Mohd__Messi's blog

By Mohd__Messi, history, 9 days ago, In English
 
 
 
 
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9 days ago, # |
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9 days ago, # |
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Auto comment: topic has been updated by Mohd__Messi (previous revision, new revision, compare).

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9 days ago, # |
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you can solve it with a one dimension dp array like this

first set the whole array to -inf execpt dp[0] = 0
for(int i = 2; i < 10001;++i)
for(int x : v) // v is the vector that hold prime numbers and primatic numbers
{
if(i — x < 0)continue;
dp[i] = min(dp[i], dp[i — x] + 1);
}
answer = dp[input]

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8 days ago, # |
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With N = 9973 (the largest prime below 10 000), your code prints 0.

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    8 days ago, # ^ |
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    Yes but why I am not able to counter that my DP solution seems OK but last few cases are giving WA. Can you please give a hint.

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8 days ago, # |
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You aren't using last prime number v[m - 1] in your dp loop, because you are going from 1..m-1 (for i), but then you are using v[i - 1] in calculation, so in reality you are using 0..m-2 prime numbers.

Second problem is array overflow, because in your for(;l<v.size();l++) for number like 10000 you will end up with l == m, which doesn't with your dp declaration.

To fix it change int dp[m][10001] to int dp[m+1][10001] and also your dp loop from for(int i=0;i<m;i++) to for(int i=0;i<=m;i++). That being said, you don't need 2 dimensional array, as pointed out by Mohamed_Saad62

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    8 days ago, # ^ |
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    Thanks a lot brother I passed the last case. I declared dp[m][10001] thats why the last prime number was not getting included. Thanks a lot for pointing out my mistake.