Permutation Descent Counts(递推)

1968: Permutation Descent Counts

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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=0,f=1; char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
inline void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
#define MX 105
//Code begin...
int dp[MX][MX];

void Init()
{
    dp[1][0]=1;
    for (int i=2;i<=100;i++)
    {
        for (int j=0;j<=i-1;j++)
        {
            dp[i][j] = dp[i-1][j]*(j+1)%MOD;
            if (j!=0)
                dp[i][j] = (dp[i][j]+dp[i-1][j-1]*(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 0;
}