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 | import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.lang.System.out;
import static java.util.Collections.*;
import java.io.*;
import java.math.*;
import java.util.*;
public class Main {
static boolean LOCAL = System.getSecurityManager() == null;
Scanner in = new Scanner(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 n, a[101];
void scan() {
scanf("%d", &n);
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
sort(a, a + n);
}
void out() {
int ans = 0, k = 0, c;
while (1) {
c = 0;
for (int i = 0; i < n; i++) {
if (a[i] != -1) {
if (a[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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, i, j, sum, p[110], q[110];
while (cin >> n) {
sum = 0;
for (i = 0; i < n; i++) cin >> p[i];
sort(p, p + n);
memset(q, 0, sizeof(q));
for (i = 0; i < n; i++)
if (!q[i]) {
sum++;
int ans = 0;
for (j = 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.util.*;
public class A {
public void solve() {
int n = in.nextInt();
int[] x = new int[n];
for (int i = 0; i < n; i++) {
x[i] = in.nextInt();
}
Arrays.sort(x);
List<Integer> curW = new ArrayList<>();
for (int i = 0; i < n; i++) {
int min = Integer.MAX_VALUE;
fo... | 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() {
int n;
cin >> n;
map<int, int> m;
for (int i = 0; i < n; i++) {
int x;
cin >> x;
m[x]++;
}
int cnt = 0;
while (!m.empty()) {
cnt++;
int tmp = 1;
int val = m.begin()->first;
m[val]--;
if (m[val] == 0) m.erase(val);... | 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()
h = [0]
A = list(map(int, input().split()))
A.sort()
for x in A:
if x < min(h):
h += [1]
else:
h[h.index(min(h))]+=1
print(len(h))
| 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;
template <class T>
inline bool read(T &x) {
int c = getchar();
int sgn = 1;
while (~c && c<'0' | c> '9') {
if (c == '-') sgn = -1;
c = getchar();
}
for (x = 0; ~c && '0' <= c && c <= '9'; c = getchar()) x = x * 10 + c - '0';
x *= sgn;
return ~c;
}
stru... | 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;
cin >> n;
int box[101];
for (int i = 0; i < n; i++) cin >> box[i];
sort(box, box + n);
bool used[101];
memset(used, false, sizeof(used));
int ai = 0, count = 0, piles = 0;
for (int i = 0; i < n; i++) {
if (!used[i]) {
ai = 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 | #include <bits/stdc++.h>
using namespace std;
int arr[2000], cap[2000];
int main() {
int a, b, c, i, j, k, n, m;
while (scanf("%d", &n) != EOF) {
for (i = 0; i < n; i++) scanf("%d", &arr[i]);
sort(arr, arr + n);
for (i = 0; i < n; i++) cap[i] = 0;
m = 0;
for (i = 0; i < n; i++) {
for (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 | def main(inp):
n = int(inp())
arr = split_inp_int(inp)
c = Counter(arr)
num_of_pile = 0
box_built = 0
while box_built < n:
current_height = 0
for i in range(101):
while i >= current_height and i in c and c[i] > 0:
c[i] -= 1
box_built +=... | 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;
template <class _T>
inline string tostr(const _T& a) {
ostringstream os("");
os << a;
return os.str();
}
int in[110];
int used[110];
int main() {
int n;
while (cin >> n) {
fill_n(used, n, 0);
for (int(i) = (0); (i) < (n); ++(i)) {
scanf("%d", in + 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 n, a[101];
bool vis[101];
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
sort(a + 1, a + n + 1);
int ans = 0;
for (int i = 1; i <= n; i++)
if (!vis[i]) {
ans++;
int now = 1;
vis[i] = 1;
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 | #include <bits/stdc++.h>
using namespace std;
int main(void) {
int n;
cin >> n;
int arr[n];
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
sort(arr, arr + n);
int ans = 1;
int height;
for (int i = 1; i < n; i++) {
height = i / ans;
while (arr[i] < height) ans++, height = i / ans;
}
cou... | 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 bisect
n = int(input())
xi = list(sorted(map(int, input().split())))
s = set(xi)
li = []
while len(xi) > 0:
li.append([xi.pop(0)])
i = 0
while i < len(xi):
if xi[i] >= len(li[-1]):
li[-1].append(xi.pop(i))
else:
i += 1
# print(li)
print(len(li)) | 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 FoxAndBoxAccumulation
{
public static void main(String[] args) throws IOException
{
Scanner in = new Scanner(System.in);
int n, t;
int[] boxes;
ArrayList<LinkedList<Integer>> piles = new 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 | n=int(input())
l=sorted(list(map(int,input().split())))
piles=[None]*100
for i in range(1,100):
piles[i]=[]
piles[0]=[l[0]]
i=1
