EXPVALUE - Editorial


Contest: Division 1
Contest: Division 2
Contest: Division 3
Contest: Division 4

Author: satyam_343
Testers: iceknight1093, yash_daga
Editorialist: iceknight1093




Linearity of expectation


Given N and p, consider the following N-step process:

  • Initialize your score S to 0.
  • In the i-th step, with probability 50\%, add p^{1-i} to S

Let S_i denote your score after i steps.
Find \mathrm{E}[S_i^2] for each 1 \leq i \leq N.


Let’s first solve this for a fixed k, i.e, compute \mathrm{E}[S_k^2].

We have the set \{1, p^{-1}, p^{-2}, \ldots, p^{1-k}\}. We choose a random subset of this set (each with probability 2^{-k} of being chosen), and we want to find the expected value of the square of the sum of this subset.

Suppose the subset we choose is \{x_1, x_2, \ldots, x_m\}.
The square of its sum is (x_1 + x_2 + \ldots + x_m)^2, which when expanded is

\sum_{i=1}^m x_i^2 + \sum_{i=1}^m \sum_{j=i+1}^m 2x_ix_j

Let’s call the first summation X and the second one Y.
That is, X = \sum_{i=1}^m x_i^2 and Y = \sum_{i=1}^m \sum_{j=i+1}^m 2x_ix_j.

By linearity of expectation, \mathrm{E}[S_k^2] = \mathrm{E}[X] + \mathrm{E}[Y], so let’s compute each one individually.

Computing E(X)

\mathrm{E}[X] is simply the expected sum of squares of the subset.

Once again, let’s use linearity of expectation: each element of the set has a 50\% chance to be in our subset, so the expected value of the sum of squares is

0.5 \cdot \left(1^2 + \left(p^{-1}\right)^2 + \left(p^{-2}\right)^2 + \ldots + \left(p^{1-k}\right)^2\right)

Notice that this is the sum of k terms of a geometric series, with first term 1 and common ratio p^{-2}.

This sum can thus be computed in \mathcal{O}(\log{MOD}) using the sum of GP formula.

Alternately, you can just maintain the sum of squares as you iterate k, each time increasing it by p^{2-2k}.

Computing E(Y)

For the third this this problem, let’s use linearity of expectation.

For each 0 \leq i\lt j \lt k, the product 2p^{-i}p^{-j} has a 25\% chance of occurring in our score (we need to pick both p^{-i} and p^{-j}, 50\% chance of each).

So, the required expected value is

0.25 \sum_{i=0}^{k-1} \sum_{j=i+1}^{k-1} 2p^{-i-j}

For a fixed i, notice that the inner summation is once again a geometric progression with first term p^{-2i-1} and common ratio p^{-1}, so it can be computed in \mathcal{O}(\log{MOD}) .

This already gives us a way to compute \mathrm{E}[Y] in \mathcal{O}(k\log{MOD}); however, doing this separately for each k is too slow.

One way to speed it up is to flip the order of summations, and look at \displaystyle \sum_{j=0}^{k-1} \sum_{i=0}^{j-1} 2p^{-i-j} instead.

Once again, for a fixed j the inner summation is a GP and can be computed quickly.
However, the nice thing about looking at it this way is that when moving from k to k+1, almost all the computations are the same!
In fact, we only need to add in the summation for one more value of j.

This allows us to move from k to k+1 in \mathcal{O}(\log{MOD}) time, since we compute only one sum of GP at each step; hence making the overall complexity \mathcal{O}(N\log {MOD}).

Even simpler, you can notice that this sum of products increases by p^{1-k} times the sum of all the powers before, so maintaining the sum so far allows this to be updated in \mathcal{O}(1) without dealing with geometric progressions.

There are also other ways to compute this part quickly: for example, the closed-form formula

\frac{ (p^{k-1} - 1) (p^{k}-1)p^{1 - 2(k-1)}}{(p+1)(p-1)^2}

Either way, \mathrm{E}[X] and \mathrm{E}[Y] can be computed in \mathcal{O}(\log{MOD}) for a fixed k, so do it separately for each k, add them up, and we’re done.


\mathcal{O}(N\log {MOD}) per testcase.


