Codeforces Round #713 (Div. 3) Editorial

#	User	Rating
1	tourist	4009
2	jiangly	3821
3	Benq	3736
4	Radewoosh	3631
5	jqdai0815	3620
6	orzdevinwang	3529
7	ecnerwala	3446
8	Um_nik	3396
9	ksun48	3388
10	gamegame	3386

#	User	Contrib.
1	cry	164
1	maomao90	164
3	Um_nik	163
4	atcoder_official	161
5	-is-this-fft-	158
6	awoo	157
7	adamant	156
8	TheScrasse	154
8	nor	154
10	Dominater069	153

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

void solve() {
    int n;
    cin >> n;
    vector<int> v(n);
    for (int &e : v) {
        cin >> e;
    }
    vector<int> a = v;
    sort(a.begin(), a.end());
    for (int i = 0; i < n; i++) {
        if (v[i] != a[1]) {
            cout << i + 1 << "\n";
        }
    }
}

int main() {
    int n;
    cin >> n;
    while (n--) {
        solve();
    }
    return 0;
}

1512B - Almost Rectangle

Author: MikeMirzayanov

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

#define forn(i, n) for (int i = 0; i < int(n); i++)

int main() {    
    int t;
    cin >> t;
    forn(tt, t) {
        int n;
        cin >> n;
        vector<string> f(n);
        vector<pair<int,int>> p;
        forn(i, n) {
            cin >> f[i];
            forn(j, n)
                if (f[i][j] == '*')
                    p.push_back({i, j});
        }

        p.push_back(p[0]);
        p.push_back(p[1]);
        if (p[0].first == p[1].first) {
            p[2].first = (p[2].first + 1) % n;
            p[3].first = (p[3].first + 1) % n;
        } else if (p[0].second == p[1].second) {
            p[2].second = (p[2].second + 1) % n;
            p[3].second = (p[3].second + 1) % n;
        } else
            swap(p[2].first, p[3].first);

        f[p[2].first][p[2].second] = '*';
        f[p[3].first][p[3].second] = '*';
        forn(i, n)
            cout << f[i] << endl;
    }
}

1512C - A-B Palindrome

Author: MikeMirzayanov

Tutorial

Solution

#include <bits/stdc++.h>
using namespace std;

void no() {
  cout << "-1" << endl;
}

void solve() {
  int a, b;
  cin >> a >> b;
  string s;
  cin >> s;
  for (int times = 0; times < 2; times++) {
    for (int i = 0; i < (int) s.size(); i++) {
      int j = (int) s.size() - i - 1;
      if (s[i] != '?') {
        if (s[j] == '?') {
          s[j] = s[i];
        } else if (s[i] != s[j]) {
          no();
          return;
        }
      }
    }
    reverse(s.begin(), s.end());
  }
  a -= count(s.begin(), s.end(), '0');
  b -= count(s.begin(), s.end(), '1');
  int ques = count(s.begin(), s.end(), '?');
  bool emptyMid = (s.size() % 2 == 1 && s[s.size() / 2] == '?');
  if (a < 0 || b < 0 || a + b != ques || (emptyMid && a % 2 == 0 && b % 2 == 0)) {
    no();
    return;
  }
  if (a % 2 == 1 || b % 2 == 1) {
    int i = s.size() / 2;
    if (s[i] != '?') {
      no();
      return;
    }
    s[i] = (a % 2 == 1 ? '0' : '1');
    if (a % 2 == 1) {
      a--;
    } else {
      b--;
    }
  }
  if (a % 2 == 1 || b % 2 == 1) {
    no();
    return;
  }
  for (int i = 0; i < (int) s.size(); i++) {
    if (s[i] == '?') {
      int j = s.size() - i - 1;
      if (a > 0) {
        a -= 2;
        s[i] = s[j] = '0';
      } else {
        b -= 2;
        s[i] = s[j] = '1';
      }
    }
  }
  cout << s << endl;
}

int main() {
  int tests;
  cin >> tests;
  while (tests-- > 0) {
    solve();
  }
  return 0;
}

