csuoj 1968 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：

要你求1~n的全排列中逆序数为 v 的个数

题解：

递推打表即可

用dp[ i ][ j ]表示 1~i 的全排列逆序数为 j 的个数

这个数来源于两个方向：dp[ i-1 ][ j-1 ] 和 dp[ i-1 ][ j ]

仔细考虑，分析已经有的逆序数对和不是逆序数对的地方插入一个数，以及头和尾考虑一下即可

代码就很简单了

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define mod 1001113
int dp[200][200];
void init()
{
for(int i=0;i<=100;i++)
dp[i][0]=1;
for(int i=1;i<=100;i++)
for(int j=1;j<i;j++)
dp[i][j]=(dp[i-1][j-1]*(i-j)%mod+dp[i-1][j]*(j+1)%mod)%mod;

}

int main()
{
int T;
init();
int t,n,k;
//freopen("in.txt","r",stdin);
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&t,&n,&k);
printf("%d %d\n",t,dp
[k]);

}
return 0;
}