Permutation Descent Counts

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一、问题：

Description

Given a positive integer, N, a permutation of order N is a one-to-one (and thus onto) function from the set of integers from 1 to N to itself. If p is such a function, we represent the function by a list of its values: [p(1)p(2). . . p(N)]

For example, [5 6 2 4 7 1 3] represents the function from {1 . . . 7} to itself which takes 1 to 5, 2 to 6, . . ., 7 to 3. For any permutation p, a descent of p is an integer k for which p(k) > p(k + 1).

For example, the permutation [5 6 2 4 7 1 3] has a descent at 2(6 > 2) and 5(7 > 1). For permutation p, des(p) is the number of descents in p. For example, des([5624713]) = 2. The identity permutation is the only permutation with des(p) = 0. The reversing permutation with p(k) = N + 1 − k is the only permutation with des(p) = N − 1. The permutation descent count (PDC) for given order N and value v is the number of permutations p of order N with des(p) = v. For example:

P DC(3, 0) = 1{[123]}

P DC(3, 1) = 4{[132], [213], [231], 312]}

P DC(3, 2) = 1{[321]}

Write a program to compute the PDC for inputs N and v. To avoid having to deal with very large numbers, your answer (and your intermediate calculations) will be computed modulo 1001113.

Input

The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. Each data set should be processed identically and independently. Each data set consists of a single line of input. It contains the data set number, K, followed by the integer order, N (2 ≤ N ≤ 100), followed by an integer value, v (0 ≤ v ≤ N − 1).

Output

For each data set there is a single line of output. The single output line consists of the data set number, K, followed by a single space followed by the PDC of N and v modulo 1001113 as a decimal integer.

Sample Input

4

1 3 1

2 5 2

3 8 3

4 99 50

Sample Output

1 4

2 66

3 15619

4 325091

 

二、题意：

第一行数字 T 代表测试样例个数。

接下来的 T 行每行第一个数 cnt 表示序号，第二个数 n，第三个数v。

求：1～n的全排列中，逆序对数为v的情况数。

 

三、思路：

dp[ i ][ j ] 表示1～ i 的全排列中，逆序对数为 j 的情况数，首先当 j =0 时，表示无逆序对，则dp[ i ] [ 0 ] = 1，即只有“顺序”一种。

其他情况，dp[ i ][ j ]的来源只有两种：

1）dp[ i-1 ][ j-1 ]，表示第 j 个数恰好也构成了逆序。

2）dp[ i-1 ][ j ] ，表示截至第 j-1 个数时逆序对数就已经为 j ，并且第 j 个不构成逆序。

状态转移方程为：dp[ i ][ j ] = dp[ i-1 ][ j-1 ] * ( i-j ) + dp[ i-1 ][ j ] * ( j+1 ) ;

由于结果可能比较大，所以还需要mod 1001113 。

 

四、代码：

#include <map>
#include <set>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define eps 1e-10
#define mod 1e9+7
#define PI acos(-1)
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
#define read(x) scanf("%d",&x)
#define mem(a,b) memset(a,0,sizeof(a))
#define fori(a,b) for(int i=a;i<=b;i++)
#define forj(a,b) for(int j=a;j<=b;j++)
#define fork(a,b) for(int k=a;k<=b;k++)
#define ifor(a,b) for(int i=a;i>=b;i--)
#define jfor(a,b) for(int j=a;j>=b;j--)
#define kfor(a,b) for(int k=a;k>=b;k--)
#define IN freopen("in.txt","r",stdin)
#define OUT freopen("out.txt","w",stdout)

using namespace std;
typedef long long LL;
const int MOD=1001113;

int main()
{
    int dp[105][105],T,cnt,n,v;
    read(T);
    mem(dp,0);
    fori(1,100)
        dp[i][0]=1;
    fori(1,100)
        forj(1,i-1)
            dp[i][j]=(dp[i-1][j-1]*(i-j)%MOD+dp[i-1][j]*(j+1)%MOD)%MOD;
    while(T--)
    {
        scanf("%d %d %d",&cnt,&n,&v);
        printf("%d %d\n",cnt,dp[n][v]);
    }
    return 0;
}

 

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