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
- MIT线性代数:18.行列式及其特性
- 大公司喜欢问的Java集合类面试题
Collection Collection是基本的集合接口,一个Collection代表一组Object,即Collection的元素(Elements).一些Collection允许相同的元素而另一 ...
- [2018-01-13] 安装Django的一些笔记
安装django pip install Django = =1.10.2 下载源码,进入根目录执行 python setup.py install 确认是否已经安装成功 python -m djan ...
- CSPS模拟 76
前序遍历,中序遍历,后序遍历 说的都是根节点在前,根节点在中,根节点在后. 长记性!
- NOIP原题 斗地主(20190804)
题目描述 牛牛最近迷上了一种叫斗地主的扑克游戏.斗地主是一种使用黑桃.红心.梅花.方片的A到K加上大小王的共54张牌来进行的扑克牌游戏.在斗地主中,牌的大小关 系根据牌的数码表示如下:3<4&l ...
- Docker常见报错解决方法记录
[问题一]OCI runtime exec failed......executable file not found in $PATH": unknown [root@localhost ...
- 说一说JVM双亲委派机制与Tomcat
双亲委派模型与JVM 类加载 讲个故事: 以前,爱捣鼓的小明突然灵机一动,写出了下面的代码 package java.lang; public class String { //...复制真正Stri ...
- python面试题2.1:如何实现栈
本题目摘自<Python程序员面试算法宝典>,我会每天做一道这本书上的题目,并分享出来,统一放在我博客内,收集在一个分类中. 2.1 如何实现栈 [阿里巴巴面试题] 难度系数:⭐⭐⭐ 考察 ...
- PHP failed to ptrace(PEEKDATA) pid 13659: Input/output error错误解决方法
PHP failed to ptrace(PEEKDATA) pid 13659: Input/output error错误解决方法 现在改linux内核文件打开限制<pre>ulimit ...
- JS 暴虐算法查找
@dd|ad|fds|d@dd|ad|fds|d@dd|ad|fds|d@ var e = []; window.onload = function () { var ...