Codeforces Round #651 (Div. 2) A. Maximum GCD (思维) 题意:在11~nn中找两个不相等的数使得他们的gcdgcd最大. 题解:水题,如果nn是偶数,那么一定取nn和n/2n/2,nn是奇数的话,取n−1n−1和(n−1)/2(n−1)/2. 代码: #include <iostream> #include <cstdio> #include <cstring> #...
最大gcd(a,b),那就是n/2向下取整的结果。因为如果gcd(a,b)越大,那么a/gcd(a,b)或者b/gcd(a,b)的值肯定越小,最小也就是2了,所以输出n/2就行 代码: View Code B. GCD Compression 题意: t组输入,你首先有一个长度为2*n的数组,你需要压缩a数组长度为n-1。你可以扔掉a数组中的两个元素,将剩...
Given array A of N elements. We are allowed to do one operation: merge two elements A[i] and A[i+1] into A[i]+A[i+1]. We need to find sequence of exactly K elements with maximum possible GCD. 1 <= K < N <= 10^5
Input: First line contains N (2<=N<=10^5). Next line contains N numbers, each of them more than 1 and less than 10^5, output: Output the largest GCD of any pair of numbers. You can iterate over all possible gcds and for each check if there exist two or more numbers divisable ...
Find the maximum possible value ofgcd(a1,a2,…,an)+gcd(b1,b2,…,bn)\gcd(a_1, a_2, \ldots, a_n) + \gcd(b_1, b_2, \ldots, b_n)gcd(a1,a2,…,an)+gcd(b1,b2,…,bn)that you can get by paying in total at mostdddcoins for swapping some ...
I was wondering if there is any faster algorithm thanO(N^2)that solves the following problem: Given a tree made ofNnodes, where each node has an integer value asociated to it, find the minimum/maximum distance between two nodes whose values are coprime(if such a pair exists). ...
CodeForces Algorithms. Contribute to DionysiosB/CodeForces development by creating an account on GitHub.
#include <bits/stdc++.h> using i64 = long long; using u64 = unsigned long long; using i128 = __int128; constexpr int N = 2E5; int LOG[N + 1], sparseTable[20][N + 1]; inline int query(int l, int r) { int t = LOG[r - l + 1]; return std::gcd(sparseTable[t][...
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a%b); } bool cmp(int a, int b) { return a > b; } ll a[maxn]; ll sum[maxn]; ll dt[maxn]; int main() { ios::sync_with_stdio(false); int n,q;
lllcm(lla,llb) {returna/gcd(a,b)*b;} llpowmod(lla,llb,llMOD) {llans=1;while(b) {if(b%2) {ans=ans*a%MOD; }a=a*a%MOD;b/=2;}returnans;} inlinevoidgetInt(int*p); constintmaxn=1000010; constintinf=0x3f3f3f3f;