prompt string | response string |
|---|---|
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.*;
/**
* Ashesh Vidyut (Drift King) *
*/
public class C {
public static void main(String[] args) {
try {
BufferedReader in = new BufferedReader(n... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | from sys import stdin,stdout,setrecursionlimit,maxint,exit
#setrecursionlimit(2*10**5)
def listInput():
return map(long,stdin.readline().split())
def printBS(li):
for i in xrange(len(li)-1):
stdout.write("%d "%li[i])
stdout.write("%d\n"%li[-1])
from math import ceil
def sin():
return stdin.readline().rstrip()
n=i... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, i, flag, piles, till_now;
int arr[100];
cin >> n;
for (i = 0; i < n; i++) {
cin >> arr[i];
}
sort(arr, arr + n);
flag = 1;
piles = 0;
while (flag != 0) {
flag = 0;
till_now = 0;
for (i = 0; i < n; i++) {
if (arr[i]... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | N = int(raw_input())
A = map(int,raw_input().split())
A.sort()
for k in xrange(1,N+1):
flag = 0
for i in xrange(N):
if A[i] < (i/k):
flag=1
break
if flag==0:
break
print k
|
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
const long long MXN = 1e6 + 1;
const long long MNN = 1e3 + 1;
const long long MOD = 1e9 + 7;
const long long INF = 1e18;
const long long MAGIC = 1000;
const long double EPS = 1e-9;
const long double PI = 3.1415926536;
const string TIME = "rovno_ushten_push_ketti";
long long... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | import java.util.*;
import java.io.*;
public class a {
public static void main(String[] args) throws IOException
{
input.init(System.in);
PrintWriter out = new PrintWriter(System.out);
int n = input.nextInt();
int[] a = new int[n];
for(int i = 0; i<n; i++) a[i] = input.nextInt();
Arrays.sor... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | import sys, math
def rs():
return sys.stdin.readline().strip()
def ri():
return int(sys.stdin.readline().strip())
def ras():
return list(sys.stdin.readline().strip())
def rai():
return map(int,sys.stdin.readline().strip().split())
def raf():
return map(float,sys.stdin.readline().strip().split())
... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
int main(void) {
int arr[100], sum;
int pile = 1;
scanf("%d", &sum);
for (int i = 0; i < sum; i++) scanf("%d", &arr[i]);
sort(arr, arr + sum);
int count = 0;
for (int i = 0; i < sum; i++) {
if (i / pile > arr[i]) {
pile++;
} else {
count++;... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
public class C {
static int compute(int[] s) {
Arrays.sort(s);
int ts=0, ret=0, j=0;
for (int i=s.length-1; i>=0; i--) {
... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
int e[100 + 1];
int main() {
int n;
cin >> n;
int a[n + 1];
for (int i = 1; i <= n; i++) cin >> a[i];
sort(a + 1, a + n + 1);
int cnt = 0;
for (int i = 1; i <= n; i++) {
if (e[i] == 1) continue;
int dem = 1;
for (int j = i + 1; j <= n; j++)
i... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | n = int(input())
arr = list(map(int, input().split()))
arr.sort()
res = 1
for i in range(n):
if arr[i] < i // res:
res += 1
print(res)
|
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | n = int(input())
l = [int(x) for x in input().split()]
ans = 0
l.sort()
while(l):
l1 = []
c = 0
for i in range(len(l)):
if (l[i]>=c):
c+=1
else:
l1.append(l[i])
l = l1[:]
ans+=1
print(ans) |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | import java.util.Arrays;
import java.util.Scanner;
public class me {
public static void main(String[] args) {
Scanner br = new Scanner(System.in);
int n = br.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; ++i) {
arr[i] = br.nextInt();
}
Arrays... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args) throws NumberFormatException, IOException {
... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | import java.util.*;
import java.io.*;
//DONT FORGET TO CHANGE CLASS NAME!
public class FoxBox {
public static void main(String[] args) throws IOException {
Scanner sc = new Scanner(System.in);
StringTokenizer st=new StringTokenizer(sc.nextLine());
int numBoxes=Integer.parseInt(st.nextToken(... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #Python 2.7
"""
Sort the array, into ascending
Then simply place it in a stack if possible, or make a new one (ie do it in reverse with the least strongest first.
