UVA 11077 Find the Permutations 递推置换

                           Find the Permutations

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 O(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
3 1
3 0
3 2
0 0
Sample Output
3
1
2

题意： 

给定一个1~n的排序，可以通过一系列的交换变成1，2，…,n, 给定n和k，统计有多少个排列至少需要交换k次才能变成有序的序列。

题解： 

每个长度为n循环需要交换n-1次才能将交换到对应的位置，例如1->2,2->4,4->1,(1,2,4)位置上对应值为(2,4,1) 相当于一个长度为3的环逆时针旋转了1格，要变换回来，需要跟原位置交换，因为成环，所以共n-1次     那么对于序列P，有x个循环节，长度为n，就需要交换n-x次     对于f[i][j]，表示交换j次能变为1~i的序列的种数     我们找到递推式：f[i][j]=f[i-1][j]+f[i-1][j-1]*(i-1)，边界是f[1][0]=1,其余为0     解释：新增的i，如果与自己构成循环，那么循环数和长度都加一，交换数不变，所以是f[i-1][j]         新增的i，如果参加到其他环中，每个数后一种，共i-1种，循环数不变，长度加一，所以是f[i-1][j-1]*(i-1)     例如1->2,2->3,3->1,{2,3,1};将4添加到1后就是1->4,4->2,2->3,3->1，序列为{4,3,1,2};其他同上

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std ;
typedef long long ll;
const int N = 105;
const int mod = 1e9 + 7;
unsigned long long  dp[N][N];
void init() {
    for(int i = 1; i <= 21; i++) {
        dp[i][0] = 1;
        for(int j = 1; j < i; j++)
            dp[i][j] = dp[i-1][j] + dp[i-1][j-1] * (i-1);
    }
}
int main () {
    init();
    int n,k;
    while(~scanf("%d%d",&n,&k)) {
        if(n == 0 && k == 0) break;
        printf("%llu
",dp[n][k]);
    }
    return 0;
}