{ "data": { "question": { "questionId": "3411", "questionFrontendId": "3145", "categoryTitle": "Algorithms", "boundTopicId": 2772487, "title": "Find Products of Elements of Big Array", "titleSlug": "find-products-of-elements-of-big-array", "content": "
A powerful array for an integer x
is the shortest sorted array of powers of two that sum up to x
. For example, the powerful array for 11 is [1, 2, 8]
.
The array big_nums
is created by concatenating the powerful arrays for every positive integer i
in ascending order: 1, 2, 3, and so forth. Thus, big_nums
starts as [1, 2, 1, 2, 4, 1, 4, 2, 4, 1, 2, 4, 8, ...]
.
You are given a 2D integer matrix queries
, where for queries[i] = [fromi, toi, modi]
you should calculate (big_nums[fromi] * big_nums[fromi + 1] * ... * big_nums[toi]) % modi
.
Return an integer array answer
such that answer[i]
is the answer to the ith
query.
\n
Example 1:
\n\nInput: queries = [[1,3,7]]
\n\nOutput: [4]
\n\nExplanation:
\n\nThere is one query.
\n\nbig_nums[1..3] = [2,1,2]
. The product of them is 4. The remainder of 4 under 7 is 4.
Example 2:
\n\nInput: queries = [[2,5,3],[7,7,4]]
\n\nOutput: [2,2]
\n\nExplanation:
\n\nThere are two queries.
\n\nFirst query: big_nums[2..5] = [1,2,4,1]
. The product of them is 8. The remainder of 8 under 3 is 2.
Second query: big_nums[7] = 2
. The remainder of 2 under 4 is 2.
\n
Constraints:
\n\n1 <= queries.length <= 500
queries[i].length == 3
0 <= queries[i][0] <= queries[i][1] <= 1015
1 <= queries[i][2] <= 105
一个整数 x
的 强数组 指的是满足和为 x
的二的幂的最短有序数组。比方说,11 的强数组为 [1, 2, 8]
。
我们将每一个正整数 i
(即1,2,3等等)的 强数组 连接得到数组 big_nums
,big_nums
开始部分为 [1, 2, 1, 2, 4, 1, 4, 2, 4, 1, 2, 4, 8, ...]
。
给你一个二维整数数组 queries
,其中 queries[i] = [fromi, toi, modi]
,你需要计算 (big_nums[fromi] * big_nums[fromi + 1] * ... * big_nums[toi]) % modi
。
请你返回一个整数数组 answer
,其中 answer[i]
是第 i
个查询的答案。
\n\n
示例 1:
\n\n输入:queries = [[1,3,7]]
\n\n输出:[4]
\n\n解释:
\n\n只有一个查询。
\n\nbig_nums[1..3] = [2,1,2]
。它们的乘积为 4 ,4 对 7 取余数得到 4 。
示例 2:
\n\n输入:queries = [[2,5,3],[7,7,4]]
\n\n输出:[2,2]
\n\n解释:
\n\n有两个查询。
\n\n第一个查询:big_nums[2..5] = [1,2,4,1]
。它们的乘积为 8 ,8 对 3 取余数得到 2 。
第二个查询:big_nums[7] = 2
,2 对 4 取余数得到 2 。
\n\n
提示:
\n\n1 <= queries.length <= 500
queries[i].length == 3
0 <= queries[i][0] <= queries[i][1] <= 1015
1 <= queries[i][2] <= 105
f(n, i)
which is the total number of numbers in [1, n]
when the ith
bit is set in O(log(n))
time.",
"Use binary search to find the last number for each query (and there might be one “incomplete” number for the query).",
"Use a similar way to find the product (we only need to save the sum of exponents of power of 2
)."
],
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