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 |
|---|---|---|---|---|---|
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | #include <bits/stdc++.h>
using namespace std;
map<unsigned long long int, unsigned long long int> m;
int main() {
unsigned long long int n;
unsigned long long int k;
unsigned long long int sol;
cin >> n;
m[1] = n;
unsigned long long int a, b;
auto it = m.begin();
while (it != m.end()) {
a = it->firs... | CPP |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | var n = parseInt(readline());
//Sol:
var r = [];
//func
var step = function( ){
//Modifica r
// Agrego un Slime
r.push(1);
//Calculo los cambios automaticos
if(r.length>1){
while(r[r.length-1]===r[r.length-2]){
r.pop();
r[r.length-1]++;
}
}
};
while(n-- > 0){
step();
}
r.map((s) => {write(s... | JAVA |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 |
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Mai... | JAVA |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | '''input
7
'''
n = input()
s = ''
res = []
for i in xrange(17, -1, -1):
# print n, pow(2, i), i
if pow(2, i) <= n:
n -= pow(2, i)
res.append(i + 1)
print ' '.join(['%s' % x for x in res]) | PYTHON |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | n = int(input())
v = []
for i in range(n):
v.append(1)
while len(v) >= 2 and v[-2] == v[-1]:
x = v.pop()
v.pop()
v.append(x+1)
print(' '.join([str(x) for x in v]))
| PYTHON3 |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | import java.io.*;
import java.util.*;
import java.lang.*;
import java.math.*;
public class abc{
public static void main(String args[]){
Scanner in=new Scanner(System.in);
int n=in.nextInt();
while(n>0){
int x=0,i=0;
while(x<=n){
x=(int)Math.pow(2,i);
i++;
}
i=i-2;
x=(int)Math.pow(2,i);
... | JAVA |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | n=bin(int(input()))[2:]
for i in range(len(n)):
if int(n[i])==1:
print(len(n)-i, end=' ') | PYTHON3 |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | #include <bits/stdc++.h>
using namespace std;
vector<long long> a;
int main() {
long long n;
cin >> n;
for (long long i = 0; i < n; i++) {
a.push_back(1);
while (a.size() > 1 && a[a.size() - 1] == a[a.size() - 2]) {
long long x = a[a.size()];
a.pop_back();
a[a.size() - 1]++;
}
}
... | CPP |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | a=[0]
for _ in range(int(input())):
a+=[1]
while a[-2]==a[-1]:
a.pop()
a[-1]+=1
print(*a[1:])
| PYTHON3 |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | #include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-9;
const int INF = 0x7f7f7f7f;
const double PI = acos(-1.0);
template <class T>
inline T _abs(T n) {
return ((n) < 0 ? -(n) : (n));
}
template <class T>
inline T _max(T a, T b) {
return (!((a) < (b)) ? (a) : (b));
}
template <class T>
inline T _min... | CPP |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | # https://codeforces.com/problemset/problem/618/A
n = int(input())
stack = [1]
for i in range(1, n):
if stack[-1] == 1:
stack[-1] = 2
while True:
if len(stack) >= 2:
last = stack.pop()
if stack[-1] == last:
stack[-1] = last + 1
... | PYTHON3 |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | #include <bits/stdc++.h>
using namespace std;
int main() {
int two_pow[17];
int max_pow;
int i;
for (i = 0; i <= 16; i++) {
two_pow[i] = pow(2, i);
}
int n, no;
cin >> n;
no = n;
for (i = 16; i >= 0; i--) {
if (two_pow[i] <= n) {
max_pow = i;
break;
}
}
while (no != 0) {
... | CPP |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | def main():
n = int(input())
a = []
for i in range(n):
a.append(1)
while len(a) > 1 and a[-1] == a[-2]:
a.pop()
a[-1] += 1
print(' '.join(str(i) for i in a))
main() | PYTHON3 |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | #include <bits/stdc++.h>
using namespace std;
vector<int> a;
int main() {
int n;
cin >> n;
for (int i = 0; true; i++) {
a.push_back(n % 2);
n /= 2;
if (n == 0) {
break;
}
}
for (int i = a.size() - 1; i >= 0; i--) {
if (a[i] == 1) {
cout << i + 1 << " ";
}
}
}
| CPP |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | #include <bits/stdc++.h>
int main() {
int n, i, s, a[100];
while (~scanf("%d", &n)) {
s = 1;
for (;; s *= 2) {
if (s > n) break;
}