"""
numStacks = 0
n = input()
boxes = raw_input().split()
boxes = [int(x) for x in boxes]
boxes.sort()
#print boxes
#array of stacks
stackArray = ... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
const int MOD = (int)(1e9) + 7;
map<int, int> M;
set<int> S;
int n, a[123], b[123][123], c[123], ans, ma;
int main() {
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a[i];
b[c[a[i]]][a[i]] = 1;
c[a[i]]++;
}
for (int j = 1; j <= 100; j++)
for (int i =... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | 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.FileReader;
import java.io.InputStreamReader;
import java.io.InputStream;
... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... |
def check(a,x):
for i in range(len(a)):
if a[i] < len(a) - i:
return False
return True
n=int(input())
a=list(map(int,input().split()))
a=sorted(a)
piles = []
for i in a:
if len(piles) == 0:
piles.append([i])
else:
flg=1
for j in range(len(piles)):
if len(piles[j])<=i:
piles[j].append(i)
f... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | import java.util.*;
import java.io.*;
import java.math.*;
public class Practice
{
static FastReader sc=new FastReader();
public static void main(String[] args)
{
int ttt =1;
//ttt=i();
int mod=(int)Math.pow(10, 9)+7;
PrintWriter out = new PrintWriter(System.out);
... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
vector<pair<string, bool>> dict;
int main() {
int N;
vector<pair<int, bool>> arr;
cin >> N;
for (int i = 0; i < N; i++) {
int q = 0;
cin >> q;
arr.push_back(make_pair(q, false));
}
sort(arr.begin(), arr.end());
int res = 0;
for (int i = 0; i < N;... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | numberofboxes = int(raw_input())
boxes = list(map(int,raw_input().split(" ")))
boxes.sort()
stacklist = [0]
for i in boxes:
for j in range(len(stacklist)):
if i >= stacklist[j]:
stacklist[j] += 1
break
else:
stacklist.append(1)
print len(stacklist) |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
int a[101];
vector<int> anda;
int main() {
int n;
cin >> n;
int ans = 0;
for (int i = 1; i <= n; i++) cin >> a[i];
sort(a + 1, a + n + 1);
for (int i = 1; i <= n; i++) {
std::vector<int>::iterator up;
sort(anda.begin(), anda.end());
up = std::upper_b... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
int n, a[101], cnt, now, p, rst, x, ans = 0, nn;
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
sort(a + 1, a + 1 + n);
rst = n;
while (rst > 0) {
now = 1;
x = a[1];
a[1] = 101;
nn = rst;
--rst;
for (int i = ... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | from collections import Counter
n=input()
x=map(int,raw_input().split())
c=Counter(x)
def dfs(d):
for k in c:
if k>=d:
c[k]-=1
if c[k]==0:
del c[k]
dfs(d+1)
break
cnt=0
while len(c):
dfs(0)
cnt+=1
print cnt |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | import java.util.*;
import java.io.*;
import java.math.BigInteger;
public class Main
{
static final int mod = (int)1e9+7;
static int N = 101;
public static void main(String[] args) throws Exception
{
FastReader in = new FastReader();
PrintWriter out = new PrintWriter(System.out);
int n = in.nextInt();
int... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
int main(int argc, char** argv) {
int N;
cin >> N;
vector<int> vec(N);
for (int i = 0; i < N; i++) cin >> vec[i];
sort(vec.begin(), vec.end());
vector<int> H;
vector<int> L;
L.push_back(vec[0]);
H.push_back(1);
for (int i = 1; i < N; i++) {
int curr ... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
int x[105], n, wtf[105], ans;
int main() {
scanf("%d", &n);
for (int i = 0; i < n; ++i) scanf("%d", &x[i]);
sort(x, x + n);
ans = 0;
for (int i = 0; i < n; ++i) {
bool br = false;
for (int j = 0; j < ans; ++j)
if (wtf[j] <= x[i]) {
++wtf[j];
... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
int a[(110)];
int n;
int b[(110)][(110)];
int pan(int x) {
memset(b, 0, sizeof(b));
for (int i = n, j = 1; i > 0; --i, j = j % x + 1) b[j][++b[j][0]] = a[i];
for (int i = 1; i <= x; ++i)
for (int j = 1; j <= b[i][0]; ++j)
if (b[i][j] < b[i][0] - j) return 0;... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
const int maxn = 222222, inf = ~0U >> 1;
int A[maxn], n, m, cur[maxn];
bool ok(int num) {
memset(cur, (0), sizeof(cur));
