LeetCode每日一题(2438. Range Product Queries of Powers)

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 answers, 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 109 + 7.

Example 1:

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.

Example 2:

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.

Constraints:

• 1 <= n <= 109
• 1 <= queries.length <= 105
• 0 <= starti <= endi < powers.length

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关键点在于 powers 实际就是 n 的二进制表达式过滤掉为 0 的位， 然后把每一位表达成一个 2 的 i 次方整数。 这一步解决了剩下的按部就班的计算每个查询的乘积就可以了

---

impl Solution {
    pub fn product_queries(n: i32, queries: Vec<Vec<i32>>) -> Vec<i32> {
        let s = format!("{:b}", n);
        let l: Vec<usize> = s
            .chars()
            .rev()
            .enumerate()
            .filter(|&(_, c)| c == '1')
            .map(|(i, _)| i)
            .collect();
        queries
            .into_iter()
            .map(|q| {
                l[q[0] as usize..=q[1] as usize]
                    .into_iter()
                    .fold(1, |mut p, &i| {
                        p *= 2i64.pow(i as u32);
                        p %= 1000000007i64;
                        p
                    }) as i32
            })
            .collect()
    }
}

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