Polygon Diagonals Solution (InterviewStreet CodeSprint Fall 2011)

Hint.md

`The required answer is : 1 / (n + k) * n-choose-r(n - 3,k) * n-choose-r(n + k, n - 1).

While I am not aware of a simple way to prove this which I can describe here, here's some practical advice on how such problems can be approached. It helps to write down a simpler solution and searching Google or OEIS (Online Encyclopedia of Integer Sequences) for small inputs like k = 3 and N = 1,2,...10. This usually gives you a formula describing the sequence, and possibly generalizations for the same.

Computing the answer using the above formula requires calculating the power of MOD (1e6 + 3) in the numerator and the demonimator. If the power of MOD is bigger in the numerator, the answer is 0. Otherwise, we can calculate all the factorials with the powers of MOD cast out. For example, if MOD = 3 and N = 11, we would calculate the product 1 * 2 * 3 / 3 * 4 * 5 * (6 / 3) * 7 * 8 * (9 / 9) * 10 * 11. This can be done easily in O(MOD). Once we have that, we can multiply the values in the numerator and mltiply with the modular inverses of the values in the denominator to get the answer.

problemsetter.cpp

#include<iostream>
#include<set>
#include<map>
#include<string>
#include<stdio.h>
#include<sstream>
#include<algorithm>
#include<queue>
#include<cmath>
#include<string.h>
using namespace std ;
#define INF (int)1e9
#define MAXN 700
int N ;
double matrix[MAXN][MAXN] ;

string normalize(string s)
{
 int mp[26],c = 0 ;
 memset(mp,255,sizeof mp) ;
 for(int i = 0;i < s.size();i++)
  if(mp[s[i] - 'a'] == -1)
   mp[s[i] - 'a'] = c++ ;
 string ret ;
 for(int i = 0;i < s.size();i++)
  ret.push_back(mp[s[i] - 'a'] + 'a') ;
 return ret ;
}

int ct = 0 ;
map<string,int> id ;
int getId(string s)
{
 if(id.count(s)) return id[s] ;
 return id[s] = ct++ ;
}

int tot = 0 ;
map<string,double> memo ;
double solve(string s)
{
 s = normalize(s) ;
 if(memo.count(s)) return memo[s] ;

 ct = 0 ;
 id.clear() ;

 int n = s.size() ;
 int swaps = n * (n - 1) / 2 ;
 
 N = 0 ;
 string t = s ;
 sort(t.begin(),t.end()) ;
 do { getId(normalize(t)) ; }
 while(next_permutation(t.begin(),t.end())) ;
 N = id.size() ;

 for(int i = 0;i <= N;i++)
  for(int j = 0;j <= N;j++)
   matrix[i][j] = 0 ;

 t = s ;
 sort(t.begin(),t.end()) ;
 do
 {
  string x = normalize(t) ;
  int id1 = getId(x) ;
  matrix[id1][id1]++ ;
    
  string tt = x ;
  reverse(tt.begin(),tt.end()) ;
  if(x == tt) continue ;  
  for(int i = 0;i < n;i++)
   for(int j = i + 1;j < n;j++)
   {
    swap(x[i],x[j]) ;
    int id2 = getId(normalize(x)) ;
    matrix[id1][id2] += -1. / swaps ;
    swap(x[i],x[j]) ;
   }
  matrix[id1][N]++ ;
 }
 while(next_permutation(t.begin(),t.end())) ;
 
 for(int i = 0;i < N;i++)
 {
  for(int j = N;j >= i;j--) matrix[i][j] /= matrix[i][i] ;
   for(int k = 0;k < N;k++) if(i != k)
    for(int j = N;j >= i;j--)
     matrix[k][j] -= matrix[k][i] * matrix[i][j] ;
 }

 sort(t.begin(),t.end()) ;
 do
 {
  string tt = normalize(t) ;
  memo[tt] = matrix[getId(tt)][N] ;
 }
 while(next_permutation(t.begin(),t.end())) ;
 
 return memo[s] ;
}

int main()
{
 int runs ;
 cin >> runs ; 
 while(runs--)
 {
  string s ;
  cin >> s ;
  double ret = solve(s) ;
  printf("%.4lf\n",ret) ;
 }
 return 0 ;
}

This code is not the answer to this problem, it is the answer to another problem: https://gist.github.com/1285176