LeetCode周赛#105 Q4 Number of Music Playlists (动态规划)

题目来源：https://leetcode.com/contest/weekly-contest-105/problems/number-of-music-playlists/

问题描述

920. Number of Music Playlists

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:

1. 0 <= K < N <= L <= 100

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题意

给定N首不同的歌，要求生成长度为L的歌单，歌单中每首歌放完后要等放过其他K首歌才能再放。求生成的不同的歌单的数量。

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思路

思路参考了清华大学计算机系算协比赛部的思路LeetCode周赛#105题解。动态规划。

dp[i][j] = 用j首不同的歌产生长度为i的歌单的种数

dp[i][j] = dp[i-1][j-1]*(N-(j-1))// 第i首歌不是前(j-1)首歌中的某一首歌

            + dp[i-1][j]*(j-K)  // 第i首歌是前j首歌中的某一首歌，与第(i-K):(i-1)首歌不同

边界条件: dp[0][0] = 1

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代码

class Solution {
public:
    const long long MOD = (long long) (1e9+7);
    int numMusicPlaylists(int N, int L, int K) {
        long long dp[105][105] = {};
        int i, j;
        dp[0][0] = 1;
        for (i=1; i<=L; i++)
        {
            for (j=1; j<=min(N,i); j++)
            {
                dp[i][j] = dp[i-1][j-1]*(N-j+1);
                dp[i][j] %= MOD;
                dp[i][j] += dp[i-1][j]*max(0, j-K);
                dp[i][j] %= MOD;
            }
        }
        return dp[L][N];
    }
};