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Commit c9f8c43

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feat: add solutions to lc problem: No.1780 (doocs#4645)
No.1780.Check if Number is a Sum of Powers of Three
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‎solution/1700-1799/1780.Check if Number is a Sum of Powers of Three/README.md‎

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我们发现,如果一个数 $n$ 可以表示成若干个"不同的"三的幂之和,那么 $n$ 的三进制表示中,每一位上的数字只能是 0ドル$ 或者 1ドル$。
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因此,我们将 $n$ 转换成三进制,然后判断每一位上的数字是否是 0ドル$ 或者 1ドル$。如果不是,那么 $n$ 就不可以表示成若干个三的幂之和,直接返回 `false`;否则 $n$ 可以表示成若干个三的幂之和,返回 `true`。
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因此,我们将 $n$ 转换成三进制,然后判断每一位上的数字是否是 0ドル$ 或者 1ドル$。如果不是,那么 $n$ 就不可以表示成若干个三的幂之和,直接返回 $\textit{false}$;否则 $n$ 可以表示成若干个三的幂之和,返回 $\textit{true}$。
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时间复杂度 $O(\log_3 n),ドル空间复杂度 $O(1)$。
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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 check_powers_of_three(n: i32) -> bool {
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let mut n = n;
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while n > 0 {
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if n % 3 > 1 {
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return false;
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}
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n /= 3;
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}
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true
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}
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}
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```
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#### C#
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```cs
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public class Solution {
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public bool CheckPowersOfThree(int n) {
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while (n > 0) {
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if (n % 3 > 1) {
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return false;
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}
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n /= 3;
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}
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return true;
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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/1700-1799/1780.Check if Number is a Sum of Powers of Three/README_EN.md‎

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We find that if a number $n$ can be expressed as the sum of several "different" powers of three, then in the ternary representation of $n,ドル each digit can only be 0ドル$ or 1ドル$.
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Therefore, we convert $n$ to ternary and then check whether each digit is 0ドル$ or 1ドル$. If not, then $n$ cannot be expressed as the sum of several powers of three, and we directly return `false`; otherwise, $n$ can be expressed as the sum of several powers of three, and we return `true`.
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Therefore, we convert $n$ to ternary and then check whether each digit is 0ドル$ or 1ドル$. If not, then $n$ cannot be expressed as the sum of several powers of three, and we directly return $\textit{false}$; otherwise, $n$ can be expressed as the sum of several powers of three, and we return $\textit{true}$.
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The time complexity is $O(\log_3 n),ドル and the space complexity is $O(1)$.
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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 check_powers_of_three(n: i32) -> bool {
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let mut n = n;
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while n > 0 {
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if n % 3 > 1 {
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return false;
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}
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n /= 3;
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}
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true
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}
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}
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```
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#### C#
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```cs
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public class Solution {
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public bool CheckPowersOfThree(int n) {
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while (n > 0) {
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if (n % 3 > 1) {
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return false;
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}
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n /= 3;
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}
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return true;
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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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public class Solution {
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public bool CheckPowersOfThree(int n) {
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while (n > 0) {
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if (n % 3 > 1) {
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return false;
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}
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n /= 3;
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}
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return true;
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}
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}
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impl Solution {
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pub fn check_powers_of_three(n: i32) -> bool {
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let mut n = n;
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while n > 0 {
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if n % 3 > 1 {
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return false;
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}
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n /= 3;
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}
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true
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}
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}

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