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 java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;
public class C {
void solve() throws Exception {
int n = nextInt();
int[] x = nextInts(n);
ArrayList[] piles = new ArrayList[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 | n = input()
boxes = map(int, raw_input().split())
s = [0 for i in range(101)]
e = [0 for i in range(102)]
for b in boxes :
s[b] += 1
e[1] = s[0]
for i in range(1, 101) :
if s[i] == 0 :
pass
else :
for j in range(1, i+1) :
# print '[%d %d] %d %d' % (i, j, s[i], e[j])
... | 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, j, T, a[200], top[200];
int main() {
scanf("%d\n", &n);
for (i = 1; i <= n; i++) scanf("%d", a + i);
sort(a + 1, a + n + 1);
top[T = 1] = 0;
for (i = 2; i <= n; i++) {
for (j = 1; j <= T; j++)
if (top[j] < a[i]) {
top[j]++;
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 | a=int(input())
z=list(map(int,input().split()))
from bisect import *
z.sort()
index=0
count=0
while(len(z)):
r=bisect_left(z,index)
if(r==len(z)):
count+=1
index=0
else:
index+=1
z.pop(r)
print(count+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>
#pragma GCC optimize("O3")
using namespace std;
int n, v[102];
bool tk[102];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
cin >> n;
for (int i = 1; i <= n; ++i) cin >> v[i];
sort(... | 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 fast{
static class Parser
{
final int BUFFER_SIZE = 1 << 16;
DataInputStream din;
byte[] buffer;
int bufferPointer, bytesRead;
public Parser(InputStream in)
{
din = new DataInputStream(in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
publ... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | N = int(raw_input())
X = [int(s) for s in raw_input().split()]
X.sort()
for res in range(1, N + 1):
for (i, x) in enumerate(X):
if i / res > x:
break
else:
print res
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 | #include <bits/stdc++.h>
using namespace std;
int piles = 1, heavy = 0, num;
int a[110], n;
void solve(int x) {
if (num == 0) {
return;
}
if (heavy <= a[x]) {
a[x] = -1;
heavy++;
num--;
solve(x + 1);
} else if (x < n) {
solve(x + 1);
} else {
for (int i = 1; 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=int(raw_input())
N=map(int,raw_input().split())
N.sort()
u=ans=0
while u<n:
ans+=1
h=0
for i in range(n):
if N[i]>=h:
N[i]= -1
h+=1
u+=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 | import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;
public class Task388A {
public static void main(String... args) throws NumberFormatException,
IOException {
Solutio... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int a[1010];
int main() {
int n, i;
cin >> n;
for (i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
int ans = 0;
int x;
for (int i = 0; i < n; i++) {
x = 1;
if (a[i] >= 0) {
ans++;
for (int j = i + 1; j < n; j++) {
if (a[j] >= 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.util.Arrays;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] boxList = new int[n];
int[] boxKeyList = new int[n];
int maxList = 0;
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 n;
int x[100 + 5];
int main() {
for (int i = 0; i < 105; ++i) {
x[i] = 0;
}
cin >> n;
int now;
for (int i = 0; i < n; ++i) {
cin >> now;
x[now]++;
}
int ans = 0, cnt = 0, level;
while (cnt < n) {
level = 0;
for (int i = 0; i <= 100; +... | 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.PrintWriter;
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
public class Abood2C {
public static void main(String[] args) throws IO... | 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, i, p, q, s, j, m, x;
int a[101] = {0};
int b[101] = {0};
cin >> n;
m = 0;
for (i = 1; i <= n; i++) {
cin >> x;
m = max(m, x);
a[x]++;
}
s = 0;
for (i = 1; i <= a[0]; i++) b[i] = 1;
if (a[0] > s) s = a[0];
