HDU3664 Permutation Counting
Permutation Counting
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1487 Accepted Submission(s): 754
Problem Description
Given a permutation a1, a2, … aN of {1, 2, …, N}, we define its E-value as the amount of elements where ai > i. For example, the E-value of permutation {1, 3, 2, 4} is 1, while the E-value of {4, 3, 2, 1} is 2. You are requested to find how many permutations of {1, 2, …, N} whose E-value is exactly k.
Input
There are several test cases, and one line for each case, which contains two integers, N and k. (1 <= N <= 1000, 0 <= k <= N).
Output
Output one line for each case. For the answer may be quite huge, you need to output the answer module 1,000,000,007.
Sample Input
3 0
3 1
Sample Output
1
4
Hint
There is only one permutation with E-value 0: {1,2,3}, and there are four permutations with E-value 1: {1,3,2}, {2,1,3}, {3,1,2}, {3,2,1}Source
题意:对于任一种N的排列A,定义它的E值为序列中满足A[i]>i的数的个数。给定N和K(K<=N<=1000),问N的排列中E值为K的个数。
dp[i][j]表示前i个数的排列中E值为j的个数,所以当插入第i+1个数时,如果放在第i+1或满足条件的j个位置时,j不变,与其余i-j个位置上的数调换时j会+1。所以
dp[i+1][j] = dp[i+1][j] + (j + 1) * dp[i][j];
dp[i+1][j+1] = dp[i+1][j+1] + (i - j) * dp[i][j];
/* ID: LinKArftc PROG: 3664.cpp LANG: C++ */ #include <map> #include <set> #include <cmath> #include <stack> #include <queue> #include <vector> #include <cstdio> #include <string> #include <utility> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define eps 1e-8 #define randin srand((unsigned int)time(NULL)) #define input freopen("input.txt","r",stdin) #define debug(s) cout << "s = " << s << endl; #define outstars cout << "*************" << endl; const double PI = acos(-1.0); const double e = exp(1.0); const int inf = 0x3f3f3f3f; const int INF = 0x7fffffff; typedef long long ll; const int maxn = 1010; const int MOD = 1000000007; ll dp[maxn][maxn]; int n, k; void init() { dp[1][0] = 1; dp[1][1] = 0; for (int i = 1; i <= 1000; i ++) { for (int j = 0; j <= 1000; j ++) { dp[i+1][j] = (dp[i+1][j] % MOD + (j + 1) * dp[i][j] % MOD) % MOD; dp[i+1][j+1] = (dp[i+1][j+1] % MOD + (i - j) * dp[i][j] % MOD) % MOD; } } } int main() { init(); while (~scanf("%d %d", &n, &k)) { printf("%d\n", dp[n][k]); } return 0; }
Read The Fucking Source Code ☺!