UVA Find the Permutations 11077 （DP&置换群）

Find the Permutations

Sorting is one of the most used operations in real life, where Computer Science comes into act. It iswell-known that the lower bound of swap based sorting is nlog(n). It means that the best possiblesorting algorithm will take at least O(nlog(n)) swaps to sort a set of n integers. However, to sort aparticular array of n integers, you can always find a swapping sequence of at most (n − 1) swaps, onceyou 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 allelements 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 sequenceis <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 willtake 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

3 1

3 0

3 2

0 0

Sample Output

3

1

2

//题意：

给出两个数n,k，对1-n的所有序列进行两两交换，统计有多少个序列至少需要进行k次操作才能变成（1,2，......n）。

//思路：

看到这个题首先就会先到置换，单元素循环是不需要交换，两个元素循环要交换一次，.......n个元素循环要交换n-1次。

这样的话，如果序列的循环节为x，总共交换的次数为n-x次。所以可以退出下面：

用dp[i][j]表示至少要进行j次操作才能将1-i，这i个数排列为（1,2,......i）这样的序列的个数。

所以就会得到方程：dp[i][j]=dp[i-1][j-1]+f[i-1][j]*(i-1)。

因为元素i要么自己形成一个循环，要么加入到前面任意一个为。

#include<stdio.h>#include<string.h>#include<algorithm>#define INF 0x3f3f3f3f#define ll unsigned long long#define N 30using namespace std;int gcd(int a,int b){return b==0?a:gcd(b,a%b);}ll dp[N][N];int init(){int i,j;memset(dp,0,sizeof(dp));dp[1][0]=1;for(i=2;i<=21;i++){for(j=0;j<i;j++){dp[i][j]=dp[i-1][j];if(j>0)dp[i][j]+=dp[i-1][j-1]*(i-1);}}}int main(){init();int n,k;while(scanf("%d%d",&n,&k),n|k){printf("%llu\n",dp[n][k]);}return 0;}