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feat: add solutions to lc problem: No.1922 (doocs#4347)
No.1922.Count Good Numbers
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‎solution/1900-1999/1922.Count Good Numbers/README.md‎

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@@ -67,7 +67,15 @@ tags:
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<!-- solution:start -->
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### 方法一
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### 方法一：快速幂
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长度为 $n$ 的好数字，偶数下标一共有 $\lceil \frac{n}{2} \rceil = \lfloor \frac{n + 1}{2} \rfloor$ 位，偶数下标可以填入 $5$ 种数字（$0, 2, 4, 6, 8$）；奇数下标一共有 $\lfloor \frac{n}{2} \rfloor$ 位，奇数下标可以填入 $4$ 种数字（$2, 3, 5, 7$）。因此长度为 $n$ 的好数字的个数为：
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$$
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ans = 5^{\lceil \frac{n}{2} \rceil} \times 4^{\lfloor \frac{n}{2} \rfloor}
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$$
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我们可以使用快速幂来计算 $5^{\lceil \frac{n}{2} \rceil}$ 和 $4^{\lfloor \frac{n}{2} \rfloor}$，时间复杂度为 $O(\log n)$，空间复杂度为 $O(1)$。
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<!-- tabs:start -->
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@@ -84,13 +92,13 @@ class Solution:
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```java
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class Solution {
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private int mod = 1000000007;
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private final int mod = (int) 1e9 + 7;
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public int countGoodNumbers(long n) {
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return (int) (myPow(5, (n + 1) >> 1) * myPow(4, n >> 1) % mod);
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return (int) (qpow(5, (n + 1) >> 1) * qpow(4, n >> 1) % mod);
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}
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private long myPow(long x, long n) {
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private long qpow(long x, long n) {
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long res = 1;
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while (n != 0) {
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if ((n & 1) == 1) {
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#### C++
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```cpp
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int MOD = 1000000007;
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class Solution {
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public:
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int countGoodNumbers(long long n) {
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return (int) (myPow(5, (n + 1) >> 1) * myPow(4, n >> 1) % MOD);
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}
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private:
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long long myPow(long long x, long long n) {
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long long res = 1;
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while (n) {
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if ((n & 1) == 1) {
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res = res * x % MOD;
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const int mod = 1e9 + 7;
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auto qpow = [](long long x, long long n) -> long long {
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long long res = 1;
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while (n) {
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if ((n & 1) == 1) {
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res = res * x % mod;
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}
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x = x * x % mod;
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n >>= 1;
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}
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x = x * x % MOD;
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n >>= 1;
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}
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return res;
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return res;
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};
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return qpow(5, (n + 1) >> 1) * qpow(4, n >> 1) % mod;
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}
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};
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```
@@ -152,6 +157,28 @@ func myPow(x, n int64) int64 {
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}
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```
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#### TypeScript
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```ts
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function countGoodNumbers(n: number): number {
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const mod = 1000000007n;
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const qpow = (x: bigint, n: bigint): bigint => {
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let res = 1n;
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while (n > 0n) {
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if (n & 1n) {
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res = (res * x) % mod;
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}
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x = (x * x) % mod;
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n >>= 1n;
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}
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return res;
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};
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const a = qpow(5n, BigInt(n + 1) / 2n);
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const b = qpow(4n, BigInt(n) / 2n);
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return Number((a * b) % mod);
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/1900-1999/1922.Count Good Numbers/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Fast Exponentiation
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For a "good number" of length $n$, the even-indexed positions have $\lceil \frac{n}{2} \rceil = \lfloor \frac{n + 1}{2} \rfloor$ digits, and these positions can be filled with $5$ different digits ($0, 2, 4, 6, 8$). The odd-indexed positions have $\lfloor \frac{n}{2} \rfloor$ digits, and these positions can be filled with $4$ different digits ($2, 3, 5, 7$). Therefore, the total number of "good numbers" of length $n$ is:
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$$
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ans = 5^{\lceil \frac{n}{2} \rceil} \times 4^{\lfloor \frac{n}{2} \rfloor}
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$$
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We can use fast exponentiation to compute $5^{\lceil \frac{n}{2} \rceil}$ and $4^{\lfloor \frac{n}{2} \rfloor}$. The time complexity is $O(\log n)$, and the space complexity is $O(1)$.
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<!-- tabs:start -->
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```java
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class Solution {
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private int mod = 1000000007;
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private final int mod = (int) 1e9 + 7;
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public int countGoodNumbers(long n) {
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return (int) (myPow(5, (n + 1) >> 1) * myPow(4, n >> 1) % mod);
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return (int) (qpow(5, (n + 1) >> 1) * qpow(4, n >> 1) % mod);
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}
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private long myPow(long x, long n) {
99+
private long qpow(long x, long n) {
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long res = 1;
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while (n != 0) {
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if ((n & 1) == 1) {
@@ -105,25 +113,22 @@ class Solution {
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#### C++
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```cpp
108-
int MOD = 1000000007;
109-
110116
class Solution {
111117
public:
112118
int countGoodNumbers(long long n) {
113-
return (int) (myPow(5, (n + 1) >> 1) * myPow(4, n >> 1) % MOD);
114-
}
115-
116-
private:
117-
long long myPow(long long x, long long n) {
118-
long long res = 1;
119-
while (n) {
120-
if ((n & 1) == 1) {
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res = res * x % MOD;
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const int mod = 1e9 + 7;
120+
auto qpow = [](long long x, long long n) -> long long {
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long long res = 1;
122+
while (n) {
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if ((n & 1) == 1) {
124+
res = res * x % mod;
125+
}
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x = x * x % mod;
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n >>= 1;
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}
123-
x = x * x % MOD;
124-
n >>= 1;
125-
}
126-
return res;
129+
return res;
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};
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return qpow(5, (n + 1) >> 1) * qpow(4, n >> 1) % mod;
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}
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};
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```
@@ -150,6 +155,28 @@ func myPow(x, n int64) int64 {
150155
}
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```
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#### TypeScript
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```ts
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function countGoodNumbers(n: number): number {
162+
const mod = 1000000007n;
163+
const qpow = (x: bigint, n: bigint): bigint => {
164+
let res = 1n;
165+
while (n > 0n) {
166+
if (n & 1n) {
167+
res = (res * x) % mod;
168+
}
169+
x = (x * x) % mod;
170+
n >>= 1n;
171+
}
172+
return res;
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};
174+
const a = qpow(5n, BigInt(n + 1) / 2n);
175+
const b = qpow(4n, BigInt(n) / 2n);
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return Number((a * b) % mod);
177+
}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/1900-1999/1922.Count Good Numbers/Solution.cpp‎

