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Commit d469b75

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feat: add solutions to lc problem: No.2787 (doocs#4623)
No.2787.Ways to Express an Integer as Sum of Powers
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‎solution/2700-2799/2787.Ways to Express an Integer as Sum of Powers/README.md‎

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@@ -60,7 +60,24 @@ tags:
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<!-- solution:start -->
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### 方法一
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### 方法一:动态规划
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我们定义 $f[i][j]$ 表示在前 $i$ 个正整数中选取一些数的 $x$ 次幂之和等于 $j$ 的方案数,初始时 $f[0][0] = 1,ドル其余均为 0ドル$。答案为 $f[n][n]$。
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对于每个正整数 $i,ドル我们可以选择不选它或者选它:
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- 不选它:此时方案数为 $f[i-1][j]$;
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- 选它:此时方案数为 $f[i-1][j-i^x]$(前提是 $j \geq i^x$)。
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因此状态转移方程为:
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$$
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f[i][j] = f[i-1][j] + (j \geq i^x ? f[i-1][j-i^x] : 0)
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$$
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注意到答案可能非常大,我们需要对 10ドル^9 + 7$ 取余。
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时间复杂度 $O(n^2),ドル空间复杂度 $O(n^2)$。其中 $n$ 是题目中给定的整数。
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<!-- tabs:start -->
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@@ -155,9 +172,7 @@ func numberOfWays(n int, x int) int {
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```ts
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function numberOfWays(n: number, x: number): number {
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const mod = 10 ** 9 + 7;
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const f: number[][] = Array(n + 1)
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.fill(0)
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.map(() => Array(n + 1).fill(0));
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const f = Array.from({ length: n + 1 }, () => Array(n + 1).fill(0));
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f[0][0] = 1;
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for (let i = 1; i <= n; ++i) {
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const k = Math.pow(i, x);
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn number_of_ways(n: i32, x: i32) -> i32 {
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const MOD: i64 = 1_000_000_007;
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let n = n as usize;
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let x = x as u32;
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let mut f = vec![vec![0; n + 1]; n + 1];
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f[0][0] = 1;
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for i in 1..=n {
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let k = (i as i64).pow(x);
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for j in 0..=n {
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f[i][j] = f[i - 1][j];
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if j >= k as usize {
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f[i][j] = (f[i][j] + f[i - 1][j - k as usize]) % MOD;
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}
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}
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}
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f[n][n] as i32
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/2700-2799/2787.Ways to Express an Integer as Sum of Powers/README_EN.md‎

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@@ -60,7 +60,24 @@ It can be shown that it is the only way to express 10 as the sum of the 2<sup>nd
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Dynamic Programming
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We define $f[i][j]$ as the number of ways to select some numbers from the first $i$ positive integers such that the sum of their $x$-th powers equals $j$. Initially, $f[0][0] = 1,ドル and all others are 0ドル$. The answer is $f[n][n]$.
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For each positive integer $i,ドル we can choose to either include it or not:
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- Not include it: the number of ways is $f[i-1][j]$;
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- Include it: the number of ways is $f[i-1][j-i^x]$ (provided that $j \geq i^x$).
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Therefore, the state transition equation is:
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$$
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f[i][j] = f[i-1][j] + (j \geq i^x ? f[i-1][j-i^x] : 0)
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$$
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Note that the answer can be very large, so we need to take modulo 10ドル^9 + 7$.
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The time complexity is $O(n^2),ドル and the space complexity is $O(n^2),ドル where $n$ is the given integer in the
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<!-- tabs:start -->
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```ts
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function numberOfWays(n: number, x: number): number {
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const mod = 10 ** 9 + 7;
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const f: number[][] = Array(n + 1)
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.fill(0)
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.map(() => Array(n + 1).fill(0));
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const f = Array.from({ length: n + 1 }, () => Array(n + 1).fill(0));
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f[0][0] = 1;
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for (let i = 1; i <= n; ++i) {
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const k = Math.pow(i, x);
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn number_of_ways(n: i32, x: i32) -> i32 {
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const MOD: i64 = 1_000_000_007;
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let n = n as usize;
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let x = x as u32;
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let mut f = vec![vec![0; n + 1]; n + 1];
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f[0][0] = 1;
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for i in 1..=n {
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let k = (i as i64).pow(x);
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for j in 0..=n {
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f[i][j] = f[i - 1][j];
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if j >= k as usize {
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f[i][j] = (f[i][j] + f[i - 1][j - k as usize]) % MOD;
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}
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}
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}
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f[n][n] as i32
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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impl Solution {
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pub fn number_of_ways(n: i32, x: i32) -> i32 {
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const MOD: i64 = 1_000_000_007;
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let n = n as usize;
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let x = x as u32;
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let mut f = vec![vec![0; n + 1]; n + 1];
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f[0][0] = 1;
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for i in 1..=n {
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let k = (i as i64).pow(x);
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for j in 0..=n {
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f[i][j] = f[i - 1][j];
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if j >= k as usize {
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f[i][j] = (f[i][j] + f[i - 1][j - k as usize]) % MOD;
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}
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}
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}
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f[n][n] as i32
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}
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}

‎solution/2700-2799/2787.Ways to Express an Integer as Sum of Powers/Solution.ts‎

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function numberOfWays(n: number, x: number): number {
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const mod = 10 ** 9 + 7;
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const f: number[][] = Array(n + 1)
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.fill(0)
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.map(() => Array(n + 1).fill(0));
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const f = Array.from({ length: n + 1 }, () => Array(n + 1).fill(0));
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f[0][0] = 1;
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for (let i = 1; i <= n; ++i) {
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const k = Math.pow(i, x);

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