1512D - Corrupted Array

Author: MikeMirzayanov

Tutorial

Solution

#include <bits/stdc++.h>
using namespace std;

void no() {
  cout << "-1" << endl;
}

void solve() {
  int n;
  cin >> n;
  vector<int> b(n + 2);
  for (int &x : b) {
    cin >> x;
  }

  multiset<int> have(b.begin(), b.end());
  long long sum = accumulate(b.begin(), b.end(), 0LL);
  for (int x : b) {
    have.erase(have.find(x));
    sum -= x;
    if (sum % 2 == 0 && sum <= 2'000'000'000 && have.find(sum / 2) != have.end()) {
      have.erase(have.find(sum / 2));
      for (int y : have) {
        cout << y << " ";
      }
      cout << endl;
      return;
    }
    sum += x;
    have.insert(x);
  }
  no();
}

int main() {
  int tests;
  cin >> tests;
  while (tests-- > 0) {
    solve();
  }
  return 0;
}

1512E - Permutation by Sum

Author: MikeMirzayanov

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

void solve() {
    int n, l, r, s;
    cin >> n >> l >> r >> s;
    l--; r--;
    for (int first = 1; first + (r - l) <= n; first++) {
        int sum = 0;
        for (int i = l; i <= r; i++) {
            sum += first + (i - l);
        }
        if (sum <= s && s - sum <= r - l + 1) {
            int needAdd = r - (s - sum) + 1;
            vector<int> ans(n);
            set<int> non_blocked;
            for (int i = 1; i <= n; i++) {
                non_blocked.insert(i);
            }
            for (int i = l; i <= r; i++) {
                ans[i] = first + (i - l);
                if (i >= needAdd) {
                    ans[i]++;
                }
                non_blocked.erase(ans[i]);
            }
            if (ans[r] > n) {
                continue;
            }
            non_blocked.erase(ans[r]);
            for (int i = 0; i < l; i++) {
                ans[i] = *non_blocked.begin();
                non_blocked.erase(non_blocked.begin());
            }
            for (int i = r + 1; i < n; i++) {
                ans[i] = *non_blocked.begin();
                non_blocked.erase(non_blocked.begin());
            }
            for (int i : ans) {
                cout << i << " ";
            }
            cout << "\n";
            return;
        }
    }
    cout << "-1\n";
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);
    int n;
    cin >> n;
    while (n--) {
        solve();
    }
}

1512F - Education

Author: sodafago

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

void solve() {
    int n, c;
    cin >> n >> c;
    vector<int> a(n);
    vector<int> b(n - 1);
    for (int &e : a) {
        cin >> e;
    }
    for (int &e : b) {
        cin >> e;
    }
    b.push_back(0);
    ll ans = 1e18;
    ll cur = 0;
    ll bal = 0;
    for (int i = 0; i < n; i++) {
        ans = min(ans, cur + max(0ll, c - bal + a[i] - 1) / a[i]);
        ll newDays = max(0ll, b[i] - bal + a[i] - 1) / a[i];
        cur += newDays + 1;
        bal += a[i] * newDays - b[i];
    }
    cout << ans << "\n";
}

int main() {
    int n;
    cin >> n;
    while (n--) {
        solve();
    }
    return 0;
}

1512G - Short Task

Author: MikeMirzayanov

Tutorial

Solution

#include <bits/stdc++.h>
using namespace std;

const int N = (int) 1e7 + 100;

long long s[N];
int d[N];
int ans[N];

int main() {
  fill(d, d + N, -1);
  d[1] = 1;
  for (int i = 2; i * i < N; i++) {
    if (d[i] == -1) {
      d[i] = i;
      for (int j = i * i; j < N; j += i) {
        if (d[j] == -1) {
          d[j] = i;
        }
      }
    }
  }
  s[1] = 1;
  for (int i = 2; i < N; i++) {
    if (d[i] == -1) {
      d[i] = i;
      s[i] = i + 1;
    } else {
      int j = i;
      s[i] = 1;
      while (j % d[i] == 0) {
        j /= d[i];
        s[i] = s[i] * d[i] + 1;
      }
      s[i] *= s[j];
    }
  }
  fill(ans, ans + N, -1);
  for (int i = N - 1; i > 0; i--) {
    if (s[i] < N) {
      ans[s[i]] = i;
    }
  }
  int tests;
  cin >> tests;
  while (tests-- > 0) {
    int c;
    cin >> c;
    cout << ans[c] << endl;
  }
  return 0;
}

MikeMirzayanov

I submitted G solution and got TLE in test case 1, but during the contest test, 1 was passed.

