AtCoder Beginner Contest 162 Announcement

5 лет назад, # ^ |

+58

Think in reverse. Try to answer question to how many sequences we will get if have gcd each from 1 to k.

So, for:

gcd = 1. -> We have N places to fill and we can use any number from 1 to k at all places(N places to fill). Number of ways = k^N.

gcd = 2 -> we will have k/2 numbers as multiples of 2. so we have k/2 options at each place. Number of ways = (k/2)^N.

And, similiarly for all gcd till k.

But wait there is overcounting happening here. Consider while calculating gcd = 2, on choosing randomly we can even get gcd = 4, 6, 8 and so on.

So what we can do is something like this, for removing overcounting,

gcd = 1 -> gcd1 — gcd2 — gcd3 — gcd4 ... — gcdk.
gcd = 2 -> gcd2 — gcd4 — gcd6 — gcd8 ... — gcdk.
gcd = 3 -> gcd3 — gcd6 — gcd9 — gcd12 ... — gcdk, and so on till k — 1 and,
gcd = k -> gcdk. (No overcounting here).

So we can just iterate from backwards to make our counting easier. My submission

I hope the explanation helped you. Ignore my english.

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one_last_chance

5 лет назад, # ^ |

Nicely explained buddy.

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Saurabh66

5 лет назад, # ^ |

I understood the solution till choosing the arrays on the basis of multiples of numbers from 1 to k, and removing the overcounting by subtracting the extra arrays counted for a particular number whose multiples are greater than the number itself but less than or equal to k which are are overcounted.

But on finding the final answer why did you multiply the count of arrays for each GCD from 1 to k with the number itself?

$$$ans+=a[i]*i$$$

where

$$$(1\le i \le k)$$$

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5 лет назад, # ^ |

Total Number of arrays with gcd = k be dp[k]. So it's contribution to answer will dp[k] * k(which is what we are asked to calculate in question)

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Saurabh66

5 лет назад, # ^ |

We had to find the total sum, that's why multiplied the count of array with its GCD to get its contribution. Missed the sum part, that's why asked such a question, my bad. By the way the above explanation of the solution to the problem is great. Did you solve the F problem?

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5 лет назад, # ^ |

Yeah glad it helped you to understand. No, I wasn't able to solve F.

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towhid1zaman

5 лет назад, # ^ |

thanks for your explanation ritik_patel05

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sh_maestro

5 лет назад, # ^ |

Nice explanation, thanks! Why do we go backwards with the calculation?

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Gats

5 лет назад, # ^ |

Because smaller numbers need to minus larger numbers, so we must first ensure the correctness of larger numbers.

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retired_

5 лет назад, # ^ |

Nice explanation!

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hardy_9795

5 лет назад, # ^ |

can you please explain how the gcd=2 terms will be k/2

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tribute_to_Ukraine_2022

5 лет назад, # ^ |

← Rev. 3 →

+15

For each $$$i$$$ there are $$$\left\lfloor\frac{k}{i}\right\rfloor^n$$$ sequences with all elements divisible by $$$i$$$. Now for any function $$$f$$$: $$$\sum\limits_{i=1}^k \left\lfloor\frac{k}{i}\right\rfloor^n f(i)$$$ is equal to sum over all arrays of $$$\sum\limits_{i | \gcd(A_1, A_2, \ldots A_N)} f(i)$$$. Now we just need to pick $$$f$$$ such that $$$\sum\limits_{i | k} f(i) = k$$$. We might know, that Euler totient function has such property, or we can fill array $$$f$$$ with intended value of this sum (in this case $$$f(i) = i$$$) and run the following code:


for (int i = 1; i <= nax; ++i)
    for (int j = i * 2; j <= nax; j += i)
        f[j] -= f[i];

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5 лет назад, # ^ |

+10

Isn't $$$O(N^2)$$$ too slow? Do you want to increase $$$j$$$ by $$$i$$$ each time?