for (i = 1; i <... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n,a,c = input(),sorted(map(int,raw_input().split())),[0]*110
for v in a:
mx = 0
for i in range(1,v+1):
if c[i]: mx = i; break
c[mx] -= 1; c[mx+1] += 1
print sum(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 | n=int(input())
a=list(map(int,input().split()))
a.sort()
piles=0
for i in range(n):
if piles*(a[i]+1)<=i:
piles+=1
print(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 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.UnsupportedEncodingException;
import java.net.URISyntaxException;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Main {
public static void main(Strin... | 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;
vector<long long> a(101);
long long n;
bool f(long long x) {
vector<vector<long long> > mas(101);
long long i, j;
for (i = 0; i < n; i++) {
long long num = i % x;
mas[num].push_back(a[i]);
}
for (i = 0; i < x; i++) {
for (j = 0; j < mas[i].size(); 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 a[110];
int h[110];
int main() {
int i, j, tail, k, n, m;
memset(h, 0, sizeof(h));
scanf("%d", &n);
for (i = 0; i < n; i++) scanf("%d", &a[i]);
tail = 0;
sort(a, a + n);
for (i = 0; i < n; i++) {
int flag = 0;
for (j = 0; j < tail; j++) {
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 | from sys import stdin
n=int(input())
a= list(map(int,stdin.readline().split()))
a.sort()
piles=0
i=0
while len(a)>0:
piles+=1
boxes=1
i=0
del a[0]
while i<len(a):
if a[i]>=boxes:
boxes+=1
del a[i]
else:
i+=1
print(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;
const int INF = INT_MAX;
int n;
vector<int> v;
bool comp(int a, int b) { return a > b; }
bool check(int mid) {
vector<int> f(mid);
for (int i = 0; i < mid; i++) f[i] = v[i];
for (int i = mid; i < n; i++) {
if (f[i % mid] > 0)
f[i % mid] = min(f[i % mid] - 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(input())
a = list(map(int, input().split()))
d = [0] * 101
for i in a:
d[i] += 1
p = 0
while True:
allZeros = True
k = 0
for i in range(101):
if d[i] != 0:
while i >= k and d[i] > 0:
d[i] -= 1
k += 1
allZeros = False
if allZeros... | 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() {
int n, x[100], height[100], ans = 0;
cin >> n;
for (int i = 0; i < n; i++) cin >> x[i];
sort(x, x + n);
for (int i = 0; i < n; i++) {
bool find = false;
for (int j = 0; j < ans; j++) {
if (height[j] <= x[i]) {
find = 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;
long long int gcd(long long int a, long long int b) {
if (!b) return a;
return gcd(b, a % b);
}
long long int power(long long int x, long long int y, long long int p) {
long long int res = 1;
x %= p;
while (y > 0) {
if (y & 1) res = (res * x) % p;
y = y >>... | 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 collections, bisect, heapq
n = int(input())
arr = list(map(int, input().split()))
arr.sort()
t = []
for a in arr:
if not t or a < t[0]:
heapq.heappush(t, 1)
else:
c = heapq.heappop(t)
heapq.heappush(t, c + 1)
print(len(t)) | PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
#pragma comment(linker, "/STACK:200000000")
using namespace std;
const int MOD = 1000000007;
int main() {
int n;
scanf("%d", &n);
int a[105];
for (int(i) = 0; (i) < (n); (i)++) scanf("%d", &a[i]);
sort(a, a + n);
int b[104] = {0};
for (int(i) = 0; (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 | import math
n = int(input())
arr = [int(z) for z in input().split()]
arr.sort()
r = 0
for i in range(n):
e = arr[i]
k = math.ceil((i+1) / (e+1))
r = max(r, k)
print(r)
| 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[105];
bool compare(int x, int y) { return x > y; }
bool ok(int x, int n) {
int y = 0;
int rem[105] = {0};
int i;
for (i = 0; i < n; ++i) {
if (i > x - 1) {
rem[y] = min(rem[y] - 1, arr[i]);
if (rem[y] < 0) {
return 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 | def arr_inp():
return [int(x) for x in stdin.readline().split()]
from sys import *
