icode_8990's blog

By icode_8990, history, 5 years ago, In English

So in this problem i am trying to solve it using dynamic programming approach. But I am unable to understand the solution since its written in poor english. I am referring to this tutorial. How is the dp[][] being developed?. it will be great if someone could explain in details since i am a newbie. Thank you in advance!

#include <iostream>
#include <stdio.h>
#include <algorithm>

using namespace std;

const int mod = 1e9 + 7;

int dp[100][2];

void add(int &a, int b)
{
    a += b;
    if(a >= mod)
        a -= mod;
}

int main(int argc, char ** argv)
{
    int n, k, d;
    cin >> n >> k >> d;
    dp[0][0] = 1;
    dp[0][1] = 0;
    
    for(int i = 1 ; i <= n ; ++i)
    {
        dp[i][0] = dp[i][1] = 0;
        
        for(int j = 1 ; j <= k ; ++j)
        {
            if(i-j < 0) break;
            
            if(j < d)
            {
                add(dp[i][0], dp[i-j][0]);
                add(dp[i][1], dp[i-j][1]);
            }
            else
            {
                add(dp[i][1], dp[i-j][0]);
                add(dp[i][1], dp[i-j][1]);
            }
        }
    }
    cout << dp[n][1] << "\n";
    return 0;
}
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5 years ago, # |
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Consider this: dp[sum to be made][do we have an edge with weight at least d]. So now you want to calculate dp[sum][false] and it will be something like this: for (int i = 1; i <= k; i++) dp[sum][false] += dp[sum-i][i >= d]; and dp[sum][true]: for (int i = 1; i <= k; i++) dp[sum][true] += dp[sum-i][true];

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5 years ago, # |
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try to solve with top — down approach . suppose we are at node x , from node x we can go to each of the k nodes just below it...

let our total sum till x node is sum now from node x , we iterate for each possible k values and each time our sum is increased by sum+i;

we also use chk variable to check whether we visited an edge with weight >=d

for more clearity check my sol .

Link

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    5 years ago, # ^ |
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    Thank you so much... it really helped

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      5 years ago, # ^ |
      Rev. 2   Vote: I like it +3 Vote: I do not like it

      how it helped. how u came across this blog. who r u? LOL.

      why u didn't submitted the solution to this problem?

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5 years ago, # |
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since i am a newbie

No you are unrated