Codeforces Round#412 Div.2
A. Is it rated?
题面
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
题意
没有!QAQ
代码
#include <bits/stdc++.h>
using namespace std;
int n;
int x,y,ans;
int a[1010];
int b[1010];
int cnt;
int sum1;
int smu2;
bool flag1;
bool flag2;
bool flag3;
int main()
{
cin>>n;
for (int i=1;i<=n;i++) cin>>a[i]>>b[i];
for (int i=1;i<=n;i++)
if (a[i]!=b[i]) return 0*puts("rated");
for (int i=1;i<n;i++) if (a[i]>=a[i+1]) ;else return 0*puts("unrated");
return 0*puts("maybe");
}
B. T-Shirt Hunt
题面
Not so long ago the Codecraft-17 contest was held on Codeforces. The top 25 participants, and additionally random 25 participants out of those who got into top 500, will receive a Codeforces T-shirt.
Unfortunately, you didn't manage to get into top 25, but you got into top 500, taking place p.
Now the elimination round of 8VC Venture Cup 2017 is being held. It has been announced that the Codecraft-17 T-shirt winners will be chosen as follows. Let s be the number of points of the winner of the elimination round of 8VC Venture Cup 2017. Then the following pseudocode will be executed:
i := (s div 50) mod 475
repeat 25 times:
i := (i * 96 + 42) mod 475
print (26 + i)
Here "div" is the integer division operator, "mod" is the modulo (the remainder of division) operator.
As the result of pseudocode execution, 25 integers between 26 and 500, inclusive, will be printed. These will be the numbers of places of the participants who get the Codecraft-17 T-shirts. It is guaranteed that the 25 printed integers will be pairwise distinct for any value of s.
You're in the lead of the elimination round of 8VC Venture Cup 2017, having x points. You believe that having at least y points in the current round will be enough for victory.
To change your final score, you can make any number of successful and unsuccessful hacks. A successful hack brings you 100 points, an unsuccessful one takes 50 points from you. It's difficult to do successful hacks, though.
You want to win the current round and, at the same time, ensure getting a Codecraft-17 T-shirt. What is the smallest number of successful hacks you have to do to achieve that?
题意
问成功hack几次,能达到一个玄学分数,该分数弄出来的名字有自己
代码
#include <bits/stdc++.h>
using namespace std;
int p,x,y;
int ans;
bool ret(int k)
{
int i = (k / 50) % 475 ;
for (int j=1;j<=25;j++)
{
i = (i * 96 + 42) % 475;
if (i+26==p) return 1;
}
return 0;
}
int main()
{
cin>>p>>x>>y;
int xx=x;
while (xx>=y) xx-=50;
y=xx+50;
for (int i=0;;i++) if (ret(y+i*50))
{
cout<<max(((y+i*50-x)/50+1)/2,0);
return 0;
}
}
C. Success Rate
题面
You are an experienced Codeforces user. Today you found out that during your activity on Codeforces you have made y submissions, out of which x have been successful. Thus, your current success rate on Codeforces is equal to x / y.
Your favorite rational number in the [0;1] range is p / q. Now you wonder: what is the smallest number of submissions you have to make if you want your success rate to be p / q?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
Each of the next t lines contains four integers x, y, p and q (0 ≤ x ≤ y ≤ 109; 0 ≤ p ≤ q ≤ 109; y > 0; q > 0).
It is guaranteed that p / q is an irreducible fraction.
Hacks. For hacks, an additional constraint of t ≤ 5 must be met.
题意
已经有y次提交,x次正确,问需要几次补充,能达到p/q的正确率?
唉,我想到枚举,为啥就想不到二分呢。。。
官方提示
二分。
代码
#include <bits/stdc++.h>
using namespace std;
int t;
using ll=long long;
ll x,y,p,q;
bool check(ll k)
{
if (k*p>=x && q*k>=y && q*k-y>=k*p-x) return 1;
else return 0;
}
ll binary(ll l,ll r)
{
if (l==r)
{
if (check(l)) return (q*l-y);
else return -1;
}
ll mid{(l+r)>>1};
if (check(mid)) return binary(l,mid);
else return binary(mid+1,r);
}
int main()
{
cin>>t;
while (t--)
{
cin>>x>>y>>p>>q;
cout<<binary(1,0x7fffffff)<<endl;
}
}
D. Dynamic Problem Scoring
题面
题意
代码
//
E. Prairie Partition
题面
题意
代码
//
F. Perishable Roads
题面
题意
代码
//
赛后总结
作为一名选手,我还是敬佩和欣赏tourist的,但是作为出题的人话。。
我真想打死他。
题目难到我的话,我其实无所谓的=A=
但是这种题意让人理解起来真的很恶心。
很令人失望的一场比赛
还有坑 日后再填
比赛链接
http://codeforces.com/contest/807
Codeforces Round#412 Div.2的更多相关文章
- Codeforces Round #412 Div. 2 第一场翻水水
大半夜呆在机房做题,我只感觉智商严重下降,今天我脑子可能不太正常 A. Is it rated? time limit per test 2 seconds memory limit per test ...
