uva 1485 - Permutation Counting(递推)
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题目链接:uva 1485 - Permutation Counting
题目大意:给定n和k,要求求一个由1~n组成的序列,要求满足ai>i的i刚好有k个的序列种数。
解题思路:dp[j][i]表示长度为i,j个位置满足的情况。
- dp[j+1][i]+=dp[j][i]∗(j+1);
1, (3), (4), 2: 括号位置代表ai>i,既满足位置,此时i = 4, j = 2.
-> 1, (3), (4), 2, 5 1 种,在最后追加
-> 1, (5), (4), 2, 5
1,(3), (5), 2, 4 j 种,与满足的位置交换,因为新追加的数肯定为当前最大的数,所以满足条件的个数不变
- dp[j+1][i+1]+=dp[j][i]∗(i−j);
-> (5), (3), (4), 2, 1
-> 1, (3), (4), (5), 2 i-j 种, 与不满足的位置交换。
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 1000;const ll MOD = 1e9+7;int N, K;ll dp[maxn+5][maxn+5];void init () { memset(dp, 0, sizeof(dp)); dp[0][0] = 1; for (int i = 0; i <= maxn; i++) { for (int j = 0; j <= i; j++) { dp[j][i+1] = (dp[j][i+1] + (j + 1) * dp[j][i]) % MOD; dp[j+1][i+1] = (dp[j+1][i+1] + (i - j) * dp[j][i]) % MOD; } }}int main () { init(); while (scanf("%d%d", &N, &K) == 2) { printf("%lld\n", dp[K][N]); } return 0;}
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