s /= 2;
for (i = 0; s != 0; i++) {
if (n >= s) {
a[i] = 1;
n = n - s;
s /= 2;
} else {
a[i] = 0;
s = s / 2;... | CPP |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | #include <bits/stdc++.h>
using namespace std;
int n;
vector<int> v;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n;
for (int i = 1; i <= n; i++) {
v.push_back(1);
while ((int)v.size() > 1) {
int sz = v.size();
if (v[sz - 1] == v[sz - 2]) {
v.pop_back()... | CPP |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;
public class A {
public static void main(String[] args) throws IOException {
br = new BufferedReader( new InputStreamRea... | JAVA |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | n = int(input())
l = [1]
for i in range(1,n):
l.append(1)
for j in range(len(l)):
if len(l)>1 and l[-1]==l[-2] :
l[-2]=l[-2]+1
l.pop(-1)
else:
break
print(*l) | PYTHON3 |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | N = input()
List=[1 for i in range(N)]
Ans=[]
for i in range(N):
Ans.append(1)
if len(Ans)>=2:
while(1):
if len(Ans)<2:
break
if Ans[-1]==Ans[-2]:
Ans[-2] += 1
del Ans[-1]
else:
break
ans = ""
for i in Ans:
ans += str(i)+" "
print ans[:-1]
| PYTHON |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | n = int(input())
L = []
for i in range(n):
L.append(1)
while len(L) >= 2 and L[-1] == L[-2]:
L.pop()
L[-1] += 1
print(' '.join(str(i) for i in L))
| PYTHON3 |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | from math import *
from Queue import *
from sys import *
n = int(raw_input())
res = [0 for i in range(n+1)]
p = 0
for i in range(n):
res[p] = 1
while (p-1 >= 0) and (res[p-1] == res[p]):
res[p] = 0
res[p-1] += 1
p -= 1
p += 1
p = 0
while res[p] != 0:
print(res[p]),
p += 1... | PYTHON |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.Collections;
import java.util.L... | JAVA |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
vector<int> v;
cin >> n;
int x = 1;
int i = 1;
if (n == 1)
cout << "1";
else {
bool dhukse = false;
while (x <= n) {
x *= 2;
if (x == n) {
cout << i + 1 << endl;
dhukse = true;
break;
} ... | CPP |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | s = input()
k = [0]
for i in xrange(s):
k.append(1)
while len(k) > 1 and k[-1] == k[-2]:
k.pop()
k[-1] = k[-1] + 1
for i in k[1: ]:
print i, | PYTHON |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | n=int(input())
d=[]
for i in range(n):
d.append(1)
if len(d)>=2:
while(d[-1]==d[-2]):
if d[-1]==d[-2]:
r=d[-1]+1
d.append(r)
d.pop(-2)
d.pop(-2)
if len(d)<2:
break
print(*d)
| PYTHON3 |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | n = int(input())
a = [2 ** i for i in range(20)]
ans = []
#print(a)
while n > 0:
for i in range(19, -1, -1):
if n >= a[i]:
print(i + 1, end=' ')
n -= a[i]
break
| PYTHON3 |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | import java.util.*;
public class slimeCombining
{
public static void main(String args[])
{
Scanner in = new Scanner(System.in);
int n;
ArrayList<Integer> al = new ArrayList<Integer>();
n=in.nextInt();
al.add(1);
for(int i=(n-1); i>0; i--)
{
al.add(1);
if(al.size()>1... | JAVA |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | n = int(input())
k = bin(n)[2::]
mas = []
l = len(k)
for a in range(l):
if k[a] == "1":
mas.append(str(l - a))
print(" ".join(mas)) | PYTHON3 |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | #include <bits/stdc++.h>
using namespace std;
struct nout {
template <class A>
const nout& operator<<(const A& a) const {
return *this;
}
} nout;
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
long long N;
cin >> N;
bool out = false;
for (long long i__count = (20), i = i__count, i__goal = (0... | CPP |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | #include <bits/stdc++.h>
using namespace std;
void reshenie() {
int n;
cin >> n;
vector<int> bin;
while (n) {
bin.push_back(n % 2);
n /= 2;
}
for (int i = bin.size() - 1; i >= 0; --i) {
if (bin[i]) {
cout << i + 1 << ' ';
}
}
cout << endl;
}
int main() {
reshenie();
return 1 - ... | CPP |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.List;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.ArrayList;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is a... | JAVA |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int next = 1;
