leetcode786 第 K 個最小的素數分數
阿新 • • 發佈:2021-08-07
思路:
二分。
實現:
1 class Solution 2 { 3 public: 4 bool check(double x, vector<int>& arr, int k) 5 { 6 int slow = 0, n = arr.size(), ans = 0; 7 for (int i = 1; i < n; i++) 8 { 9 if (arr[slow] > x * arr[i]) continue; 10 int tmp = slow;11 while (slow < i and arr[slow] <= x * arr[i]) slow++; 12 ans += (slow - tmp) * (n - i); 13 } 14 return ans >= k; 15 } 16 vector<int> kthSmallestPrimeFraction(vector<int>& arr, int k) 17 { 18 int n = arr.size(); 19double l = 0, r = 1, res = -1; 20 for (int i = 0; i < 30; i++) 21 { 22 double m = (l + r) / 2; 23 if (check(m, arr, k)) 24 { 25 res = m; r = m; 26 } 27 else l = m; 28 } 29 double minn = 0x3f3f3f3f; 30 int a = -1, b = -1; 31 for (int i = 0; i < n; i++) 32 { 33 double x = res * arr[i]; 34 int p = upper_bound(arr.begin(), arr.begin() + i, x) - arr.begin(); 35 if (p) 36 { 37 p--; 38 double tmp = (double)arr[p] / arr[i]; 39 if (abs(res - tmp) < minn) 40 { 41 minn = abs(res - tmp); 42 a = p; b = i; 43 } 44 } 45 } 46 return vector<int>{arr[a], arr[b]}; 47 } 48 };