Codeforces Round #357 (Div. 2) A
1 second
256 megabytes
standard input
standard output
Codeforces user' handle color depends on his rating — it is red if his rating is greater or equal to 2400; it is orange if his rating is less than 2400 but greater or equal to 2200, etc. Each time participant takes part in a rated contest, his rating is changed depending on his performance.
Anton wants the color of his handle to become red. He considers his performance in the rated contest to be good if he outscored some participant, whose handle was colored red before the contest and his rating has increased after it.
Anton has written a program that analyses contest results and determines whether he performed good or not. Are you able to do the same?
The first line of the input contains a single integer n (1 ≤ n ≤ 100) — the number of participants Anton has outscored in this contest .
The next n lines describe participants results: the i-th of them consists of a participant handle namei and two integers beforei and afteri ( - 4000 ≤ beforei, afteri ≤ 4000) — participant's rating before and after the contest, respectively. Each handle is a non-empty string, consisting of no more than 10 characters, which might be lowercase and uppercase English letters, digits, characters «_» and «-» characters.
It is guaranteed that all handles are distinct.
Print «YES» (quotes for clarity), if Anton has performed good in the contest and «NO» (quotes for clarity) otherwise.
3
Burunduk1 2526 2537
BudAlNik 2084 2214
subscriber 2833 2749
YES
3
Applejack 2400 2400
Fluttershy 2390 2431
Pinkie_Pie -2500 -2450
NO
In the first sample, Anton has outscored user with handle Burunduk1, whose handle was colored red before the contest and his rating has increased after the contest.
In the second sample, Applejack's rating has not increased after the contest, while both Fluttershy's and Pinkie_Pie's handles were not colored red before the contest.
题意:存在红名(rate>=2400的)并且涨分的输出YES 否则输出NO
题解:水题
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
int n;
char a[];
int be,af;
int main()
{
scanf("%d",&n);
int flag=;
getchar();
for(int i=;i<=n;i++)
{
scanf("%s %d %d",a,&be,&af);
if(be>=&&af>be)
flag=;
}
if(flag)
printf("YES\n");
else
printf("NO\n");
return ;
}
Codeforces Round #357 (Div. 2) A的更多相关文章
- Codeforces Round #357 (Div. 2) E. Runaway to a Shadow 计算几何
E. Runaway to a Shadow 题目连接: http://www.codeforces.com/contest/681/problem/E Description Dima is liv ...
- Codeforces Round #357 (Div. 2) D. Gifts by the List 水题
D. Gifts by the List 题目连接: http://www.codeforces.com/contest/681/problem/D Description Sasha lives i ...
- Codeforces Round #357 (Div. 2) C. Heap Operations 模拟
C. Heap Operations 题目连接: http://www.codeforces.com/contest/681/problem/C Description Petya has recen ...
- Codeforces Round #357 (Div. 2) B. Economy Game 水题
B. Economy Game 题目连接: http://www.codeforces.com/contest/681/problem/B Description Kolya is developin ...
- Codeforces Round #357 (Div. 2) A. A Good Contest 水题
A. A Good Contest 题目连接: http://www.codeforces.com/contest/681/problem/A Description Codeforces user' ...
- Codeforces Round #357 (Div. 2) E 计算几何
传说中做cf不补题等于没做 于是第一次补...这次的cf没有做出来DE D题的描述神奇 到现在也没有看懂 于是只补了E 每次div2都是hack前2~3题 终于打出一次hack后的三题了...希望以后 ...
- Codeforces Round #357 (Div. 2) 优先队列+模拟
C. Heap Operations time limit per test 1 second memory limit per test 256 megabytes input standard i ...
- Codeforces Round #357 (Div. 2) C
C. Heap Operations time limit per test 1 second memory limit per test 256 megabytes input standard i ...
- Codeforces Round #357 (Div. 2) B
B. Economy Game time limit per test 1 second memory limit per test 256 megabytes input standard inpu ...
随机推荐
- SQLSERVER存储过程基本语法使用
一.定义变量 --简单赋值 declare @a int print @a --使用select语句赋值 ) select @user1='张三' print @user1 ) print @user ...
- Thinkphp 取消Url默认模块的现实
例子http://www.tp.com/home/index/index 想要现实的效果是:http://www.tp.com/index/index 1是通过配置路由来达到目的 2通过配置首页的入口 ...
- 页面刷新 方法总结 JSP刷新
1) <meta http-equiv="refresh"content="10;url=跳转的页面"> 10表示间隔10秒刷新一次 2) < ...
- HTML基本教程,及一些基本常用标签。
HTML基本结构,及常用标签 <DOCTYPE html> <html> <head> <meta charset="UTF-8" /&g ...
- 43_2.VUE学习之--不使用组件computed计算属性超简单的实现美团购物车原理
<!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8&quo ...
- MySQL查询优化 对not in 、in 的优化
因为 not in不走索引,所以不在不得已情况下,就不要使用not in 下面使用 join 来替代not in 做查询 select ID from A where ID not in (selec ...
- 设置默认以管理员运行的WinForm
右键工程名, 属性; 选择"安全性"; 勾选"启用ClickOnce安全设置"与"这是完全可信的应用程序"; 退出该页面, app.mani ...
- 20145202马超 《Java程序设计》第五周学习总结
异常:程序在运行的时候出现不正正常的情况 由来:问题也是可以通过java对不正常情况进行描述后的对象的体现. 问题的划分:(1).严重的问题,java通过error类进行描述,对于error一般不编写 ...
- Javascript 属性高级写法
http://www.cnblogs.com/YuanSong/p/3899287.html
- shell脚本递归删除空文件夹
有时我们需要递归删除空文件夹,网上找了一下,没有发现比较好的脚本,于是自己动手写了一个 脚本 #!/bin/bash # author: 十年后的卢哥哥(http://www.cnblogs.com/ ...