After final testing, I submitted Previous G's solution and got Accepted.

can you check it?

#include <bits/stdc++.h> using namespace std; #define lli long long int #define ff first #define ss second #define endl '\n' int main() { int t; cin>>t; while(t--) { int a,b; cin>>a>>b; string s; cin>>s; int n=s.size(); int flag=0; if((a+b)!=n) { cout<<"-1"<<endl; continue; } for(int i=0; i<(n/2); i++) { if(s[i]=='0') { if(s[n-1-i]=='0') a-=2; else if(s[i]=='1') { flag++; break; } else { s[n-1-i]='0'; a-=2; } } if(s[i]=='1') { b-=2; if(s[n-1-i]=='?') { s[n-1-i]='1'; } if(s[n-1-i]=='0') { flag++; break; } } if(s[i]=='?') { if(s[n-1-i]=='0') { s[i]='0'; a-=2; } else if(s[n-1-i]=='1') { s[i]='1'; b-=2; } } } if((n&1)==1) { if(s[n/2]=='0') a--; else if(s[n/2]=='1') b--; } if((a<0)||(b<0)||(flag!=0)) { cout<<"-1"<<endl; continue; } for(int i=0; i<n; i++) { if(s[i]=='?') { if(((n&1)==1)&&(i==n/2)) // for middle element if n is odd { if((a&1)==1) { s[i]='0'; a--; } else { s[i]='1'; b--; } } else if(a>1) { s[i]='0'; s[n-1-i]='0'; a-=2; } else { s[i]='1'; s[n-1-i]='1'; b-=2; } } } if((a<0)||(b<0)) { cout<<"-1"<<endl; continue; } cout<<s<<endl; } return 0; }

#include<bits/stdc++.h> using namespace std; long long q, j, x, n, a[10000001], f[10000001]; main(){ n=10000000; for (int i=1; i<=n; i++){ for(j=i; j<=n; j+=i)a[j]+=i; if(a[i]<=n&&!f[a[i]])f[a[i]]=i; } cin>>q; while(q--){ cin>>x; cout<<f[x]-(!f[x])<<endl; } }

Comments (109)

Show archived | Write comment?

RainFall

4 years ago, # |

← Rev. 2 →

Is there any simpler solution for B?

→ Reply

jineshparakh

4 years ago, # ^ |

For both the *'s in the matrix at least one of these 3 conditions will hold:
1. They will form a horizontal edge(both having the same row)
2. They will form a vertical edge(both having the same column)
3. They will form a diagonal

After that, a rectangle can be made using some basic casework.
You can refer to My Submission

lawarganayra

I did that , but got wrong answer, could you please help! 112540024

bhavyaganatra

I think error is in below line:

poin1 = make_pair(0,first_x);

It should be like this: poin1 = make_pair(first_x,0);

Here is the test case in which your code will fail:

1
3
...
..*
..*

starkspi

Can someone check my problem C submission and tell the test case I am failing because I tried very hard but not able to figure it out even jury is not showing the test case I failed. Kindly help me I am an absolute beginner your support means a lot to me

Coder_Arjun

Please post your code so that we can debug it...

aditya_01

you can do stress testing between your code and solution code

stefdasca

+26

Videos out for E, F, G

spookywooky

+21

We can do D in O(n) by observing two cases:

One of the biggest two numbers was choosen as x, and the other one is the sum of all smaller numbers.
Any smaller number was choosen for x and the biggest number is the sum of all other numbers.

