B. Parade
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Very soon there will be a parade of victory over alien invaders in Berland. Unfortunately, all soldiers died in the war and now the army consists of entirely new recruits, many of whom do not even know from which leg they should begin to march. The civilian population also poorly understands from which leg recruits begin to march, so it is only important how many soldiers march in step.

There will be n columns participating in the parade, the i-th column consists of li soldiers, who start to march from left leg, and ri soldiers, who start to march from right leg.

The beauty of the parade is calculated by the following formula: if L is the total number of soldiers on the parade who start to march from the left leg, and R is the total number of soldiers on the parade who start to march from the right leg, so the beauty will equal |L - R|.

No more than once you can choose one column and tell all the soldiers in this column to switch starting leg, i.e. everyone in this columns who starts the march from left leg will now start it from right leg, and vice versa. Formally, you can pick no more than one index i and swap values li and ri.

Find the index of the column, such that switching the starting leg for soldiers in it will maximize the the beauty of the parade, or determine, that no such operation can increase the current beauty.

Input

The first line contains single integer n (1 ≤ n ≤ 105) — the number of columns.

The next n lines contain the pairs of integers li and ri (1 ≤ li, ri ≤ 500) — the number of soldiers in the i-th column which start to march from the left or the right leg respectively.

Output

Print single integer k — the number of the column in which soldiers need to change the leg from which they start to march, or 0 if the maximum beauty is already reached.

Consider that columns are numbered from 1 to n in the order they are given in the input data.

If there are several answers, print any of them.

Examples
Input
3
5 6
8 9
10 3
Output
3
Input
2
6 5
5 6
Output
1
Input
6
5 9
1 3
4 8
4 5
23 54
12 32
Output
0
Note

In the first example if you don't give the order to change the leg, the number of soldiers, who start to march from the left leg, would equal 5 + 8 + 10 = 23, and from the right leg — 6 + 9 + 3 = 18. In this case the beauty of the parade will equal |23 - 18| = 5.

If you give the order to change the leg to the third column, so the number of soldiers, who march from the left leg, will equal 5 + 8 + 3 = 16, and who march from the right leg — 6 + 9 + 10 = 25. In this case the beauty equals |16 - 25| = 9.

It is impossible to reach greater beauty by giving another orders. Thus, the maximum beauty that can be achieved is 9.

/*
找出最大的差就可以
*/
#include<bits/stdc++.h>
using namespace std;
int l[],r[];
int main()
{
//freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
int n;
int L=,R=;
scanf("%d",&n);
int cur=;
for(int i=;i<n;i++)
{
scanf("%d%d",&l[i],&r[i]);
L+=l[i];
R+=r[i];
}
int Max=abs(L-R);
int Maxi=;
for(int i=;i<n;i++)
{
L-=l[i];
L+=r[i];
R-=r[i];
R+=l[i];
if(abs(L-R)>Max)
{
Max=abs(L-R);
Maxi=i+;
}
L+=l[i];
L-=r[i];
R+=r[i];
R-=l[i];
}
printf("%d\n",Maxi);
return ;
}

Codeforces 378B. Parade的更多相关文章

  1. CodeForces 733B Parade

    B. Parade time limit per test1 second memory limit per test256 megabytes inputstandard input outputs ...

  2. [Codeforces 35E] Parade

    Link: Codeforces 35E 传送门 Brief Intro: 给定$n$个矩形,求出轮廓线的所有顶点 Solution: 对于此类可拆分成多个事件点的题目,使用扫描线的方式 将每个矩形分 ...

  3. Codeforces 35E Parade 扫描线

    题意: 给出\(n\)个底边在\(x\)轴上的矩形,求外面的轮廓线顶点. 分析: 将每个矩形拆成两个事件:\(\\\{ l, y, + \\\}\)和\(\\\{ r, y, - \\\}\)分别表示 ...

  4. Codeforces 35E Parade 扫描线 + list

    主题链接:点击打开链接 意甲冠军:特定n矩阵(总是接近底部x轴) 然后找到由上面的矩阵所包围的路径,的点 给定n 以下n行给定 y [x1, x2] 表示矩阵的高度和2个x轴坐标 思路: 扫描线维护每 ...