vector<int> ans;
while (n) {
if (n & 1) ans.push_back(next);
next++;
n >>= 1;
}
for (auto i = ans.rbegin(); i != ans.rend(); ++i) cout << *i << " ";
return 0;
}
| CPP |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | import sys
import math
def good(n):
while n%2==0:
n/=2
if n==1:
return True
else :
return False
arr={1:[1],}
t=2
while t<100005:
arr[t]=[]
if good(t):
arr[t]=[int(math.log2(t))+1]
else:
tmp=2**(int(math.log2(t)))
arr[t]=arr[tmp]+arr[t-tmp]
t=t+1
for line in sys.stdin:
n=line.split()
a=int(n[0])
... | PYTHON3 |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | import java.io.*;
import static java.lang.Math.*;
import java.util.*;
import java.util.function.*;
import java.lang.*;
public class Main {
final static boolean debug = false;
final static String fileName = "";
final static boolean useFiles = false;
public static void main(String[] args) throws FileNot... | JAVA |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | n=int(input())
l=[2**i for i in range(20)]
while n>0:
if n in l:
i=l.index(n)
print(i+1)
n=0
else:
l.append(n)
l.sort()
i=l.index(n)
print(i,end=" ")
l.remove(n)
n-=2**(i-1) | PYTHON3 |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | L=[1];
for i in range(2,int(input())+1):
L.append(1);
a=len(L)-1;
while(a>0):
if L[a]==L[a-1]:
L[a-1]=L[a]+1;
L[a:a+1]=[];
else:
break
a=len(L)-1;
if a==0:
break
for i in L:
print(i,end=' ') | PYTHON3 |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | n = input()
res = []
for x in xrange(30, -1, -1):
if (n & (1 << x)):
res += [x+1]
print " ".join(map(str, res)) | PYTHON |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | import math
def rlist(t):
return map(t, raw_input().split())
def read_int_list():
return rlist(int)
x = input()
ans = []
while x:
pow = 1
i = 0
while pow <= x:
pow *= 2
i += 1
x -= 2**(i-1)
ans.append(str(i))
print " ".join(ans) | PYTHON |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Stack;
import java.util.StringTokenizer;
public class A {
static void solve(InputReader in, PrintWriter out) {
int N = i... | JAVA |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | #!/usr/bin/env python
import math, sys, collections, heapq
cache = [1]
for i in range(20):
cache.append(cache[-1]*2)
def twotopow(i):
for j in range(19):
if cache[j] <= i < cache[j+1]: return j
return -1
n = int(raw_input())
res = []
while n:
r = twotopow(n)
res.append(r+1)
n %= 2**r
... | PYTHON |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | n = int(input())
k = []
for x in range(n):
k.append(1)
while len(k) >= 2 and k[-2] == k[-1]:
k.pop()
k[-1] += 1
print(" ".join(map(str, k))) | PYTHON3 |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int n;
cin >> n;
for (int i = 30; ~i; i--)
if (n >> i & 1) cout << i + 1 << ' ';
return 0;
}
| CPP |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | n = int(input())
ans = list()
for i in range(n):
ans.append(1)
while len(ans) > 1 and ans[-1] == ans[-2]:
t = ans.pop()
ans[-1] = t + 1
print(' '.join(map(str, ans)))
| PYTHON3 |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | n = int(input())
s = str(bin(n))[2:]
for i in range(len(s)):
if s[i] == '1':
print(len(s) - i, end=' ') | PYTHON3 |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | #include <bits/stdc++.h>
using namespace std;
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
const int maxn = 100010;
const double eps = 1e-8;
int a[maxn], sz = 0;
int main() {
int n;
scanf("%d", &n);
a[sz++] = 1;
n--;
while (n--) {
a[sz++] = 1;
while (sz >= 2 and a[sz - 1] == a[sz - ... | CPP |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | n = int(input())
i = 1;
arr = [];
while True:
if (n & 1 == 1):
arr.append(i);
n >>= 1;
i = i + 1;
if ( n == 0 ):
break;
outstr = "";
for elm in arr:
outstr = str(elm) + " " + outstr;
print(outstr); | PYTHON3 |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n, i, j, p, ans[1003];
;
cin >> n;
j = 1;
p = n;
while (p > 0) {
int k = p % 2;
ans[j++] = k;
p = p / 2;
}
for (i = j - 1; i >= 1; i--) {
if (ans[i] == 1) {
cout << i << " ";
}
}
return 0;
}
| CPP |
618_A. Slime Combining | Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.
