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
- 差异:后缀数组(wzz模板理解),单调栈
因为涉及到对模板的理解,所以就着代码看会好一些. 让那些坚决不颓代码的人受委屈了. 我是对着wzz的板子默写的,可能不完全一样啊. 还有代码注释里都是我个人的理解,不保证正确,但欢迎指正. 可以有选择 ...
- JS- 封装、继承、多态
http://www.cnblogs.com/silence516/articles/1509456.html
- P2114 [NOI2014]起床困难综合症
#include<iostream> #include<cstdio> using namespace std; ; ]; long long n,m; long long t ...
- php imagick 文字居中的方法
php imagick 文字居中的方法<pre> public function getwenzinfo($nickName) { $nickNamelen = mb_strlen($ni ...
- jquery序列帧播放(支持视频自动播放和不是全屏播放)
jquery序列帧播放 这个弊端就是到时候需要升级下带宽 至少10MB 保证不卡.. ae导出序列真的时候 每秒10帧 就是代码每秒播放10张图片 尺寸适当的可以压小点<pre> < ...
- Linux软件包管理和磁盘管理实践
一.自建yum仓库,分别为网络源和本地源 本地yum仓库的搭建就是以下三个步骤: 创建仓库目录结构 上传相应的包到目录下,或者直接挂载光盘也行,如果挂载光盘,第三步就可以省略,因为光盘默认里有repo ...
- Python 基础之 线程与进程
Python 基础之 线程与进程 在前面已经接触过了,socket编程的基础知识,也通过socketserver 模块实现了并发,也就是多个客户端可以给服务器端发送消息,那接下来还有个问题,如何用多线 ...
- 关于设备与canvas画不出来的解决办法
连续四天解决一个在三星手机上面画canvas的倒计时饼图不出来的问题,困惑了很久,用了很多办法,甚至重写了那个方法,还是没有解决,大神给的思路是给父级加 "overflow: visible ...
- SpringBoot 源码解析 (六)----- Spring Boot的核心能力 - 内置Servlet容器源码分析(Tomcat)
Spring Boot默认使用Tomcat作为嵌入式的Servlet容器,只要引入了spring-boot-start-web依赖,则默认是用Tomcat作为Servlet容器: <depend ...
- JQury自动切换图片
[标签]Jquery图片自动切换<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "ht ...