C++ code O(n)
Intuition\n Describe your first thoughts on how to solve this problem. \ntreat the part between x and y as a ring (on which the shortest path cannot exceed half
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\ntreat the part between x and y as a ring (on which the shortest path cannot exceed half of the ring length). Numerous edge cases to consider. \n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nusing ll = long long;\nclass Solution {\npublic:\n    vector<long long> countOfPairs(int n, int x, int y) {\n        vector<ll> ans;\n        if(y<x) swap(x,y);\n        for(int i = 1; i < n; i++){\n            ll cur = 0;\n            cur += max(2*(x-i), 0);\n            cur += max(2*(n-y+1 - i), 0);\n            if(i>=2 && i <= n - (y-x)) {\n                int tmp = min(i-1-min(y-x,1), x-1);\n                tmp = min(tmp, n-y);\n                tmp = min(tmp, n - max(y-x, 1) - i+1);\n                cur += 2*tmp;\n            }\n            if(i <= (y-x+1)/2) {\n                int tmp = 2* (y-x+1);\n                if((y-x+1)%i==0 && (y-x+1)/i==2) tmp /=2;\n                cur += tmp;\n            }\n            int lr = y - x +1;\n            int mx_d = lr/2;\n            if(i<=mx_d + x-1&& i>1) {\n                cur += 4* min(min(min((x-1), mx_d), mx_d +x - i), i-1);\n                if(x>1 && lr%2==0 && lr/2<=mx_d+x-1 && i > lr/2) cur -=2;\n            }\n            if(i<=mx_d + n-y&& i>1) {\n                cur += 4* min(min(min(n-y, mx_d), mx_d +n-y - i+1), i-1);\n                if(n-y>=1 && lr%2==0 && lr/2<=mx_d+n-y && i > lr/2) cur -=2;\n            }\n            ans.push_back(cur);\n        }\n        ans.push_back(0);\n        return ans;\n    }\n};\n```
['C++']