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feat: add solutions to lc problem: No.1969 (doocs#1269)
No.1969.Minimum Non-Zero Product of the Array Elements
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‎solution/1900-1999/1969.Minimum Non-Zero Product of the Array Elements/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：贪心 + 快速幂**
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我们注意到，每一次操作，并不会改变元素的和，而在元素和不变的情况下，要想使得乘积最小，应该尽可能最大化元素的差值。
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由于最大的元素为 $2^p - 1$，无论与哪个元素交换，都不会使得差值变大，因此我们不需要考虑与最大元素交换的情况。
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对于其它的 $[1,..2^p-2]$ 的元素，我们依次将首尾元素两两配对，即 $x$ 与 $2^p-1-x$ 进行配置，那么经过若干次操作过后，每一对元素都变成了 $(1, 2^p-2)$，那么最终的乘积为 $(2^p-1) \times (2^p-2)^{2^{p-1}-1}$。
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时间复杂度 $O(p)$，空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def minNonZeroProduct(self, p: int) -> int:
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mod = 10**9 + 7
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return (2**p - 1) * pow(2**p - 2, 2 ** (p - 1) - 1, mod) % mod
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int minNonZeroProduct(int p) {
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final int mod = (int) 1e9 + 7;
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long a = ((1L << p) - 1) % mod;
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long b = qmi(((1L << p) - 2) % mod, (1L << (p - 1)) - 1, mod);
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return (int) (a * b % mod);
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}
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long qmi(long a, long k, long p) {
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long res = 1;
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while (k != 0) {
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if ((k & 1) == 1) {
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res = res * a % p;
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}
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k >>= 1;
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a = a * a % p;
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}
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return res;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int minNonZeroProduct(int p) {
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const int mod = 1e9 + 7;
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long long a = ((1LL << p) - 1) % mod;
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long long b = qmi(((1LL << p) - 2) % mod, (1L << (p - 1)) - 1, mod);
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return a * b % mod;
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}
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long long qmi(long long a, long long k, int p) {
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long long res = 1;
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while (k != 0) {
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if ((k & 1) == 1) {
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res = res * a % p;
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}
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k >>= 1;
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a = a * a % p;
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}
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return res;
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}
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};
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```
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### **Go**
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```go
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func minNonZeroProduct(p int) int {
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const mod int = 1e9 + 7
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a := ((1 << p) - 1) % mod
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b := qmi(((1<<p)-2)%mod, (1<<(p-1))-1, mod)
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return a * b % mod
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}
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func qmi(a, k, p int) int {
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res := 1
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for k != 0 {
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if k&1 == 1 {
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res = res * a % p
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}
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k >>= 1
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a = a * a % p
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}
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return res
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}
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```
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### **...**

‎solution/1900-1999/1969.Minimum Non-Zero Product of the Array Elements/README_EN.md

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### **Python3**
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```python
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class Solution:
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def minNonZeroProduct(self, p: int) -> int:
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mod = 10**9 + 7
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return (2**p - 1) * pow(2**p - 2, 2 ** (p - 1) - 1, mod) % mod
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```
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### **Java**
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```java
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class Solution {
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public int minNonZeroProduct(int p) {
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final int mod = (int) 1e9 + 7;
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long a = ((1L << p) - 1) % mod;
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long b = qmi(((1L << p) - 2) % mod, (1L << (p - 1)) - 1, mod);
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return (int) (a * b % mod);
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}
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long qmi(long a, long k, long p) {
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long res = 1;
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while (k != 0) {
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if ((k & 1) == 1) {
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res = res * a % p;
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}
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k >>= 1;
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a = a * a % p;
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}
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return res;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int minNonZeroProduct(int p) {
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const int mod = 1e9 + 7;
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long long a = ((1LL << p) - 1) % mod;
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long long b = qmi(((1LL << p) - 2) % mod, (1L << (p - 1)) - 1, mod);
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return a * b % mod;
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}
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long long qmi(long long a, long long k, int p) {
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long long res = 1;
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while (k != 0) {
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if ((k & 1) == 1) {
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res = res * a % p;
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}
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k >>= 1;
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a = a * a % p;
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}
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return res;
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}
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};
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```
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### **Go**
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```go
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func minNonZeroProduct(p int) int {
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const mod int = 1e9 + 7
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a := ((1 << p) - 1) % mod
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b := qmi(((1<<p)-2)%mod, (1<<(p-1))-1, mod)
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return a * b % mod
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}
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func qmi(a, k, p int) int {
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res := 1
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for k != 0 {
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if k&1 == 1 {
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res = res * a % p
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}
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k >>= 1
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a = a * a % p
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}
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return res
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}
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```
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### **...**

‎solution/1900-1999/1969.Minimum Non-Zero Product of the Array Elements/Solution.cpp

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class Solution {
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public:
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int minNonZeroProduct(int p) {
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const int mod = 1e9 + 7;
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long long a = ((1LL << p) - 1) % mod;
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long long b = qmi(((1LL << p) - 2) % mod, (1L << (p - 1)) - 1, mod);
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return a * b % mod;
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}
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long long qmi(long long a, long long k, int p) {
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long long res = 1;
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while (k != 0) {
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if ((k & 1) == 1) {
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res = res * a % p;
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}
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k >>= 1;
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a = a * a % p;
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}
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return res;
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}
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};

‎solution/1900-1999/1969.Minimum Non-Zero Product of the Array Elements/Solution.go

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func minNonZeroProduct(p int) int {
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const mod int = 1e9 + 7
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a := ((1 << p) - 1) % mod
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b := qmi(((1<<p)-2)%mod, (1<<(p-1))-1, mod)
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return a * b % mod
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}
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func qmi(a, k, p int) int {
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res := 1
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for k != 0 {
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if k&1 == 1 {
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res = res * a % p
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}
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k >>= 1
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a = a * a % p
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}
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return res
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}

‎solution/1900-1999/1969.Minimum Non-Zero Product of the Array Elements/Solution.java

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class Solution {
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public int minNonZeroProduct(int p) {
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final int mod = (int) 1e9 + 7;
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long a = ((1L << p) - 1) % mod;
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long b = qmi(((1L << p) - 2) % mod, (1L << (p - 1)) - 1, mod);
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return (int) (a * b % mod);
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}
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long qmi(long a, long k, long p) {
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long res = 1;
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while (k != 0) {
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if ((k & 1) == 1) {
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res = res * a % p;
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}
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k >>= 1;
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a = a * a % p;
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}
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return res;
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}
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}

‎solution/1900-1999/1969.Minimum Non-Zero Product of the Array Elements/Solution.py

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class Solution:
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def minNonZeroProduct(self, p: int) -> int:
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mod = 10**9 + 7
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return (2**p - 1) * pow(2**p - 2, 2 ** (p - 1) - 1, mod) % mod

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