int sumRegion(int row1, int col1, int row2, int col2) { if (row1 == 0 && col1 == 0) return pre_sum[row2][col2]; else if (row1 == 0) return pre_sum[row2][col2] - pre_sum[row2][col1-1]; else if (col1 == 0) return pre_sum[row2][col2] - pre_sum[row1-...
buffering → Codeforces Global Round 26 NaZaR.IO → KEP Contest #6 Cryomenace → (Problem Solved) Sublime Text Editor taking too much time to compile a C/C++ code Alma_s → Codeforces practice grind rating range 1600-1800 i_love_penguins → Codeforces Round #932 (Div. 2) Editorial ...
void solve() { int n, k; std::cin >> n >> k; std::vector a(n + 1), sum(n + 1), min(n + 1); for (int i = 1; i <= n; i ++) std::cin >> a[i]; auto check = [&](int x) { for (int i = 1; i <= n; i ++) sum[i] = sum[i - 1] + (a[i] ...
Codeforces blog 树状数组解法 所有的奇数位置的数字和原数组对应位置的相同,偶数位置是原数组若干位置之和,若干是根据坐标的最低位 Low Bit 来决定的 (x&-x) [i, j] 区间和:sum[j]-sum[i-1] 1classNumArray {2public:3NumArray(vector<int>&nums) {4data.resize(nums.size());5bit.resize(nums.si...
CF1771F Hossam and Range Minimum Query 题解 合集- codeforces题解集1(37) 1.CF1916E Happy Life in University 题解2023-12-312.CF763E Timofey and our friends animals题解01-033.CF1270G Subset with Zero Sum01-054.CF1045G AI robots题解01-055.CF940F Machine Learning题解01-066.CF678F ...
CF1458F Range Diameter Sum 一棵树,定义\(diam(l,r)\)表示区间\([l,r]\)中的点的直径。求\(\sum_{l<r} diam(l,r)\)。 \(n\le 10^5\) 题解有详细证明:https://codeforces.com/blog/entry/85750 先将所有的边拆成两条,中间插个虚点。
Output Print a single integer —∑1≤l≤r≤ndiam(l,r)\sum_{1 \leq l \leq r \leq n} \mathrm{diam}(l, r)∑1≤l≤r≤ndiam(l,r). Samples 输入数据 1 4 1 2 2 4 3 2 Copy 输出数据 1 10 Copy 输入数据 2 10 1 8 2 9 5 6 4 8 4 2 7 9 3 6 10 4 3 9 ...
DEFINE : RSQ[i]= X[1] + X[2] + .. + X[i] = rangeSumQuery(1,i) Now say we have a query which tells us to adduto all elements in the range [l,r] then if I perform a point update and make X[l] = X[l] +uthink what happens to RSQ[i] for different values of i. ...
In 1-D we can find sum in range [l, r] using sum[r] - sum[l - 1]. In 2-D we can find sum in range from (x1, y1) to (x2, y2) using sum[x2][y2] - sum[x2][y1] - sum[x1][y2] + sum[x1][y1] where sum[] or sum[][] is ...
一棵树,定义$diam(l,r)\(表示区间\)[l,r]\(中的点的直径。求\)\sum_{l<r} diam(l,r)$。 \(n\le 10^5\) 题解有详细证明:https://codeforces.com/blog/entry/85750 先将所有的边拆成两条,中间插个虚点。 定义