diff --git a/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/README.md b/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/README.md index 4e4dbed8681eb..ad49c17f04cfc 100644 --- a/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/README.md +++ b/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/README.md @@ -90,32 +90,142 @@ tags: -### 方法一 +### 方法一：动态规划 + +我们定义 $f[i][j]$ 表示前 $i$ 个表演者安排到 $j$ 个节目的方案数。初始时 $f[0][0] = 1$，其余 $f[i][j] = 0$。 + +对于 $f[i][j]$，其中 $1 \leq i \leq n$, $1 \leq j \leq x$，我们考虑第 $i$ 个表演者： + +- 如果被安排到了一个已经有表演者的节目，一共有 $j$ 种选择，即 $f[i - 1][j] \times j$； +- 如果被安排到了一个没有表演者的节目，一共有 $x - (j - 1)$ 种选择，即 $f[i - 1][j - 1] \times (x - (j - 1))$。 + +所以状态转移方程为： + +$$ +f[i][j] = f[i - 1][j] \times j + f[i - 1][j - 1] \times (x - (j - 1)) +$$ + +对于每个 $j$，一共有 $y^j$ 种选择，所以最终答案为： + +$$ +\sum_{j = 1}^{x} f[n][j] \times y^j +$$ + +注意，由于答案可能很大，我们需要对 $10^9 + 7$ 取模。 + +时间复杂度 $O(n \times x)$，空间复杂度 $O(n \times x)$。其中 $n$ 和 $x$ 分别为表演者的数量和节目的数量。 #### Python3 ```python - +class Solution: + def numberOfWays(self, n: int, x: int, y: int) -> int: + mod = 10**9 + 7 + f = [[0] * (x + 1) for _ in range(n + 1)] + f[0][0] = 1 + for i in range(1, n + 1): + for j in range(1, x + 1): + f[i][j] = (f[i - 1][j] * j + f[i - 1][j - 1] * (x - (j - 1))) % mod + ans, p = 0, 1 + for j in range(1, x + 1): + p = p * y % mod + ans = (ans + f[n][j] * p) % mod + return ans ``` #### Java ```java - +class Solution { + public int numberOfWays(int n, int x, int y) { + final int mod = (int) 1e9 + 7; + long[][] f = new long[n + 1][x + 1]; + f[0][0] = 1; + for (int i = 1; i <= n; ++i) { + for (int j = 1; j <= x; ++j) { + f[i][j] = (f[i - 1][j] * j % mod + f[i - 1][j - 1] * (x - (j - 1) % mod)) % mod; + } + } + long ans = 0, p = 1; + for (int j = 1; j <= x; ++j) { + p = p * y % mod; + ans = (ans + f[n][j] * p) % mod; + } + return (int) ans; + } +} ``` #### C++ ```cpp - +class Solution { +public: + int numberOfWays(int n, int x, int y) { + const int mod = 1e9 + 7; + long long f[n + 1][x + 1]; + memset(f, 0, sizeof(f)); + f[0][0] = 1; + for (int i = 1; i <= n; ++i) { + for (int j = 1; j <= x; ++j) { + f[i][j] = (f[i - 1][j] * j % mod + f[i - 1][j - 1] * (x - (j - 1) % mod)) % mod; + } + } + long long ans = 0, p = 1; + for (int j = 1; j <= x; ++j) { + p = p * y % mod; + ans = (ans + f[n][j] * p) % mod; + } + return ans; + } +}; ``` #### Go ```go +func numberOfWays(n int, x int, y int) int { + const mod int = 1e9 + 7 + f := make([][]int, n+1) + for i := range f { + f[i] = make([]int, x+1) + } + f[0][0] = 1 + for i := 1; i <= n; i++ { + for j := 1; j <= x; j++ { + f[i][j] = (f[i-1][j]*j%mod + f[i-1][j-1]*(x-(j-1))%mod) % mod + } + } + ans, p := 0, 1 + for j := 1; j <= x; j++ { + p = p * y % mod + ans = (ans + f[n][j]*p%mod) % mod + } + return ans +} +``` +#### TypeScript + +```ts +function numberOfWays(n: number, x: number, y: number): number { + const mod = BigInt(10 ** 9 + 7); + const f: bigint[][] = Array.from({ length: n + 1 }, () => Array(x + 1).fill(0n)); + f[0][0] = 1n; + for (let i = 1; i <= n; ++i) { + for (let j = 1; j <= x; ++j) { + f[i][j] = (f[i - 1][j] * BigInt(j) + f[i - 1][j - 1] * BigInt(x - (j - 1))) % mod; + } + } + let [ans, p] = [0n, 1n]; + for (let j = 1; j <= x; ++j) { + p = (p * BigInt(y)) % mod; + ans = (ans + f[n][j] * p) % mod; + } + return Number(ans); +} ``` diff --git a/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/README_EN.md b/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/README_EN.md index 9b2ce3b1998ff..d23cee3b2cfbf 100644 --- a/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/README_EN.md +++ b/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/README_EN.md @@ -87,32 +87,142 @@ tags: -### Solution 1 +### Solution 1: Dynamic Programming + +We define $f[i][j]$ to represent the number of ways to arrange the first $i$ performers into $j$ programs. Initially, $f[0][0] = 1$, and the rest $f[i][j] = 0$. + +For $f[i][j]$, where $1 \leq i \leq n$ and $1 \leq j \leq x$, we consider the $i$-th performer: + +- If the performer is assigned to a program that already has performers, there are $j$ choices, i.e., $f[i - 1][j] \times j$; +- If the performer is assigned to a program that has no performers, there are $x - (j - 1)$ choices, i.e., $f[i - 1][j - 1] \times (x - (j - 1))$. + +So the state transition equation is: + +$$ +f[i][j] = f[i - 1][j] \times j + f[i - 1][j - 1] \times (x - (j - 1)) +$$ + +For each $j$, there are $y^j$ choices, so the final answer is: + +$$ +\sum_{j = 1}^{x} f[n][j] \times y^j +$$ + +Note that since the answer can be very large, we need to take the modulo $10^9 + 7$. + +The time complexity is $O(n \times x)$, and the space complexity is $O(n \times x)$. Here, $n$ and $x$ represent the number of performers and the number of programs, respectively. #### Python3 ```python - +class Solution: + def numberOfWays(self, n: int, x: int, y: int) -> int: + mod = 10**9 + 7 + f = [[0] * (x + 1) for _ in range(n + 1)] + f[0][0] = 1 + for i in range(1, n + 1): + for j in range(1, x + 1): + f[i][j] = (f[i - 1][j] * j + f[i - 1][j - 1] * (x - (j - 1))) % mod + ans, p = 0, 1 + for j in range(1, x + 1): + p = p * y % mod + ans = (ans + f[n][j] * p) % mod + return ans ``` #### Java ```java - +class Solution { + public int numberOfWays(int n, int x, int y) { + final int mod = (int) 1e9 + 7; + long[][] f = new long[n + 1][x + 1]; + f[0][0] = 1; + for (int i = 1; i <= n; ++i) { + for (int j = 1; j <= x; ++j) { + f[i][j] = (f[i - 1][j] * j % mod + f[i - 1][j - 1] * (x - (j - 1) % mod)) % mod; + } + } + long ans = 0, p = 1; + for (int j = 1; j <= x; ++j) { + p = p * y % mod; + ans = (ans + f[n][j] * p) % mod; + } + return (int) ans; + } +} ``` #### C++ ```cpp - +class Solution { +public: + int numberOfWays(int n, int x, int y) { + const int mod = 1e9 + 7; + long long f[n + 1][x + 1]; + memset(f, 0, sizeof(f)); + f[0][0] = 1; + for (int i = 1; i <= n; ++i) { + for (int j = 1; j <= x; ++j) { + f[i][j] = (f[i - 1][j] * j % mod + f[i - 1][j - 1] * (x - (j - 1) % mod)) % mod; + } + } + long long ans = 0, p = 1; + for (int j = 1; j <= x; ++j) { + p = p * y % mod; + ans = (ans + f[n][j] * p) % mod; + } + return ans; + } +}; ``` #### Go ```go +func numberOfWays(n int, x int, y int) int { + const mod int = 1e9 + 7 + f := make([][]int, n+1) + for i := range f { + f[i] = make([]int, x+1) + } + f[0][0] = 1 + for i := 1; i <= n; i++ { + for j := 1; j <= x; j++ { + f[i][j] = (f[i-1][j]*j%mod + f[i-1][j-1]*(x-(j-1))%mod) % mod + } + } + ans, p := 0, 1 + for j := 1; j <= x; j++ { + p = p * y % mod + ans = (ans + f[n][j]*p%mod) % mod + } + return ans +} +``` +#### TypeScript + +```ts +function numberOfWays(n: number, x: number, y: number): number { + const mod = BigInt(10 ** 9 + 7); + const f: bigint[][] = Array.from({ length: n + 1 }, () => Array(x + 1).fill(0n)); + f[0][0] = 1n; + for (let i = 1; i <= n; ++i) { + for (let j = 1; j <= x; ++j) { + f[i][j] = (f[i - 1][j] * BigInt(j) + f[i - 1][j - 1] * BigInt(x - (j - 1))) % mod; + } + } + let [ans, p] = [0n, 1n]; + for (let j = 1; j <= x; ++j) { + p = (p * BigInt(y)) % mod; + ans = (ans + f[n][j] * p) % mod; + } + return Number(ans); +} ``` diff --git a/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/Solution.cpp b/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/Solution.cpp