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 ;
}
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