  5. 线段树详解 (原理,实现与应用)(转载自:http://blog.csdn.net/zearot/article/details/48299459)

    原文地址:http://blog.csdn.net/zearot/article/details/48299459(如有侵权,请联系博主,立即删除.) 线段树详解    By 岩之痕 目录: 一:综述 ...

  6. 【非常高%】【codeforces 733B】Parade

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

  7. Codeforces Beta Round #35 (Div. 2) E. Parade(扫描线)

    题目链接 只要会做,周长并,这题肯定有思路. 有个小地方敲错了,细心啊,扫描线,有一段时间没写过了,还有注意排序的问题,很重要. #include <iostream> #include ...

  8. CodeForces - 1250J The Parade 二分

    题目 题意: 一共n种身高,每一个士兵有一个身高.你需要把他们安排成k行(士兵不需要全部安排),每一行士兵身高差距小于等于1.你要找出来最多能安排多少士兵 题解: 这道题很容易就能看出来就是一道二分, ...

  9. Codeforces水题集合[14/未完待续]

    Codeforces Round #371 (Div. 2) A. Meeting of Old Friends |B. Filya and Homework A. Meeting of Old Fr ...

随机推荐

  1. Apache Spark 2.2.0 中文文档 - Spark SQL, DataFrames and Datasets Guide | ApacheCN

    Spark SQL, DataFrames and Datasets Guide Overview SQL Datasets and DataFrames 开始入门 起始点: SparkSession ...

  2. Floyd算法(最短路)

    如题,这是最短路算法Floyd. Floyd,是只有五行的代码. 简单,易懂.O(N的三方)的时间也可以. 遇到简单的就这么用. #include<iostream> #include&l ...

  3. Java虚拟机-运行时数据区域

    Java虚拟机管理的内存包括如图所示的运行时数据区域: 下面分别进行介绍: 1)程序计数器(Program Counter Register) 占用的内存空间比较小,主要作用就是标识当前线程执行的字节 ...

  4. yum软件管理器,及yum源配置

    说到yum源就必须说到linux系统中特有的依赖关系问题,yum就是为了解决依赖关系而存在的.yum源就相当是一个目录项,当我们使用yum机制安装软件时,若需要安装依赖软件,则yum机制就会根据在yu ...

  5. 认识jQuery的Promise

    先前了解了ES6的Promise对象,来看看jQuery中的Promise,也就是jQuery的Deferred对象. 打开浏览器的控制台先. <script> var defer = $ ...

  6. 童话故事 --- 通信协议之 HDLC 浅析

    高飞狗: "高飞的白鹭浮水的鹅,唐诗里有画-" 布鲁托: "高飞狗,又在做你的高飞梦哪!" 高飞狗: "哈罗,布鲁托,这几天好郁闷呐!" 布 ...

  7. Cmder 软件中修改λ符号方法

    以前的版本 网上都有,我就不介绍了,  只介绍现在的 1. 打开Cmder软件安装位置 2. 打开vendor文件夹 profile.ps1文件 3. 找到第77行  Write-Host " ...

  8. ThinkJS框架入门详细教程(二)新手入门项目

    一.准备工作 参考前一篇:ThinkJS框架入门详细教程(一)开发环境 安装thinkJS命令 npm install -g think-cli 监测是否安装成功 thinkjs -v 二.创建项目 ...

  9. Quartz.NET实现作业调度

    一.Quartz.NET介绍 Quartz.NET是一个强大.开源.轻量的作业调度框架,是 OpenSymphony 的 Quartz API 的.NET移植,用C#改写,可用于winform和asp ...

  10. 用C#实现字符串相似度算法(编辑距离算法 Levenshtein Distance)

    在搞验证码识别的时候需要比较字符代码的相似度用到"编辑距离算法",关于原理和C#实现做个记录. 据百度百科介绍: 编辑距离,又称Levenshtein距离(也叫做Edit Dist ...