for (int i = 0; i < n; i++) {
int x = A[i];
int no = 0;
for (int j = 0; j < num; j++)
if (cur[j] <= x && cur[j] >= cur[no]) no = j;... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
int a[105], x[105];
int main() {
int n, i, cn = 0, j, w, wn, ck;
scanf("%d", &n);
memset(a, 0, sizeof a);
for (i = 0; i < n; i++) {
scanf("%d", &w);
a[w]++;
}
while (1) {
wn = 0;
ck = 0;
for (i = 0; i < 101; i++) {
while (a[i] && i >= w... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | n=int(input())
arr=[int(i) for i in input().split()]
arr.sort()
for i in range(1,n+1):
temp=[]
l=n-1
for j in range(i):
temp.append(arr[l])
l-=1
boo=True
while(l>=0 and boo):
boo=False
for j in range(i):
if temp[j]>0:
boo=True
... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | import java.util.Arrays;
import java.util.Scanner;
public class FoxAndBoxAccumaltion {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | n, required, searched = int(input()), 0, []
boxes = list(sorted([int(i) for i in input().split()]))
while True:
stack = 0
for i in range(n):
if i not in searched and boxes[i] >= stack:
searched.append(i)
stack += 1
if stack:
required += 1
else:
bre... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
bool R[107];
int ar[107];
int find(int val, int pos, int n) {
for (int i = pos; i < n; i++)
if (ar[i] >= val and !R[i]) return i;
return n;
}
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) cin >> ar[i];
sort(ar, ar + n);
bool did = true;
int r... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | from collections import deque
def getLine():
return list(map(int,input().split()))
n = int(input())
arr = getLine()
arr.sort()
d = deque()
d.append(arr[0])
l = [d]
for i in range(1,len(arr)):
found = False
for d in l:
if len(d) <= arr[i]:
found=True
d.append(arr[i])
... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
int n, x;
cin >> n;
priority_queue<int, vector<int>, greater<int> > Q;
for (int i = 0; i < n; i++) {
cin >> x;
Q.push(x);
}
int ans = 0;
auto V = vector<int>();
int cnt = 0, t;
while (n) {
t = Q.top()... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
int n, m, a[1005], p[1005];
int main() {
scanf("%d", &n);
for (int i = 0; i < n; ++i) scanf("%d", a + i);
sort(a, a + n);
for (int i = 0; i < n; ++i) {
int w = -1;
for (int j = 0; j < m; ++j)
if (p[j] <= a[i]) {
++p[j], w = j;
break;
... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
int n, a[101];
bool vis[101];
int main() {
cin >> n;
for (int i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
int sum = 0;
int t = n;
int flag = 0;
while (t) {
flag = 0;
for (int i = 0; i < n; i++) {
if (flag <= a[i] && !vis[i]) {
vis[i] ... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;
public class A {
public static void main(String[] args) throws IOException {
InputReader in = new InputReader();
int n = in.nextInt();
int[] a = new int[n];
f... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
template <typename T>
string NumberToString(T Number) {
ostringstream ss;
ss << Number;
return ss.str();
}
template <typename T>
T StringToNumber(const string &Text) {
istringstream ss(Text);
T result;
return ss >> result ? result : 0;
}
int v[100];
int main() {... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | import java.util.*;
public class FoxAndBoxAccumulation
{
public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int[] boxes = new int[n];
for(int x = 0; x < n; x++)
{
boxes[x] = in.nextInt();
}
Arrays.sort(boxes);
outer:
for(int y = 1; y <... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | import java.util.Arrays;
import java.util.Scanner;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int[] v = new int[n];
boolean[] f = new boolean[n];
for (int i = 0; i < n; i++)
v[i] = in.nextInt();
Arrays.sort(v);
in... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
int n, t, m = 0;
cin >> n;
vector<int> s;
for (int i = 0; i < n; i++) {
cin >> t;
s.push_back(t);
}
sort(s.begin(), s.end());
while (true) {
t = -1;
for (int i = 0; i < n; i++) {
if (s[i] !... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | import java.lang.reflect.Array;
import java.util.*;
import java.io.*;
import java.math.*;
import java.lang.*;
/**
* Created by vishakha on 07-11-2016.