You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all a... | 2 | 7 | import java.util.ArrayList;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner=new Scanner(System.in);
int n=0;
n=scanner.nextInt();
ArrayList arrayList = new ArrayList();
while(n>0){
arrayList.add(1);
... | JAVA |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
template <class T>
inline void rread(T& num) {
num = 0;
T f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') num = num * 10 + ch - '0', ch = getchar();
num *= f;
}
inline i... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.Arrays;
import java.io.BufferedWriter;
import java.util.InputMismatchException;
import java.util.HashMap;
import java.util.Set;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.Pri... | JAVA |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
import java.util.Scanner;
public class PogChamp {
public static void main(String[] args) {
new PogChamp().bUrself(Syst... | JAVA |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int read() {
int ans = 0, f = 1;
char c = getchar();
while (c > '9' || c < '0') {
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
ans = ans * 10 + c - '0';
c = getchar();
}
return ans * f;
}
const int N = 2e5 + 5;
int n, d... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int M = 200001;
struct Edge {
int u, v;
int ot(int x) { return u ^ x ^ v; }
} e[M];
vector<vector<int>> adj;
bool done[M];
int deg[M];
int num[M];
const int N = 200001;
vector<int> sol[N];
void dfs(int u, int p = -1, int n = -1) {
int cur = 1;
for (auto el : a... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | import java.io.*;
import java.math.*;
import java.util.*;
import static java.lang.System.out;
public class solver {
public static List<pair>[] g;
public static List<Integer>[] arr;
static void dfs(int from, int v, int day) throws Exception{
int d = 0; if(d == day) ++d;
for(pair i : g[v]){
if... | JAVA |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
cout << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
const char* comma = strchr(names + 1, ',');
cout.write(nam... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int LEN = 200200;
vector<int> ed[LEN];
vector<pair<int, int> > res[LEN];
int n, a, b, use[LEN], k;
map<int, int> q[LEN];
void dfs(int v, int d) {
int u, k = 1;
use[v] = 1;
for (int i = 0; i < ed[v].size(); i++) {
u = ed[v][i];
if (use[u]) continue;
i... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 200010;
bool used[MAXN];
int c[MAXN];
int father[MAXN];
vector<pair<int, int> > color[MAXN];
vector<int> g[MAXN];
map<pair<int, int>, int> e;
int res;
void dfs(int v, int pr, int num) {
father[v] = pr;
used[v] = 1;
int i = 0;
if (v != 1) {
if (c... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > G[200100];
vector<int> rs[200100];
void dfs(int x, int y, int p) {
int pos = 0;
int n = G[x].size();
for (int i = 0; i < n; ++i) {
int to = G[x][i].first;
if (to == y) continue;
if (pos == p) pos++;
rs[pos].push_back(G[x][i].sec... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9;
const double PI = 3.1415926535897932384626433832795;
int n, st[200007];
vector<pair<int, int>> v[200007];
vector<int> ans[200007];
void deep(int c, int day) {
int i, now = 0;
if (day == 0) now++;
st[c] = 1;
for (i = 0; i < v[c].size();... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
std::vector<std::vector<int>> g;
std::vector<std::vector<int>> rn;
std::vector<std::pair<int, int>> d;
std::vector<std::vector<int>> plan;
void calc_dp(int v, int p) {
int mx = 0;
int children = 0;
for (int n : g[v]) {
if (n != p) {
calc_dp(n, v);
mx =... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
vector<vector<int> > g;
map<pair<int, int>, int> MAP;
map<int, vector<int> > colors;
int mx;
void dfs(int v, int c, int p = -1) {
int cur(0);
for (int i(0), _l((int)(((int)g[v].size())) - 1); i <= _l; ++i) {
int to(g[v][i]);
if (to == p) continue;
int k(MAP[... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | // practice with kaiboy
import java.io.*;
import java.util.*;
public class CF638C extends PrintWriter {
CF638C() { super(System.out); }
