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Dynamic Programming solving an algorithm


Dynamic Programming solving an algorithm

By : NBStudios
Date : November 17 2020, 11:55 AM
To fix the issue you can do Your task is similiar to fibonnaci :) First i will explain you fibonacci.
F(1) = 1
code :
#include <cstdio>
#include <cstdlib>

using namespace std;

int f(int n) {

    //printf("n = %d\n", n);
    if(n == 1 || n == 2) // base case
        return 1;
    return f(n - 1) + f(n - 2); // recurrence relation

}

int main() {

    int n; scanf("%d", &n);
    printf("%d\n", f(n));

    return 0;
}
#include <cstdio>
#include <cstdlib>
#include <cstring>

using namespace std;
const int N = 50;

long long memo[N];

long long f(int n) {
    if(memo[n] != -1) // if we already computed the value of f(N), then return that value
        return memo[n];
    return memo[n] = f(n - 1) + f(n - 2); // else compute the value, and save it into the table
}

int main() {

    memset(memo, -1, sizeof(memo));

    memo[1] = memo[2] = 1; // add answer for base case to the table

    int n; scanf("%d", &n);
    printf("%lld\n", f(n));

    return 0;
}
#include <cstdio>
#include <cstdlib>
#include <cstring>

using namespace std;
const int N = 25;

int check, memo[N];

int f(int x) {
    if(memo[x] != check) // if f(n) was already computed
        return memo[x]; // return computed value
    return memo[x] = f(x - 2) - f(x - 1) + 2; // else compte given value and add it to a table
}

int main() {

    memset(memo, 63, sizeof(memo)); // very big number, if the value of h(n) is different then that very big number, then we know we have computed the value for h(n)
    check = memo[0];

    memo[1] = 2; // base case
    memo[2] = 3; // base case

    int n; scanf("%d", &n);
    printf("%d\n", f(n));

    return 0;
}


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What is dynamic programming? (in solving technique)

What is dynamic programming? (in solving technique)


By : Wycliff Bironga
Date : March 29 2020, 07:55 AM
like below fixes the issue Maybe your friend means to find T1, T2, etc. with idle time in between and try to shorten this? In general dp means to eliminate variables that are unecessary for the solution thus speed up the overall compute time. Here is a good link: difference between back tracking and Dynamic programming
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By : Ajayi Justina
Date : March 29 2020, 07:55 AM
This might help you I think you can refer to the geometry, especially the analytic geometry. Simplex algorithm actually means that the optimal result always stay in the vertex instead of in the line or in the face, it's very intuitive.
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By : Amy Joscelyn
Date : March 29 2020, 07:55 AM
This might help you The magic words you need to google are "Simplex Algorithm" for linear programming constraint maximisation. This slide deck looks like it might be what you want. http://www.cs.nccu.edu/~melikyan/mat_fm/lec/lec5_4.pdf
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By : Bjørn Bjerkeli
Date : March 29 2020, 07:55 AM
I wish this helpful for you Instead of using let with fn's in it, try letfn. doseq doseq -> looks like it would likely be better as a for comprehension Your cond / zero? / = 1 code would be easier to read (and faster) with case. My spidey-sense tells me that the loop/recurs here should be some kind of map call instead I strongly suspect that this would be far faster with primitive arrays (and possibly cleaner in some places) You might like to use or look at the source for Incanter
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By : Afdel
Date : March 29 2020, 07:55 AM
this will help I don't know much about DP but I have a few observations that I hope will contribute to the conversation.
First, your example Scala code doesn't appear to solve the LIS problem. I plugged in the Van der Corput sequence as found on the Wikipedia page and did not get the designated result.
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