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tribute_to_Ukraine_2022

5 лет назад, # ^ |

Ooooops. It should iterate over all multiples of $$$i$$$ and take $$$O(n \log n)$$$ time

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5 лет назад, # ^ |

+13

I understood what you meant, just making it clear for readers. Best of luck for div1 round!

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5 лет назад, # ^ |

← Rev. 3 →

can we solve E like this.

for each i we want to find how many subsequences have gcd exactly i. So every value will be multiple of i (1*i,2*i,3*i..etc)

let x=k/i

for any such subsequence there should atleast one value which is equal to i otherwise if everyvalue is >=2*i then gcd will be >i.

so count of such subsequences = total — none of those values is equal to i

so count=x**n-(x-1)**n

will this work i'm getting wrong answer for sample testcases , am i overcounting something?

upd: Found my mistake

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hotshot

5 лет назад, # ^ |

what was the mistake?

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towhid1zaman

5 лет назад, # ^ |

Thank you, sir. Now I got it

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yuppy1234

5 лет назад, # ^ |

Can you please explain your solution through Euler totient function a little more I didn't understand it.

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5 лет назад, # |

I was trying bruteforce of F in O(N*N). But it gave me the wrong answer.

Can anyone tell why this solution is giving wrong answer. Link

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akulenok

5 лет назад, # ^ |

Wont N^2 give TL?

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QOEINDIEDIS

5 лет назад, # ^ |

I use dp to solve F

Spoiler

void solve(){
	int n;
	cin>>n;
	VI a(n+1);
	rep(i,1,n+1){
		cin>>a[i];
	}
	vector<ll> dp(n+1,0);
	dp[1] = 0;
	vector<ll> prefix(n+1);
	prefix[1] = a[1];
	for(int i=3;i<=n;i+=2){
		prefix[i] = prefix[i-2]+a[i];
	}
	for(int i=2;i<=n;i++){
		if(i&1){
			dp[i]=max(dp[i-1],dp[i-2]+a[i]);
		}else{
			dp[i] = a[i] + dp[i-2];
			dp[i] = max(dp[i],prefix[i-1]);
		}
	}
	cout<<dp[n];
}

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5 лет назад, # ^ |

← Rev. 2 →

we can maintain dp[i][j], j = {0,1}

Initially dp[2][1] = arr[2] and dp[2][0] = arr[1] (Base Case)

1. dp[i][1] denotes maximum sum when we are at i and we are picking ith element and floor(i / 2) elements(subset of 1...i such that adjacent element are not picked).

2. dp[i][0] denotes maximum sum when we are at i and we are NOT picking ith element and floor(i / 2) elements (subset of 1...(i — 1) such that adjacent element are not picked)

With some work we can fill dp upto n.

Our final answer will be max(dp[n][1],dp[n][0]) My Submission

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donpablo

5 лет назад, # ^ |

for odd i

dp[i][0] may be just dp[i-1][1]

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5 лет назад, # ^ |

may be!!!! designing transition can be different

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donpablo

5 лет назад, # ^ |

I mean for this dp[i][0] = max({arr[i — 1] + dp[i — 2][0], arr[i — 1] + dp[i — 3][1], arr[i — 1] + dp[i — 3][0], arr[i — 2] + dp[i — 3][0]});

you can just write dp[i][0]=dp[i-1][1]

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5 лет назад, # ^ |

what about if we can get max answer with arr[i — 1]

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Gats

5 лет назад, # ^ |

too short to understand,can you give some explanation? :D

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Video tutorials for this contest

5 лет назад, # ^ |

since we know already how many elements we have to take i.e N/2 so instead of using N^2 dynamic programming solution we can do it in O(N) by maintaining at each i that we have taken i/2 elements upto now and have maximum possible sum till now, since at each i we have two possiblity either pick that element or don't pick it.

We have two cases to handle when i is odd and i is even

when i is odd and we want to pick ith element we can build a current state from (i — 2)th and (i -3)rd state because we know we have already calculated answer for (i/2) — 1 element up to (i — 2) and (i — 3) so that after adding total elements become i/2.

