LeetCode每日一题(1862. Sum of Floored Pairs)_sum of floor division of all pairs from given arra-优快云博客

本文探讨了一个特定的算法问题：给定一个整数数组，如何计算所有元素对(i, j)的floor(nums[i]/nums[j])之和，并确保结果在10^9+7的模意义下计算。通过分析不同整数比值范围内的取整规律及使用前缀和优化查找过程，给出了一种高效解决方案。

Given an integer array nums, return the sum of floor(nums[i] / nums[j]) for all pairs of indices 0 <= i, j < nums.length in the array. Since the answer may be too large, return it modulo 109 + 7.

The floor() function returns the integer part of the division.

Example 1:

Input: nums = [2,5,9]
Output: 10

Explanation:
floor(2 / 5) = floor(2 / 9) = floor(5 / 9) = 0
floor(2 / 2) = floor(5 / 5) = floor(9 / 9) = 1
floor(5 / 2) = 2
floor(9 / 2) = 4
floor(9 / 5) = 1
We calculate the floor of the division for every pair of indices in the array then sum them up.

Example 2:

Input: nums = [7,7,7,7,7,7,7]
Output: 49

Constraints:

1 <= nums.length <= 105
1 <= nums[i] <= 105

对于 n, m, 如果 m < n, 则 floor(m / n) = 0, 如果 n <= m < 2n, 则 floor(m / n) = 1, 以此类推当 i _ n <= m < (i + 1) _ n 时， floor(m / n) = i。我们再反过来想，我们遍历 i, 查找处于[i _ n, (i+1) _ n)的 m 的数量就可以得到以 n 作为分母部分的答案。我们把所有以 n 为分母的答案加和起来就是最终答案。这里要快速的查出 m 的数量，我们要对 nums 做一个 prefix sum, prefix_sum[(i+1)n - 1] - prefix_sum[in-1]就是我们要的 m 的数量。为了对应大量重复数字的情况，我们对数字进行计数


use std::collections::HashMap;

const MOD: i64 = 1000000007;

impl Solution {
    pub fn sum_of_floored_pairs(nums: Vec<i32>) -> i32 {
        let mut max = 0i64;
        let mut freq = vec![0i64; 100000 * 2 + 1];
        let mut counts = HashMap::new();
        for &n in &nums {
            max = max.max(n as i64);
            freq[n as usize] += 1;
            *counts.entry(n as i64).or_insert(0i64) += 1;
        }
        for i in 1..freq.len() {
            freq[i] += freq[i - 1];
        }
        let mut ans = 0;
        for (n, c) in counts {
            let mut m = 1i64;
            while n * m <= max {
                ans +=
                    (freq[((m + 1) * n - 1) as usize] - freq[(n * m - 1) as usize]) * m * c % MOD;
                m += 1;
            }
        }
        (ans % MOD) as i32
    }
}