Hello Codeforces .
Plase help.
Given n, array a(a1, a2,,, an). n, a[i] <= 100.
How can i make dp[i][j].
Corectly dp[i][j] = 1, if we can get sum j without a[i].
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Hello Codeforces .
Plase help.
Given n, array a(a1, a2,,, an). n, a[i] <= 100.
How can i make dp[i][j].
Corectly dp[i][j] = 1, if we can get sum j without a[i].
Название |
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How can you say its easy if you are stuck in it and asking for help? :/
it is easy for some people
I have thought of a solution.
Take dpt[j] be the number of ways you can make the sum j using any elements. So, for each sum from 1 to 10000, you take each of the a[i] into consideration and do a dpt[sum-a[i]]++.
Now, dp[i][j] is 1 if dpt[j]>=2 or else its 0.
What are the constraints on n ?
If you don't mind, can you put a link to the exact question. This is kind of hard to understand.
"n, a[i] <= 100"
Calculate DP for prefixes and suffixes. Now, DP[i][j] = 1 if there exists A and B such that A+B=j and A is obtained as sum from prefix and B is obtained as sum from suffix.