1968: Permutation Descent Counts
Submit Page Summary Time Limit: 1 Sec Memory Limit: 128 Mb Submitted: 123 Solved: 96
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！
1 # include <cstdio>
2 # include <cstring>
3 # include <cstdlib>
4 # include <iostream>
5 # include <vector>
6 # include <queue>
7 # include <stack>
8 # include <map>
9 # include <bitset>
10 # include <set>
11 # include <cmath>
12 # include <algorithm>
13 using namespace std;
14 #define lowbit(x) ((x)&(-x))
15 #define pi acos(-1.0)
16 #define eps 1e-8
17 #define MOD 1001113
18 #define INF 0x3f3f3f3f
19 #define LL long long
20 inline int scan() {
21 int x=0,f=1; char ch=getchar();
22 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘) f=-1; ch=getchar();}
23 while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘; ch=getchar();}
24 return x*f;
25 }
26 inline void Out(int a) {
27 if(a<0) {putchar(‘-‘); a=-a;}
28 if(a>=10) Out(a/10);
29 putchar(a%10+‘0‘);
30 }
31 #define MX 105
32 //Code begin...
33 int dp[MX][MX];
34
35 void Init()
36 {
37 dp[1][0]=1;
38 for (int i=2;i<=100;i++)
39 {
40 for (int j=0;j<=i-1;j++)
41 {
42 dp[i][j] = dp[i-1][j]*(j+1)%MOD;
43 if (j!=0)
44 dp[i][j] = (dp[i][j]+dp[i-1][j-1]*(i-j))%MOD;
45 }
46 }
47 }
48
49 int main()
50 {
51 Init();
52 int t = scan();
53 while (t--)
54 {
55 int c = scan();
56 int n = scan();
57 int m = scan();
58 printf("%d %d\n",c,dp[n][m]);
59 }
60 return 0;
61 }