Adi_Max

i tried that but some test cases r failed ,please let me know my mistakes ,my solution 112546189

lazy_and_slow

you took everything long long except the sum variable lol 112575126

wysoserious

Hi, I did the same, however got wrong answer. I apologize if it's too much to ask, but can you please review my code? I compared your code with mine, and it appears I did all the necessary things and yet code fails on the second test case. My submission — 112881623

It is testcase 1 of second testcase, you answer -1 where the answer is 1:

1
1
1 1 1

Letscode_sundar

3 years ago, # ^ |

can you please find the mistake in this 116030741

i am following the same method that you mentioned.

parthsharma6321

Can someone point out the mistake in this? tried so many times but cant find it. the case is also not visible to debug. https://codeforces.me/contest/1512/submission/112547774 Verdict: wrong answer jury found answer, but participant did not (test case 88)

Newton

1
8 2
?????0000?
correct ans is 1000000001

Geothermal

← Rev. 4 →

+67

Is something wrong with the input validator for A? Perhaps I'm being dumb, but I can't find the error in my hacked solution, and it seems strange to me that the top twenty or so participants from the original ranklist have all been hacked. Moreover, a friend of mine reported that the test case 1 3 1 1 1 gives an "unsuccessful hacking attempt" verdict, when it should give invalid input, leading me to suspect that invalid test cases are being used to hack solutions. (UPD: I'm also hearing that the case 1 3 1 2 100 is successfully hacking solutions, which definitely shouldn't be happening, as there are more than two distinct elements in the array.)

+41

After being hacked and dropping to 344th place, it's nice to be back on top B)

code_warrior

How did so many good coders like you got hacked on A?

See my post above--the validator is incorrect, and the inputs used for hacking do not satisfy the constraints of the problem. I'm assuming a fix will be issued sometime later today and everyone's solutions will be restored.

Mong_Mang

Can you please tell all the extensions name on the last 3 rows? How can I get those extension. Please reply.

kaiyu

+11

Here is the link to the extension : https://chrome.google.com/webstore/detail/carrot/gakohpplicjdhhfllilcjpfildodfnnn?hl=en

swapnil159

Can you tell us which extension are you using to see the rating change on right?? UPD: Sorry for asking again, just saw the comment above.

chilly_willy

+39

just git gud lmao

+13

Yep, sorry for it. It is fixed now: https://codeforces.me/blog/entry/89476#comment-778860

btme

← Rev. 3 →

CODE

ll c; cin>>c;

//for(int i=0;i<=c;i++) {cout<<seive[i]<<" ";}cout<<endl;
if(c==1) ret(1);

//assert(c<
for(int i=2;i<=c;i++){
    map<ll,ll> mp;
    ll n=i;
    while(n!=1){
       if(seive[n]==1) {mp[n]++;break;}
       mp[seive[n]]++;
       n/=seive[n];
    }
    ll sum=1;
    for(auto x:mp){
       sum*=(((ll)pow(x.first,x.second+1)-1)/(x.first-1));
    }
    //deb(i,sum);
    if(sum==c) ret(i);
}

ret(-1);

why its giving TLE for problem G?

burak

Do not use pow, I do not know your intended solution but pow can result in both precision error and linear calculation (which makes your whole algorithm $$$O(n\;maxB)$$$ if you iterate through $$$n$$$ values and use pow to calculate some power $$$a^b$$$ in $$$O(b)$$$ time instead of $$$O(log\;b)$$$ time with binary exponentiation.)

AB.devil

Well, it should TLE. For every query, you are looping from 2 to c to find appropriate n.
Instead, you can for all values of c store the minimum n and then answer queries in constant time.

Darshan_Bhangale

Use Binary Exponentiation to calculate the value of a raised to b. pow function has a time complexity of O(n). while Binary Exponentiation has a time complexity of O(log(n)).

fortune10

hold on can u declare an array of size 1e7 ?? wouldn't that give run time error

You can because it is roughly ~40MB (in case of ints) and in CF the memory limit for most of the problems is 256MB.

vineet16

Time limits were too tight for problem G. Java and Python users are bound to get TLE, quite unfair.