new file mode 100644 index 0000000000000..faf5a763f56a3 --- /dev/null +++ b/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/Solution.cpp @@ -0,0 +1,20 @@ +class Solution { +public: + int numberOfWays(int n, int x, int y) { + const int mod = 1e9 + 7; + long long f[n + 1][x + 1]; + memset(f, 0, sizeof(f)); + f[0][0] = 1; + for (int i = 1; i <= n; ++i) { + for (int j = 1; j <= x; ++j) { + f[i][j] = (f[i - 1][j] * j % mod + f[i - 1][j - 1] * (x - (j - 1) % mod)) % mod; + } + } + long long ans = 0, p = 1; + for (int j = 1; j <= x; ++j) { + p = p * y % mod; + ans = (ans + f[n][j] * p) % mod; + } + return ans; + } +}; diff --git a/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/Solution.go b/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/Solution.go new file mode 100644 index 0000000000000..9744c5bf8c6dc --- /dev/null +++ b/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/Solution.go @@ -0,0 +1,19 @@ +func numberOfWays(n int, x int, y int) int { + const mod int = 1e9 + 7 + f := make([][]int, n+1) + for i := range f { + f[i] = make([]int, x+1) + } + f[0][0] = 1 + for i := 1; i <= n; i++ { + for j := 1; j <= x; j++ { + f[i][j] = (f[i-1][j]*j%mod + f[i-1][j-1]*(x-(j-1))%mod) % mod + } + } + ans, p := 0, 1 + for j := 1; j <= x; j++ { + p = p * y % mod + ans = (ans + f[n][j]*p%mod) % mod + } + return ans +} diff --git a/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/Solution.java b/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/Solution.java new file mode 100644 index 0000000000000..e78bec1754949 --- /dev/null +++ b/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/Solution.java @@ -0,0 +1,18 @@ +class Solution { + public int numberOfWays(int n, int x, int y) { + final int mod = (int) 1e9 + 7; + long[][] f = new long[n + 1][x + 1]; + f[0][0] = 1; + for (int i = 1; i <= n; ++i) { + for (int j = 1; j <= x; ++j) { + f[i][j] = (f[i - 1][j] * j % mod + f[i - 1][j - 1] * (x - (j - 1) % mod)) % mod; + } + } + long ans = 0, p = 1; + for (int j = 1; j <= x; ++j) { + p = p * y % mod; + ans = (ans + f[n][j] * p) % mod; + } + return (int) ans; + } +} diff --git a/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/Solution.py b/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/Solution.py new file mode 100644 index 0000000000000..4f131d2cf3de7 --- /dev/null +++ b/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/Solution.py @@ -0,0 +1,13 @@ +class Solution: + def numberOfWays(self, n: int, x: int, y: int) -> int: + mod = 10**9 + 7 + f = [[0] * (x + 1) for _ in range(n + 1)] + f[0][0] = 1 + for i in range(1, n + 1): + for j in range(1, x + 1): + f[i][j] = (f[i - 1][j] * j + f[i - 1][j - 1] * (x - (j - 1))) % mod + ans, p = 0, 1 + for j in range(1, x + 1): + p = p * y % mod + ans = (ans + f[n][j] * p) % mod + return ans diff --git a/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/Solution.ts b/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/Solution.ts new file mode 100644 index 0000000000000..90ce88d41053b --- /dev/null +++ b/solution/3300-3399/3317.Find the Number of Possible Ways for an Event/Solution.ts @@ -0,0 +1,16 @@ +function numberOfWays(n: number, x: number, y: number): number { + const mod = BigInt(10 ** 9 + 7); + const f: bigint[][] = Array.from({ length: n + 1 }, () => Array(x + 1).fill(0n)); + f[0][0] = 1n; + for (let i = 1; i <= n; ++i) { + for (let j = 1; j <= x; ++j) { + f[i][j] = (f[i - 1][j] * BigInt(j) + f[i - 1][j - 1] * BigInt(x - (j - 1))) % mod; + } + } + let [ans, p] = [0n, 1n]; + for (let j = 1; j <= x; ++j) { + p = (p * BigInt(y)) % mod; + ans = (ans + f[n][j] * p) % mod; + } + return Number(ans); +}