Codeforces 622B The Time 【水题】】的更多相关文章

Problem G. Grave 题目连接: http://codeforces.com/gym/100531/attachments Description Gerard develops a Halloween computer game. The game is played on a rectangular graveyard with a rectangular chapel in it. During the game, the player places new rectangul…
题目链接: A. Beru-taxi 题意: 问那个taxi到他的时间最短,水题; AC代码: #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <bits/stdc++.h> #include <stack> #include <map> using n…
题目链接: B. Inventory time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Companies always have a lot of equipment, furniture and other things. All of them should be tracked. To do this, there is…
A. SwapSort time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output In this problem your goal is to sort an array consisting of n integers in at most n swaps. For the given array find the sequence…
题目链接: A. Opponents time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Arya has n opponents in the school. Each day he will fight with all opponents who are present this day. His opponents have…
题意:给定两个速度,一个一初速度,一个末速度,然后给定 t 秒时间,还每秒速度最多变化多少,让你求最长距离. 析:其实这个题很水的,看一遍就知道怎么做了,很明显就是先从末速度开始算起,然后倒着推. 代码如下: #include <bits/stdc++.h> using namespace std; typedef long long LL; const int maxn = 1e5 + 5; const int INF = 0x3f3f3f3f; vector<int> ans;…
Problem I. iSharpTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=86821#problem/I Description You are developing a new fashionable language that is not quite unlike C, C++, and Java. Since your langua…
题目链接:http://codeforces.com/problemset/problem/705/A 从第三个输出中可看出规律, I hate that I love that I hate it I hate I love 是来回循环的,不到最后一个每一次循环后面都输出 that,如果到最后则输出 it #include<bits/stdc++.h> using namespace std; int main() { ][]= {"I love ","I ha…
题意:给定 n 个杯子,里面有不同体积的水,然后问你要把所有的杯子的水的体积都一样,至少要倒少多少个杯子. 析:既然最后都一样,那么先求平均数然后再数一下,哪个杯子的开始的体积就大于平均数,这是一定要倒的. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #inclu…
*这种题好像不用写题解... 题意: 一个人要改动别人的实验记录,实验记录记录是一个集合 实验记录本身满足:$max(X)-min(X)<=2$ 改动结果要求: 1.新的集合平均值和之前的一样 2.新的集合,$max(Y)<=max(X),min(Y)>=min(X)$ 求新一个和之前相同数值最少的新记录 题解: 首先考虑不同情况, 如果$max-min<=1$ :因为要保证平均值且值域受限制不变,无法改变值(增加一个值之后,要相应的把另外一值减小,而数值只有2/1种,改动没有意义…