AdityaG

time limits were too tight for G bcz bruteforce with slight optimization gives AC in c++ . submission link

__algebra__

Problem G with bigger constraint INVDIV

dhritishman22

Div 3 contests were great when Vohuh was a problem setter.

rohitssingh17

Problem C : https://ideone.com/dZeurz can someone please tell me a testcase where my solution fails i am not able to figure it out.

thank you for helping ..

try to seperate if(s[i]='?'&&s[n-i-1]='?') case from rest of the other case and do in separate loop

Zealotrahl

Can anybody provide knapsack dp solution for problem E;

jpratik15

It will give TLE , I guess

amar277

Here is my submission Take a look at this https://codeforces.me/contest/1512/submission/112546632

dedsec_29

Can you explain your dp state, what does dp[i][j] denote?

i means no. of elements which add upto j i.e. sum

ChemicalWarfare

Hi! Thanks for the editorial! Can anyone here check my submission for problem C and tell me in which test cases, my program failed? I am trying to find a corner case for so long.

If you don't have time to read my code, could you at least give me some counterexamples so I can test my code on them?

Thank you!

Here's my code: https://ideone.com/TZb6bR

faraz16iqbal

Solution for C failing. Can someone help?

YareYare

Try this test

Test

1
9 6
??????0??0????1

Answer isn't -1

JaxYoung

Why should we check whether the sum is below 2e9?

// the solution code
if (sum % 2 == 0 && sum <= 2'000'000'000 && have.find(sum / 2) != have.end()) {

I mean that b[i] has a constraint 1 <= b[i] <= 1e9. So if sum > 2e9 then sum / 2 > 1e9, as b[i] <= 1e9, we can't find sum / 2 in b[i]. Thus sum <= 2e9 should be redundant. But I got a wrong answer without it, and passed for adding it.

Appreciate for anyone who could explain it.

My WA submission: 112590439 My AC submission: 112590535

Continued_Story

same question

If there exists an array a, the sum of the elements must be in array b.
Since the maximum sum can be 1e9 in that case(b_i <= 1e9), so the sum of the elements of a + this sum (again which appears in b) can be at most 2e9.

Thank you for your reply, but I still can't understand.

I see that. But as I said before, if sum > 2e9, have.find(sum / 2) != have.end() should be false. That's to say the if statement result remains the same without evaluating sum <= 2e9.

I am so confused that the only possible exception I would say is b[i] has element larger than 1e9.

The main problem was you were using a multiset of integers.
I changed it to long long and it worked.
AC submission https://codeforces.me/contest/1512/submission/112652296
I guess have multiset when given arbitrarily large values does not point to end().

soutrikdas

Can Someone help me with B , 112546789 I checked the test cases, and its still hard to decipher, perhaps someone has faced the same problem as me ?

In that block one variable name has a typo:

Spoiler

	else
		{
			x3=x1;
			y3=y2;
 
			x4=x2;
			y3=y1;
		}

Thank you very much for going through my code, I tried with the corrected code and it got accepted.

arpitP

 if (sum % 2 == 0 && sum <= 2'000'000'000 && have.find(sum / 2) != have.end())

Can someone please tell me what is the meaning of this condition?

Candidate_noob

If the sum is even and less than equal to 2e9 && it is present in the set named "have".

digishek

Can someone explain how binary search solution of F works ?

Noctambulant

Can anyone help me by sharing the code of O(n) complexity sieve of problem G!!

codinginveins

https://cp-algorithms.com/algebra/prime-sieve-linear.html

18ucs134

112599399 This is my submission of problem F. Please someone tell me whats wrong with this..

nitrousoxide

The convention might by platform-dependent, but sometimes in C++ (-1)/2 = 0. As a result, if at a point you have already made enough money to buy the computer/take the course for the next level, you won't need to spend an extra day to make the money.

Thank you very much.. I was initially adding ai for each day at least once. But as you said if already we have enough money we need not spend a day.

mushfiq13

Contest time code: https://codeforces.me/contest/1512/submission/112549540

After final testing: https://codeforces.me/contest/1512/submission/112601755

mnishhh

Same with me. In final testing my solution for G gave tle on test 1 which was accepted at the time of contest. Please help:

They tested again, Accepted :)

dv.jakhar

In G's code can someone explain this line plz:

s[i] = s[i] * d[i] + 1;

Like how are we calculating when two same prime divisors comes (for eg. 4=2*2) and why/how it works?

