Question

1155. 掷骰子等于目标和的方法数

English Version

题目描述

这里有 n 个一样的骰子，每个骰子上都有 k 个面，分别标号为 1 到 k 。

给定三个整数 n、k 和 target，请返回投掷骰子的所有可能得到的结果（共有 kn 种方式），使得骰子面朝上的数字总和等于 target。

由于答案可能很大，你需要对 109 + 7 取模。

 

示例 1：

输入：n = 1, k = 6, target = 3
输出：1
解释：你掷了一个有 6 个面的骰子。
得到总和为 3 的结果的方式只有一种。

示例 2：

输入：n = 2, k = 6, target = 7
输出：6
解释：你掷了两个骰子，每个骰子有 6 个面。
有 6 种方式得到总和为 7 的结果: 1+6, 2+5, 3+4, 4+3, 5+2, 6+1。

示例 3：

输入：n = 30, k = 30, target = 500
输出：222616187
解释：返回的结果必须对 109 + 7 取模。

 

提示：

• 1 <= n, k <= 30
• 1 <= target <= 1000

解法

方法一：动态规划

我们定义 $f[i][j]$ 表示使用 $i$ 个骰子，和为 $j$ 的方案数。那么我们可以得到状态转移方程：

$$ f[i][j] = \sum_{h=1}^{\min(j, k)} f[i-1][j-h] $$

其中 $h$ 表示第 $i$ 个骰子的点数。

初始时 $f[0][0] = 1$，最终的答案即为 $f[n][target]$。

时间复杂度 $O(n \times k \times target)$，空间复杂度 $O(n \times target)$。

我们注意到，状态 $f[i][j]$ 只和 $f[i-1][]$ 有关，因此我们可以使用滚动数组的方式，将空间复杂度优化到 $O(target)$。

class Solution:
  def numRollsToTarget(self, n: int, k: int, target: int) -> int:
    f = [[0] * (target + 1) for _ in range(n + 1)]
    f[0][0] = 1
    mod = 10**9 + 7
    for i in range(1, n + 1):
      for j in range(1, min(i * k, target) + 1):
        for h in range(1, min(j, k) + 1):
          f[i][j] = (f[i][j] + f[i - 1][j - h]) % mod
    return f[n][target]

class Solution {
  public int numRollsToTarget(int n, int k, int target) {
    final int mod = (int) 1e9 + 7;
    int[][] f = new int[n + 1][target + 1];
    f[0][0] = 1;
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= Math.min(target, i * k); ++j) {
        for (int h = 1; h <= Math.min(j, k); ++h) {
          f[i][j] = (f[i][j] + f[i - 1][j - h]) % mod;
        }
      }
    }
    return f[n][target];
  }
}

class Solution {
public:
  int numRollsToTarget(int n, int k, int target) {
    const int mod = 1e9 + 7;
    int f[n + 1][target + 1];
    memset(f, 0, sizeof f);
    f[0][0] = 1;
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= min(target, i * k); ++j) {
        for (int h = 1; h <= min(j, k); ++h) {
          f[i][j] = (f[i][j] + f[i - 1][j - h]) % mod;
        }
      }
    }
    return f[n][target];
  }
};

func numRollsToTarget(n int, k int, target int) int {
  const mod int = 1e9 + 7
  f := make([][]int, n+1)
  for i := range f {
    f[i] = make([]int, target+1)
  }
  f[0][0] = 1
  for i := 1; i <= n; i++ {
    for j := 1; j <= min(target, i*k); j++ {
      for h := 1; h <= min(j, k); h++ {
        f[i][j] = (f[i][j] + f[i-1][j-h]) % mod
      }
    }
  }
  return f[n][target]
}

