diff --git a/solution/2000-2099/2081.Sum of k-Mirror Numbers/README.md b/solution/2000-2099/2081.Sum of k-Mirror Numbers/README.md index 53bad43650637..224a410e3d6a1 100644 --- a/solution/2000-2099/2081.Sum of k-Mirror Numbers/README.md +++ b/solution/2000-2099/2081.Sum of k-Mirror Numbers/README.md @@ -85,26 +85,112 @@ tags: -### 方法一 +### 方法一：折半枚举 + 数学 + +对于一个 k 镜像数字，我们可以将其分为两部分：前半部分和后半部分。对于偶数长度的数字，前半部分和后半部分完全相同；对于奇数长度的数字，前半部分和后半部分相同，但中间的数字可以是任意数字。 + +我们可以通过枚举前半部分的数字，然后根据前半部分构造出完整的 k 镜像数字。具体步骤如下： + +1. **枚举长度**：从 1 开始枚举数字的长度，直到找到满足条件的 k 镜像数字。 +2. **计算前半部分的范围**：对于长度为 $l$ 的数字，前半部分的范围是 $[10^{(l-1)/2}, 10^{(l+1)/2})$。 +3. **构造 k 镜像数字**：对于每个前半部分的数字 $i$，如果长度为偶数，则直接将 $i$ 作为前半部分；如果长度为奇数，则将 $i$ 除以 10 得到前半部分。然后将前半部分的数字反转并添加到后半部分，构造出完整的 k 镜像数字。 +4. **检查 k 镜像数字**：将构造出的数字转换为 k 进制，检查其是否是回文数。 +5. **累加结果**：如果是 k 镜像数字，则将其累加到结果中，并减少计数器 $n$。当计数器 $n$ 减至 0 时，返回结果。 + +时间复杂度主要取决于枚举的长度和前半部分的范围。由于 $n$ 的最大值为 30，因此在实际操作中，枚举的次数是有限的。空间复杂度 $O(1)$，因为我们只使用了常数级别的额外空间。 +#### Python3 + +```python +class Solution: + def kMirror(self, k: int, n: int) -> int: + def check(x: int, k: int) -> bool: + s = [] + while x: + s.append(x % k) + x //= k + return s == s[::-1] + + ans = 0 + for l in count(1): + x = 10 ** ((l - 1) // 2) + y = 10 ** ((l + 1) // 2) + for i in range(x, y): + v = i + j = i if l % 2 == 0 else i // 10 + while j > 0: + v = v * 10 + j % 10 + j //= 10 + if check(v, k): + ans += v + n -= 1 + if n == 0: + return ans +``` + #### Java ```java class Solution { public long kMirror(int k, int n) { long ans = 0; - for (int l = 1;; ++l) { + for (int l = 1;; l++) { int x = (int) Math.pow(10, (l - 1) / 2); int y = (int) Math.pow(10, (l + 1) / 2); for (int i = x; i < y; i++) { long v = i; - for (int j = l % 2 == 0 ? i : i / 10; j > 0; j /= 10) { + int j = (l % 2 == 0) ? i : i / 10; + while (j > 0) { + v = v * 10 + j % 10; + j /= 10; + } + if (check(v, k)) { + ans += v; + n--; + if (n == 0) { + return ans; + } + } + } + } + } + + private boolean check(long x, int k) { + List s = new ArrayList<>(); + while (x > 0) { + s.add((int) (x % k)); + x /= k; + } + for (int i = 0, j = s.size() - 1; i < j; ++i, --j) { + if (!s.get(i).equals(s.get(j))) { + return false; + } + } + return true; + } +} +``` + +#### C++ + +```cpp +class Solution { +public: + long long kMirror(int k, int n) { + long long ans = 0; + for (int l = 1;; ++l) { + int x = pow(10, (l - 1) / 2); + int y = pow(10, (l + 1) / 2); + for (int i = x; i < y; ++i) { + long long v = i; + int j = (l % 2 == 0) ? i : i / 10; + while (j > 0) { v = v * 10 + j % 10; + j /= 10; } - String ss = Long.toString(v, k); - if (check(ss.toCharArray())) { + if (check(v, k)) { ans += v; if (--n == 0) { return ans; @@ -114,14 +200,113 @@ class Solution { } } - private boolean check(char[] c) { - for (int i = 0, j = c.length - 1; i < j; i++, j--) { - if (c[i] != c[j]) { +private: + bool check(long long x, int k) { + vector s; + while (x > 0) { + s.push_back(x % k); + x /= k; + } + for (int i = 0, j = s.size() - 1; i < j; ++i, --j) { + if (s[i] != s[j]) { return false; } } return true; } +}; +``` + +#### Go + +```go +func kMirror(k int, n int) int64 { + check := func(x int64, k int) bool { + s := []int{} + for x > 0 { + s = append(s, int(x%int64(k))) + x /= int64(k) + } + for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 { + if s[i] != s[j] { + return false + } + } + return true + } + + var ans int64 = 0 + for l := 1; ; l++ { + x := pow10((l - 1) / 2) + y := pow10((l + 1) / 2) + for i := x; i < y; i++ { + v := int64(i) + j := i + if l%2 != 0 { + j = i / 10 + } + for j > 0 { + v = v*10 + int64(j%10) + j /= 10 + } + if check(v, k) { + ans += v + n-- + if n == 0 { + return ans + } + } + } + } +} + +func pow10(exp int) int { + res := 1 + for i := 0; i < exp; i++ { + res *= 10 + } + return res +} +``` + +#### TypeScript + +```ts +function kMirror(k: number, n: number): number { + function check(x: number, k: number): boolean { + const s: number[] = []; + while (x > 0) { + s.push(x % k); + x = Math.floor(x / k); + } + for (let i = 0, j = s.length - 1; i < j; i++, j--) { + if (s[i] !