1000101

I got it. they are at first dividing the number by some prime which divides it. d[x] for the number x. (Remember we got the prime using the sieve.)

Now, Notice that for $$${prime}^{r}$$$ we will have the divisor function equals

$$${prime}^{r} + {prime}^{r-1} + {prime}^{r-2}+ ..... + 1.$$$

If you think about it for some time then you will realize that that's what the expression $$$s[i]=s[i]*d[i]+1$$$ when ran in a loop till it is divisible with "j" is doing.

the remaining part "j" (which gives us s[j]) has gcd equals one with this prime exponent so we can using the above expresssion simply multiply it.

and get $$$s[i]=s[i]*s[j].$$$

Where does this comes from. Is there some there theorem? Seems like some property like that of phi function but I can not figure out why.

Use the multiplicativity of the function d(n):

d(a⋅b)=d(a)⋅d(b) if gcd(a,b)=1.

Let me explain this...

suppose some number has prime divisors a, b, c and all are prime so using above formula we can write

d(a.b.c)=d(a).d(b).d(c)

Now d(a)=a+1, d(b)=b+1, d(c)=c+1

so d(a.b.c)=(a+1)(b+1)(c+1)

If we expand the R.H.S we get,

d(a.b.c) = a.b.c+a.b+a.c+b.c+a+b+c+1

and that is what d(a.b.c) would be if you calculate, so both are equal.

Let's take 30 as an example.

=> 30 = 2*3*5

so using above formula d(30)=2*3*5+2*3+2*5+3*5+2+3+5+1

=> d(30)=72 & also d(30)=(2+1)*(3+1)*(5+1)=72.

Ask me if you have any doubt.

PhantomAssassin

because if a decomposition of A of this form is p1 ^ a1 * p2 ^ a2 ... and a decomposition of B of this form is q1 ^ b1 * q2 ^ b2 ... then the sum of divisors of A equals to: (1 + p1 + p1 ^ 2 + .... p1 ^ a1) * (1 + p2 + p2 ^ 2 + .... p2 ^ a2) ... -> because no matter how we take numbers from each the parentheses are divisors of A and they are all present.

Next-> d (a) = (1 + p1 + p1 ^ 2 + .... p1 ^ a1) * (1 + p2 + p2 ^ 2 + .... p2 ^ a2), d (b) = (1 + q1 + q1 ^ 2 + .... q1 ^ b1) * (1 + q2 + q2 ^ 2 + .... q2 ^ b2) i.e. gcd (a, b) = 1 then no p and q coincide. hence d (a * b) = d (a) * d (b)

9Curd1Cream

nice bro, got the intuition now.

Zahid_Hasan_Sahin

If java coder use Arrays.sort() then We got TLE in problem D :) , Please solve this Supermagzzz

112527550

dhruv7888

I don't know much java , but as much as I have heard Arrays.sort() is implemented using quick Sort and is hence O(N^2)

IshtiaQ99

[Problem G] Same Code getting TLE and AC verdict. TLE Code — link AC Code — link Is it a joke?!

chirag_h

Can someone help me find the loophole in this approach for Problem C?

It keeps failing Test 2

112630518

Try this

john9925

how to calculate d(a*b) when a%b==0 and b is a prime by using linear sieve of Eratosthenes? Thanks.

Problem G

Maxis

Any proof for problem F ? why is it enough to calculate the number of days we should spend on a certain job a_i to get >= c and repeat this process taking the minimum for all jobs taking into consideration how many days will we stay in all the previous jobs ?

Consider a pair of days today and tomorrow.

If we got enough money to to an exam today it is allways benefical to do it today, never tomorrow.

This means by induction that for a given number of exams, there is no better solution than doing the exams as early as possible.

sharingan_coder

Hello, i am very new to codeforces(this was my first contest)

My code for problem C works perfectly well on my compiler, running it with valgrind gives no memory errors, but the diagnostics fail multiple test cases(which ran successfully on my compiler). I don't understand this!!

