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Add solution and test-cases for problem 2438 #1284

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6boris merged 1 commit into 6boris:main from 0xff-dev:2438
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34 changes: 20 additions & 14 deletions leetcode/2401-2500/2438.Range-Product-Queries-of-Powers/README.md
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@@ -1,28 +1,34 @@
# [2438.Range Product Queries of Powers][title]

> [!WARNING|style:flat]
> This question is temporarily unanswered if you have good ideas. Welcome to [Create Pull Request PR](https://github.com/kylesliu/awesome-golang-algorithm)

## Description
Given a positive integer `n`, there exists a **0-indexed** array called `powers`, composed of the **minimum** number of powers of `2` that sum to `n`. The array is sorted in **non-decreasing** order, and there is **only one** way to form the array.

You are also given a **0-indexed** 2D integer array `queries`, where `queries[i] = [lefti, righti]`. Each `queries[i]` represents a query where you have to find the product of all `powers[j]` with `lefti <= j <= righti`.

Return an array `answer`, equal in length to `queries`, where `answers[i]` is the answer to the `ith` query. Since the answer to the `ith` query may be too large, each `answers[i]` should be returned **modulo** `10^9 + 7`.

**Example 1:**

```
Input: a = "11", b = "1"
Output: "100"
Input: n = 15, queries = [[0,1],[2,2],[0,3]]
Output: [2,4,64]
Explanation:
For n = 15, powers = [1,2,4,8]. It can be shown that powers cannot be a smaller size.
Answer to 1st query: powers[0] * powers[1] = 1 * 2 = 2.
Answer to 2nd query: powers[2] = 4.
Answer to 3rd query: powers[0] * powers[1] * powers[2] * powers[3] = 1 * 2 * 4 * 8 = 64.
Each answer modulo 109 + 7 yields the same answer, so [2,4,64] is returned.
```

## 题意
> ...
**Example 2:**

## 题解

### 思路1
> ...
Range Product Queries of Powers
```go
```

Input: n = 2, queries = [[0,0]]
Output: [2]
Explanation:
For n = 2, powers = [2].
The answer to the only query is powers[0] = 2. The answer modulo 109 + 7 is the same, so [2] is returned.
```

## 结语

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package Solution

func Solution(x bool) bool {
return x
import "sort"

func construct(n int, powers [33]int) []int {
if n&(n-1) == 0 {
return []int{n}
}
index := sort.Search(33, func(i int) bool {
return powers[i] > n
})
set := []int{powers[index-1]}
left := construct(n-powers[index-1], powers)
set = append(set, left...)
return set
}

func modPow(x int64, n int64, m int64) int64 {
result := int64(1)
x %= m
for n > 0 {
if n&1 == 1 {
result = (result * x) % m
}
x = (x * x) % m
n >>= 1
}
return result
}

func modInverse(x int64, m int64) int64 {
return modPow(x, m-2, m)
}

const mod = 1000000007

func Solution(n int, queries [][]int) []int {

powers := [33]int{}
powers[0] = 1
for i := 1; i < 33; i++ {
powers[i] = powers[i-1] * 2
}
arr := construct(n, powers)
for s, e := 0, len(arr)-1; s < e; s, e = s+1, e-1 {
arr[s], arr[e] = arr[e], arr[s]
}
ans := make([]int, len(queries))
prefix := make([]int64, len(arr)+1)
index := 0
prefix[index] = 1
for i := 0; i < len(arr); i++ {
prefix[i+1] = (prefix[index] * int64(arr[i])) % mod
index++
}
for i, q := range queries {
l, r := q[0], q[1]
res := (prefix[r+1] * modInverse(prefix[l], mod)) % mod
ans[i] = int(res)
}
return ans
}
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Expand Up @@ -9,31 +9,31 @@ import (
func TestSolution(t *testing.T) {
// 测试用例
cases := []struct {
name string
inputs bool
expect bool
name string
n int
queries [][]int
expect []int
}{
{"TestCase", true, true},
{"TestCase", true, true},
{"TestCase", false, false},
{"TestCase1", 15, [][]int{{0, 1}, {2, 2}, {0, 3}}, []int{2, 4, 64}},
{"TestCase2", 2, [][]int{{0, 0}}, []int{2}},
}

// 开始测试
for i, c := range cases {
t.Run(c.name+" "+strconv.Itoa(i), func(t *testing.T) {
got := Solution(c.inputs)
got := Solution(c.n, c.queries)
if !reflect.DeepEqual(got, c.expect) {
t.Fatalf("expected: %v, but got: %v, with inputs: %v",
c.expect, got, c.inputs)
t.Fatalf("expected: %v, but got: %v, with inputs: %v %v",
c.expect, got, c.n, c.queries)
}
})
}
}

//压力测试
//压力测试
func BenchmarkSolution(b *testing.B) {
}

//使用案列
//使用案列
func ExampleSolution() {
}
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