from collections import deque
from bisect import *
n, a, ans, all = int(input()), sorted(arr_inp()), 0, []
for i in range(n):
if not a[i]:
ans += 1
insort_right(all, 1)
else:
ix = bisect_right(all, ... | 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.Arrays;
import java.util.StringTokenizer;
public class CopyOfC {
public static void main(String[] args) {
MyScanner in = new MyScanner();
int n = in.nextInt();
int[] x = 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.util.*;
import java.io.*;
public class ultrafast
{
static class InputReader {
private InputStream stream;
private byte[] inbuf = new byte[1024];
private int start= 0;
private int end = 0;
public InputReader(InputStream stream) {
this.stream = stream;
... | 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 dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
int dx2[] = {1, -1, -1, 1, 0, 0, -1, 1};
int dy2[] = {1, -1, 1, -1, 1, -1, 0, 0};
int kmx[] = {-1, -1, 1, 1, 2, -2, 2, -2};
int kmy[] = {2, -2, 2, -2, -1, -1, 1, 1};
class Timer {
public:
clock_t T;
Timer() { T = cloc... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.IOException;
import java.io.InputStream;
import java.util.InputMismatchException;
public class Main {
public static void main(String[] args) {
InputReader jin = new InputReader(System.in);
int n = jin.readInt();
int[] a = new int[n];
int[] count = new int[101];
int[] piles = new 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;
long long pow(long long x, long long y, long long mod) {
long long temp;
if (y == 0) return 1;
temp = (pow(x, y / 2, mod)) % mod;
if (y % 2 == 0)
return (temp * temp) % mod;
else
return (((x * temp) % mod) * temp) % mod;
}
void solve(long long tno) {
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 | n = int(raw_input())
c = [ 0 for i in range(200) ]
l = map(int , raw_input().split())
for i in l:
c[i] += 1
ans = 0
RANGE = range(101);
while True:
flag = False
for i in RANGE:
if c[i]:
flag = True
cnt = 1
c[i] -= 1
ans += 1
for j in RANGE:
while j >= cnt and c[j]:
c[j] -= 1
cnt += 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 | import sys
n = int(sys.stdin.readline()[:-1])
a = [ int(i) for i in sys.stdin.readline().split()]
a = sorted(a)
M = []
for i in range(len(a)):
if M == []:
newar = [a[i]]
M.append(newar)
else:
minsize = min(M, key = lambda x: len(x))
index = M.index(minsize)
if len(minsize) <= a[i]:
M[index].append(a[i... | PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.*;
import java.io.*;
public class A{
public static void main(String[] args) throws IOException {
Scanner sc=new Scanner(System.in);
PrintWriter pw=new PrintWriter(System.out);
int n=sc.nextInt();
int ans=0;
int []cnt=new int[110];
for(int j=0;j<n;j++)
cnt[sc.nextInt()]++;
int []pi... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n=int(raw_input())
arr=sorted([int(i) for i in raw_input().split()])
l=[]
for x in arr:
k=True
for i in range(len(l)):
if l[i]<=x:
l[i]+=1
k=False
break
if k:
l.append(1)
print len(l)
| PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | raw_input()
A = sorted(map(int, raw_input().split()))
ans = 0
while len(A) > 0:
c = 0
for i in xrange(len(A)):
if A[i] >= c:
A[i] = -1
c += 1
A = filter(lambda x: x != -1, A)
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 | import java.io.*;
import java.util.*;
import java.math.*;
public class Main {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
InputStream inputstream = System.in;
OutputStream outputstream = System.out;
InputReader in = new InputReader(inputstream);
Outp... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
public class CFTest104 {
static BufferedReader br;
public static void main(String[] args) {
br = new BufferedReader(new InputStreamReader(System.in));
try {
int n = readInt();
int[] 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 | def solve():
for i in range(1, 101):
pile = [0] * i
ind = 0