for j in range(1,n):
flag=0
for k in range(0,i):
if l[j] >= len(piles[k]):
piles[k].append(l[j])
flag=1
break
if flag==0:
pi... | 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,input().split()));print(max([((ind+1)//(i+1)+((ind+1)%(i+1)!=0)) for ind,i in enumerate(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 | n = int(input())
a = list(map(int, input().split()))
a.sort()
counter = 0
while len(a) != 0:
numbers = {-1}
for elements in a:
numbers.add(elements)
numbers.remove(-1)
for values in numbers:
a.remove(values)
j = 0
s = 0
for values in numbers:
if len(a) == 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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Solution {
static class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
... | 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;
struct pairvar {
int a;
int b;
int c;
int max;
};
bool isPrime(long long x) {
if (x == 1) {
return false;
}
long long a = (long long)x;
for (int i = 2; i <= (long long)sqrt((double)a); i++) {
if (a % i == 0) {
return false;
}
}
return t... | 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 n, m, x, y, z, k, sol, sum, ans, l, r, xx, yy, a[1000000], b[1000000];
vector<long long> v;
vector<long long> v1;
vector<pair<long long, long long>> v2;
pair<long long, pair<long long, long long>> pp[1000000];
pair<long long, long long> p[1000000];
map<long long, ... | 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[105];
int cnt[105];
int find(int x) {
for (int i = x - 1; i >= 0; i--)
if (cnt[i]) return i;
return -1;
}
int main() {
int n;
memset(cnt, 0, sizeof cnt);
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
cnt[a[i]]++;
}
boo... | 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 solve2() {
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
sort(a.begin(), a.end());
;
vector<bool> use(n, false);
int lo = 0;
int ans = 0;
while (lo < n) {
ans++;
int top = 0;
for (int i = 0; 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 | import math
def ok(pile):
for i in range(len(pile)):
if(len(pile) - i-1 > pile[i]):
return False
return True
def solve(sbox, p):
ap = [];
for i in range(p):
pile = [sbox[0]]
sbox.pop(0);
ap.append(pile);
currentPile = 0;
full = [False]*p
while s... | 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 | from sys import stdin
inp = [l.split() for l in stdin.readlines()]
p = [int(a) for a in inp[1]]
p = sorted(p)
s = [0]*(len(p)+5)
#print [i for i in p]
for i in range(len(p)):
for j in range(len(p)):
if p[i] >= s[j]:
s[j] +=1
break
print s.index(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.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.BitSet;
import java.util.Collections;
import java.util.Comparator;
impor... | 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 a[100 + 10];
bool used[100 + 10];
map<int, int> cc;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
int rs = 0;
for (int i = 0; i < n; i++) {
if (used[i]) continue;
... | 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 static java.lang.System.in;
import static java.lang.System.out;
import java.io.*;
import java.util.*;
public class Main {
static final double EPS = 1e-10;
static final double INF = 1 << 31;
static final double PI = Math.PI;
public static Scanner sc = new Scanner(in);
StringBuilder sb = new StringBuilder(... | 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.io.PrintWriter;
import java.util.StringTokenizer;
/**
* @author Don Li
*/
public class FoxBoxAccumulation {
void solve() {
int n = in.nextInt();
int[] x = new int[n];
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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, t;
while (scanf("%d", &n) == 1) {
int a[105];
bool vis[105] = {false};
memset(a, 0, sizeof(a));
for (int i = 0; i < n; i++) {
scanf("%d", &a[i]);
}
sort(a, a + n);
int ans = 0;
for (int i = 0; i < n; 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 | //http://codeforces.com/contest/389/problem/C
import java.util.*;
public class FoxBox {
/*
* find top box, find smallest box that can go below that, keep going until 1 stack is done
* recurse on remaining list of boxes
*/
public static void magic(ArrayList<Integer> v, int c) {
if (v.size() == 0) {
Syst... | 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.Arrays;
import java.util.Scanner;
public class C228 {
/**
* @param args
*/
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int arr[] = new int[n];
for (int i = 0; i < n; ... | 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 N = 1e5 + 5;
int n, ans;
multiset<int> setik;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n;