And if we don't pick ith element we have to construct answer so that total element chosen is i/2 and sum is maximum we can construct answer by taking arr[i — 1] and help from states (i — 1),(i — 2),(i — 3) (for arr[i — 1] only these are possibility can you see why???) and also by using arr[i — 2] and help from state (i — 3) (for arr[i — 2] only this is possible). You don't have to consider other elements and states because you cannot build current state having i/2 elements.

Similarly for even but possiblity for even i's are less than that of odd i's you can think about it.

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Gats

5 лет назад, # ^ |

thanks a lot!

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Mucosolvan

5 лет назад, # ^ |

Your solution might underflow, check the testcase where every value is negative.

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stefdasca

5 лет назад, # |

+33

Also, F is this JOI task, which helped me a lot to win the contest

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lemongrab

5 лет назад, # ^ |

Thanks a lot for your regular videos, just one feedback that audio quality is can improve, but still very helpful!

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stefdasca

5 лет назад, # ^ |

+16

Yeah, I didn't expect to record now, so my audio quality wasn't the best due to circumstances not under my control.

But I'll take that in account from now on.

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Invaders

5 лет назад, # ^ |

Can you share your solution idea to the JOI problem you mentioned(Candies) ?

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inksamurai

5 лет назад, # |

how to E ?

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The_Dark_Knight_Rises

5 лет назад, # ^ |

I used Mobius Inversion, don't know if there was an easier solution using symmetry.

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https://atcoder.jp/contests/abc162/submissions/11854513

5 лет назад, # ^ |

wdym by modbius can your share your AC code?

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Bisnu

5 лет назад, # ^ |

[user:The _Dark_Knight_Rises] Can you please share your idea?

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zxcv890

5 лет назад, # |

← Rev. 3 →

I solved problem D in last 2 minutes. Spent almost 1 hour debugging it and found there was an overflow error in my $$$nC2$$$ and $$$nC3$$$ functions. I got huge rush when I saw AC on it!! Wonderful questions!!

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5 лет назад, # ^ |

← Rev. 2 →

Did you really need combinatorics for D? I would say, there was easy and simpler solution. Iterate over $$$i$$$ and $$$k$$$ ( such that $$$s[i] != s[k]$$$ ) and find number of $$$j$$$ between them such that ( $$$s[j] = \{'R','G','B'\} \setminus \{s[i],s[k]\}$$$ ).

Link — My solution.

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supermatthew

5 лет назад, # ^ |

Agreed, no combinatorics needed. Just needed to precompute prefixes to answer the "number of j between them" queries.

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akulenok

5 лет назад, # |

How to solve E and F? In E I had an idea to check all possible gcds and find answer, in F I have a dp approach but it works in O(n^2).

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5 лет назад, # ^ |

Read these:

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heath_cliff

5 лет назад, # ^ |

← Rev. 2 →

gupta_samarth I just went through your solution for E, phi[i] gives the count of numbers from 1 to i which are coprime with i. now = expo( k/g, n ) means inserting all the multiples of g in the array of size n . What I don't understand is that why did you do phi[g]*now ?

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5 лет назад, # ^ |

← Rev. 2 →

Read the blogs ( the mobius inversion one, the other is pre requisite ), they explain the problem for $$$2$$$ numbers taking values upto $$$n$$$.

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arjun_9426

5 лет назад, # ^ |

for each possible gcd try to maintain the number of possible arrays , dp[k+1] such that dp[i] means number of arrays which produce gcd=i, now if you try to find dp[i] means each number must be divisible by i(as this should be the gcd), so each number can be [i*1, i*2,i*3.....i*(k/i)] number of possibilities are k/i but these possibilities may also produce GCDs which are multiple of i (for eg if you took each element as i*2) so to get the exact number of sequences which produce 'i' as gcd you mustremove the count of sequences which exactly produce i*2, i*3 and so on. you can do this using dp just iterate the dp backwards

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.-O_O-.