*/
public class Problem {
public static void main(String[] args) {
FastReader1 sc = new FastReader1(System.in);
int n = sc.nextInt();
Arr... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
int n, a[105];
bool ok(int k) {
for (int i = 1; i <= n; ++i)
if ((i - 1) / k > a[i]) return false;
return true;
}
int main() {
cin >> n;
for (int i = 1; i <= n; ++i) cin >> a[i];
sort(a + 1, a + n + 1);
for (int i = 1; i <= n; ++i)
if (ok(i)) {
cou... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | import java.io.*;
import java.util.*;
public class Test{
static final int p = 1000000007;
public static void main(String [] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
StringTokenizer st = ne... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Collection;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Map;
import java.util.Scanner;
public class Main {
public static void main(String args[]) {
Scann... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Collections;
import java.util.LinkedList;
import java.util.StringTokenizer;
public class C {
static class Scanner{
BufferedReader br=null;
StringTokenizer tk=null;
public Sca... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> t(n);
bool c[n];
for (int i = 0; i < n; i++) {
cin >> t[i];
c[i] = true;
}
sort(t.begin(), t.end());
int cnt = 0, ans = 0;
while (cnt != n) {
int cie = 0;
for (int i = 0; i < n; i++) {
if (c[i... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #double underscore makes a class variable or a class method private
mod = 1000000007
ii = lambda : int(input())
si = lambda : input()
dgl = lambda : list(map(int, input()))
f = lambda : map(int, input().split())
il = lambda : list(map(int, input().split()))
ls = lambda : list(input())
n = ii()
l = il()
vis=[0]*n
l.sort... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
#pragma comment(linker, "/STACK:36777216")
using namespace std;
template <class T>
inline T &RD(T &);
template <class T>
inline void OT(const T &);
inline long long RD() {
long long x;
return RD(x);
}
inline double &RF(double &);
inline double RF() {
double x;
return RF(x);
}
inline cha... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | from itertools import *
n = int(raw_input())
x = map(int, raw_input().split())
d = {}
for e in x:
if e not in d:
d[e] = 0
d[e]+=1
k = max(d.values())
x = sorted(x, reverse=True)
for p in xrange(1,k+1):
bins = [200 for _ in xrange(p)]
for e in x:
_,i = max(zip(bins,xrange(p)), key=lambda a: a[0])
... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | n = int(raw_input())
r = list(sorted(int(c) for c in raw_input().split()))
piles = [0] * 100
np = 0
for i in r:
for j in xrange(100):
if piles[j] > i:
continue
piles[j] += 1
np = max(np,j+1)
break
print np |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
int a[110];
int main() {
int i, N, x;
cin >> N;
for (i = 0; i < N; i++) {
cin >> x;
++a[x];
}
int ans, p, need;
for (ans = 0;; ++ans) {
for (p = 0; p <= 100; p++)
if (a[p]) break;
if (p > 100) break;
p = 0;... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
const int co = 1000;
int a[co + 10];
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) scanf("%d", a + i);
sort(a, a + n);
int used = 0, ans = 0;
while (used < n) {
ans++;
int h = 0;
for (int i = 0; i < n; i++)
if (a[i] >= h) {
... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
int b[103];
int w[103] = {0};
int ans = 1;
int bb[103];
int main() {
int n, i, j;
cin >> n;
for (i = 1; i <= n; i++) cin >> b[i];
for (i = 1; i <= n; i++) bb[i] = -1;
sort(b + 1, b + 1 + n);
for (i = 1; i <= n; i++) {
for (j = 1;; j++) {
if (b[i] >= bb... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int arr[n];
