PERMUT1 - Permutations editorial -spoj - UCS

 

#dynamic-programming

This is question for  PERMUT1 - Permutations editorial - spoj

Let A = [a₁,a₂,...,a_n] be a permutation of integers 1,2,...,n. A pair of indices (i,j), 1<=i<=j<=n, is an inversion of the permutation A if a_i>a_j. We are given integers n>0 and k>=0. What is the number of n-element permutations containing exactly k inversions?
For instance, the number of 4-element permutations with exactly 1 inversion equals 3.

Task

Write a program which for each data set from a sequence of several data sets:

• reads integers n and k from input,
• computes the number of n-element permutations with exactly k inversions,
• writes the result to output.

Input

The first line of the input file contains one integer d, 1<=d<=10, which is the number of data sets. The data sets follow. Each data set occupies one line of the input file and contains two integers n (1<=n<=12) and k (0<=k<=98) separated by a single space.

Output

The i-th line of the output file should contain one integer - the number of n-element permutations with exactly k inversions.

Example

Sample input:
1 
4 1 

Sample output:
3 

Problem Link for PERMUT1 - Permutations is:

http://www.spoj.com/problems/PERMUT1/ 

The solution for PERMUT1 - Permutations is explained from here.

Let us see the code or solution to PERMUT1 - Permutations first.

 #include <iostream>
using namespace std;
long  long dp[15][100];
long long count(long long n,long long k)
{


// Solution to  PERMUT1 - Permutations

      if(n==0)

      return 0;

      if(k==0)

      return 1;

      if(dp[n][k]!=-1)

      return dp[n][k];

      long long val=0;

      for(int i=0;i<n&&k-i>=0;i++)

      {

            val+=count(n-1,k-i);

      }

      return (dp[n][k]=val);

}

int main() {


//Main method starts for  PERMUT1 - Permutations problem

    long long t,i,j,k,n;

    cin>>t;

    for(i=0;i<t;i++)

    {

//First take input here dp is used.

          cin>>n>>k;

          for(int l=0;l<=n;l++)

          {

           for(int m=0;m<=k;m++)

           {

                 dp[l][m]=-1;

           }

          }

         cout<< count(n,k)<<endl;

    }

    return 0;

}

What is inversion permutation in problem PERMUT1 - Permutations is that 

suppose consider any pair of indices (i,j), 1<=i<=j<=n, is an inversion of the permutation A if a_i>a_j. 

According to problem PERMUT1 - Permutations if n equals to zero then there are 0 inversions of permutations are possible.

In the same way, According to problem PERMUT1 - Permutations if k equals to zero then number of inversion of permutations is one.This is simply the permutation is in ascending order.

These two are base cases in problem PERMUT1 - Permutations problem solution.

The other DP approach is used is implemented in the code and the final value or answer of our problem is present in dp[n][k]. 

This is solution to the problem PERMUT1 - Permutations.

Happy Coding.........

Feel free to ask queries in comments...........