for j in range(n):
if pile[ind % i] <= L[j]:
pile[ind % i] += 1
ind += 1
else:
break
else:
return i
n = input()
L = sorted(map(int, raw_inp... | 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 = 100;
int x[N], n;
bool good(int u) {
for (int i = 0; i < n; i++)
if (x[i] < i / u) return 0;
return 1;
}
int main() {
scanf("%d", &n);
for (int i = 0; i < n; i++) scanf("%d", &x[i]);
sort(x, x + n);
int l = 1, r = n;
while (l < r) {
int m... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int n, x[101], f[101];
int comp(const void *a, const void *b) { return *(int *)a - *(int *)b; }
int main() {
ios::sync_with_stdio(false);
cin >> n;
x[0] = 0;
for (int i = 1; i <= n; i++) cin >> x[i];
for (int i = 1; i <= n; i++) f[i] = 1;
qsort(x, n + 1, sizeof(... | 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 Solution {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] s = br.readLine().split("\\s");
int N = Integer.parseInt(s[0]);
// int Q = Inte... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n = int(input())
x = sorted(list(map(int, input().split())))
ans = 1
for i in range(n):
if(x[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 time,math,bisect,sys
from sys import stdin,stdout
from collections import deque
from fractions import Fraction
from collections import Counter
from collections import OrderedDict
pi=3.14159265358979323846264338327950
def II(): # to take integer input
return int(stdin.readline())
def IO(): # to take string in... | 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 INF = 2147483647;
int inf = -2147483648;
int dir[8][2] = {-1, 0, 1, 0, 0, -1, 0, 1, -1, -1, 1, 1, 1, -1, -1, 1};
const double PI = acos(-1.0);
int a;
int mapp[1005];
int main() {
ios::sync_with_stdio(false);
int n;
int cnt = 0;
int sum = 0;
int sum1 = 0;
cin... | 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().strip())
x = map(int, raw_input().strip().split())
x.sort()
y = []
for item in x:
put = False
for i in xrange(len(y)):
if y[i] <= item:
y[i] += 1
put = True
break
if not put:
y.append(1)
print len(y)
| PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long int a[400];
long long int h[400];
multiset<long long int> st;
int main() {
ios_base ::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
long long int n;
cin >> n;
long long int maxm = 0, l = -1, cnt = 0, lll = 5;
for (int i = 0; i < n; i++) cin >> 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() {
int n;
cin >> n;
int a[n];
for (int i = 0; i < n; i++) {
cin >> a[i];
}
sort(a, a + n);
int count = 1;
for (int i = 1; i < n; i++) {
if (a[i] < (i / count)) count++;
}
cout << count;
}
| 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;
int ans;
struct NODE {
int w, s;
} x[105];
int cmp(const NODE &a, const NODE &b) {
if (a.s < b.s)
return 1;
else if (a.s == b.s) {
if (a.w < b.w)
return 1;
else
return 0;
} else
return 0;
}
int main() {
int i, j, flag;
scanf("%... | 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[120], b[120];
int main(void) {
int n;
scanf("%d", &n);
for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
sort(a, a + n);
int sum = 0;
while (1) {
int st = 0;
while (st < n && b[st] != 0) ++st;
if (st >= n) break;
int d = 0;
for (int i = s... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | # Target - Expert on CF
# Be Humblefool
import sys
inf = float("inf")
# sys.setrecursionlimit(10000000)
# abc='abcdefghijklmnopqrstuvwxyz'
# abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, '... | 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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.OutputStream;
import java.util.Arrays;
import java.io.InputStreamReader;
import java.io.FileNotFoundException;
import java.io.File;
import java.util.StringTokenizer;
import java.io.Writer;
imp... | 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 Can(vector<int> v, int x) {
for (int i = 0; i < ((int)(v).size()); i++)
