69 lines
2.0 KiB
C++
69 lines
2.0 KiB
C++
// Source : https://leetcode.com/problems/sum-of-floored-pairs/
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// Author : Hao Chen
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// Date : 2021-05-22
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/*****************************************************************************************************
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*
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* Given an integer array nums, return the sum of floor(nums[i] / nums[j]) for all pairs of indices 0
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* <= i, j < nums.length in the array. Since the answer may be too large, return it modulo 10^9 + 7.
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*
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* The floor() function returns the integer part of the division.
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*
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* Example 1:
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*
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* Input: nums = [2,5,9]
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* Output: 10
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* Explanation:
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* floor(2 / 5) = floor(2 / 9) = floor(5 / 9) = 0
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* floor(2 / 2) = floor(5 / 5) = floor(9 / 9) = 1
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* floor(5 / 2) = 2
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* floor(9 / 2) = 4
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* floor(9 / 5) = 1
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* We calculate the floor of the division for every pair of indices in the array then sum them up.
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*
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* Example 2:
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*
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* Input: nums = [7,7,7,7,7,7,7]
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* Output: 49
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*
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* Constraints:
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*
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* 1 <= nums.length <= 10^5
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* 1 <= nums[i] <= 10^5
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******************************************************************************************************/
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class Solution {
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public:
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int sumOfFlooredPairs(vector<int>& nums) {
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const int MAX_NUM = 100001;
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int cnt[MAX_NUM] = {0};
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int maxn = 0;
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for(auto& n : nums) {
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cnt[n]++;
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maxn = max(maxn, n);
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}
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vector<vector<int>> stats;
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for(int i=1; i<MAX_NUM; i++) {
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if (cnt[i] > 0) {
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stats.push_back({i, cnt[i]});
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}
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cnt[i] += cnt[i-1];
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}
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const int MOD = 1e9+7;
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int result = 0;
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for(int i=0; i < stats.size(); i++) {
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int n = stats[i][0];
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int c = stats[i][1];
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for(int x=2; x <= maxn/n+1; x++) {
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int pre = (x-1) * n - 1;
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int cur = min( x * n - 1, MAX_NUM-1);
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result = (result + (cnt[cur] - cnt[pre]) * long(x-1) * c) % MOD;
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}
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}
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return result;
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}
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};
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