Educational Codeforces Round 34 (Rated for Div. 2) A B C D
Educational Codeforces Round 34 (Rated for Div. 2)
A Hungry Student Problem
题目链接:
http://codeforces.com/contest/903/problem/A
思路:
直接模拟
代码:
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
scanf("%d",&n);
while(n--) {
int num,flag=0;
scanf("%d",&num);
for(int i=0;i<=num/3;++i) for(int j=0;j<=num/7;++j) if(3*i+7*j==num) flag=1;
if(flag) printf("YES\n");
else printf("NO\n");
}
return 0;
}
B The Modcrab
题目链接:
http://codeforces.com/contest/903/problem/B
思路:
模拟打怪兽的过程,需要注意的是,能够尽量打的情况坚决不舔包。就是说在一个回合中,怪兽能把你打死,但是你也能打死怪兽,这个时候先下手为强。其余情况下,保证自己活到下一回合。
代码:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e7+5;
ll d[maxn];
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
ll h1,h2,a1,a2,c1,tot=0;
cin>>h1>>a1>>c1;
cin>>h2>>a2;
while(!(h2<=0)) {
if(h2-a1<=0) {
d[tot]=1;
h2=h2-a1;
} else if(h1-a2>0) {
d[tot]=1;
h2=h2-a1;
} else {
d[tot]=0;
h1=h1+c1;
}
tot=tot+1;
h1=h1-a2;
}
cout<<tot<<endl;
for(int i=0;i<tot;i=i+1) {
if(d[i]) {
cout<<"STRIKE"<<endl;
} else {
cout<<"HEAL"<<endl;
}
}
return 0;
}
C Boxes Packing
题目链接:
http://codeforces.com/contest/903/problem/C
思路:
找到某一个数的数量,且该数的数量是全部数里面最大的,就是答案。多此一举的离散化了一下。(¦3」∠)
代码:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll maxn = 1e9+5;
ll data[5005],ans[5005],res[5005];
int main() {
ll n,maxnum=0;
scanf("%I64d",&n);
for(int i=0;i<n;++i) scanf("%I64d",&data[i]),ans[i]=data[i];
sort(data,data+n);
int tot=unique(data,data+n)-data;
for(int i=0;i<n;++i) {
ans[i]=lower_bound(data,data+tot,ans[i])-data;
res[ans[i]]++;
}
for(int i=0;i<tot;++i) maxnum=max(maxnum,res[i]);
printf("%I64d\n",maxnum);
return 0;
}
D Almost Difference
题目链接:
http://codeforces.com/contest/903/problem/D
思路:
爆了long long,所以使用long double。另外是用c++14提交的,c++11提交就是过不了,读入数据部分就会和本地不一样。
代码:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 200005;
ll a[maxn];
int n;
long double sum=0;
map<ll,int> mp;
int main() {
scanf("%d",&n);
for(int i=1;i<=n;++i) {
scanf("%I64d",&a[i]);
sum+=(long double)(i-1)*(long double)a[i];
sum-=(long double)(n-i)*(long double)a[i];
}
for(int i=1;i<=n;++i) {
mp[a[i]]++;
sum-=(long double)mp[a[i]-1];
sum+=(long double)mp[a[i]+1];
}
printf("%.0Lf\n",sum);
return 0;
}
Educational Codeforces Round 34 (Rated for Div. 2) A B C D的更多相关文章
- Educational Codeforces Round 34 (Rated for Div. 2) D - Almost Difference(高精度)
D. Almost Difference Let's denote a function You are given an array a consisting of n integers. You ...
- Educational Codeforces Round 34 (Rated for Div. 2) C. Boxes Packing
C. Boxes Packing time limit per test 1 second memory limit per test 256 megabytes input standard inp ...
- Educational Codeforces Round 34 (Rated for Div. 2)
A. Hungry Student Problem time limit per test 1 second memory limit per test 256 megabytes input sta ...
- Educational Codeforces Round 34 (Rated for Div. 2) B题【打怪模拟】
B. The Modcrab Vova is again playing some computer game, now an RPG. In the game Vova's character re ...
- Educational Codeforces Round 71 (Rated for Div. 2)-F. Remainder Problem-技巧分块
Educational Codeforces Round 71 (Rated for Div. 2)-F. Remainder Problem-技巧分块 [Problem Description] ...
- Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship
Problem Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship Time Limit: 2000 mSec P ...
- Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems(动态规划+矩阵快速幂)
Problem Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems Time Limit: 3000 mSec P ...
- Educational Codeforces Round 43 (Rated for Div. 2)
Educational Codeforces Round 43 (Rated for Div. 2) https://codeforces.com/contest/976 A #include< ...
- Educational Codeforces Round 35 (Rated for Div. 2)
Educational Codeforces Round 35 (Rated for Div. 2) https://codeforces.com/contest/911 A 模拟 #include& ...
随机推荐
- 通俗易懂了解Vuex
1.前言 在使用Vue进行开发的时候,关于vue组件通信的方式,除了通俗易懂了解Vue组件的通信方式这篇博文谈到三种通信方式,其实vue更提倡我们使用vuex来进行组件间的状态管理以及通信问题.Vue ...
- 爬虫学习--Urllib库基本使用 Day1
一.Urllib库详解 1.什么是Urllib Python内置的HTTP请求库 urllib.request 请求模块(模拟实现传入网址访问) urllib.error ...
- F#周报2019年第46期
新闻 使用Pulumi和.NET Core创建现代云应用 宣告.NET Core 3.1预览版3 ML.NET模型构建器升级 .NET Framework修复工具 Mac上的Visual Studio ...
- Python Excel 绘制柱形图
本文主要讲述如何使用Python操作Excel绘制柱形图. 相关代码请参考 https://github.com/RustFisher/python-playground 本文链接:https://w ...
- Matlab 文件格式化/Matlab Source File Formator
由于需要使用到别人编写的Matlab代码文件,但是呢不同的人有不同的风格,有的写得就比较糟糕了. 为了更好地理解代码的内容,一个比较美观的代码会让人身心愉悦. 但是在网上并没有找到一个比较好的实现,此 ...
- MathType转Word公式(OMML)
背景 由于之前个人喜欢在Word里做笔记,而有很多笔记里存在着大量的公式.在早期,由于对Word自身的公式的不理解,所以便使用了MathType这个工具来编写公式.但是现在本人已经转战到LatTeX了 ...
- RabbitMQ学习笔记之五种模式及消息确认机制
本文详细介绍简单模式Simple.工作模式Work.发布订阅模式Publish/Subscribe.Topic.Routing. Maven依赖引用 <dependencies> < ...
- 力扣(LeetCode)2的幂 个人题解
给定一个整数,编写一个函数来判断它是否是 2 的幂次方. 示例 1: 输入: 1 输出: true 解释: 20 = 1 示例 2: 输入: 16 输出: true 解释: 24 = 16 示这题是考 ...
- Project Euler 62: Cubic permutations
立方数\(41063625 (345^3)\)的各位数重新排列形成另外两个立方数\(6623104 (384^3)\)和\(66430125 (405^3)\).事实上,\(41063625\)是满足 ...
- centos7 设置连接无线wifi
安装系统后,首先要联网. 1.首先使用网线连接,之后尝试ping www.baidu.com我的是自动通的 2.需要查看网卡型号,先安装工具 yum -y install pciutils* 3.查看 ...