Scanner sc = new Scanner(System.in);
public static void main(String[] $) {
CF638C o = new CF638C(); o.main(); o.flush();
}
int n;
int[] eo; int[][] eh;
int[] ij;
void appen... | JAVA |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
vector<pair<int, int>> v[200200];
vector<int> ans[200200];
bool vis[200200];
int k = 0;
void dfs(int curr, int c) {
vis[curr] = true;
int x = 0;
for (auto child : v[curr]) {
if (!vis[child.first]) {
if (++x == c) x++;
ans[x].push_back(child.second);
... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 7;
vector<pair<int, int> > E[maxn];
vector<int> ans[maxn];
int tot = 0;
void dfs(int x, int fa, int te) {
int now = 0;
for (int i = 0; i < E[x].size(); i++) {
int v = E[x][i].first;
if (v == fa) continue;
now++;
if (now == te) now+... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | import java.io.*;
import java.util.*;
public final class road_improvement
{
static FastScanner sc=new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
static PrintWriter out=new PrintWriter(System.out);
static ArrayList<Node>[] al;
static ArrayList<Integer>[] days;
static int max=1;
static... | JAVA |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
inline int next_int() {
int32_t x;
scanf("%d", &x);
return x;
}
const int maxn = 2e5 + 10;
vector<pair<int, int> > g[maxn];
int res[maxn];
int ans;
inline void dfs(int node, int father = -1, int preCol = -1) {
int curCol = 0;
for (auto u : g[node]) {
if (u.fir... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
vector<int> neigh[200005];
int ptr[200005];
vector<int> children[200005];
int order[200005];
int par[200005];
int get(int v) {
if (par[v] == v) return v;
return par[v] = get(par[v]);
}
void del(int v) { par[get(v)] = par[get(v + 1)]; }
int N;
void init() {
vector<bool... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
vector<int> v[200200], ans[200200];
int color[200200], a[200200], b[200200], n, pat, p[200200], m, j;
void dfs(int x) {
int m = v[x].size(), i, curr = 0;
for (i = 0; i < m; i++) {
if (v[x][i] == 1) continue;
if (color[v[x][i]] != 0) continue;
p[v[x][i]] = x;... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
unordered_map<int, vector<pair<int, int> > > ans;
vector<bool> used;
void rec(vector<vector<int> > &v, int pos, int color) {
int act_color = 0;
for (int act : v[pos])
if (!used[act]) {
act_color += (act_color == color);
used[act] = true;
ans[act_co... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
#pragma comment(linker, "/STACK:100000000000,100000000000")
const int INF = 1e9;
const double cp = 2 * asin(1.0);
const double eps = 1e-9;
const long long mod = 1000000007;
using namespace std;
int n;
vector<vector<pair<int, int> > > g;
int used[200020];
int res;
vector<int> ans[200020];
void d... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > g[200005];
vector<vector<int> > ans;
int n;
bool u[200005];
void dfs(int v, int w) {
int it = 0;
if (it == w) it++;
for (int i = 0; i < g[v].size(); i++)
if (!u[g[v][i].second]) {
ans[it].push_back(g[v][i].second);
u[g[v][i].sec... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
vector<pair<long long int, long long int>> v[200007];
vector<long long int> ans[200007];
long long int cc = 0;
void dfs(long long int x, long long int y, long long int z) {
long long int c = 0;
for (long long int i = 0; i < v[x].size(); i++) {
long long int val = v[... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
using Graph = vector<vector<int32_t> >;
using Plan = map<int32_t, int32_t>;
using Roads = map<pair<int32_t, int32_t>, int32_t>;
Graph G;
Plan plan;
Roads roads;
void dfs_path(int32_t curr, int32_t last, int32_t pair_day) {
int32_t day = 0;
for (auto to : G[curr]) {
... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
struct edge {
int v, id, day;
};
vector<edge> g[200100];
vector<int> d[200100];
bool used[200200];
int max_day = 0;
void dfs(int v, int pr_day) {
int day = 1;
for (int i = 0; i < g[v].size(); i++) {
if (day == pr_day) day++;
if (!used[g[v][i].v]) {
used[... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
template <class T>
T sqr(T x) {
return x * x;
}
template <class T>
T gcd(T a, T b) {