5 лет назад, # ^ |

you can optimize O(N^2) solution to O(N*sqrt(N))....

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Not-Afraid

5 лет назад, # ^ |

A problem similiar to E. The only difference is we have one array here and not $$$k$$$^$$$n$$$.

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rohit_goyal

5 лет назад, # |

← Rev. 5 →

Solutions : D : Iterate on the L,R boundaries of the sequence, maintaining the count of each character in the boundary and if s[L] != s[R], add number of indices that have characters different from both of them to the answer and subtract 1 from it if(L == R(mod 2) and s[(L+R)]/2 == required character ). Time Complexity : (N^2)

E : Simple implementation of sieve. For each i, say number of sequences of length n is f(i) that has gcd = i, then f(i) = (floor(k/i))^n — (f(2*i) + f(3*i) + f(4*i) .... f(i*(floor(k/i))), calculate it from backwards.Time Complexity : O(KlogK)

F : It is easy to observe that gap between consecutive elements cannot exceed 2 or 3, but to be on safe side, lets make it 10. So we can make a dp[i][j] denoting the maximum answer if the sequence ends at index i(taking it) and (ceil(i+1)/2 — j) is the length of the sequence for this answer.It is easy to calculate it via contribution technique.Say before i we take j (j>i-11), and at j we take state k, then a possible transition is dp[i][ceil(i/2) — (ceil(j/2) -k)+(i-j-1)] = max(dp[i][ceil(i/2) — (ceil(j/2) -k)+(i-j-1)] ,dp[j][k] + arr[i]).Don't forget to initialize a possible starting state for each i, ie, dp[i][i] = arr[i] for all possible i. Time Complexity : (n*fac*fac), where fac is around 3-4 for optimized solutions.

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pranitchawla

5 лет назад, # ^ |

In problem E, does it make a difference if we calculate it from backwards? I was not able to get the right answer by summing from 1 to k.

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rohit_goyal

5 лет назад, # ^ |

Yes because for f(i) we already need to know f(2*i) and so on already.If you get confused at any time about iteration, just do memoization dp.

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arjun_9426

5 лет назад, # |

my solution for F, don't know what is wrong , for me this should work, could anyone help me in figuring out where i am wrong https://atcoder.jp/contests/abc162/submissions/11855696

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TheEpicCowOfLife

5 лет назад, # |

Instant editorial:

A: Just implement. I scanned in as a string to make it easier B: Implement . Use the modulo operator to check for divisibility, a % b == 0 if a is divisible by b C: Implement, just brute force using a triple nested for loop, using Euclid's algorithm to find gcd(i,gcd(j,k)) in O(n^3 logn) time D: Keep a prefix sum of the number of appearances of 'R' in subarray [1,i] for all 1<=i<=n. Do the same for 'G' and 'B'. Now brute force over all (j,k), and one can easily calculate all triples.

E: Let a[i] be the number of tuples with gcd i. Observe that if a tuple has gcd a multiple of i, then all elements of the tuple must also be multiples of i. Let p(i) = number of multiples of i that are <= k. The number of tuples with gcd i can be simply calculated by p(i)^n — a[i*2] — a[i*3]...

We can calculate each a[i] from i = k to 1 in that order.

Calculating each a[i] takes O(k/i) time. Via the harmonic series upper bound, this means that across all i from 1 to k, this method will take O(klogk) time.

F: This is some nasty case bash. First we should calculate a prefix sum for all odd indexed numbers and even indexed numbers.

If n is a even: then either you take all odd indexed numbers, even indexed numbers, or there is a single gap of two. It is enough to consider these cases and to brute force all possible gaps.

If n is odd, then you either don't take either endpoint, you take both endpoints, or you take one of them.

If you don't take either endpoint then there is only one choice: all the "middle" elements.