if (v[i] < i / x) return false;
return true;
}
int main() {
int n;
scanf("%d", &n);
vector<int> v(n);
for (int i = 0; i < n; i++) scanf("%d", v.begin() + i);
sort(v.begin(), v.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 | n=int(input())
l=list(map(int,input().split()))
l=sorted(l)
l=l[::-1]
l1=[0]*n
k=0
for i in range(n) :
if l1[i]!=1 :
t=l[i]
p=1
r=0
l1[i]==1
V=[t]
for j in range(n) :
if l1[j]==0 and l[j]<t :
t=l[j]
l1[j]=1
V... | 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 = 105;
int n, a[N];
bool vis[N];
int main(void) {
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &a[i]);
}
sort(a, a + n);
int ans = 0;
bool flag = true;
while (flag) {
ans++;
int now = 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 | n = int(raw_input())
a = map(int, raw_input().split())
a.sort()
b = []
for x in a:
for i, y in enumerate(b):
if x >= y:
b[i] += 1
break
else:
b.append(1)
print len(b)
| PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | def check(mid , lis):
pile=[10000000000]*mid
c=0
for i in range(n-1,-1,-1):
pile[i%mid]=min(pile[i%mid]-1,lis[i])
# print(pile,mid)
if min(pile)<0:
return 0
else:
return 1
n = int(input())
lis = sorted(map(int,input().split()))
l=1
r=n
while l<=r:
mid = l + (r-... | 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(raw_input())
a=map(int,raw_input().split())
a.sort()
for ans in range(1,n+1):
OK=1
b=[1 for i in range(ans)]
j=0
for i in range(ans,n):
if a[i]<b[j]:
OK=0
break
b[j]+=1
j+=1
if j==ans:
j=0
if OK:
print ans
break
| 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.lang.reflect.Array;
import java.text.SimpleDateFormat;
import java.util.Arrays;
import java.util.Locale;
import java.util.Map;
imp... | 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 check_append(list, elem):
if elem >= len(list):
# list.append(elem)
return True
else:
return False
n = int(input())
a = [int(x) for x in raw_input().split()]
a.sort()
main = []
for x in a:
if len(main) == 0:
main.append([x])
else:
flag = True
for t in... | 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<int> pile[102], v;
int main() {
ios_base::sync_with_stdio(false);
int n, x;
cin >> n;
for (int i = 0; i < n; i++) cin >> x, v.push_back(x), pile[i].clear();
sort(v.begin(), v.end());
int ans = 0, box = 0;
for (int i = 0; i < n; i++) {
x = v[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 | n = int(input())
boxes = [int(item) for item in input().split(' ')]
boxes.sort()
stacks = 0
while len(boxes):
stacks += 1
n_boxes = []
current_weight = 0
for box in boxes:
if box >= current_weight:
current_weight += 1
else:
n_boxes.append(box)
boxes = n_box... | 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())
l=sorted(map(int,input().split()))
pile,box=0,n
visited=[0]*n
while box!=0:
t=0
for i in range(n):
if l[i]>=t and visited[i]==0:
visited[i]=1
t+=1
box-=1
if t>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 | # Description of the problem can be found at http://codeforces.com/problemset/problem/388/A
n = int(input())
l_b = list(map(int, input().split()))
l_b.sort()
p = 0
s_d = list()
while len(l_b) > 0:
t = 0
for i, b in enumerate(l_b):
if b >= t:
s_d.append(i)
t += 1
for i in... | 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(raw_input())
a = map(int, raw_input().split())
a.sort()
count = [0] * 200
for elem in a:
i = 0
while count[i] > elem:
i += 1
count[i] += 1
answer = 0
i = 0
while count[i] > 0:
i += 1
print i
| PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.*;
import java.io.*;
public class Main388A
{
static PrintWriter out=new PrintWriter(System.out);
public static void main(String[] args) throws IOException
{
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int[] a=sc.nextIntArray(n);
Arrays.sort(a);
int cnt=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 | __author__ = 'Shailesh'
n = int(raw_input())
data = sorted(map(int, raw_input().split()))
height = 1
