翻转对
class Solution {
public:int findpairs(vector<int> &nums, int l, int r) {int mid = (l + r) >> 1;int i = l, j = mid + 1;int res = 0;for (; i <= mid; ++i) {while (j <= r && (1LL * nums[i] > 2LL * nums[j])) {res += mid - i + 1;++j;}}return res;}int merge(vector<int> &nums, int nums2[], int l, int r) {if (l >= r) return 0;int res = 0;int mid = (l + r) >> 1;res = merge(nums, nums2, l, mid) + merge(nums, nums2, mid + 1, r) + findpairs(nums, l, r);int i = l, j = mid + 1;int id = l;while (i <= mid && j <= r ) {if (nums[i] <= nums[j]) nums2[id++] = nums[i++];else nums2[id++] = nums[j++];}while (i <= mid) nums2[id++] = nums[i++];while (j <= r) nums2[id++] = nums[j++];for(int ind = l;ind <= r;ind++) nums[ind] = nums2[ind];return res;}int reversePairs(vector<int>& nums) {int n = nums.size();int nums2[n];if (n == 0) return 0;return merge(nums, nums2, 0, n - 1);}
};
小和问题
#include<bits/stdc++.h>
using namespace std;
#define maxn 100005
#define ll long longint a[maxn];
int b[maxn];ll find(int a[], int l, int r) {if (l == r) return 0;int mid = (l + r) >> 1;int i = l, j = mid + 1;ll res = 0;for (; j <= r; ++j) {while (i <= mid && a[i] <= a[j]) {res += 1LL * (r - j + 1) * a[i];i++;}}return res;
}ll merge(int a[], int b[], int l, int r) {if (l >= r) return 0;int mid = (l + r) >> 1;int i = l, j = mid + 1, k = l;ll res = merge(a, b, l, mid) + merge(a, b, mid + 1, r) + find(a, l, r);while (i <= mid && j <= r) {if (a[i] <= a[j]) b[k++] = a[i++];else b[k++] = a[j++];}while (i <= mid) b[k++] = a[i++];while (j <= r) b[k++] = a[j++];for (int i = l; i < k; ++i) a[i] = b[i];return res;
}int main() {int n;cin >> n;for (int i = 0; i < n; ++i) cin >> a[i];cout << merge(a, b, 0, n - 1);return 0;
}