Can someone please help me?

This is my code:

Thank you

mstallion

Test your code on this testcase.

3 1

1??0

2 1

???

Hope it helps

itti_da

In problem E, can someone help me understand this:

 ans = min(ans, cur + max(0ll, c — bal + a[i] — 1) / a[i]);
        ll newDays = max(0ll, b[i] — bal + a[i] — 1) / a[i];
        cur += newDays + 1;
        bal += a[i] * newDays — b[i];

I got the idea of what this is trying to do but i can't understand how exactly the mathematical operations are working.

Drstrange_

In problem G since O(NlogN) is sufficient can someone please tell why can't we do something of type

int fact[1000003];

for(int i=1;i<N;i++)
    for(int j=i;j<N;j+=i)
    fact[j]+=i;

and store the first occurences of every value and rest will be -1.

Agrawal31

This approach will give TLE, due to the second loop. Since starting j from i*i in the sieve of eranthoses prunes a lot of iterations.

punk_tech

It seems like solution of D from editorial doesn't fit in time.

UPD Oh, I'm wrong

in problem G the multiplicativity of the function d(n) : d(a⋅b)=d(a)⋅d(b) if gcd(a,b)=1. is their any proof for it?

OleschY

← Rev. 9 →

Write $$$n=p_1^{a_1} \cdot p_2^{a_2} \cdot ... \cdot p_m^{a_m}$$$ with $$$p_i$$$ primes. Then $$$d(n)=(1+p_1^1+...+p_1^{a_1})\cdot(1+p_2^1+...+p_2^{a_2})\cdot...\cdot(1+p_n^1+...+p_n^{a_n})$$$ (This was a wrong formula first, thanks to put_peace for correcting it!)

Let {$$$q_i$$$} and {$$$r_i$$$} be sets of primes with empty intersection. Let $$$Q=q_1^{b_1} \cdot q_2^{b_2} \cdot ... \cdot q_m^{b_m}$$$ and $$$R=r_1^{c_1} \cdot r_2^{c_2} \cdot ... \cdot r_k^{c_k}$$$. Then $$$gcd(Q, R)=1$$$ and $$$d(Q \cdot R) = d(Q) \cdot d(R)$$$ using the formula for $$$d$$$.

This isnt a formal proof, but it outlines the idea for a proof.

See also https://en.wikipedia.org/wiki/Euler%27s_totient_function the totient function is similar to $$$d$$$ here.

put_peace

$$$d(n)=(a1+1)⋅(a2+1)⋅...⋅(am+1)$$$, I think this is not correct.

Oh, you are totally right. What I wrote was the amount of divisors not the sum of divisors!

Your comment below is right. If I replace my error with $$$d(n)=(1+p_1^1+...+p_1^{a_1})\cdot(1+p_2^1+...+p_2^{a_2})\cdot...\cdot(1+p_n^1+...+p_n^{a_n})$$$ then it's correct. Thanks for the heads up!

Just to see how it is correct, $$$d(p^{a}).d(q^{b}).d(r^{c}) = (1 +p + p^{2} + ... + p^{a}).(1 +q + q^{2} + ... + q^{b}).(1 + r + r^{2} + ... + r^{c})$$$
if you expand the RHS, you can verify that it will give $$$d(p^{a}q^{b}r^{c})$$$

yes got it,
we can even think of it like
d(p^a) as sigma(p^ai) where 0<=ai<=a
similarly, d(p^a.q^b.r^c) = sigma(p^ai.q^bi.r^ci) for all 0<=ai<=a 0<=bi<=b 0<=ci<=c
its like all combination of powers till a,b,c which can be written as
d(p^a.q^b.r^c) = (1+p+p^2+...+p^a).(1+q+q^2+...+q^b).(1+r+r^2+...+r^c)
here the rhs its same as choosing power of p and power of q and power of r (i.e same as getting all possible combinations for d(p^a.q^b.r^c)) same as d(p^a).d(q^b).d(r^c)

waltz47

Why is E not doable with knapsack? https://codeforces.me/contest/1512/submission/113093207 Passed sample cases but WA on TC2

shivansh1102

Can anyone please tell me why I am getting Wrong answer on test 2(test case 392) for problem 1512C (A-B Pallindrome). I tried lot but unable to find a case where it fails.

