Your music player contains N
different songs and she wants to listen to L
(not necessarily different) songs during your trip. You create a playlist so that:
- Every song is played at least once
- A song can only be played again only if
K
other songs have been played
Return the number of possible playlists. As the answer can be very large, return it modulo 10^9 + 7
.
Example 1:
Input: N = 3, L = 3, K = 1 Output: 6 Explanation: There are 6 possible playlists. [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1].
Example 2:
Input: N = 2, L = 3, K = 0 Output: 6 Explanation: There are 6 possible playlists. [1, 1, 2], [1, 2, 1], [2, 1, 1], [2, 2, 1], [2, 1, 2], [1, 2, 2]
Example 3:
Input: N = 2, L = 3, K = 1 Output: 2 Explanation: There are 2 possible playlists. [1, 2, 1], [2, 1, 2]
Note:
0 <= K < N <= L <= 100
思路:设i为目前歌单的长度,j为已听歌曲数(不重复的数目),dp[i][j]为排列的方案数,最终求dp[L][N]。
1)若将要听的下一首歌为新歌,那么状态转移方程为:dp[i][j] = dp[i-1][j-1]*(N-(j-1))
2) 若将要听的下一首歌为旧歌,且目前已听歌曲数(不重复的数目)>=k,那么就可以选取旧歌了,此时状态转移方程为:dp[i][j] = dp[i-1][j-1]*(N-(j-1)) + dp[i-1][j]*(j-k)
注意dp[0][0]=1
class Solution {
public:
const long long MOD = (long long) (1e9+7);
int numMusicPlaylists(int N, int L, int K) {
long long dp[104][104] = {};
dp[0][0]=1;
for(int i=1;i<=L;i++){
for(int j=1;j<=N;j++){
dp[i][j] = (dp[i-1][j-1]*(N-(j-1))) % MOD;
if(j>=K){
//注意dp[i][j] += dp[i-1][j]*(j-K)) % MOD 会溢出,还是分开写吧
dp[i][j] = (dp[i][j]+(dp[i-1][j]*(j-K)) % MOD) % MOD;
}
}
}
return (int)dp[L][N];
}
};