CodeForces-1370A.Maximum GCD 原题如下: A.Maximum GCD time limit per test: 1 second memory limit per test: 256 megabytes input: standard input output: standard output Let’s consider all integers in the range from 1 to n (inclusi......
Here's a solution: First, we'll look at it from a different angle: we need to choosekdistinct subarrays (continuous), taking a total of the whole array, so that the gcd of all the subarray sums is maximal. Now, by this observation, we can see that since all the subarrays' sums ...
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> #...
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 ...
Codeforces Round #651 (Div. 2) A Maximum GCD、B GCD Compression、C Number Game、D Odd-Even Subsequence A. Maximum GCD 题意: t组输入,然后输入一个n,让你在区间[1,n]之间找出来两个不相等的数a,b。求出来gcd(a,b)(也就是a,b最大公约数)。让你求出来最大的gcd(a,b)是多少。
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 ...
给你两个长度为n的数列a和b,要求你最多翻转一次a序列的一段连续的区间,使得n∑i=1ai∗bi最大。 分析 首先暴力方法是先枚举区间长度再枚举起点,整个过程是O(n2)的,然后就是如何O(1)求出翻转后的值,先去找了一下每个翻转策略之间有没有冗余部分,发现没有啥可以优化的,想不出怎么快速求值,于是否掉了这个...
Output the resulting graph or report that no solution exists. Input The first line contains a single integer n (3≤n≤500) — the number of vertices in the graph. The second line contains n integers a1,a2,…,an (1≤ai≤n−1) — the upper limits to vertex degrees. ...
There can be onlydifferent gcd values among suffixes of an array (due to Euclide's algorithm complexity). So you can add characters one by one keeping all possible gcd's of current array's suffixes. By the way, does anyone know solution for the original problem now, 2 years later?
https://www.luogu.org/problemnew/solution/P5170 1#include<bits/stdc++.h>2usingnamespacestd;3typedeflonglongll;45ll f(ll n,ll a,ll b,ll c)6{if(n<0)return0;7if(!n)returnb/c;8if(a>=c || b>=c)returnf(n,a%c,b%c,c)+n*(n+1)/2*(a/c)+(n+1)*(b/c);9ll m=(...