(最大流)CodeForces - 499E Array and Operations You have written on a piece of paper an array ofnpositive integersa[1], a[2], ..., a[n] andmgood pairs of integers(i1, j1), (i2, j2), ..., (im, jm). Eachgood pair(ik, jk) meets the following condit...
所以我们对于每一个质因数建图,原图中的边的两个点都存在这个这个质因数的连边,边权为两个点这个质因数的数量中较小的那一个,奇数点与源建边,边权为它存在这个质因数的个数,偶数点同理,与汇建边,然后跑最大流。 因为通过筛素数,我们也就能筛出105以内的素数,对于大于这个边界的我们可以考虑,一个数(109以...
Codeforces 498C Array and Operations(最大流) 题目是给一些数和<数对>的下标,然后进行操作:对某个<数对>中的两个数同时除以一个都能被它们整除且不等于1的数,要求的就是最多能进行多少次操作。 除数一定是素数,就是要决定某素数要除哪些<数对>使除的次数最多,ik + jkis an odd number 可以想到这个...
Codeforces Round #284 (Div. 1) C. Array and Operations 二分图匹配,因为只有奇偶之间有操作,可以看出是二分图,然后拆质因子,二分图最大匹配求答案就好啦。
Greg has an array a = a1, a2, ..., an and m operations. Each operation looks as: li, ri, di, (1 ≤ li ≤ ri ≤ n). To apply operation i to the array means to increase all array elements with numbers li, li + 1, ..., ri by value ...
Ifk=3k=3, it's impossible to make allaiaiequal to00. After performing the first operation, we'll get three elements equal to00and one equal to44. After that, all elimination operations won't change anything, since at least one chosen element will always be equal to...
简介:codeforces 315 B.Sereja and Array 地址 B. Sereja and Array time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Sereja has got an array, consisting of n integers, a1, a2, ..., an. Sereja is an active boy, so he is...
First solution, n adds and 1 mod First, let's make ai = x * n + i (for some x). Then, let's mod the whole array with n (making ai = i). If the "add update" changed one index, we can just add i + n - ai%n to index i. The problem is,...
Minimum operations required to remove an array Given an array of N integers where N is even. There are two kinds of operations allowed on the array. Increase the value of any element A[i] by 1. If two adjacent elements in the array are consecutive prime number, delete both the element....
CodeForces 1353 D. Constructing the Array 2. 解读 用优先队列 priority_queue 根据区间长度构造大顶堆。 每次取出堆顶的区间元素进行赋值,每次赋值完以后,将左子区间和右子区间放入堆中。和分治法的思想比较类似。 重复以上步骤直到堆为空。 3. 代码 #include <algorithm> #include <iostream> #include <queue...