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voidsolve(){intn, m; cin >> n >> m;vector<int>a(n);for(auto& x : a){ cin >> x; x %= m; } set<int> sett;for(constauto& x : a){ set<int> cur;for(constauto& it : sett){ cur.insert((it + x) % m); } sett.insert(cur.begin(), cur.end()); sett.insert(x)...
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}else{ maxn =max<longlong>(maxn, t); } }if(nums.empty() || nums.size() == n) { cout <<"0\n";return; }sort(nums.begin(), nums.end());intm = (int)nums.size();for(inti =0; i < m; ++i) {if(maxn < nums[i]) { cout << m +1<<'\n';return; } maxn += ...
We can show that the tree consists of O(k)O(k) vertices at most. Notice that the answer must be leaf. So we can use dp to find furthest leaf from each vertex to keep track of that. Now the answer depends on which node's subtree it is in. Let's say there is some leaf at ...
LeetCode 834.树中距离之和 Sum of distance in tree 题目虽然是hard级的,但是感觉难度其实不高。属于较为基础的树状DPDPDP 首先题目是计算以各点开始的所有距离之和。如果直接计算两点距离。最快的方案是弗洛伊德算法,这是O(N3)O(N^3)O(N3)的,直接挂 下面分析一下思路: 假设某个节点rootrootroot,令dp[ro...
If the values compared in the merge() function are equal then any of them could produce a larger answer later on while building a tree. I didn't think of it before. It might be the problem in this approach. →Reply rnsiehemt
2019-12-12 08:00 − 原题链接在这里:https://leetcode.com/problems/range-sum-of-bst/ 题目: Given the root node of a binary search tree, return the sum of values of all nod... Dylan_Java_NYC 0 487 Codeforces Round #608 (Div. 2) D Portals 2019-12-19 17:03 − #include...
树状数组部分有点同于求区间数的种数。 last记录每个数前一次出现的位置。 走到i时,如果a[i]出现过,那么把他上次出现的位置异或掉,再在i位置上异或上a[i]。 然后对所有以i结尾的讯问区间进行操作,这里就是异或计算了。 /*** Author :devil ***/#include<cstdio>#include<cstring>#include<iostream...
Using these operations, he adds the numbers in binary. So, according to his calculations,3 (011) + 7 (111) = 4 (100) Now you are given an array of n integers, indexed from 0 to n-1, you have to find two indices i j in the array(0 ≤ i ≤ j < n)such that the summation...