Permutation Descent Counts(递推)

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([5 6 2 4 7 1 3]) = 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:

PDC(3, 0) = 1 { [ 1 2 3 ] }
PDC(3, 1) = 4 { [ 1 3 2 ], [ 2 1 3 ], [ 2 3 1 ], 3 1 2 ] }
PDC(3, 2) = 1 { [ 3 2 1 ] }`

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

Hint

Source

2017湖南多校第十三场

//题意：给出 n，v 求 1 -- n 的排列中，相邻的数，出现 v 次前面数比后面数大的种数。

题解：假如设 dp[i][j] 为 1 -- i 的排列，出现 j 次前面数比后面数大的情况的种数，那么

递推，有两个来源，dp[i-1][j] 和 dp[i-1][j-1] ，只要考虑 i 放置的位置即可，分清楚情况讨论清楚即可！

比赛时没想清楚唉！

 # include <cstdio>

 # include <cstring>

 # include <cstdlib>

 # include <iostream>

 # include <vector>

 # include <queue>

 # include <stack>

 # include <map>

 # include <bitset>

 # include <set>

 # include <cmath>

 # include <algorithm>

 using namespace std;

 #define lowbit(x) ((x)&(-x))

 #define pi acos(-1.0)

 #define eps 1e-8

 #define MOD 1001113

 #define INF 0x3f3f3f3f

 #define LL long long

 inline int scan() {

     int x=,f=; char ch=getchar();

     while(ch<''||ch>''){if(ch=='-') f=-; ch=getchar();}

     while(ch>=''&&ch<=''){x=x*+ch-''; ch=getchar();}

     return x*f;

 }

 inline void Out(int a) {

     if(a<) {putchar('-'); a=-a;}

     if(a>=) Out(a/);

     putchar(a%+'');

 }

 #define MX 105

 //Code begin...

 int dp[MX][MX];

 void Init()

 {

     dp[][]=;

     for (int i=;i<=;i++)

     {

         for (int j=;j<=i-;j++)

         {

             dp[i][j] = dp[i-][j]*(j+)%MOD;

             if (j!=)

                 dp[i][j] = (dp[i][j]+dp[i-][j-]*(i-j))%MOD;

         }

     }

 }

 int main()

 {

     Init();

     int t = scan();

     while (t--)

     {

         int c = scan();

         int n = scan();

         int m = scan();

         printf("%d %d\n",c,dp[n][m]);

     }

     return ;

 }