likelihood 12, he leaves his present club, then, at that point, makes another club himself and goes along with it. There is just a single understudy (himself) in the new club he makes. With likelihood 12, he doesn't make new clubs. For this situation, he changes his club to another one (perhaps a similar club he is in right now) with likelihood relative to the number of understudies
Correct answer will be upvoted else downvoted. Computer science.
With likelihood 12, he leaves his present club, then, at that point, makes another club himself and goes along with it. There is just a single understudy (himself) in the new club he makes.
With likelihood 12, he doesn't make new clubs. For this situation, he changes his club to another one (perhaps a similar club he is in right now) with likelihood relative to the number of understudies in it. Officially, assume there are k clubs and there are bi understudies in the I-th club for 1≤i≤k (before the understudy blows up). He leaves his present club, and afterward joins the I-th club with likelihood canister.
We note that when a club becomes unfilled, understudies won't ever go along with this is on the grounds that any understudy who blows up will get a vacant club together with likelihood 0 as per the above assertion.
Homer ponders the normal number of days until each understudy is in a similar club interestingly.
We can demonstrate that the appropriate response can be addressed as an objective number pq with gcd(p,q)=1. In this way, you are requested to find the worth from pq−1mod998244353. It very well may be shown that qmod998244353≠0 under the given limitations of the issue.
Input
The main line contains an integer m (1≤m≤1000) — the number of clubs at first.
The subsequent line contains m integers a1,a2,… ,am (1≤
Output
Print one integer — the normal number of days until each understudy is in a similar club interestingly, modulo 998244353.
Step by step
Solved in 3 steps with 1 images