return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
return (a / gcd<T>(a, b) * b);
}
template <class T>
inline T bigmod(T p, T e, T M) {
if (e == 0) retur... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int stat[200010];
int fixe[200010];
int loop;
int coun;
vector<pair<int, int>> grafh[200010];
vector<int> day[200010];
int main() {
int m;
scanf("%d", &m);
m--;
for (int i = 0; i < m; i++) {
int a, b;
scanf("%d %d", &a, &b);
grafh[a].push_back({b, i + 1}... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
vector<vector<int> > g(2e5 + 10), d;
map<pair<int, int>, int> mp;
queue<pair<int, int> > q;
int main() {
ios_base::sync_with_stdio();
int n;
cin >> n;
int ans = 0;
for (int i = 0; i < n - 1; i++) {
int x, y;
scanf("%d%d", &x, &y);
g[x].push_back(y);
... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int N = 200000 + 1000;
vector<pair<int, int> > g[N];
vector<int> ans[N];
int n, sum, tot, u, v;
void dfs(int x, int fa, int son_t) {
int l = g[x].size();
int sum = 0;
for (int i = 0; i < l; i++) {
int v = g[x][i].first;
if (v == fa) continue;
sum++;
... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long inf = (1ll << 62) - 1;
const long long MOD = 1e9 + 7;
const int MAX = 2 * 1e9 + 10;
const int N = 2e5 + 5;
using namespace std;
int bPow(int a, int b) {
int res = 1;
while (b) {
if (b & 1ll) {
res = 1ll * res * a % MOD;
}
b >>= 1;
a... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
vector<vector<pair<int, int>>> vi;
vector<vector<int>> days;
long long result;
long long vis[10000000];
void dfs(int node, int lastday) {
if (vis[node]) return;
vis[node] = 1;
long long day = 1;
for (int nodes = 0; nodes < vi[node].size(); nodes++) {
int n = vi[... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int n, x, y, zc;
vector<pair<int, int> > a[200020];
vector<int> z[200020];
void dfs(int x, int y, int c) {
int j = 0;
for (pair<int, int> i : a[x]) {
if (i.first != y) {
++j;
if (j == c) {
++j;
}
z[j].push_back(i.second);
zc = m... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 55;
vector<pair<int, int>> v[N];
vector<int> days[N];
bool visited[N];
int mx = 0;
void dfs(int s, int m) {
int l = 1;
if (visited[s]) return;
visited[s] = 1;
for (auto x : v[s]) {
if (visited[x.first]) continue;
if (m == l) l++;
days... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
const double PI = 3.141592653589793;
using namespace std;
std::vector<int> v[200500];
std::vector<int> ans[200500];
int was[200500];
int st[200500];
map<std::pair<int, int>, int> m;
vector<std::pair<int, int> > r;
void dfs(int x, int c) {
was[x] = 1;
int cnt = 1;
for (int y : v[x]) {
... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > ed[(int)(200005)];
vector<int> ans[(int)(200005)];
int n, cvp;
void dfs(int cur, int back, int back_ed) {
int top = 0;
for (auto i : ed[cur]) {
if (i.first == back) continue;
top++;
if (top == back_ed) top++;
cvp = max(cvp, top);
... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int inf = (int)1e9;
const double eps = (double)1e-8;
const int mod = (int)1000000007;
const int maxn = (int)2 * 1e5 + 5;
int n, x, y, mx;
int u[maxn], pr[maxn];
pair<int, int> t[maxn];
vector<pair<int, int> > a[maxn];
vector<int> ans;
list<int> q;
list<int>::iterator ... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 200002;
struct NODE {
int v;
int id;
};
vector<NODE> G[MAXN];
vector<int> res[MAXN];
int n, ans = 0;
void DFS(int u, int p, int pcnt) {
int cnt = 0;
for (int i = 0; i < G[u].size(); i++) {
int v = G[u][i].v, id = G[u][i].id;
if (v != p) {
... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | import java.io.*;
import java.lang.reflect.Array;
import java.util.*;
public class Codeforces {
ArrayList<Pair>[]graph;
ArrayList<Integer>[]days;
public void dfs(int from,int to,int day){
int d = 0;
if(day == d)d++;
for(Pair p:graph[to]){
if(p.first == from)continue;
... | JAVA |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 7;
vector<pair<int, int> > e[MAXN];
vector<int> ans[MAXN];
void fixRoads(int u, int p, int c) {
int i, edgeIndex, v, d = 0;
for (auto it : e[u]) {
v = it.first;
edgeIndex = it.second;
if (v != p) {