Here is my code :

DeepBisman

Can someone help and tell me why my solution is failing at test case 13 for prob D

https://codeforces.me/contest/1512/submission/115064448

kumasento

G is such a great question!

Calanosay

why such a brute force solution could solve problem G? 1e6*1e6=1e12,isn't it?

oh..just forget what i said ,hahaha

aim100

3 years ago, # |

Can someone please explain me the solution for E (editorial one), I have been trying for a long time but unable to understand how this condition came :

(s — sum <= r — l + 1).

nikhilnagrale2

my solution is fairly straightforward and understandable. you can check out here

Thanks, It helped.

Hidder

null

gnudgnaoh

Could someone explain for me the logic behind this part in the editorial code for problem G, please?

that part

      int j = i;
      s[i] = 1;
      while (j % d[i] == 0) {
        j /= d[i];
        s[i] = s[i] * d[i] + 1;
      }
      s[i] *= s[j];

I knew it was related to the multiplicativity of d(n) but I couldn't fully understand.

And this is

the whole code

Thanks in advance!

MartinRistov

Hello I am new to programming. Just finished "1512A — Spy Detected!". And now I read the tutorial. I am wondering is my way more efficient than the tutorial and code showed?

My idea was to assume the first element of array is spy. Then compare first element until you hit different element. In that moment we have two options first element is spy or the one we just discovered. if (a[0]!=a[n-1] && a[0]!=[n-2]) is simple way to find out who is spy.

I would like a feedback on this idea, is this good approach or not? **I wrote my code in C

tyagi_abhishek

if( arr[0] !=arr[1]) 0 or 1 is different element , if(arr[0]==arr[2]) answer is 1 else 0 . if( arr[0]==arr[1]) element !=arr[0] is the answer !

mostafaabdelaal_03

22 months ago, # |

← Rev. 6 →

I solved E using simpler approach in O(N)

if we have $$$s$$$ and $$$r-l+1=k$$$, so $$$\frac{(k)(k+1)}{2}$$$must be greater or equal to$$$s$$$

First we will subtract the k Values from s (we will Know later Why) $$$m=s-\frac{(k)(k+1)}{2}$$$ and if we try to distribute m oranges in k boxes, then every box will have m/k orange(floor division), and remainder will be r oranges, r<k

k Boxes: $$$b_1, b_2, b_3, b_4, ....b_{k-1}, b_k$$$

k Boxes: $$$\frac{m}{k},\frac{m}{k}, \frac{m}{k}, \frac{m}{k}, ....\frac{m}{k}, \frac{m}{k}$$$

so we will distribute remainder and each box will take 1 orange until we have 0 remainder(I know that there is some boxes that we won't take orange).we will distribute remainder from right to left

after we distribute remainder we will distribute k values that we subtracted in the first: 1, 2, 3, .....k;

example will clarify:

suppose that we have $$$r-l+1=k=4, s=13, m=13-\frac{(4)(4+1)}{2}=3$$$

we will make each box have $$$\frac{m}{k}=\frac{3}{4}=0$$$ and we have $$$r=3$$$

k Boxes: $$$b_1, b_2, b_3, b_4$$$

distribute quotiont: $$$0,\ 0,\ 0,\ 0$$$

distribute remainders: $$$0,\ 1,\ 1,\ 1$$$

distribute K values that we subtract in the first : $$$1,\ 2,\ 3,\ 4$$$

the summation of three rows above: $$$0+0+1,\ 0+1+2,\ 0+1+3,\ 0+1+4$$$

the summation of three rows above: $$$1,\ 3,\ 4,\ 5$$$ $$$= 13$$$

mysubmission

redraven

4 months ago, # |

In the solution for problem D, if (sum % 2 == 0 && sum <= 2'000'000'000 && have.find(sum / 2) != have.end()) , why we check if sum less than 2x1E9?

Supermagzzz's blog

MikeMirzayanov