B. Maximum of Maximums of Minimums time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given an array a1, a2, ..., an consisting of n integers, and an integer k. You have to split the a…
C. Maximum splitting time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given several queries. In the i-th query you are given a single positive integer ni. You are to represent ni as…
A. Laptops time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the…
题意: 我们在研究罗马数字.罗马数字只有4个字符,I,V,X,L分别代表1,5,10,100.一个罗马数字的值为该数字包含的字符代表数字的和,而与字符的顺序无关.例如XXXV=35,IXI=12. 现在求问一个长度为 nnn 的罗马数字可以有多少种不同的值. n<=109n<=10^9n<=109. 题解: 我们可以用暴力的方法求出前20项的值,其中前111111 项采用打表的方式,而从第12项开始答案始终 = 前一项的答案 + 49,即 292+(n−11)∗49292+(n-11)*…
Codeforces Round #440 (Div. 2) codeforces 870 A. Search for Pretty Integers(水题) 题意:给两个数组,求一个最小的数包含两个数组各至少一个数. #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int n, m; ], b[]; int main() { , y = , s = ; scanf(…
A. Search for Pretty Integers 题目链接:http://codeforces.com/contest/872/problem/A 题目意思:题目很简单,找到一个数,组成这个数的数字即在A数组中出现过,也在B数组中出现过,问这个数最小是多少. 题目思路:首先要么一个数两个数组都出现过直接输出来,要么分别取两个数组中最小的数组合一下输出. 代码: //Author: xiaowuga #include <bits/stdc++.h> using namespace st…
A. Search for Pretty Integers time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given two lists of non-zero digits. Let's call an integer pretty if its (base 10) representation has at…
A. Search for Pretty Integers You are given two lists of non-zero digits. Let's call an integer pretty if its (base 10) representation has at least one digit from the first list and at least one digit from the second list. What is the smallest positi…
B. Maximum of Maximums of Minimums You are given an array a1, a2, ..., an consisting of n integers, and an integer k. You have to split the array into exactly k non-empty subsegments. You'll then compute the minimum integer on each subsegment, and ta…
// Codeforces Round #365 (Div. 2) // C - Chris and Road 二分找切点 // 题意:给你一个凸边行,凸边行有个初始的速度往左走,人有最大速度,可以停下来,竖直走. // 问走到终点的最短时间 // 思路: // 1.贪心来做 // 2.我觉的二分更直观 // 可以抽象成:一条射线与凸边行相交,判断交点.二分找切点 #include <bits/stdc++.h> using namespace std; #define LL long lon…