== s[j]) { + return false; + } + } + return true; + } + + let ans = 0; + for (let l = 1; ; l++) { + const x = Math.pow(10, Math.floor((l - 1) / 2)); + const y = Math.pow(10, Math.floor((l + 1) / 2)); + for (let i = x; i < y; i++) { + let v = i; + let j = l % 2 === 0 ? i : Math.floor(i / 10); + while (j > 0) { + v = v * 10 + (j % 10); + j = Math.floor(j / 10); + } + if (check(v, k)) { + ans += v; + n--; + if (n === 0) { + return ans; + } + } + } + } } ``` diff --git a/solution/2000-2099/2081.Sum of k-Mirror Numbers/README_EN.md b/solution/2000-2099/2081.Sum of k-Mirror Numbers/README_EN.md index ec3ab6bf5b1bd..2238182ec1e96 100644 --- a/solution/2000-2099/2081.Sum of k-Mirror Numbers/README_EN.md +++ b/solution/2000-2099/2081.Sum of k-Mirror Numbers/README_EN.md @@ -86,26 +86,112 @@ Their sum = 1 + 2 + 4 + 8 + 121 + 151 + 212 = 499. -### Solution 1 +### Solution 1: Half Enumeration + Mathematics + +For a k-mirror number, we can divide it into two parts: the first half and the second half. For numbers with even length, the first and second halves are exactly the same; for numbers with odd length, the first and second halves are the same, but the middle digit can be any digit. + +We can enumerate the numbers in the first half, and then construct the complete k-mirror number based on the first half. The specific steps are as follows: + +1. **Enumerate Lengths**: Start enumerating the length of the numbers from 1, until we find enough k-mirror numbers that meet the requirements. +2. **Calculate the Range of the First Half**: For a number of length $l$, the range of the first half is $[10^{(l-1)/2}, 10^{(l+1)/2})$. +3. **Construct k-Mirror Numbers**: For each number $i$ in the first half, if the length is even, use $i$ directly as the first half; if the length is odd, divide $i$ by 10 to get the first half. Then reverse the digits of the first half and append them to form the complete k-mirror number. +4. **Check k-Mirror Numbers**: Convert the constructed number to base $k$ and check whether it is a palindrome. +5. **Accumulate the Result**: If it is a k-mirror number, add it to the result and decrease the counter $n$. When $n$ reaches 0, return the result. + +The time complexity mainly depends on the length being enumerated and the range of the first half. Since the maximum value of $n$ is 30, the number of enumerations is limited in practice. The space complexity is $O(1)$, since only a constant amount of extra space is +#### Python3 + +```python +class Solution: + def kMirror(self, k: int, n: int) -> int: + def check(x: int, k: int) -> bool: + s = [] + while x: + s.append(x % k) + x //= k + return s == s[::-1] + + ans = 0 + for l in count(1): + x = 10 ** ((l - 1) // 2) + y = 10 ** ((l + 1) // 2) + for i in range(x, y): + v = i + j = i if l % 2 == 0 else i // 10 + while j > 0: + v = v * 10 + j % 10 + j //= 10 + if check(v, k): + ans += v + n -= 1 + if n == 0: + return ans +``` + #### Java ```java class Solution { public long kMirror(int k, int n) { long ans = 0; - for (int l = 1;; ++l) { + for (int l = 1;; l++) { int x = (int) Math.pow(10, (l - 1) / 2); int y = (int) Math.pow(10, (l + 1) / 2); for (int i = x; i < y; i++) { long v = i; - for (int j = l % 2 == 0 ? i : i / 10; j > 0; j /= 10) { + int j = (l % 2 == 0) ? i : i / 10; + while (j > 0) { + v = v * 10 + j % 10; + j /= 10; + } + if (check(v, k)) { + ans += v; + n--; + if (n == 0) { + return ans; + } + } + } + } + } + + private boolean check(long x, int