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Original file line numberDiff line numberDiff line change
@@ -1,21 +1,18 @@
1-
int MOD = 1000000007;
2-
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class Solution {
42
public:
53
int countGoodNumbers(long long n) {
6-
return (int) (myPow(5, (n + 1) >> 1) * myPow(4, n >> 1) % MOD);
7-
}
8-
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private:
10-
long long myPow(long long x, long long n) {
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long long res = 1;
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while (n) {
13-
if ((n & 1) == 1) {
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res = res * x % MOD;
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const int mod = 1e9 + 7;
5+
auto qpow = [](long long x, long long n) -> long long {
6+
long long res = 1;
7+
while (n) {
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if ((n & 1) == 1) {
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res = res * x % mod;
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}
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x = x * x % mod;
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n >>= 1;
1513
}
16-
x = x * x % MOD;
17-
n >>= 1;
18-
}
19-
return res;
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return res;
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};
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return qpow(5, (n + 1) >> 1) * qpow(4, n >> 1) % mod;
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}
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};

‎solution/1900-1999/1922.Count Good Numbers/Solution.java‎

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class Solution {
2-
private int mod = 1000000007;
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private final int mod = (int) 1e9 + 7;
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public int countGoodNumbers(long n) {
5-
return (int) (myPow(5, (n + 1) >> 1) * myPow(4, n >> 1) % mod);
5+
return (int) (qpow(5, (n + 1) >> 1) * qpow(4, n >> 1) % mod);
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}
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8-
private long myPow(long x, long n) {
8+
private long qpow(long x, long n) {
99
long res = 1;
1010
while (n != 0) {
1111
if ((n & 1) == 1) {

‎solution/1900-1999/1922.Count Good Numbers/Solution.ts‎

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,17 @@
1+
function countGoodNumbers(n: number): number {
2+
const mod = 1000000007n;
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const qpow = (x: bigint, n: bigint): bigint => {
4+
let res = 1n;
5+
while (n > 0n) {
6+
if (n & 1n) {
7+
res = (res * x) % mod;
8+
}
9+
x = (x * x) % mod;
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n >>= 1n;
11+
}
12+
return res;
13+
};
14+
const a = qpow(5n, BigInt(n + 1) / 2n);
15+
const b = qpow(4n, BigInt(n) / 2n);
16+
return Number((a * b) % mod);
17+
}

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