CF-807A
A. Is it rated?time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.
InputThe first line contains a single integer n (2 ≤ n ≤ 1000) — the number of round participants.
Each of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 4126) — the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
OutputIf the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examplesinput6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884outputratedinput4
1500 1500
1300 1300
1200 1200
1400 1400outputunratedinput5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699outputmaybeNoteIn the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
题意:
对于给出的数字对,若前后不相等,则为rated,若相等且按降序排列则为maybe,否则为unrated。
附AC代码:
#include<bits/stdc++.h>
using namespace std; int a[],c[]; int main(){
int n,b,flag=;
cin>>n;
for(int i=;i<n;i++){
cin>>a[i]>>b;
if(a[i]!=b)
flag=;
c[i]=a[i];
}
if(flag){
cout<<"rated"<<endl;
return ;
}
flag=;
sort(c,c+n,greater<int>());//对C数组降序排列
for(int i=;i<n;i++){
//cout<<a[i]<<" "<<c[i]<<endl;
if(a[i]!=c[i]){
flag=;
break;
}
}
if(flag){
cout<<"unrated"<<endl;
return ;
}
cout<<"maybe"<<endl;
return ;
}
CF-807A的更多相关文章
- ORA-00494: enqueue [CF] held for too long (more than 900 seconds) by 'inst 1, osid 5166'
凌晨收到同事电话,反馈应用程序访问Oracle数据库时报错,当时现场现象确认: 1. 应用程序访问不了数据库,使用SQL Developer测试发现访问不了数据库.报ORA-12570 TNS:pac ...
- 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..... ...
- cf Round 613
A.Peter and Snow Blower(计算几何) 给定一个点和一个多边形,求出这个多边形绕这个点旋转一圈后形成的面积.保证这个点不在多边形内. 画个图能明白 这个图形是一个圆环,那么就是这个 ...
- ARC下OC对象和CF对象之间的桥接(bridge)
在开发iOS应用程序时我们有时会用到Core Foundation对象简称CF,例如Core Graphics.Core Text,并且我们可能需要将CF对象和OC对象进行互相转化,我们知道,ARC环 ...
- [Recommendation System] 推荐系统之协同过滤(CF)算法详解和实现
1 集体智慧和协同过滤 1.1 什么是集体智慧(社会计算)? 集体智慧 (Collective Intelligence) 并不是 Web2.0 时代特有的,只是在 Web2.0 时代,大家在 Web ...
- CF memsql Start[c]UP 2.0 A
CF memsql Start[c]UP 2.0 A A. Golden System time limit per test 1 second memory limit per test 256 m ...
- CF memsql Start[c]UP 2.0 B
CF memsql Start[c]UP 2.0 B B. Distributed Join time limit per test 1 second memory limit per test 25 ...
- CF #376 (Div. 2) C. dfs
1.CF #376 (Div. 2) C. Socks dfs 2.题意:给袜子上色,使n天左右脚袜子都同样颜色. 3.总结:一开始用链表存图,一直TLE test 6 (1)如果需 ...
- CF #375 (Div. 2) D. bfs
1.CF #375 (Div. 2) D. Lakes in Berland 2.总结:麻烦的bfs,但其实很水.. 3.题意:n*m的陆地与水泽,水泽在边界表示连通海洋.最后要剩k个湖,总要填掉多 ...
- CF #374 (Div. 2) D. 贪心,优先队列或set
1.CF #374 (Div. 2) D. Maxim and Array 2.总结:按绝对值最小贪心下去即可 3.题意:对n个数进行+x或-x的k次操作,要使操作之后的n个数乘积最小. (1)优 ...
随机推荐
- start-dfs.sh 和 start-all.sh的区别
start-dfs.sh 只启动namenode 和datanode, start-all.sh还包括yarn的resourcemanager 和nodemanager 之前就所以因为只启动了star ...
- 【TensorFlow-windows】(三) 多层感知器进行手写数字识别(mnist)
主要内容: 1.基于多层感知器的mnist手写数字识别(代码注释) 2.该实现中的函数总结 平台: 1.windows 10 64位 2.Anaconda3-4.2.0-Windows-x86_64. ...
- poj1125--Floyd
题解: 有N个股票经济人能够互相传递消息.他们之间存在一些单向的通信路径.如今有一个消息要由某个人開始传递给其它全部人.问应该由哪一个人来传递,才干在最短时间内让全部人都接收到消息. 显然,用Floy ...
- android handler looper
http://www.cnblogs.com/plokmju/p/android_Handler.html
- EasyDarwin Streaming Server对Task的调用方法
我们在EasyDarwin流媒体服务器的二次开发过程中,经常会需要定义自己的Task类,例如在EasyDarwin中,RTSPSessioin.HTTPSession.RTCPTask等,都是Task ...
- java设计模式之综述
一.什么是设计模式 设计模式是一套被反复使用的.多数人知晓的.经过分类编目的.代码设计经验的总结.使用设计模式是为了重用代码.让代码更容易被他人理解.保证代码可靠性. 毫无疑问,设计模式于己于他人于系 ...
- 九度OJ 1109:连通图 (最小生成树)
时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:2783 解决:1432 题目描述: 给定一个无向图和其中的所有边,判断这个图是否所有顶点都是连通的. 输入: 每组数据的第一行是两个整数 n ...
- AndroidPageObjectTest_TimeOutManagement.java
以下代码使用ApiDemos-debug.apk进行测试 //这个脚本用于演示PageFactory的功能:设置timeout时间. package com.saucelabs.appium; imp ...
- spring cloud服务注册与发现无法发现的可能原因
1.注册中心服务端默认90秒检测一次,看服务是否还存活,不存活则删除掉服务,还存活则继续注册上去 2. spring: profiles: dev cloud: config: name: clean ...
- HDU 1878(1Y) (判断欧拉回路是否存在 奇点个数为0 + 一个联通分量 *【模板】)
欧拉回路 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submis ...