B. Divisiblity of Differences time limit per test 1 second memory limit per test 512 megabytes input standard input output standard output You are given a multiset of n integers. You should select exactly k of them in a such way that the difference b…

B. Divisiblity of Differencestime limit per test1 secondmemory limit per test512 megabytesinputstandard inputoutputstandard outputYou are given a multiset of n integers. You should select exactly k of them in a such way that the difference between an…

B. Divisiblity of Differences time limit per test 1 second memory limit per test 512 megabytes input standard input output standard output You are given a multiset of n integers. You should select exactly k of them in a such way that the difference b…

B. Divisiblity of Differences You are given a multiset of n integers. You should select exactly k of them in a such way that the difference between any two of them is divisible by m, or tell that it is impossible. Numbers can be repeated in the origi…

http://codeforces.com/contest/876/problem/B 题意: 给出n个数,要求从里面选出k个数使得这k个数中任意两个的差能够被m整除,若不能则输出no. 思路: 差能够被m整除,其实就是对m取余的余数相同.那么就统计n个数的余数丢到一个map里面,最后判断是否有某个数的数量大于等于k. 代码: #include <stdio.h> #include <map> #include <vector> using namespace std;…

题目链接:http://codeforces.com/contest/876/problem/B 题意: 给你n个数a[i],让你找出一个大小为k的集合,使得集合中的数两两之差为m的倍数. 若有多解,输出任意一个集合即可. 题解: 若一个集合中的数,两两之差为m的倍数,则他们 mod m 的值均相等. 所以O(N)扫一遍,对于每个数a:vector v[a%m].push_back(a) 一旦有一个集合大小为k,则输出. AC Code: #include <iostream> #includ…

题意:给定n个数,从中选取k个数,使得任意两个数之差能被m整除,若能选出k个数,则输出,否则输出“No”. 分析: 1.若k个数之差都能被m整除,那么他们两两之间相差的是m的倍数,即他们对m取余的余数是相同的. 2.记录n个数对m取余的余数,计算出数量最多的余数ma. 3.ma>=k,才能选出,并输出即可. #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #in…

A. Trip For Meal 题目链接:http://codeforces.com/contest/876/problem/A 题目意思:现在三个点1,2,3,1-2的路程是a,1-3的路程是b,2-3的路程是c,从1点开始,小熊维尼在1点吃过一次蜂蜜了,但是他要吃n次蜂蜜,每次他离开一个地方以后这个地方的蜂蜜就会自动补充,问最少需要走多少距离. 题目思路:如果n=1,那么就是0,如果n等于2,那么答案说就是min(a,b),如果n>2,答案就是min(a,b)+(n-2)*min(a,b,…

Codeforces Round #441 (Div. 2) codeforces 876 A. Trip For Meal(水题) 题意:R.O.E三点互连,给出任意两点间距离,你在R点,每次只能去相邻点,要走过n个点,求走过的最短距离. #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int main() { int n, a, b, c; scanf("…

Codeforces Round #441 (Div. 2) A. Trip For Meal 题目描述:给出\(3\)个点,以及任意两个点之间的距离,求从\(1\)个点出发,再走\(n-1\)个点的最短路径. solution 当\(n=1\)时,答案为\(0\),当\(n=2\)时,答案等于与开始点相连的两条边的最小值,当\(n>2\)时,答案等于与开始点相连的两条边的最小值+三条边最小值*\((n-2)\) 时间复杂度:\(O(1)\) B. Divisiblity of Differen…