k) { + List s = new ArrayList<>(); + while (x > 0) { + s.add((int) (x % k)); + x /= k; + } + for (int i = 0, j = s.size() - 1; i < j; ++i, --j) { + if (!s.get(i).equals(s.get(j))) { + return false; + } + } + return true; + } +} +``` + +#### C++ + +```cpp +class Solution { +public: + long long kMirror(int k, int n) { + long long ans = 0; + for (int l = 1;; ++l) { + int x = pow(10, (l - 1) / 2); + int y = pow(10, (l + 1) / 2); + for (int i = x; i < y; ++i) { + long long v = i; + int j = (l % 2 == 0) ? i : i / 10; + while (j > 0) { v = v * 10 + j % 10; + j /= 10; } - String ss = Long.toString(v, k); - if (check(ss.toCharArray())) { + if (check(v, k)) { ans += v; if (--n == 0) { return ans; @@ -115,14 +201,113 @@ class Solution { } } - private boolean check(char[] c) { - for (int i = 0, j = c.length - 1; i < j; i++, j--) { - if (c[i] != c[j]) { +private: + bool check(long long x, int k) { + vector s; + while (x > 0) { + s.push_back(x % k); + x /= k; + } + for (int i = 0, j = s.size() - 1; i < j; ++i, --j) { + if (s[i] != s[j]) { return false; } } return true; } +}; +``` + +#### Go + +```go +func kMirror(k int, n int) int64 { + check := func(x int64, k int) bool { + s := []int{} + for x > 0 { + s = append(s, int(x%int64(k))) + x /= int64(k) + } + for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 { + if s[i] != s[j] { + return false + } + } + return true + } + + var ans int64 = 0 + for l := 1; ; l++ { + x := pow10((l - 1) / 2) + y := pow10((l + 1) / 2) + for i := x; i < y; i++ { + v := int64(i) + j := i + if l%2 != 0 { + j = i / 10 + } + for j > 0 { + v = v*10 + int64(j%10) + j /= 10 + } + if check(v, k) { + ans += v + n-- + if n == 0 { + return ans + } + } + } + } +} + +func pow10(exp int) int { + res := 1 + for i := 0; i < exp; i++ { + res *= 10 + } + return res +} +``` + +#### TypeScript + +```ts +function kMirror(k: number, n: number): number { + function check(x: number, k: number): boolean { + const s: number[] = []; + while (x > 0) { + s.push(x % k); + x = Math.floor(x / k); + } + for (let i = 0, j = s.length - 1; i < j; i++, j--) { + if (s[i] !== s[j]) { + return false; + } + } + return true; + } + + let ans = 0; + for (let l = 1; ; l++) { + const x = Math.pow(10, Math.floor((l - 1) / 2)); + const y = Math.pow(10, Math.floor((l + 1) / 2)); + for (let i = x; i < y; i++) { + let v = i; + let j = l % 2 === 0 ? i : Math.floor(i / 10); + while (j > 0) { + v = v * 10 + (j % 10); + j = Math.floor(j / 10); + } + if (check(v, k)) { + ans += v; + n--; + if (n === 0) { + return ans; + } + } + } + } } ``` diff --git a/solution/2000-2099/2081.Sum of k-Mirror Numbers/Solution.cpp b/solution/2000-2099/2081.Sum of k-Mirror Numbers/Solution.cpp new file mode 100644 index 0000000000000..f0378bdb0147e --- /dev/null +++ b/solution/2000-2099/2081.Sum of k-Mirror Numbers/Solution.cpp @@ -0,0 +1,39 @@ +class Solution { +public: + long long kMirror(int k, int n) { + long long ans = 0; + for (int l = 1;; ++l) { + int x = pow(10, (l - 1) / 2); + int y = pow(10, (l + 1) / 2); + for (int i = x; i < y; ++i) { + long long v = i; + int j = (l % 2 == 0) ? i : i / 10; + while (j > 0) { + v = v * 10 + j % 10; + j /= 10; + } + if (check(v, k)) { + ans += v; + if (--n == 0) { + return ans; + } + } + } + } + } + +private: + bool check(long long x, int k) { + vector s; + while (x > 0) { + s.push_back(x % k); + x /= k; + } + for (int i = 0, j = s.size() - 1; i < j; ++i, --j) { + if (s[i] != s[j]) { + return false; + } + } + return true; + } +}; diff --git a/solution/2000-2099/2081.Sum of k-Mirror Numbers/Solution.go b/solution/2000-2099/2081.Sum of k-Mirror Numbers/Solution.go new file mode 100644 index 0000000000000..b99c8b2fea28d --- /dev/null +++ b/solution/2000-2099/2081.Sum of k-Mirror Numbers/Solution.go @@ -0,0 +1,47 @@ +func kMirror(k int, n int) int64 { + check := func(x int64, k int) bool { + s := []int{} + for x > 0 { + s = append(s, int(x%int64(k))) + x /= int64(k) + } + for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 { + if s[i] != s[j] { + return false + } + } + return true + } + + var ans int64 = 0 + for l := 1; ; l++ { + x := pow10((l - 1) / 2) + y := pow10((l + 1) / 2) + for i := x; i < y; i++ { + v := int64(i) + j := i + if l%2 != 0 { + j = i / 10 + } + for j > 0 { + v = v*10 + int64(j%10) + j /= 10 + } + if check(v, k) { + ans += v + n-- + if n == 0 { + return ans + } + } + } + } +} + +func pow10(exp int) int { + res := 1 + for i := 0; i < exp; i++ { + res *= 10 + } + return res +} diff --git a/solution/2000-2099/2081.Sum of k-Mirror Numbers/Solution.java b/solution/2000-2099/2081.Sum of k-Mirror Numbers/Solution.java index c08776873b60b..f9dae28dfd03c 100644 --- a/solution/2000-2099/2081.Sum of k-Mirror Numbers/Solution.java +++ b/solution/2000-2099/2081.Sum of k-Mirror Numbers/Solution.java @@ -1,18 +1,20 @@ class Solution { public long kMirror(int k, int n) { long ans = 0; - for (int l = 1;; ++l) { + for (int l = 1;; l++) { int x = (int) Math.pow(10, (l - 1) / 2); int y = (int) Math.pow(10, (l + 1) / 2); for (int i = x; i < y; i++) { long v = i; - for (int j = l % 2 == 0 ? i : i / 10; j > 0; j /= 10) { + int j = (l % 2 == 0) ? i : i / 10; + while (j > 0) { v = v * 10 + j % 10; + j /= 10; } - String ss = Long.toString(v, k); - if (check(ss.toCharArray())) { + if (check(v, k)) { ans += v; - if (--n == 0) { + n--; + if (n == 0) { return ans; } } @@ -20,12 +22,17 @@ public long kMirror(int k, int n) { } } - private boolean check(char[] c) { - for (int i = 0, j = c.length - 1; i < j; i++, j--) { - if (c[i] != c[j]) { + private boolean check(long x, int k) { + List s = new ArrayList<>(); + while (x > 0) { + s.add((int) (x % k)); + x /= k; + } + for (int i = 0, j = s.size() - 1; i < j; ++i, --j) { + if (!s.get(i).equals(s.get(j))) { return false; } } return true; } -} \ No newline at end of file +} diff --git a/solution/2000-2099/2081.Sum of k-Mirror Numbers/Solution.py b/solution/2000-2099/2081.Sum of k-Mirror Numbers/Solution.py new file mode 100644 index 0000000000000..df18cf29d7316 --- /dev/null +++ b/solution/2000-2099/2081.Sum of k-Mirror Numbers/Solution.py @@ -0,0 +1,24 @@ +class Solution: + def kMirror(self, k: int, n: int) -> int: + def check(x: int, k: int) -> bool: + s = [] + while x: + s.append(x % k) + x //= k + return s == s[::-1] + + ans = 0 + for l in count(1): + x = 10 ** ((l - 1) // 2) + y = 10 ** ((l + 1) // 2) + for i in range(x, y): + v = i + j = i if l % 2 == 0 else i // 10 + while j > 0: + v = v * 10 + j % 10 + j //= 10 + if check(v, k): + ans += v + n -= 1 + if n == 0: + return ans diff --git a/solution/2000-2099/2081.Sum of k-Mirror Numbers/Solution.ts b/solution/2000-2099/2081.Sum of k-Mirror Numbers/Solution.ts new file mode 100644 index 0000000000000..4d5e04035acec --- /dev/null +++ b/solution/2000-2099/2081.Sum of k-Mirror Numbers/Solution.ts @@ -0,0 +1,36 @@ +function kMirror(k: number, n: number): number { + function check(x: number, k: number): boolean { + const s: number[] = []; + while (x > 0) { + s.push(x % k); + x = Math.floor(x / k); + } + for (let i = 0, j = s.length - 1; i < j; i++, j--) { + if (s[i] !== s[j]) { + return false; + } + } + return true; + } + + let ans = 0; + for (let l = 1; ; l++) { + const x = Math.pow(10, Math.floor((l - 1) / 2)); + const y = Math.pow(10, Math.floor((l + 1) / 2)); + for (let i = x; i < y; i++) { + let v = i; + let j = l % 2 === 0 ? i : Math.floor(i / 10); + while (j > 0) { + v = v * 10 + (j % 10); + j = Math.floor(j / 10); + } + if (check(v, k)) { + ans += v; + n--; + if (n === 0) { + return ans; + } + } + } + } +}