Codeforces Round #131 (Div. 2)】的更多相关文章

题目链接: http://codeforces.com/problemset/problem/213/B B. Numbers time limit per test 2 secondsmemory limit per test 256 megabytes 问题描述 Furik loves writing all sorts of problems, especially such that he can't solve himself. You've got one of his proble…
题目链接: http://codeforces.com/problemset/problem/214/B Hometask time limit per test:2 secondsmemory limit per test:256 megabytes 问题描述 Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark fo…
题目链接: http://codeforces.com/problemset/problem/214/E Relay Race time limit per test4 secondsmemory limit per test256 megabytes 问题描述 Furik and Rubik take part in a relay race. The race will be set up on a large square with the side of n meters. The gi…
A. System of Equations \(a\)的范围在\(\sqrt n\)内,所以暴力枚举即可. B. Hometask 需要被2.5整除,所以末位必然为0,如果0没有出现,则直接返回-1. 在不考虑被3整除的情况下,为了凑出最大数,必然是从9到0的顺序填数. 设位数和模3的值为\(sum\),如果\(sum=0\),显然当前数即答案,否则考虑去除一些数使得数被3整除. 因为\(sum \ne 0\),所以\(sum=1\ or\ sum = 2\),显然去除的数不会超过2个.优先考…
首先能被2,5整除的数结尾必须是0: 如果没有0肯定不行: 然后判断他们的和ans%3: 如果==0,直接从大到小输出就行: 如果==1,要么删除它们之间最小的那个%3==1的,要么删除两个小的并且%3==2的: 如果==2,要么删除它们之中最小的那个%2==2的,要么删除两个小的并且%3==1的: 代码: #include<iostream> using namespace std; ],a,u; main() { cin>>n; while(n--)cin>>a,+…
A. Game time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Furik and Rubik love playing computer games. Furik has recently found a new game that greatly interested Rubik. The game consists of…
Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate it n = int(raw_input()) s = "" a = ["I hate that ","I love that ", "I hate it","I love it"] for i in ran…
Codeforces Round #354 (Div. 2) Problems     # Name     A Nicholas and Permutation standard input/output 1 s, 256 MB    x3384 B Pyramid of Glasses standard input/output 1 s, 256 MB    x1462 C Vasya and String standard input/output 1 s, 256 MB    x1393…
直达–>Codeforces Round #368 (Div. 2) A Brain’s Photos 给你一个NxM的矩阵,一个字母代表一种颜色,如果有”C”,”M”,”Y”三种中任意一种就输出”#Color”,如果只有”G”,”B”,”W”就输出”#Black&White”. #include <cstdio> #include <cstring> using namespace std; const int maxn = 200; const int INF =…
 cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅.....       其实这个应该是昨天就写完的,不过没时间了,就留到了今天.. 地址:http://codeforces.com/contest/651/problem/A A. Joysticks time limit per test 1 second memory limit per test 256…