My submission:- https://pastebin.com/aL9vCYWu
Please anyone tell what is wrong in this code
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My submission:- https://pastebin.com/aL9vCYWu
Please anyone tell what is wrong in this code
Название |
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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]
With
N = 9973
(the largest prime below 10 000), your code prints0
.You aren't using last prime number
v[m - 1]
in your dp loop, because you are going from1..m-1 (for i)
, but then you are usingv[i - 1]
in calculation, so in reality you are using0..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 withl == m
, which doesn't with yourdp
declaration.To fix it change
int dp[m][10001]
toint dp[m+1][10001]
and also your dp loop fromfor(int i=0;i<m;i++)
tofor(int i=0;i<=m;i++)
. That being said, you don't need 2 dimensional array, as pointed out by Mohamed_Saad62