题目：
题解：
class Solution {
public:vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {int m = nums1.size();int n = nums2.size();auto count = [&](int target){long long cnt = 0;int start = 0;int end = n - 1;while (start < m && end >= 0) {if (nums1[start] + nums2[end] > target) {end--;} else {cnt += end + 1;start++;}}return cnt;};/*二分查找第 k 小的数对和的大小*/int left = nums1[0] + nums2[0];int right = nums1.back() + nums2.back();int pairSum = right;while (left <= right) {int mid = left + ((right - left) >> 1);if (count(mid) < k) {left = mid + 1;} else {pairSum = mid;right = mid - 1;}}vector<vector<int>> ans;int pos = n - 1;/*找到小于目标值 pairSum 的数对*/for (int i = 0; i < m; i++) {while (pos >= 0 && nums1[i] + nums2[pos] >= pairSum) {pos--;}for (int j = 0; j <= pos && k > 0; j++, k--) {ans.push_back({nums1[i], nums2[j]});}}/*找到等于目标值 pairSum 的数对*/pos = n - 1;for (int i = 0; i < m && k > 0; i++) {int start1 = i;while (i < m - 1 && nums1[i] == nums1[i + 1]) {i++;}while (pos >= 0 && nums1[i] + nums2[pos] > pairSum) {pos--;}int start2 = pos;while (pos > 0 && nums2[pos] == nums2[pos - 1]) {pos--;}if (nums1[i] + nums2[pos] != pairSum) {continue;}int count = (int) min((long) k, (long) (i - start1 + 1) * (start2 - pos + 1));for (int j = 0; j < count && k > 0; j++, k--) {ans.push_back({nums1[i], nums2[pos]});}}return ans;}long min(long num1, long num2) {return num1 <= num2 ? num1 : num2;}
};