin = new BufferedReader(new InputStreamReader(System.in));
stati... | 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;
int C[101];
int main() {
cin >> N;
for (int i = 0; i < N; i++) {
int x;
cin >> x;
C[x]++;
}
int k = 0;
while (true) {
bool f = false;
int c = -1;
for (int x = 100; x >= 0; x--) {
if (C[x] > 0) {
C[x]--;
c = x;
... | 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 N = 4000009;
int main() {
int n;
cin >> n;
int a[111];
int x, ans = 0;
int xep[111] = {0};
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
sort(a + 1, a + n + 1);
for (int i = 1; i <= n; i++) {
x = 0;
if (!xep[i]) {
x++;
ans... | 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, a[200];
int dp[200];
int main() {
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
dp[0] = 1;
for (int i = 1; i < n; i++) {
int ma = -1, v = -1;
for (int j = 0; j < i; j++)
if (dp[j] <= a[i] && dp[j] > ma) ma = dp[j], v = 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.*;
import java.math.BigInteger;
import java.util.*;
import java.text.*;
public class cf389c {
static BufferedReader br;
static Scanner sc;
static PrintWriter out;
public static void initA() {
try {
br = new BufferedReader(new InputStreamReader(System.in));
... | 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 n;
cin >> n;
vector<int> x;
int ans = 0;
for (int i = 0; i < n; i++) {
int t;
cin >> t;
x.push_back(t);
}
sort(x.begin(), x.end());
while (!x.empty()) {
for (int i = 0; i <= n; i++) {
if (lower_bound(x.begin(), x.end(),... | 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.*;
public class BruteForce{
public static Scanner in=new Scanner(System.in);
public static void main(String[]args){
int n=in.nextInt();
int[]z=new int[n];
for(int i=0;i<n;i++)
z[i]=in.nextInt();
Arrays.sort(z);
int count=0;
int given... | 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=[int(i) for i in input().split()]
l.sort()
piles=[]
for x in l:
if x==0:
piles.append(1)
continue
f=0
for i in range(len(piles)):
if piles[i]<=x:
piles[i]+=1
f=1
break
if not f:
piles.append(1)
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 | from sys import maxint
raw_input()
box_caps = map(int, raw_input().split(' '))
box_caps.sort(key=lambda x: -x)
stacks = []
while len(box_caps) != 0:
current_cap = box_caps.pop()
# Select for the current element a stack that it can hold with the lowest surplus capacity
selected_stack = None
min_surplus = maxin... | 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 | /***** BY MURAD ******/
/*Everyone has a different way of thinking, so God Created us*/
/*Hope You Respect My Way..,Thank You*/
import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
Out... | 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.BufferedWriter;
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.Map;
import java.util.Scanner;
import... | 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 n;
cin >> n;
vector<int> v(n);
for (int i = 0; i < n; i++) {
cin >> v[i];
}
sort(v.rbegin(), v.rend());
int ans = n;
for (int i = 1; i < n; i++) {
vector<vector<int> > piles(i);
for (int j = 0; j < n; j++) piles[j % i].push_back(... | 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.*;
import java.io.*;
//DONT FORGET TO CHANGE CLASS NAME!
public class FoxBox {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
StringTokenizer st=new StringTokenizer(sc.nextLine());
int numBoxes=Integer.parseInt(st.nextToken(... | 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;
int x[107];
bool cmpr(int a, int b) { return a > b; }
int main() {
cin >> n;
for (int i = 0; i < n; ++i) cin >> x[i];
sort(x, x + n, cmpr);
for (int ans = 1; ans <= n; ++ans) {
bool ok = true;
for (int i = 0; i < ans && ok; ++i) {
for (int j = 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 class cf {
static class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner(Reader in) {
br = new BufferedReader(in);
}
public FastScanner() {
this(new InputStreamReade... | 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::sync_with_stdio(false);
cin.tie(0);
long long n;
cin >> n;
vector<long long> vec(n);
for (int i = 0; i < n; i++) cin >> vec[i];
sort(vec.begin(), vec.end());
vector<long long> vis(n, 0);
long long tot = 0, ans = 0;
while (tot < 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 | def solve(a):
a.sort()
res = 0
n = len(a)
while n > 0:
r = []
for i, x in enumerate(a):
if x == -1: continue
if x >= len(r):
r.append(x)
n -= 1
a[i] = -1
res += 1
return res
n = int(raw_input())
x = map(int, raw_input().split())
print solve(x)
| 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 i, j, k, x, y, z, m, n, ans, p, q, r;
int ara[400];
int main() {
cin >> n;
for (i = 0; i < n; i++) cin >> ara[i];
sort(ara, ara + n);
ans = 0;
for (i = 0; i < n; i++) {
if (ara[i] == -1) continue;
x = ara[i];
ara[i] = -1;
for (j = i + 1, x = 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;
const int maxn = 1e2 + 10;
int dp[maxn];
int n;
int a[maxn];
int mp[maxn];
int mo[maxn];
int v(int u) {
int res = 0;
for (int i = u; i < maxn; i++) {
res += mo[i];
}
return res;
}
void of(int k, int f) {
for (int i = k; i < maxn && f > 0; i++) {
if (mo[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.*;
import static java.lang.Math.*;
public class Main {
public static void main(String args[]) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
int[] data = new 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;
void solve() {