int c = 0, z = 0;
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
sort(arr, arr + n);
for (int i = 0; i < n; i++) {
if (c * (arr[i] + 1) <= i) c++;
}
cout << c << endl;
return 0;
}
|
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | import java.io.*;
import java.util.*;
public class C {
static BufferedReader stdin = new BufferedReader(new InputStreamReader(
System.in));
static StringTokenizer st = new StringTokenizer("");
public static void main(String[] args) throws Exception {
int n = readInt();
int[] w ... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | var num = parseInt(readline());
var arr = readline().split(" ");
var map = {};
var col = 0;
var cols = [];
for(var i in arr) {
arr[i] = parseInt(arr[i]);
}
arr.sort(unSort);
for(var i in arr) {
var
value = arr[i],
isNew = true
;
for(var j in cols) {
if (cols[j].length <= valu... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | n=input()
arr=map(int,raw_input().split())
arr=sorted(arr)
an=[1]
for i in range(1,len(arr)):
a=arr[i]
found=False
for j in range(len(an)):
b=an[j]
if b<=a:
an[j]+=1
found=True
break
if found==False:
an.append(1)
an=sorted(an,reverse=True)
... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | def solve():
n = input()
L = sorted(map(int, raw_input().split()))
for i in range(1, n + 1):
pile = [0] * i
ind = 0
for j in range(n):
if pile[ind % i] <= L[j]:
pile[ind % i] += 1
ind += 1
else:
break
els... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | from sys import stdin,stdout
nmbr=lambda:int(stdin.readline())
lst = lambda: list(map(int,input().split()))
for _ in range(1):#nmbr()):
n=nmbr()
a=sorted(lst())
ans=0
while 1:
left=[]
cap=0
for v in a:
if v>=cap:
cap+=1
else:left+=[v]
... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | def input_int_array():
return [int(string) for string in input().split()]
# input part
n = input_int_array()[0]
boxes = input_int_array()
counts = [0]*101
for hardess in boxes:
counts[hardess] += 1
# bingo
columns = []
for hardness in range(0, 101) :
if counts[hardness] <= 0 :
continue
... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
map<int, int> cnt;
vector<int> ara;
int main() {
int n, x, ans = 0;
cin >> n;
for (int i = 0; i < (n); i++) {
cin >> x;
ara.push_back(x);
}
sort(ara.begin(), ara.end());
for (int i = 0; i < (ara.size()); i++) {
x = ara[i];
cnt[x] = i + 1;
x =... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | n = int(input())
a = list(map(int,input().split()))
r = 0
flags = [False]*n
a.sort()
count = 0
for i in range(n):
if(count==n):
break
r+=1
last = -1
for j in range(n):
if(flags[j]):
continue
if(a[j]==last):
continue
flags[j] = True
count+=1
last+=1
print(r) |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | n=int(input())
a=list(map(int,input().split()))
sm=1
b=[0 for i in range(101)]
for i in range(n):
b[a[i]]+=1
x=[0]*n
z=0
zm=0
for i in range(101):
while b[i]!=0:
if x[z]>i:
z+=1
zm=max(zm,z)
if x[z]<=i and b[i]>0:
x[z]+=1
b[i]-=1
z=0
pr... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | #include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5, M = 1e9 + 7;
int main() {
ios ::sync_with_stdio(0);
cin.tie(0);
int n;
cin >> n;
vector<int> a;
for (int i = 1; i <= n; ++i) {
int x;
cin >> x;
a.push_back(x);
}
int ret = 0;
while (!a.empty()) {
++ret;
vector<int... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... | n = int( input() )
a = sorted( list( map( int, input().split() ) ) )
used = [False]*(n+1)
ans = 0
while True:
pos = -1
for i in range( n ):
if used[i] == False:
pos = i
break
if pos == -1:
break
ans += 1
total = 1
used[pos] = True
for j in range(... |
Problem: Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box).
Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. F... |
import java.util.*;
import java.io.*;
public class Codeforces {
static int arr[];
static boolean used[];
public static void main(String[] args) throws IOException {
Reader.init(System.in);
int n = Reader.nextInt();
arr = new int[n];
used = new boolean[n];
for (in... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in PYTHON): | i = [0, 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1, 52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4, 267, 1, 4, 1, 5, 1, 4, 1, 50, 1, 2, 3, 4, 1, 6, 1, 52, 15, 2, 1, 15, 1, 2, 1, 12, 1, 10, 1, 4, 2 ]
print(i[input()... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in PYTHON): | oes=[ 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51, 1,2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1, 52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4, 267, 1, 4, 1, 5, 1, 4, 1, 50, 1, 2, 3, 4, 1, 6, 1, 52, 15, 2, 1, 15, 1, 2, 1, 12, 1, 10, 1, 4, 2]
value = int(raw_in... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in CPP): | #include <bits/stdc++.h>
using namespace std;
int s[65] = {0, 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14,
1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51, 1,
2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1, 52, 2, 5,
1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2,... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in JAVA): | import java.io.InputStreamReader;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static ... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in CPP): | #include <bits/stdc++.h>
int a[1020] = {0, 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1,
14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1,
51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1,
52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4,
... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in PYTHON3): | print([0,1,1,1,2,1,2,1,5,2,2,1,5,1,2,1,14,1,5,1,5,2,2,1,15,2,2,5,4,1,4,1,51,1,2,1,14,1,2,2,14,1,6,1,4,2,2,1,52,2,5,1,5,1,15,2,13,2,2,1,13,1,2,4,267,1,4,1,5,1,4,1,50,1,2,3,4,1,6,1,52,15,2,1,15,1][int(input())])
|
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in CPP): | #include <bits/stdc++.h>
using namespace std;
int n;
int main() {
cin >> n;
switch (n) {
case 0:
cout << "0";
break;
case 1:
cout << "1";
break;
case 2:
cout << "1";
break;
case 3:
cout << "1";
break;
case 4:
cout << "2";
break;
cas... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in CPP): | #include <bits/stdc++.h>
using namespace std;
int main() {
int num[] = {1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14,
1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51,
1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1, 52,
2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1,... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in PYTHON3): | #include<iostream>
a = [0, 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1, 52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4, 267, 1, 4, 1, 5, 1, 4, 1, 50, 1, 2, 3, 4, 1, 6, 1, 52, 15, 2, 1, 15, 1, 2, 1, 12, 1, 10, 1, 4, ... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in JAVA): | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class F {
public static void main(String[] args) throws IOException {
FastScanner scan = new FastScanner();
PrintWriter out = new PrintWriter(System.ou... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in CPP): | #include <bits/stdc++.h>
const int a[] = {0, 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1,
14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1,
51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1,
52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4,
... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in PYTHON): |
a = [ 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1,
15, 2, 2, 5, 4, 1, 4, 1, 51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1,
52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4, 267, 1, 4, 1, 5, 1, 4, 1,
50, 1, 2, 3, 4, 1, 6, 1, 52, 15, 2, 1, 15, 1, 2, 1, 12, 1, 10, 1, 4, 2]
n = input... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in CPP): | #include <bits/stdc++.h>
using namespace std;
int main() {
int a[105] = {0, 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1,
14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1,
51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1,
52, 2, 5, 1, 5, 1, 15, 2, ... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in PYTHON3): | a=[0,1,1,1,2,1,2,1,5,2,2,1,5,1,2,1,14,1,5,1,5,2,2,1,15,2,2,5,4,1,4,1,51,1,2,1,14,1,2,2,14,1,6,1,4,2,2,1,52,2,5,1,5,1,15,2,13,2,2,1,13,1,2,4,267,1,4,1,5,1,4,1,50,1,2,3,4,1,6,1,52,15,2,1,15,1,2,1,12,1,10,1,4,2]
n=int(input())
print(a[n]) |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in CPP): | #include <bits/stdc++.h>
using namespace std;
const double PI = 3.14159;