piles = 1
i = 1
while i < len(data):
if data[i] < height:
piles += 1
i += height
continue
i += piles
height += 1
print piles
| PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n = int(input())
l = [int(i) for i in 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 | input();print-~max(x/-~f for x,f in enumerate(sorted(map(int,raw_input().split())))) | 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 inf = 0x7FFFFFFF;
struct point_int {
int x, y;
point_int() {}
point_int(int a, int b) { x = a, y = b; }
};
struct point_double {
double x, y;
point_double() {}
point_double(double a, double b) { x = a, y = b; }
};
struct Node {
int v, w;
Node() {}
... | 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 CodeforcesRound228Div2ProblemC {
static class Problem {
Scanner reader;
PrintWriter writer;
Problem() {
reader = new Scanner(System.in);
writer = new PrintWriter(System.out);
}
Problem(File inputFile,... | 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, ans = 0, c = 0;
int a[123];
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
for (int i = 0; i < n; i++) {
if (a[i] != -1) {
a[i] = -1;
c++;
for (int j = i + 1; j < n; j++) {
if (a[j] >= 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 | n = int(input())
a = list(map(int, input().split()))
a = sorted(a)
k = 0
for i in range(n):
if k * (a[i] + 1) <= i:
k += 1
print(k)
| PYTHON3 |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:16777216")
const double EPS = 1e-7;
double iabs(const double a) { return (a < -EPS) ? -a : a; }
double imin(const double a, const double b) { return (a - b > EPS) ? b : a; }
double imax(const double a, const double b) { return (a - b > EPS) ?... | 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>
const int INF = 0x7F7F7F7F;
const double EPS = 1e-10;
const long long mod7 = 1e9 + 7;
const long long mod9 = 1e9 + 9;
using namespace std;
inline long long rit() {
long long f = 0, key = 1;
char c;
do {
c = getchar();
if (c == '-') key = -1;
} while (c < '0' || c > '9');
do {
... | 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 C389C {
public static void main(String[] args) {
Scanner ah = new Scanner(System.in);
int n = ah.nextInt() , i , j , k , s = 0, a [] = new int[n];
for(i = 0;i < n;i ++)a[i] = ah.nextInt();
Arrays.sort(a);
boolean b[] = new boolean[n];
for(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 | import java.util.Arrays;
import java.util.Scanner;
public class BoxAccumulation {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int[] a = new int[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 | import java.util.*;
import java.io.*;
import java.lang.*;
import java.math.*;
public class C {
public static void main(String[] args) throws Exception {
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
// Scanner scan = new Scanner(System.in);
// PrintWriter out = 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 | #include <bits/stdc++.h>
using namespace std;
const long long N = 1e10 + 0;
deque<long long> v, vc;
long long cnt, ans, sum, n, a;
map<long long, long long> mp;
void DNM() {
cin >> n;
for (long long i = 0; i < n; i++) {
cin >> a;
v.push_back(a);
}
sort(v.begin(), v.end());
for (long long i = 0; i < v.... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | n=int(input())
arr=list(map(int,input().split()))
arr.sort()
for k in range(1,n+1):
i=0
v=0
ok=True
while i<n:
for j in range(i,min(i+k,n)):
if arr[j]<v:
ok=False
break
i+=k
v+=1
if ok:
print(k)
exit()
| 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.awt.Point;
import java.io.*;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;
import static java.lang.Math.*;
public class Solution implements Runnable {
BufferedReader in;
PrintWriter out;
StringTokenizer tok = new StringTokenizer("");
... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.*;
// import java.lang.*;
import java.io.*;
// THIS TEMPLATE MADE BY AKSH BANSAL.