if (d < c) {
i = d;
... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
void fast_io() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
vector<int> g[200005];
int n;
vector<int> day[200005];
vector<bool> vis(200005, 0);
map<pair<int, int>, int> mp;
int max_cnt = 0;
void dfs(int node, int cnt) {
vis[node] = 1;
max_cnt = max(max_cn... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long n, a;
const long long MAXN = 200000;
vector<pair<long long, long long> > graph[MAXN];
vector<long long> times[MAXN];
long long maxD = 0;
void dfs(long long v, long long p, long long t) {
long long timer = 0;
long long deg = (p == -1 ? 0 : 1);
for (long long ... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
long long bfs(long long n, vector<vector<pair<long long, long long> > > &g,
vector<vector<long long> > &path) {
queue<long long> q;
queue<long long> qs;
vector<bool> used(n + 1);
long long s;
s = 1;
long long k;
q.push(s);
qs.push(0);
qs.push... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int n;
int useday[N];
vector<int> g[N];
int a[N], b[N], fl[N];
int getnxt(int t, int v) {
if (t + 1 != useday[v]) return t + 1;
return t + 2;
}
void dfs(int v, int p) {
int t = getnxt(0, v);
for (int i : g[v]) {
int u = a[i] + b[i] - v;
... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
std::vector<pair<int, int> > adjList[200010];
int up_time[200010];
std::vector<int> ans_list[200010];
int U[200010], V[200010];
int ans = 0;
void dfs(int curr, int par) {
std::vector<pair<int, int> >::iterator it;
pair<int, int> temp;
int upd = 1;
it = adjList[curr]... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int MAX = 2000005;
int n, x, y, mxday;
map<pair<int, int>, int> mp2;
vector<vector<int> > adj(MAX);
vector<pair<int, int> > edge;
void dfs(int node, int parent, int pDay) {
int cnt = 1;
for (int i(0); i < int(adj[node].size()); i++) {
int ch = adj[node][i];
... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 100;
struct Edge {
int v, nxt;
int id;
Edge() {}
Edge(int vv, int next) {
v = vv;
nxt = next;
}
} edge[N << 1];
int idx;
int head[N];
void addEdge(int u, int v, int id) {
edge[++idx] = Edge(v, head[u]);
edge[idx].id = id;
head[u] ... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
vector<vector<int> > a, b, ans;
int i, j, n, mx;
int x, y, used[200005];
void paint(int v, int parent_color) {
used[v] = 1;
int cur_color = 0;
for (int i = 0; i < a[v].size(); i++) {
if (!used[a[v][i]]) {
if (cur_color == parent_color) cur_color++;
ans... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > V[200000];
int colors[200000];
int w[200000];
list<int> q;
map<int, vector<int> > answ;
void DFS() {
while (!q.empty()) {
int index = q.front();
q.pop_front();
colors[index] = 1;
vector<pair<int, int> >::iterator it = V[index].begin... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > graph[200001];
vector<int> ans[200001];
int vis[200001] = {0}, v1;
void dfs(int u, int prev) {
vis[u] = 1;
int v, ind, k = 1;
for (int j = 0; j < graph[u].size(); j++) {
v = graph[u][j].first;
ind = graph[u][j].second;
if (!vis[v]) ... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 400000;
const int MAXM = 200000;
const int INF = 1000000010;
const long long int MOD = 1000000007;
const long long int P = 31;
const double EPS = 1e-6;
int N;
vector<pair<int, int> > e;
bool rep[MAXN] = {false};
set<int> g[MAXN];
vector<int> gd[MAXN];
set<i... | CPP |
638_C. Road Improvement | In Berland there are n cities and n - 1 bidirectional roads. Each road connects some pair of cities, from any city you can get to any other one using only the given roads.
In each city there is exactly one repair brigade. To repair some road, you need two teams based in the cities connected by the road to work simulta... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int a[int(1e6 + 6)], i, m, ans, k, l, j, q, x, y, n, ma, mi;
set<int> s;
vector<int> v[int(1e6 + 6)];
vector<int> g[int(1e6 + 6)];
vector<int> an[int(1e6 + 6)];
void go(int x, int p, int raf) {
int beg = 0;
for (int i = 0; i < v[x].size(); i++) {
int to = v[x][i];
... | CPP |
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