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…

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…

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…

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…

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;…

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…