public class Solution {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
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 N, S[105];
bool feasible(int pile) {
priority_queue<int> pq;
for (int i = 0; i < pile; ++i) pq.push(S[N - i - 1]);
for (int i = N - pile - 1; i >= 0; --i) {
int lim = pq.top();
pq.pop();
if (lim <= 0) return false;
pq.push(min(lim - 1, S[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 | def check(s, k):
for i in xrange(len(s)):
if s[i] < i/k:
return False
return True
def f(n):
s = sorted(map(int, raw_input().split()))
start, end = 1, n
while start < end:
mid = (start+end)/2
if check(s, mid):
end = mid
else:
start ... | PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.util.*;
public class C {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n = s.nextInt();
Vector<Pile> list = new Vector<Pile>();
for(int i=0; i<n; i++)
list.addElement(new Pile(1, s.nextInt()));
Collections.sort(list);
int count = 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 | import heapq
#br = open('a.in')
f = lambda: map(int, raw_input().strip().split())
n, a, b = f()[0], sorted(f()), []
for i in a:
h = next((j for j, k in enumerate(b) if k <= i), None)
if h is None:
heapq.heappush(b, 1)
else:
b[h] += 1
print len(b)
| PYTHON |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | //Author: net12k44
import java.io.*;
import java.util.*;
public
class Main{//}
static PrintWriter out;
static void solve() {
int n = in.nextInt();
int a[] = new int [n];
int d[] = new int [n];
for(int i = 0; i < n ;++i) a[i] = in.nextInt();
Arrays.sort(a);
int kq = 0;
for(int i = 0; i < n; ++i) {
boolean... | 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 scan = new Scanner(System.in);
int n,count=1;
n=scan.nextInt();
int[] boxes = new int[n];
for(int i=0;i<n;i++){
boxes[i]=scan.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 | import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;
public class A228{
void solve()
{
int n = ni();
int[] a = ia(n);
Arrays.sort(a);
for(int i=0;i<n;i++)
{
if(a[i] < 0)
continue;
int j = i+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 main() {
int n;
cin >> n;
vector<int> vec(n);
int res = 0;
for (int i = 0; i < n; i++) cin >> vec[i];
sort(vec.begin(), vec.end());
int num = 0;
bool vis[110] = {0};
for (int a = 0; a < n; a++) {
bool ok = 0;
num = 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 | #include <bits/stdc++.h>
using namespace std;
int ri() {
int r;
cin >> r;
return r;
}
unsigned int rui() {
unsigned int r;
cin >> r;
return r;
}
long long rl() {
long long r;
cin >> r;
return r;
}
unsigned long long rul() {
unsigned long long r;
cin >> r;
return r;
}
double rd() {
double r;
... | 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.OutputStream;
import java.util.Arrays;
import java.io.IOException;
import java.util.Random;
import java.io.UncheckedIOException;
import java.util.AbstractMap;
import java.util.TreeMap;
import java.io.Closeable;
import ja... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int a[41111], n, was[41111], cnt;
int main(void) {
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
int ans = 0;
while (cnt < n) {
int x = 0;
for (int i = 0; i < n; i++)
if (was[i] == 0 && x <= a[i]) x++, was[i] = 1;
cnt += x, a... | CPP |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | def go():
n = int(input())
a = [int(i) for i in input().split(' ')]
a.sort()
o = 1
for i in range(n):
if(a[i] < i // o):
o += 1
return o
print(go())
| 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 static java.lang.Math.*;
import static java.lang.System.currentTimeMillis;
import static java.lang.System.exit;
import static java.lang.System.arraycopy;
import static java.util.Arrays.sort;
import static java.util.Arrays.binarySearch;
import static java.util.Arrays.fill;
import java.util.*;
import java.io.*;
p... | JAVA |
389_C. Fox and Box Accumulation | Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For exampl... | 2 | 9 | import java.io.*;
import java.util.*;
import java.lang.*;
public class Accumulation {
public static void main(String[] args) throws java.lang.Exception {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWr... | JAVA |
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