long long n;
cin >> n;
long long a[n];
for (long long int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
vector<long long int> v;
v.push_back(0);
for (long long int i = 0; i < n; i++) {
long long j = 0;
while (1) {
if (j == (lon... | 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 maxn = 3e5, mod = 1e9 + 7;
multiset<int> q, v;
int n, a[maxn], cnt, ans;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
sort(a + 1, a + n + 1);
int cur = 1;
while (cur <= 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 | # coding: utf-8
n = int(input())
ans = 0
li1 = [int(i) for i in input().split()]
li2 = []
while li1:
li1.sort()
n = len(li1)
i = 0
while i < n:
if li1[i] < i:
li2.append(li1[i])
del(li1[i])
n -= 1
else:
i += 1
ans += 1
li1 = li2
... | 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 a[101];
bool b[101];
int main() {
int n, i, j, nrb, rasp = 0;
cin >> n;
for (i = 1; i <= n; i++) cin >> a[i];
sort(a + 1, a + n + 1);
for (i = 1; i <= n; i++)
if (b[i] == 0) {
nrb = 1;
b[i] = 1;
j = i + 1;
while (j <= n) {
f... | 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.Scanner;
public class C {
public static void main(String args[]){
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int a[] = new int[500];
for(int i=0;i<n;i++){
int b = in.nextInt();
a[b]++;
}
... | 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 int maxn = 111;
int a[maxn];
int b[maxn];
int main() {
int n;
while (~scanf("%d", &n)) {
memset(b, 0, sizeof(b));
for (int i = 0; i < n; i++) {
scanf("%d", &a[i]);
b[a[i]]++;
}
sort(a, a + n);
int x = a[n - 1];
int t = n;
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 | n=input()
a=map(int,raw_input().split())
a.sort()
p=[0]*n
c=0
for i in a:
for j in range(n):
if p[j]<=i:
p[j]+=1
c=max(c,j)
break
print c+1 | 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];
int used[101];
int p = 0;
int main() {
int n;
scanf("%d", &n);
int i, j;
for (i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
sort(a + 1, a + 1 + n);
for (i = 1; i <= n; i++) {
if (used[i] == 0) {
int t = 1;
p++;
used[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 | //package codeforces;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;
public class FoxandBoxAccumulation {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
int N = in.nextIn... | 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 n, m, a[2000], s, k, l, i, j, b[2000], p;
int main() {
cin >> n;
for (i = 1; i <= n; i++) cin >> a[i];
sort(a + 1, a + 1 + n);
while (p == 0) {
p = 1;
l = 0;
for (i = 1; i <= n; i++)
if (l <= a[i] && b[i] == 0) {
l++;
b[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;
inline int init() {
int now = 0, ju = 1;
char c;
bool flag = false;
while (1) {
c = getchar();
if (c == '-')
ju = -1;
else if (c >= '0' && c <= '9') {
now = now * 10 + c - '0';
flag = true;
} else if (flag)
return now * ju;
... | 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;
void solve();
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int test = 1;
while (test--) {
solve();
}
return 0;
}
void solve() {
int n;
cin >> n;
multiset<int> st;
for (int i = 0; i < n; i++) {
int x;
cin >> x;
st.... | 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 | num = int(input())
boxes = list(map(int, 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))
# 1536683664960
| 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 | from sys import stdin
inFile = stdin
tokens = []
tokens_next = 0
def next_str():
global tokens, tokens_next
while tokens_next >= len(tokens):
tokens = inFile.readline().split()
tokens_next = 0
tokens_next += 1
return tokens[tokens_next - 1]
def nextInt():
return int(next_str())
de... | 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 | input()
X=sorted(map(int,raw_input().split()))
r=0
while X:
r+=1;X=X[1:];t=0
while 1:
l=[i for i in X if i>t]
if[]==l:break
X.remove(l[0]);t+=1
print r
| 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.*;
import java.util.*;
public class FoxAndBoxAccumulation {
public static void main(String[] args){
int n = ni();
int[] a = na(n);
Arrays.sort(a);
int i, group = 0, count;
for(i=0; i<n; i++){
if(a[i]!=-1){
group++;
count = 1;
for(int j=i+1; j<n; j++){
if(a[j]!=-1){
... | 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.*;
public class problem389C {
public static void main (String[]args)throws IOException{
BufferedReader x = new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(x.readLine());
StringTokenizer st=new StringTokenizer(x.readLine());
int[]values=new 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 | import java.io.*;
import java.util.*;
public class FoxAndBoxAccumulation
{
public static void main(String[] args) throws IOException
{
Scanner in = new Scanner(System.in);
StringBuilder out = new StringBuilder();
int n, ind, t;
int[] boxes;
ArrayList<LinkedList<Integer>>... | 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
import collections
case = int(sys.stdin.readline());
test = sys.stdin.readline().split(" ");
num = [int(x) for x in test]
stats = collections.defaultdict(int);
for i in num:
stats[i]+=1;
key = sorted(stats)
total = 0;
while sum(stats.values())>0:
stack = [];
for i in key:
if(stats[i]>0... | 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.InputStreamReader;
import java.io.IOException;
import java.util.Arrays;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public cla... | JAVA |
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