- Codeforces Round #412 Div. 2 补题 D. Dynamic Problem Scoring
D. Dynamic Problem Scoring time limit per test 2 seconds memory limit per test 256 megabytes input s ...
- Codeforces Round #412 (Div. 2)ABCD
tourist的剧毒contest,题干长到让人不想做... A.看不太懂题意直接看下面input output note n组里有两数不一样的一组就rated 否则单调不增为maybe,否则unra ...
- Codeforces Round #366 (Div. 2) ABC
Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate ...
- Codeforces Round #354 (Div. 2) ABCD
Codeforces Round #354 (Div. 2) Problems # Name A Nicholas and Permutation standard input/out ...
- Codeforces Round #368 (Div. 2)
直达–>Codeforces Round #368 (Div. 2) A Brain’s Photos 给你一个NxM的矩阵,一个字母代表一种颜色,如果有”C”,”M”,”Y”三种中任意一种就输 ...
- cf之路,1,Codeforces Round #345 (Div. 2)
cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅..... ...
- Codeforces Round #279 (Div. 2) ABCDE
Codeforces Round #279 (Div. 2) 做得我都变绿了! Problems # Name A Team Olympiad standard input/outpu ...
- Codeforces Round #262 (Div. 2) 1003
Codeforces Round #262 (Div. 2) 1003 C. Present time limit per test 2 seconds memory limit per test 2 ...
随机推荐
- ubuntu下sudo apt-get update Sources 404 Not Found 解决方法
刚安装了ubuntu之后的主要安装命令无非就是apt-get install了,然而很多都在这里就夭折了. 使用apt-get install ***需要先执行apt-get update 加载文件包 ...
- DOS下的安全空间
我们需要直接向一段内存中写入内容: 这段内存空间不应存放系统或其他程序的数据或代码,否则写入操作很可能引发错误: DOS方式下,一般情况,0:200~0:2ff空间中没有系统或其他程序的数据或代码; ...
- SideBar 选择城市时右侧边上的 选择bar
需要定义一个SideBar的视图类 在布局文件中引用 同时在布局中设置一个textView默认不可见 当触摸时才显示 在调用的Activity中 sideBar.setOnTouchingL ...
- jquery关于attr和prop的差异
转自:http://www.jb51.net/article/88068.htm 处理像checkbox,radio和select这样的元素时,经常会发现明明使用了attr设置了selected或ch ...
- Error starting daemon: error initializing graphdriver: driver not supported
Error starting daemon: error initializing graphdriver: driver not supported systemctl stop docker rm ...
- PopupWindow与Edittext结合使用所遇到的坑
PopupWindow与Edittext结合使用一起实现目的:既可以编辑输入想要的内容,还可以通过下拉列表来实现内容的选择. 我就是这样的一个目的,结果很简单的目的却遇到了很大的坑,下面我将把我遇到的 ...
- BOM心得-定时器
写在前面的话:之前一直以为定时器的返回值是Object类型,所以timer初始化也是写null,今天发现返回值是number,进而发觉这个返回值代表的是定时器的索引,指代这是第几个定时器 个人觉得只用 ...
- (O)JS高阶函数应用——函数节流
在一些函数需被频繁调用的场景,如:window.onresize.mousemove.scroll滚动事件.上传进度等等,操作频繁导致性能消耗过高,而造成浏览器卡顿现象,我们可以通过函数节流的方式解决 ...
- 论坛:获取当前原始请求中的远程IP地址
topic.setIpAddr(ServletActionContext.getRequest().getRemoteAddr());//当前原始请求中的远程IP地址
- sqli-labs:5-6,盲注
思考1:当# --+都被过滤时,只能考虑闭合处理 思考2:union联合注入时必须先判断字段长度 eg. id=1' order by 3 and '1'='1 sqli5: 首先判断出对id经过了' ...