B. Vasya and Wrestling
time limit per test 2 seconds
memory limit per test 256 megabytes
input standard input
output standard output

Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.

When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.

If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.

Input

The first line contains number n — the number of techniques that the wrestlers have used (1 ≤ n ≤ 2·105).

The following n lines contain integer numbers ai (|ai| ≤ 109, ai ≠ 0). If ai is positive, that means that the first wrestler performed the technique that was awarded with ai points. And if ai is negative, that means that the second wrestler performed the technique that was awarded with ( - ai) points.

The techniques are given in chronological order.

Output

If the first wrestler wins, print string "first", otherwise print "second"

Sample test(s)
input
5
1
2
-3
-4
3
output
second
input
3
-1
-2
3
output
first
input
2
4
-4
output
second
Note

Sequence x  =  x1x2... x|x| is lexicographically larger than sequence y  =  y1y2... y|y|, if either |x|  >  |y| and x1  =  y1,  x2  =  y2, ... ,  x|y|  =  y|y|, or there is such number r (r  <  |x|, r  <  |y|), that x1  =  y1,  x2  =  y2,  ... ,  xr  =  yr and xr  +  1  >  yr  +  1.

We use notation |a| to denote length of sequence a.

模拟

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#define ll long long
using namespace std;
ll i,j,t,n,m,l,r,k,z,y,x;
ll a[200005],b[200005];
ll suma,sumb;
int main()
{
scanf("%I64d",&n);
a[0]=b[0]=0;
suma=sumb=0;
for (i=1;i<=n;i++)
{
scanf("%I64d",&x);
if (x>0)
{
a[++a[0]]=x;
suma+=x;
}
if (x<0)
{
b[++b[0]]=-x;
sumb+=-x;
}
t=x;
}
if (suma>sumb) printf("first\n");
else if (suma<sumb) printf("second\n");
else
{
for (i=1;i<=max(a[0],b[0]);i++) if (a[i]!=b[i]) break;
if (a[i]>b[i]) printf("first\n");
else if (a[i]<b[i]) printf("second\n");
else
{
if (t>0) printf("first\n");
else printf("second\n");
}
}
return 0;
}

cf493B Vasya and Wrestling的更多相关文章

  1. Codeforces Round #281 (Div. 2) B. Vasya and Wrestling 水题

    B. Vasya and Wrestling 题目连接: http://codeforces.com/contest/493/problem/B Description Vasya has becom ...

  2. CodeForces 493B Vasya and Wrestling 【模拟】

    B. Vasya and Wrestling time limit per test 2 seconds memory limit per test 256 megabytes input stand ...

  3. codeforces 493B.Vasya and Wrestling 解题报告

    题目链接:http://codeforces.com/problemset/problem/493/B 题目意思:给出 n 个 techniques,每个 technique 的值为 ai. ai & ...

  4. Codeforces Round #281 (Div. 2)

    题目链接:http://codeforces.com/contest/493 A. Vasya and Football Vasya has started watching football gam ...

  5. codeforces493B

    Vasya and Wrestling CodeForces - 493B Vasya has become interested in wrestling. In wrestling wrestle ...

  6. Codeforces Round #281 (Div. 2) B 模拟

    B. Vasya and Wrestling time limit per test 2 seconds memory limit per test 256 megabytes input stand ...

  7. Milliard Vasya's Function-Ural1353动态规划

    Time limit: 1.0 second Memory limit: 64 MB Vasya is the beginning mathematician. He decided to make ...

  8. CF460 A. Vasya and Socks

    A. Vasya and Socks time limit per test 1 second memory limit per test 256 megabytes input standard i ...

  9. 递推DP URAL 1353 Milliard Vasya's Function

    题目传送门 /* 题意:1~1e9的数字里,各个位数数字相加和为s的个数 递推DP:dp[i][j] 表示i位数字,当前数字和为j的个数 状态转移方程:dp[i][j] += dp[i-1][j-k] ...

随机推荐

  1. Apache HttpClient组件封装工具类

    package com.mengyao.spider.utils; import java.util.ArrayList;import java.util.HashMap;import java.ut ...

  2. Python模拟登录实战(一)

    今天,学习了模拟登录新浪微博.模拟登录主要有两种方式,一.利用Cookie:二.模仿浏览器的请求,发送表单. 法一: Cookie:指某些网站为了辨别用户身份而储存在用户本地终端上的数据(通常经过加密 ...

  3. NetAnalyzer笔记 目录

    目录 NetAnalyzer笔记 之 一 开篇语 NetAnalyzer笔记 之 二 简单的协议分析 NetAnalyzer笔记 之 三 用C++做一个抓包程序 NetAnalyzer笔记 之 四 C ...

  4. 正则表达式:网页爬虫:从TXT中获取邮箱地址(获取的练习,缺点:一行只能匹配一个)

    import java.util.regex.*; import java.io.*; class L { public static void main(String[] args) throws  ...

  5. CSU1659: Graph Center(最短路)

    Description The center of a graph is the set of all vertices of minimum eccentricity, that is, the s ...

  6. Python3.2官方文档翻译--继承

    6.5 继承 当然,一门语言特性假设不支持继承那么名称类就失去了价值.子类继承父类的方法例如以下: class DerivedClassName(BaseClassName): <stateme ...

  7. 收集的URL

    *******************************************看文章的好地方************************************** http://www. ...

  8. JavaScript异步编程 ( 一 )

    1. 异步编程 Javascript语言的执行环境是"单线程"(single thread).所谓"单线程",就是指一次只能完成一件任务.如果有多个任务,就必须 ...

  9. linux中文乱码问题及locale详解

    一.修改系统默认语言及中文乱码问题记录系统默认使用语言的文件是/etc/sysconfig/i18n,如果默认安装的是中文的系统,i18n的内容如下: LANG="zh_CN.UTF-8&q ...

  10. django: startproject

    python 的 django 框架的安装教程很多,这里不列举安装过程,直接开始记开发应用过程. 1  startprojec,新建项目 $ django-admin.py startproject ...