Setter's code (C++)
//#pragma GCC target ("avx2")
#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")

#include <bits/stdc++.h>   
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;   
using namespace std;  
#define ll long long
const ll INF_MUL=1e13; 
const ll INF_ADD=1e18; 
#define pb push_back                 
#define mp make_pair          
#define nline "\n"                           
#define f first                                          
#define s second                                             
#define pll pair<ll,ll> 
#define all(x) x.begin(),x.end()     
#define vl vector<ll>             
#define vvl vector<vector<ll>>    
#define vvvl vector<vector<vector<ll>>>          
#ifndef ONLINE_JUDGE    
#define debug(x) cerr<<#x<<" "; _print(x); cerr<<nline;
#define debug(x);    
void _print(ll x){cerr<<x;}   
void _print(char x){cerr<<x;}     
void _print(string x){cerr<<x;}    
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());   
template<class T,class V> void _print(pair<T,V> p) {cerr<<"{"; _print(p.first);cerr<<","; _print(p.second);cerr<<"}";}
template<class T>void _print(vector<T> v) {cerr<<" [ "; for (T i:v){_print(i);cerr<<" ";}cerr<<"]";}
template<class T>void _print(set<T> v) {cerr<<" [ "; for (T i:v){_print(i); cerr<<" ";}cerr<<"]";}
template<class T>void _print(multiset<T> v) {cerr<< " [ "; for (T i:v){_print(i);cerr<<" ";}cerr<<"]";}
template<class T,class V>void _print(map<T, V> v) {cerr<<" [ "; for(auto i:v) {_print(i);cerr<<" ";} cerr<<"]";} 
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
typedef tree<ll, null_type, less_equal<ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_multiset;
typedef tree<pair<ll,ll>, null_type, less<pair<ll,ll>>, rb_tree_tag, tree_order_statistics_node_update> ordered_pset;
const ll MOD=998244353;           
const ll MAX=200200;
ll binpow(ll a,ll b,ll MOD){
    ll ans=1;
    return ans;  
ll inverse(ll a,ll MOD){
    return binpow(a,MOD-2,MOD);  
ll getv(ll p,ll n){
    ll num=binpow(p,n,MOD)-1+MOD;
    ll den=inverse(p+MOD-1,MOD);
    return num; 
void solve(){
    ll n,p; cin>>n>>p;  
    for(ll i=1;i<=n;i++){
        ll ans=(getv(p,i)*getv(p,i)+getv(p*p,i))%MOD;
        cout<<ans<<" \n"[i==n]; 
int main()                                                                                                         
    #ifndef ONLINE_JUDGE                          
    freopen("input.txt", "r", stdin);                                                    
    freopen("output.txt", "w", stdout);      
    freopen("error.txt", "w", stderr);                          
    ll test_cases=1;                
Tester's code (C++)
#include <bits/stdc++.h>                   
#define int long long     
using namespace std;
const int mod=998244353;

int power(int a, int b, int p) {
    return 0;
    int res=1;
    while(b>0) {
    return res;

int32_t main() {
    int n, p;
    assert(n>=1 && n<=300000 && p>=2 && p<=100000);
    int exp_sq=0, exp_sum=0, exp_prod=0, inv2=power(2, mod-2, mod);
    for(int i=0;i<n;i++) {
        int cur=power(power(p, i, mod), mod-2, mod);
        exp_sq=(exp_sq + (cur*cur))%mod;
        exp_prod=(exp_prod + (cur*exp_sum))%mod;
        exp_sum=(exp_sum + cur)%mod;
        int ans=((exp_sq + exp_prod)*inv2)%mod;
        cout<<ans<<" ";
Editorialist's code (Python)
def pw(a, n, mod):
	n %= mod-1
	r = 1
	while n > 0:
		if n%2 == 1: r = (r * a)%mod
		a = (a * a)%mod
		n //= 2
	return r

mod = 998244353
def gp_sum(a, r, n):
	ret = (pw(r, n, mod) - 1) * pw(r-1, mod-2, mod)
	return (a * ret) % mod
def what(p, n):
	ret = pw(p, 1 - 2*n, mod) * (pw(p, n, mod) - 1) * (pw(p, n+1, mod) - 1)
	ret %= mod
	ret *= pw((p - 1)**2, mod-2, mod)
	ret *= pw(p+1, mod-2, mod)
	return ret%mod;

n, p = map(int, input().split())
half = (mod+1)//2
r = pow(p*p, mod-2, mod)
for i in range(1, n+1):
	x = gp_sum(1, r, i) * half
	y = what(p, i-1) * half
	print((x + y)%mod, end = ' ')
1 Like

What’s the purpose of editorial when cheaters can solve before majority of actual solvers. they are about 5 times more than actual solvers. They are fearless because they know they will get away from this.