int main() {
ios::sync_with_stdio(false);
int a[64] = {1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14,
1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51,
1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1, 52,... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in JAVA): | import java.util.*;
public class f {
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
int[] res = new int[]{1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1, 52, 2, 5, 1, 5, 1, 15, 2, ... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in JAVA): | import java.math.BigInteger;
import java.io.BufferedWriter;
import java.io.File;
import java.io.PrintWriter;
import java.util.Scanner;
// yeah another submission for this problem
public class Main {
private void solve() throws Exception {
int groupsOfOrderN[] = { 0, 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1,... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in CPP): | #include <bits/stdc++.h>
using namespace std;
int a[64] = {1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14,
1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51,
1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1, 52,
2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4, 267};
int main... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in PYTHON): | print[0,1,1,1,2,1,2,1,5,2,2,1,5,1,2,1,14,1,5,1,5,2,2,1,15,2,2,5,4,1,4,1,51,1,2,1,14,1,2,2,14,1,6,1,4,2,2,1,52,2,5,1,5,1,15,2,13,2,2,1,13,1,2,4,267,1,4,1,5,1,4,1,50,1,2,3,4,1,6,1,52,15,2,1,15,1,2,1,12,1,10,1,4,2][input()] |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in CPP): | #include <bits/stdc++.h>
using namespace std;
int main() {
int a[] = {0, 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1,
14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1,
51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1,
52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in PYTHON3): | a=[0,1,1,1,2,1,2,1,5,2,2,1,5,1,2,1,14,1,5,1,5,2,2,1,15,2,2,5,4,1,4,1,51,1,2,1,14,1,2,2,14,1,6,1,4,2,2,1,52,2,5,1,5,1,15,2,13,2,2,1,13,1,2,4,267,1,4,1,5,1,4,1,50,1,2,3,4,1,6,1,52,15,2,1,15,1,2,1,12,1,10,1,4,2]
b=int(input())
print(a[b]) |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in PYTHON3): | n = int(input())
res=[0, 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1, 52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4, 267, 1, 4, 1, 5, 1, 4, 1, 50, 1, 2, 3, 4, 1, 6, 1, 52, 15, 2, 1, 15, 1, 2, 1, 12, 1, 10, 1, 4, 2]... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in CPP): | #include <bits/stdc++.h>
using namespace std;
int ans[] = {0, 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1,
14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1,
51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1,
52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in PYTHON): | lst = [ 0, 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1, 52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4, 267, 1, 4, 1, 5, 1, 4, 1, 50, 1, 2, 3, 4, 1, 6, 1, 52, 15, 2, 1, 15, 1, 2, 1, 12, 1, 10, 1, 4, 2]
print lst[in... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in CPP): | #include <bits/stdc++.h>
int a[100] = {0, 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1,
14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1,
51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1,
52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4,
2... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in PYTHON): | import math
print[0,1,1,1,2,1,2,1,5,2,2,1,5,1,2,1,14,1,5,1,5,2,2,1,15,2,2,5,4,1,4,1,51,1,2,1,14,1,2,2,14,1,6,1,4,2,2,1,52,2,5,1,5,1,15,2,13,2,2,1,13,1,2,4,267,1,4,1,5,1,4,1,50,1,2,3,4,1,6,1,52,15,2,1,15,1,2,1,12,1,10,1,4,2][input()] |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in PYTHON3): | OeisA000001 = [0, 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1, 52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4, 267, 1, 4, 1, 5, 1, 4, 1, 50, 1, 2, 3, 4, 1, 6, 1, 52, 15, 2, 1, 15, 1, 2, 1, 12, 1, 10, 1, 4, 2]
print(... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in CPP): | #include <bits/stdc++.h>
int n, a[110] = {0, 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1,
14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1,
51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1,
52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4,
... |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in JAVA): | var a = [1,1,1,2,1,2,1,5,2,2,1,5,1,2,1,14,1,5,1,5,2,2,1,15,2,2,5,4,1,4,1,51,1,2,1,14,1,2,2,14,1,6,1,4,2,2,1,52,2,5,1,5,1,15,2,13,2,2,1,13,1,2,4,267];
var n = readline();
print(a[n-1]); |
Problem:
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Solution (in JAVA): | import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class r217_b {
static StringTokenizer st;
static BufferedReader br;
static PrintWriter ... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.