for (int i = 1; i <= n; i++) {
int x;
cin >> x;
setik.insert(x);
}
while (setik.size()) {
int cur = 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>
int comp(const void* a, const void* b) { return *(int*)a - *(int*)b; }
int main() {
int n, i, j, t, count, p;
int s[105];
while (scanf("%d", &n) != EOF) {
t = n;
for (i = 0; i < n; i++) scanf("%d", &s[i]);
qsort(s, n, sizeof(int), comp);
while (t) {
for (i = 0, p = 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 ans = 0;
int n, tmp;
bool vis[106];
int a[106];
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &a[i]);
vis[i] = true;
}
sort(a, a + n);
for (int i = 0; i < n; i++) {
if (vis[i]) {
ans++;
tmp = 1;
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 | #include <bits/stdc++.h>
using namespace std;
int main() {
int a[105] = {0};
int n;
ios_base::sync_with_stdio(false);
cin.tie(0);
cin >> n;
int temp, mx = 0;
for (int i = 0; i < n; ++i) {
cin >> temp;
a[temp]++;
mx = max(mx, temp);
}
for (int i = 1; i < mx + 1; ++i) a[i] += a[i - 1];
for... | 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()]
v = [True]*n
a.sort()
ans = 0
prev = 0
i = 0
cnt = 0
while(cnt<n):
if a[i]>=prev and v[i]:
cnt += 1
v[i] = False
prev += 1
i += 1
if i==n:
i = 0
prev = 0
ans += 1
if prev != 0:
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 | #include <bits/stdc++.h>
using namespace std;
int box_strength[111], piles[111];
int main() {
int num, i, j;
cin >> num;
for (i = 0; i <= num - 1; i++) cin >> box_strength[i];
sort(box_strength, box_strength + num);
int pile_index = 0, min = INT_MAX, min_index;
for (i = 0; i <= num - 1; i++) {
min = 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.util.*; //Scanner;
import java.io.PrintWriter; //PrintWriter
public class R228_Div2_C
{
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
solve(in, out);
out.close();
in.close();
}
public static void solve(Scann... | 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 | /**
* Created with IntelliJ IDEA.
* User: den
* Date: 2/3/14
* Time: 6:12 PM
* To change this template use File | Settings | File Templates.
*/
import java.io.*;
import java.util.Arrays;
import java.util.Collections;
import java.util.StringTokenizer;
public class TaskC extends Thread {
public static int an... | 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.OutputStreamWriter;
import java.io.BufferedWriter;
import java.util.Comparator;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.Writer;
import java.io.IOException;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.NoSuchElementException;
import jav... | 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;
import java.util.LinkedList;
import java.util.Queue;
import java.util.StringTokenizer;
public class C228 {
public void run() throws NumberFormatException, IOException {
BufferedReader bReader = ne... | 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.Collections;
import java.util.LinkedList;
import java.util.List;
import java.util.ListIterator;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
List<Integer> list = new LinkedList<>();
for(... | 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, strength[104], count = 0;
bool strikeoff[104];
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &strength[i]);
strikeoff[i] = false;
}
sort(strength, strength + n);
int weight = 0, start_at = 0;
bool start_set = true;
... | 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 + 5;
int a[maxn];
int bo[maxn];
int fin[105];
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
bo[a[i]]++;
}
sort(a + 1, a + n + 1);
int maxx = 1;
int i = 0;
while (n) {
int k = 1;
while (bo[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 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
int n, a[105];
vector<int> v;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a[i];
}
sort(a, a + n);
for (int i = 0; i < n; i++) {
int j;
for (j = 0; j < (int)v.size(); j++) {
if (v[j] <= 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 | #include <bits/stdc++.h>
using namespace std;
const int INF = 1000000000;
const int maxn = 110;
int n;
int s[maxn];
bool used[maxn];
int main() {
scanf("%d", &n);
for (int i = 0; i < n; i++) scanf("%d", &s[i]);
sort(s, s + n);
for (int i = 0; i < n; i++) used[i] = false;
int cnt = 0;
for (int i = 0; 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 | /******************************************************************************
Online Java Compiler.
Code, Compile, Run and Debug java program online.
Write your code in this editor and press "Run" button to execute it.