They’ll not hear you can’t do anything in this case.

1 Like

What happened? People are cheating in these competitions!!!

To learn how to solve something you were unable to solve in-contest (or perhaps learn an alternate solution/better implementation, if you did solve it).

I understand being frustrated with cheaters but they have nothing to do with the existence of the editorial (and other people cheating isn’t preventing you from learning, at the very least).

1 Like

I am having trouble understanding the question. Please help
Lets take P=2 as taken in first testcase
After 2 moves we have scores

  • 0 with probability 1/4

  • 1/2 with probability 1/4

  • 1 with probability 1/4

  • 3/2 with probability 1/4

This gives expected score of 1/8 + 1/4 + 3/8 = 5/8

But how can we F(2) in this case? In question its given 7/8

You need to find the expected value of the squares, so it will be (1/16)+(1/4)+(9/16) = 7/8

1 Like

Thanks for clearing, I misunderstood the question :sweat_smile:

I understand you too, but there are thousands of unsolved problems. Why we give contest then?

By mistake the video editorial is of problem ZEROSTRING and not of this question. Please change it.

I don’t know, why do you participate in contests? That’s a question only you can answer for yourself.

I do it as a hobby because I find it fun — what about you?

Who cares? Maybe cheaters doesn’t matter to you. But I went from 100ish rank to about 400 and lost points. (A Disclaimer before contests might help - You should treat it as fun, don’t taking ratings seriously).
If I invest 3 long hours, the minimum I would want is a approach to curtail cheaters( I have never seen my ranks go up after plagiarism check unlike codeforces, so All your claims doesn’t even matter).

You asked me a question, so I assume you did.
At any rate it seems you don’t share my view on participating in contests for fun, that’s ok. Not my place to tell you how seriously to take them.

There’s a plagiarism thread whose first post mentions the number of people penalized after each contest. At the moment it goes back to Starters 74.
I don’t know the exact procedure or how effective it is but claiming that nothing is being done is decidedly false.

CF recomputes ratings a few days after each contest, the plag thread here says they do it every month.

It doesn’t look like you participated in yesterday’s contest but you’re talking about your rank in it, so I’ll assume you used an alt.
Either way, make note of your rank currently, then come back sometime next month and see if it’s changed. (Alts are against the code of conduct btw, for whatever that page is worth)

This is the last message about cheating I’m making on this page, a problem’s editorial page isn’t the place for it.

1 Like

Ya, now target the alts, maybe alts are not your biggest problem here( you got 70% of people cheating). Anyways that shows the attitude you guys have towards cheating, justifying in a way.

Do whatever you want to do, one thing that is certain is I’ll never join any contests here anymore. #PRO_CHEATERS.

1 Like

yes you are right

1 Like

I am having a really hard time understanding this editorial. Why there is a 0.25% chance of getting 2p^-i p ^-j ? Can anyone please explain?

I am not able to understand why for calculating Y we have used mod-2

In the given “EQUATION” what was variable ‘m’ ?

Think about when you get a 2p^{-i}p^{-j} term. It only happens when both p^{-i} and p^{-j} are chosen, right?
The chance of choosing p^{-i} is 1/2. The chance of choosing p^{-j} is 1/2.
These choices are completely independent, so the chance of choosing both of them is 1/2 \times 1/2 = 1/4, which is 25\%.

The fact that you can do this separately for each (i, j) pair (where i \neq j) is a consequence of linearity of expectation, which is used liberally in this problem.

Also, notice that I’m making a distinction between different ways to write the same power: in our case, 2p^{-2-3} and 2p^{-1-4} are treated separately, and each have a 25\% chance of appearing, even though they’re technically both 2p^{-5}.
It might help your understanding if you took a concrete example. Take the set \{1, p^{-1}, p^{-2}, p^{-3}\}. Write out all its subsets, see what happens when you square their sums.
For each 0 \leq i \lt j \leq 3, see what the probability of getting 2p^{-i-j} is.

If you’re unfamiliar with linearity of expectation in general, maybe this blog will help.

Taking something to the power mod-2 means you’re dividing by it in modulo. I recommend reading about modular multiplicative inverses.

Oh that should’ve been k, not m. I’ve fixed the typo, thanks for pointing out.

1 Like

I got your idea @iceknight1093, thanks. But I think I should try easy EV problems first in order to better understand the notion of EV and linearity of EV in general.