If you take exactly one of the endpoints, there will be one gap of two or three. Brute force over all the O(n) possible gaps

If you take both endpoints, then you may have two gaps of two or one gap of three. Brute forcing over the O(n) possible gaps of 3 is fine, but there are O(n^2) possible configurations of two gaps of two. However, suppose our gaps are at location i and j, writing down the expression for the resulting sum f(i,j) will look something like f(i,j) = g(i) + h(j) + c for some constant c, and some functions g and h. The exact details depend on implementation details I won't bore you with. By looping over it in a specific way, you can keep a running max of f(i) for some prefix i as you iterate over j, keeping in mind whatever precise restrictions you have on i and j.

Thus, an O(n) algorithm is achieved.

More details upon request.

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Lossser

5 лет назад, # |

Problem in Q.D what is the problem with this code in QUESTION D ? I am getting WA in 2 test cases

int n;
   cin>>n;
   string s;
   cin>>s;
   if(n==1)
   {cout<<0<<endl;continue;}
   if(n==2)
    {cout<<0<<endl;continue;}

   int a[3]={0};
   for(int i=0;i<n;i++)
   {
       if(s[i]=='R')
        a[0]++;
       else if(s[i]=='G')
        a[1]++;
       else
        a[2]++;
   }
   ll s1=0;
   s1=a[0]*a[1]*a[2];
   for(int i=0;i<n;i++)
   {
       for(int j=1;j<=min(i,n-i-1);j++)
       {
           if(s[i]!=s[i-j] && s[i]!=s[i+j] && s[i-j]!=s[i+j])
           {
               s1--;
           }
       }
   }
   cout<<s1<<endl;

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Wrynn

5 лет назад, # ^ |

Check out the code below and watch out for mathematical overflow.

s1=a[0]*a[1]*a[2];

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Lossser

5 лет назад, # ^ |

I got it. Thanks

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knighthood

5 лет назад, # |

Problem E, can someone help me find the bug in my logic? for i : 1 -> k

i = 1,

number of array such that there is at least one 1. -> k^n — (k-1)^n for all such arrays, gcd will be 1.

Now our Integer set reduced to {2,3,4...k}

i = 2,

number of array such that there is at least one 2 out of our Integer set {2,3,4...k} -> T = (k-1)^n — (k-2)^n. for all such arrays, gcd can be either 1 or 2. Number of arrays such that gcd is 2 will be where all elements are power of 2, -> C2 = (k/i)^n-(k/i — 1)^n : Gcd contribution = 2 All the other arrays will have gcd 1 and number of such array is = T-C2.

Continue like this . .

i = k

in the last iteration our integer set will {k}. then number of arrays with this is 1 and gcd contribution is k.

Unable to find my mistake. :( Please help!!

ll fans = 0;
for(int i = 1; i <= k; ++i) 
{
    ll cnt = (pwr(k-i+1,n) - pwr(k-i,n) + mod) % mod; 
    ll a = (k/i);
    ll cnt2 = pwr(a,n) - pwr(a-1,n); // i
    ll cnt3 = (cnt-cnt2+mod)%mod; // 1
    ll tans = (i*cnt2 + 1*cnt3) % mod;
    fans = (fans + tans) % mod;
}

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5 лет назад, # ^ |

← Rev. 3 →

I was thinking in a similar way but look

consider i=2 case according to your formula count of such subseq which give gcd=2 is (k-1)^n-(k-2)^n.

consider this subsequence {2,2,...,2,3} all 2's except the last one being 3 , this will be counted under (k-1)^n and this subseq won't be counted under (k-2)^n as it won't consider 2 as it's element now is the gcd of {2,2,...,2,3} equal to 2? no right it's 1 so you are counting some extra subseq whose gcd isn't i.

read my previous comment.

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knighthood

5 лет назад, # ^ |

For i=2 case, the total number of arrays possible are T = (k-1)^n -(k-2)^n, out of which some arrays will have gcd 1 and some 2. Your example will be counted in case where gcd is 1 for i=2. Out of T arrays, gcd will be 2 if we have an array with multiples of 2. (2,2,...3) will fall in case where all are not multiple of 2 hence gcd 1.

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