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. Struts+Spring搭建

    前言 本文以Tomcat为j2ee容器,数据库为Sqlserver2005进行说明.Struts版本为2.3.15.3,Spring版本为3.2.5 Spring简介 Spring也是appache下 ...

  2. ListView OnItemClickListener position 索引不正确

    在使用ListView添加如下代码时 listview.setOnItemClickListener(new OnItemClickListener() { @Override public void ...

  3. 我对Laravel ThinkPHP Yii symfony2 CI cakephp 的看法

    这是我的真心体会,在尝试使用Laravel.ThinkPHP.Yii.symfony2.CI.cakephp.Yii2 之后的真实想法(default7#zbphp.com). 1)ThinkPHP ...

  4. rac 10g 加入节点具体解释

    目标: 当前我环境中是有两个节点RAC1和RAC2 节点.如今添加一个RAC3节点.   概要:为现有的Oracle10g RAC 加入节点大致包含下面步骤: 1. 配置新的server节点上的硬件及 ...

  5. Adobe/Flash Media Server 5.0 linux 64位系统下的安装

    一.下载 Adobe/Flash MS5.0下载地址: http://fs1.d-h.st/download/00036/VOt/adobemediaserver_5_ls1_linux64.tar. ...

  6. SQL数值函数

    /*abs(n)返回参数n所指定数值的绝对值(如果参数值为NULL,则返回结果为NULL,下同).*/--SELECT ABS(-3.14) FROM DUAL; --3.14 /*round(n[, ...

  7. SVN 代码下载,上传

    代码下载,如: svn co https://99.99.16.1:8080/svn/pavenas/webpy --username bg 代码上传,如: svn commit -m "备 ...

  8. sql server 系统表系统视图 及作用说明

    sql server 系统视图,可分为: 目录视图   兼容性视图    动态管理视图和函数    信息架构视图    复制视图 系统表: sysaltfiles主数据库 保存数据库的文件 sysch ...

  9. NPM与调试工具的使用

    1)NPM 2)nodemon 自动监视文件的变化并重启应用 3)pm2 启动node,资源共享 4)node-inspector node调试工具 5)Chrome Developer Tools

  10. Oracle之事务

    一,oracle的事务: 是指对数据操作的一系列动作的统称.即:事务的任务便是使数据库从一种状态变换成为另一种状态,这不同于文件系统,它是数据库所特用的. 事务有四大特性(ACID): 1,原子性(a ...