function numRollsToTarget(n: number, k: number, target: number): number {
  const f = Array.from({ length: n + 1 }, () => Array(target + 1).fill(0));
  f[0][0] = 1;
  const mod = 1e9 + 7;
  for (let i = 1; i <= n; ++i) {
    for (let j = 1; j <= Math.min(i * k, target); ++j) {
      for (let h = 1; h <= Math.min(j, k); ++h) {
        f[i][j] = (f[i][j] + f[i - 1][j - h]) % mod;
      }
    }
  }
  return f[n][target];
}

impl Solution {
  pub fn num_rolls_to_target(n: i32, k: i32, target: i32) -> i32 {
    let _mod = 1_000_000_007;
    let n = n as usize;
    let k = k as usize;
    let target = target as usize;
    let mut f = vec![vec![0; target + 1]; n + 1];
    f[0][0] = 1;

    for i in 1..=n {
      for j in 1..=target.min(i * k) {
        for h in 1..=j.min(k) {
          f[i][j] = (f[i][j] + f[i - 1][j - h]) % _mod;
        }
      }
    }

    f[n][target]
  }
}

方法二

class Solution:
  def numRollsToTarget(self, n: int, k: int, target: int) -> int:
    f = [1] + [0] * target
    mod = 10**9 + 7
    for i in range(1, n + 1):
      g = [0] * (target + 1)
      for j in range(1, min(i * k, target) + 1):
        for h in range(1, min(j, k) + 1):
          g[j] = (g[j] + f[j - h]) % mod
      f = g
    return f[target]

class Solution {
  public int numRollsToTarget(int n, int k, int target) {
    final int mod = (int) 1e9 + 7;
    int[] f = new int[target + 1];
    f[0] = 1;
    for (int i = 1; i <= n; ++i) {
      int[] g = new int[target + 1];
      for (int j = 1; j <= Math.min(target, i * k); ++j) {
        for (int h = 1; h <= Math.min(j, k); ++h) {
          g[j] = (g[j] + f[j - h]) % mod;
        }
      }
      f = g;
    }
    return f[target];
  }
}

class Solution {
public:
  int numRollsToTarget(int n, int k, int target) {
    const int mod = 1e9 + 7;
    vector<int> f(target + 1);
    f[0] = 1;
    for (int i = 1; i <= n; ++i) {
      vector<int> g(target + 1);
      for (int j = 1; j <= min(target, i * k); ++j) {
        for (int h = 1; h <= min(j, k); ++h) {
          g[j] = (g[j] + f[j - h]) % mod;
        }
      }
      f = move(g);
    }
    return f[target];
  }
};

func numRollsToTarget(n int, k int, target int) int {
  const mod int = 1e9 + 7
  f := make([]int, target+1)
  f[0] = 1
  for i := 1; i <= n; i++ {
    g := make([]int, target+1)
    for j := 1; j <= min(target, i*k); j++ {
      for h := 1; h <= min(j, k); h++ {
        g[j] = (g[j] + f[j-h]) % mod
      }
    }
    f = g
  }
  return f[target]
}

function numRollsToTarget(n: number, k: number, target: number): number {
  const f = Array(target + 1).fill(0);
  f[0] = 1;
  const mod = 1e9 + 7;
  for (let i = 1; i <= n; ++i) {
    const g = Array(target + 1).fill(0);
    for (let j = 1; j <= Math.min(i * k, target); ++j) {
      for (let h = 1; h <= Math.min(j, k); ++h) {
        g[j] = (g[j] + f[j - h]) % mod;
      }
    }
    f.splice(0, target + 1, ...g);
  }
  return f[target];
}

impl Solution {
  pub fn num_rolls_to_target(n: i32, k: i32, target: i32) -> i32 {
    let _mod = 1_000_000_007;
    let n = n as usize;
    let k = k as usize;
    let target = target as usize;
    let mut f = vec![0; target + 1];
    f[0] = 1;

    for i in 1..=n {
      let mut g = vec![0; target + 1];
      for j in 1..=target {
        for h in 1..=j.min(k) {
          g[j] = (g[j] + f[j - h]) % _mod;
        }
      }
      f = g;
    }

    f[target]
  }
}