*****************************************************... | 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>
int n, MaxStr, i, x, f[105], ans;
int max(int a, int b) { return a > b ? a : b; }
void solve(int str, int now) {
if (str > MaxStr) return;
if ((now > str) || (!f[str]))
solve(str + 1, now);
else {
f[str]--;
solve(str, now + 1);
}
}
int main() {
scanf("%d", &n);
memset(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.*;
import java.io.*;
public class Fox_And_Fox_Accumulation {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
... | 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 Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int x[] = new int[n];
boolean taken[] = new boolean[n];
for(int i = 0; i < n; 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 | """
greedy alg
properties : len(stack) < x[i], if not, new stack
"""
number_of_items = int(raw_input())
items_strength = map(int, raw_input().split(" "))
items_strength.sort()
stacks = []
new_stack = []
stacks.append(new_stack)
for item in items_strength:
flag = True
for stack in stacks:
if len(stack) <= item:... | 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.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.Arrays;
import java.util.InputMismatchException;
public class CF {
public static void main(String[] args) {
FasterScanner sc = new FasterScanner();
PrintWriter out = new PrintWriter(Sy... | 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 = 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 sys
n=int(raw_input())
ar=map(int,raw_input().split())
ar.sort()
p=0
c_ar=ar[::]
while ar!=[]:
## print ar
nb=0
p+=1
for i in ar:
if nb<=i:
nb+=1
c_ar.remove(i)
ar=c_ar[::]
print 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 | #include <bits/stdc++.h>
using namespace std;
int n, i, a[10000], t, q, kol, ans[10000];
int main() {
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
for (int i = 0; i < n; i++) {
if (ans[i] == 0) {
t = 1;
q = a[i];
kol++;
ans[i] = -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;
int main() {
bool used[101] = {};
int n;
int a[101] = {};
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
int ans = 0;
while (1) {
int numbx = 0;
bool ff = false;
for (int i = 0; i < n; i++) {
if (!used[i] && a[i] >= numb... | 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())
li = sorted([int(x) for x in input().split()])
res = 1
for i in range(n):
if (li[i] < i // res):
res += 1
print(res)
| 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 | # -*- coding: utf-8 -*-
"""
Created on Fri Jun 21 00:21:11 2019
@author: sj
"""
n=int(input())
s=input().split(" ")
for i in range(0,n):
s[i]=int(s[i])
s=sorted(s)
a=[]
tt=0
for i in range(0,n):
c=0
for j in range(tt,len(a)):
if s[i]>a[j]:
a[j]+=1
c=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 | #include <bits/stdc++.h>
using namespace std;
int main(void) {
int n, a[105], k = 0;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a[i];
}
sort(a, a + n);
for (int i = 0; i < n; i++) {
if (k * a[i] + k <= i) k++;
}
cout << 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 | /**
* Created with IntelliJ IDEA.
* User: igarus
* Date: 13.12.13
* Time: 18:48
* To change this template use File | Settings | File Templates.
*/
import java.util.Arrays;
import java.util.Scanner;
public class CF228_C {
public static void main(String[] args) {
Scanner scanner = new Scanner(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() {
std::ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
;
int n, i;
cin >> n;
vector<int> a(n);
for (i = 0; i < n; i++) {
cin >> a[i];
}
sort(a.begin(), a.end());
int count = 0;
i = 0;
while (n) {
i = 0;
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.InputStreamReader;
import java.util.*;
public class Main {
public static void main(String[] args) throws IOException {
//Scanner sc = new Scanner(System.in);
BufferedReader br = new BufferedReader(new InputStreamReader(Syste... | 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 itertools import combinations
# from bisect import bisect_left
from collections import Counter
I = lambda: list(map(int, input().split()))
n, a = I(), I()
a.sort()
h = [1]
for el in a[1:]:
mn = min(h)
if el >= mn:
h[h.index(mn)] += 1
else:
h.append(1)
print(len(h)) | 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())
ar = list(map(int,input().split(' ')))
ar.sort()
s=[]
for i in ar:
s.sort(reverse=True)
for j in range(len(s)):
if i>=s[j]:
s[j]+=1
break
else:
s.append(1)
print(len(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 | def isPossible(arr,pile):
arr.sort(reverse=True)
size = [10**18]*pile
#print(arr)
c = 0
for item in arr:
if size[c%pile] >= 1:
size[c%pile] = min(size[c%pile]-1,item)
else:
return False
c = c + 1
return True
n = int(input())
arr = [int(num) ... | 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())
nums = sorted(list(map(int, input().split(' '))))
v = []
b = [0] * 105
cnt = 0
k = 0
while cnt < n:
k += 1
for i in range(n):
if nums[i] >= len(v) and b[i] == 0:
v.append(nums[i])
b[i] = 1
cnt += 1
#print(list(reversed(v)))
v = []
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;
const int N = 1e3;
int m[N], n, tmp, ans, z;
int main() {
cin >> n;
for (int i = 0; i < n; i++) {
cin >> tmp;
m[tmp]++;
z += (tmp == 0);
}
for (int i = 0; i < 103; i++) {
if (m[i]) {
while (m[i]) {
m[i]--;
ans++;
tmp = 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;
int n, a[111];
vector<int> L[111];
bool ok;
int main() {
cin >> n;
for (int i = 0; i < n; ++i) cin >> a[i];
sort(a, a + n);
int cnt = 0;
for (int i = 0; i < n; ++i) {
ok = 1;
for (int j = 0; j < cnt; ++j) {
if (a[i] >= L[j].size()) {
L[j].pus... | 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 Main{
public static void main(String args[]) {
Scanner in= new Scanner(System.in);
int n=in.nextInt(),num;
int []a=new int[n];
for(int i=0;i<n;i++)
a[i]=in.nextInt();
Arrays.sort(a);
for (num=1;;num++) {
int flag=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;
bool cmp(int a, int b) { return (a > b); }
bool check(int num, int x[], int N) {
int cap[100];
int totalCap = 0;
for (int i = 0; i < num; i++) {
cap[i] = x[i];
totalCap += cap[i];
}
if (totalCap < N - num) return false;
for (int i = num; i < N; 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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Built using CHelper... | 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;
bool visited[10000];
int main() {
int n;
cin >> n;
int a[n];
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
int str = 0, count1 = 0;
for (int i = 0; i < n; i++) {
if (visited[i]) continue;
count1++;
str = 0;
for (int j = i + 1; j < 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 | n = int(input())
a = list(map(int,input().split()))
a.sort()
res = 0
for i in range(n):
cnt = i+1
lvl = a[i]+1
res = max(res , (cnt+lvl-1)//lvl)
print(res) | 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 cols[200], a[200], sizes[200];
int main() {
int n, active = 0;
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
for (int i = 0; i < n; i++) {
int l = 0;
while (l < active) {
if (cols[l] <= a[i] && sizes[l] <= a[i]) {
c... | 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 power(long long a, long long b) {
long long result = 1;
while (b > 0) {
if (b % 2 == 1) {
result *= a;
}
a *= a;
b /= 2;
}
return result;
}
long long gcd(long long x, long long y) {
long long r;
while (y != 0 && (r = x % y) != 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 n, a[110], i, hsh[110][3], j, cnt, k, ans, span;
int main() {
ios::sync_with_stdio(0);
cin >> n;
for (i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
for (i = 0, k = 0; i < n; i = j) {
cnt = 1;
for (j = i + 1; j < n; j++) {
if (a[j] != a[i]) brea... | 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[101], piller;
int main() {
cin >> n;
for (auto i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
for (auto i = 0; i < n; i++) {
if (piller * a[i] + piller <= i) piller++;
}
cout << piller;
}
| 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[105], f, ans;
int main() {
cin >> n;
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
sort(a, a + n);
while (n) {
++ans;
for (int i = 0; i < n; ++i) {
if (a[i] >= f) {
++f;
a[i] = 999;
}
}
sort(a, a + n);
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 | import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
public class A525 {
public static void main(String[] Args) throws Exception{
FastScanner sc = new FastScanner(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 v[1000], x[1000];
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) scanf("%d", &x[i]);
sort(x, x + n);
int ans = n;
int high = n;
int low = 1;
while (high >= low) {
int flag = 1;
int med = (high + low) / 2;
int i = n - 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 | n = int(raw_input())
x_s = sorted([int(i) for i in raw_input().split()])
num_piles = n
for k in range(1, n+1):
flag = True
for i in range(n):
if x_s[i] < i/k:
flag = False
break
if flag:
num_piles = k
break
print num_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 double PI = acos(-1);
const long long OO = 0x3f3f3f3f3f3f3f3f;
const int oo = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
const int N = 1e6 + 5;
void run_case() {
int n, ans = 0;
cin >> n;
vector<int> a(n);
