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 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 Input
5
1
2
-3
-4
3
Sample Output
second
Hint
题意
有n个分数,然后输入,如果是正数,那么就属于第一个人,如果是负数,那么就属于第二个人。
谁分数大,谁胜利。
如果分数一样,那么看字典序大小。
如果字典序大小一样,看最后的那个数属于谁
题解:
读题比做题难……
模拟一下就好了
代码
#include<bits/stdc++.h>
using namespace std;
vector<int> a,b;
int judge()
{
for(int i=0;i<a.size()&&i<b.size();i++)
{
if(a[i]>b[i])return 1;
if(b[i]>a[i])return 2;
}
return 0;
}
int main()
{
int n;
long long sum1=0,sum2=0;
scanf("%d",&n);
int flag = 0;
for(int i=0;i<n;i++)
{
int x;scanf("%d",&x);
if(x>0)
{
a.push_back(x);
sum1+=x;
flag = 1;
}
else
{
b.push_back(-x);
sum2+=-x;
flag = 2;
}
}
if(sum1>sum2)return puts("first"),0;
if(sum2>sum1)return puts("second"),0;
if(judge()==1)return puts("first"),0;
else if(judge()==2)return puts("second"),0;
if(flag==1)return puts("first"),0;
else return puts("second"),0;
}
Codeforces Round #281 (Div. 2) B. Vasya and Wrestling 水题的更多相关文章
- Codeforces Round #322 (Div. 2) A. Vasya the Hipster 水题
A. Vasya the Hipster Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/581/p ...
- Codeforces Round #281 (Div. 2) D. Vasya and Chess 水
D. Vasya and Chess time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- Codeforces Round #297 (Div. 2)A. Vitaliy and Pie 水题
Codeforces Round #297 (Div. 2)A. Vitaliy and Pie Time Limit: 2 Sec Memory Limit: 256 MBSubmit: xxx ...
- Codeforces Round #290 (Div. 2) A. Fox And Snake 水题
A. Fox And Snake 题目连接: http://codeforces.com/contest/510/problem/A Description Fox Ciel starts to le ...
- Codeforces Round #373 (Div. 2) B. Anatoly and Cockroaches 水题
B. Anatoly and Cockroaches 题目连接: http://codeforces.com/contest/719/problem/B Description Anatoly liv ...
- Codeforces Round #368 (Div. 2) A. Brain's Photos 水题
A. Brain's Photos 题目连接: http://www.codeforces.com/contest/707/problem/A Description Small, but very ...
- Codeforces Round #359 (Div. 2) A. Free Ice Cream 水题
A. Free Ice Cream 题目连接: http://www.codeforces.com/contest/686/problem/A Description After their adve ...
- Codeforces Round #355 (Div. 2) A. Vanya and Fence 水题
A. Vanya and Fence 题目连接: http://www.codeforces.com/contest/677/problem/A Description Vanya and his f ...
- Codeforces Round #281 (Div. 2) C. Vasya and Basketball 二分
C. Vasya and Basketball time limit per test 2 seconds memory limit per test 256 megabytes input stan ...
随机推荐
- bzoj千题计划190:bzoj4300: 绝世好题
http://www.lydsy.com/JudgeOnline/problem.php?id=4300 f[i] 表示第i位&为1的最长长度 #include<cstdio> # ...
- linux下编译make文件报错“/bin/bash^M: 坏的解释器,使用grep快速定位代码位置
一.linux下编译make文件报错“/bin/bash^M: 坏的解释器 参考文章:http://blog.csdn.net/liuqiyao_01/article/details/41542101 ...
- 【BZOJ】2111: [ZJOI2010]Perm 排列计数 计数DP+排列组合+lucas
[题目]BZOJ 2111 [题意]求有多少1~n的排列,满足\(A_i>A_{\frac{i}{2}}\),输出对p取模的结果.\(n \leq 10^6,p \leq 10^9\),p是素数 ...
- Codeforces 238 div2 B. Domino Effect
题目链接:http://codeforces.com/contest/405/problem/B 解题报告:一排n个的多米诺骨牌,规定,若从一边推的话多米诺骨牌会一直倒,但是如果从两个方向同时往中间推 ...
- caoha
- mac lsof使用查看端口
安装 brew install lsof 在Mac OS系统中,无法使用netstat来查看端口占用情况,可以使用lsof来代替,这种方式在Linux下也适用. sudo lsof -nP -iTCP ...
- 如何将IOS版本的更新下载文件指向到自己的服务器
针对那些使用企业签名但是没有发布到AppSotre的IOS版本APP自动更新问题解决方案: 在apicloud中是这样说明的: 因为要填写plist地址所以不能向安卓那样直接填写服务器文件地址,但是直 ...
- elasticsearch RTF版本介绍
说明:elastic search官方版本没有集成中文分词以及各种插件,需要手动配置,手动编译jar,对Windows用户很不友好.下载地址:https://github.com/medcl/elas ...
- 005_ss-link.info的ping探测工具
用小工具ping.py测试距离您最快的节点 #!/usr/bin/env python # coding: utf-8 """ A pure python ping im ...
- Android 拍摄(横\竖屏)视频的懒人之路
想一想,我们聊过AudioReord,AudioTrack,MediaPlayer,那多媒体四大金刚,就剩下了MediaRecorder了(SoundPool?我这里信号不好···).其实MediaR ...