UVA 11077 Find the Permutations DP

Find the Permutations

Time Limit: 3000MSMemory Limit: Unknown64bit IO Format: %lld & %llu

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Description

Problem D
Find the Permutations
Input: Standard Input

Output: Standard Output

 

Sorting is one of the most used operations in real life, where Computer Science comes into act. It is well-known that the lower bound of swap based sorting is nlog(n). It means that the best possible sorting algorithm will take at least W(nlog(n)) swaps to sort a set of n integers. However, to sort a particular array of n integers, you can always find a swapping sequence of at most (n-1) swaps, once you know the position of each element in the sorted sequence. For example – consider four elements <1 2 3 4>. There are 24 possible permutations and for all elements you know the position in sorted sequence.

 

If the permutation is <2 1 4 3>, it will take minimum 2 swaps to make it sorted. If the sequence is <2 3 4 1>, at least 3 swaps are required. The sequence <4 2 3 1> requires only 1 and the sequence <1 2 3 4> requires none. In this way, we can find the permutations of N distinct integers which will take at least K swaps to be sorted.

Input

Each input consists of two positive integers N (1≤N≤21) and K (0≤K<N) in a single line. Input is terminated by two zeros. There can be at most 250 test cases.

Output

For each of the input, print in a line the number of permutations which will take at least K swaps.

Sample Input                             Output for Sample Input

3 1

3 0

3 2

0 0

3

1

2

 

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Problemsetter: Md. Kamruzzaman

Special Thanks: Abdullah-al-Mahmud

 

 

/* ***********************************************Author        :CKbossCreated Time  :2015年05月06日 星期三 22时13分56秒File Name     :UVA11077.cpp************************************************ */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <set>#include <map>using namespace std;typedef unsigned long long int LL;LL n,k;LL dp[30][30];void init(){dp[1][0]=1;for(int i=2;i<=21;i++){dp[i][0]=1;for(int j=0;j<=i;j++){dp[i][j]=dp[i-1][j]+dp[i-1][j-1]*(i-1);}}}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);init();while(cin>>n>>k){if(n==0&&k==0) break;cout<<dp[n][k]<<endl;}        return 0;}