for (int &it : a) cin >> it;
sort(a.begin(), a.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 |
var n=+readline();
var x=readline().split(' ').map(function(v){return+v;});
x.sort(function(a,b){return a-b;});
var ans=[];
while(true){
var q=[]
ans.push(q);
for(var j=0;j<x.length;j++){
if(x[j]>=q.length){
q.push(x[j]);
x.splice(j,1);
j--;
}
}
if(x.length===0){
break;
}
}
print(ans.length); | 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[105];
int main() {
ios_base::sync_with_stdio(false);
scanf("%d", &n);
for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
sort(a, a + n);
for (int k = 1; k <= n; ++k) {
int ok = true;
for (int i = 0; i < n; ++i)
if (a[i] < i / k) ok = false;
... | 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
n = int(input())
boxes = sorted(list(map(int, input().split())))
piles = [1]
for b in boxes[1:]:
if piles[0] <= b:
heapq.heappush(piles, heapq.heappop(piles)+1)
else:
heapq.heappush(piles, 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 | #include <bits/stdc++.h>
using namespace std;
int n, a[101], c = 1, s, used[101];
int mi(int s2) {
int minn = 5154541, minnn = -1;
for (int i = 1; i <= n; i++)
if (a[i] >= s2 && a[i] < minn && !used[i]) minn = a[i], minnn = i;
used[minnn] = 1;
return minnn;
}
bool allused(void) {
for (int i = 1; i <= n; 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;
const int N = 1e5;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout << fixed;
cout.precision(10);
;
int n;
cin >> n;
vector<int> arr(n);
for (auto &i : arr) cin >> i;
sort(arr.begin(), arr.end());
int temp[101]{};
int cnt = 0;
for... | 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=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 | #include <bits/stdc++.h>
using namespace std;
vector<vector<int> > piles;
vector<int> inp;
int n;
void process(int w) {
bool can_reuse = false;
for (int i = 0; i < piles.size(); i++)
if (piles[i].size() <= w) {
can_reuse = true;
piles[i].push_back(w);
return;
}
if (can_reuse == false) {
... | 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()
h = [0]
A = sorted(map(int, input().split()))
for x in A:
if x < min(h):
h += [1]
else:
h[h.index(min(h))]+=1
print(len(h)) | 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 n;
int a[110];
priority_queue<int> f;
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
sort(a + 1, a + 1 + n);
f.push(-1);
for (int i = 2; i <= n; i++) {
a[i] = -a[i];
if (f.top() < a[i])
f.push(-1);
else
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 sys
input = sys.stdin.readline
read_tuple = lambda _type: map(_type, input().split(' '))
def is_correct(seq):
if len(seq) == 1:
return True
_len = 1
res = True
for idx in range(1, len(seq)):
res = res and seq[idx] >= _len
_len += 1
return res
def solve():
n = 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 | n = int(input())
a = [int(x) for x in input().split()]
a.sort()
pile,tc=0,n
visited = [0]*n
while tc != 0:
tt=0
for i in range(0,n):
if a[i]>=tt and visited[i] != 1:
visited[i]=1
tt+=1
tc-=1
if(tt>0):
pile+=1
print(pile)
| 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
2
50 50
'''
from sys import stdin
def get_freq(arr):
freq = [0] * (101)
for i in arr:
freq[i] += 1
return freq
# main starts
n = int(stdin.readline().strip())
arr = list(map(int, stdin.readline().split()))
freq = get_freq(arr)
i = 0
c = 0
while i <= 100:
if freq[i] == 0:
i += 1
else:
freq[i] -= 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.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int[] boxes = new int[100+1];
for (int i=0 ; i<n ; i++) boxes[in.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;
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 | #include <bits/stdc++.h>
using namespace std;
int a[101], n, res;
bool can(int x) {
vector<int> b(a, a + x);
int k = 0;
for (int j = x; j < n; ++j) {
bool f = 0;
int c = 0;
while (c != x) {
if (k == x) k = 0;
if (b[k] > 0) {
b[k] = min(a[j], b[k] - 1);
k++;
f = 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;
int main(void) {
vector<int> piles[105];
vector<int> boxes;
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
int buff;
scanf("%d", &buff);
boxes.push_back(buff);
}
sort(boxes.begin(), boxes.end());
for (int i = 1; i <= n; i++) {
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 | #include <bits/stdc++.h>
using namespace std;
int p[100100];
int main() {
int n, a, b, t, cn = 1;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a;
p[i] = a;
}
sort(p, p + n);
int ans = 0;
vector<int> v;
for (int i = 0; i < n; i++) {
int j = 0;
for (j = 0; j < v.size(); j++) {
if (... | CPP |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.