help me in this problem idea or intuition

Правка en4, от singh23saurav, 2024-11-10 12:00:41

can anyone resolve better than O(N*N) Problem Breakdown 1. Find LCM for Each Pair: * For each unique pair (a[i], a[j]) in the array a, compute the Least Common Multiple (LCM). * The pairs include (a[i], a[i]) for each single number, as well as every combination of distinct numbers (a[i], a[j]) where i ≠ j. 2. Calculate Minimum Multiplier: * For each pair (a[i], a[j]), find the smallest integer m

such that: LCM(a[i],a[j])×m≥k\text{LCM}(a[i], a[j]) \times m \geq kLCM(a[i],a[j])×m≥k * This product should also be a multiple of kkk. 3. Calculate the Cost: * For each pair, compute the cost of the smallest multiplier mmm, which is simply the value of mmm. * Sum up the costs for all pairs to get the total answer. eg: [4,5,6] k=12 ans=>3+12+2+2*(3+2+1)=29, give a complexity less than O(N2) less than 108 as n<=10**5

as lcm of(4,4)->4*3=12(multiple of k) (5,5)->5*12=60(mutplit of k) similarly, (6,5)->30*2=60(multiple of k) and (5,6)=2 same answer

(4,5)=(5,4)=3 that's why foe different pairs we multiple by 2 for answer

История

 
 
 
 
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en6 Английский singh23saurav 2024-11-10 12:45:20 76
en5 Английский singh23saurav 2024-11-10 12:05:20 727
en4 Английский singh23saurav 2024-11-10 12:00:41 6 Tiny change: ' integer mmm such that' -> ' integer m\n\n such that'
en3 Английский singh23saurav 2024-11-10 12:00:05 1032
en2 Английский singh23saurav 2024-11-10 11:57:28 139 Tiny change: 'can anyon resolve b' -> 'can anyone resolve b'
en1 Английский singh23saurav 2024-11-10 11:55:46 77 Initial revision (published)