Educational Codeforces Round 34 (Rated for Div. 2) D - Almost Difference(高精度)
D. Almost Difference
Let's denote a function
You are given an array a consisting of n integers. You have to calculate the sum of d(ai, aj) over all pairs (i, j) such that 1 ≤ i ≤ j ≤ n.
Input
The first line contains one integer n (1 ≤ n ≤ 200000) — the number of elements in a.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — elements of the array.
Output
Print one integer — the sum of d(ai, aj) over all pairs (i, j) such that 1 ≤ i ≤ j ≤ n.
input
5
1 2 3 1 3
output
4
input
4
6 6 5 5
output
0
input
4
6 6 4 4
output
-8
Note
In the first example:
- d(a1, a2) = 0;
- d(a1, a3) = 2;
- d(a1, a4) = 0;
- d(a1, a5) = 2;
- d(a2, a3) = 0;
- d(a2, a4) = 0;
- d(a2, a5) = 0;
- d(a3, a4) = - 2;
- d(a3, a5) = 0;
- d(a4, a5) = 2.
哇,好不容易写到第四题,突然弹出消息说这题爆long long,然后就懵逼了,看了下状态AC的全是Python。赛后发现Hacker在疯狂Hack C++,血赚场?
怎么全世界都会long double,不过瞄到qls也被Hack了,窝q(小纠结.JPG)
#include <bits/stdc++.h>
using namespace std;
map<double,double>a;
int main()
{
int n;
scanf("%d",&n);
long double ans=;
for (int i=;i<n;i++)
{
double x;scanf("%lf",&x);
a[x]++;
ans= ans+ a[x+]-a[x-]+x*(i+-n+i);
}
cout << fixed << setprecision() << ans << endl;
return ;
}
q巨赛后补题代码:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN=;
const ll BASE=1000000000000000000LL;
int a[MAXN];
int main()
{
int n;
scanf("%d",&n);
ll high=,low=;
for(int i=;i<=n;i++)
{
scanf("%d",&a[i]);
low+=1LL*(*(i-)-(n-))*a[i];
while(low>=BASE)low-=BASE,high++;
while(low<)low+=BASE,high--;
}
map<int,int> mp;
for(int i=;i<=n;i++)
{
low-=mp[a[i]-],low+=mp[a[i]+];
while(low>=BASE)low-=BASE,high++;
while(low<)low+=BASE,high--;
mp[a[i]]++;
}
if(high>=- && high<=)printf("%lld\n",high*BASE+low);
else if(high>)printf("%lld%018lld\n",high,low);
else printf("%lld%018lld\n",high+(low>),(low>)*BASE-low);
return ;
}
Educational Codeforces Round 34 (Rated for Div. 2) D - Almost Difference(高精度)的更多相关文章
- Educational Codeforces Round 34 (Rated for Div. 2) A B C D
Educational Codeforces Round 34 (Rated for Div. 2) A Hungry Student Problem 题目链接: http://codeforces. ...
- Educational Codeforces Round 34 (Rated for Div. 2) C. Boxes Packing
C. Boxes Packing time limit per test 1 second memory limit per test 256 megabytes input standard inp ...
- Educational Codeforces Round 34 (Rated for Div. 2)
A. Hungry Student Problem time limit per test 1 second memory limit per test 256 megabytes input sta ...
- Educational Codeforces Round 34 (Rated for Div. 2) B题【打怪模拟】
B. The Modcrab Vova is again playing some computer game, now an RPG. In the game Vova's character re ...
- Educational Codeforces Round 71 (Rated for Div. 2)-F. Remainder Problem-技巧分块
Educational Codeforces Round 71 (Rated for Div. 2)-F. Remainder Problem-技巧分块 [Problem Description] ...
- Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship
Problem Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship Time Limit: 2000 mSec P ...
- Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems(动态规划+矩阵快速幂)
Problem Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems Time Limit: 3000 mSec P ...
- Educational Codeforces Round 43 (Rated for Div. 2)
Educational Codeforces Round 43 (Rated for Div. 2) https://codeforces.com/contest/976 A #include< ...
- Educational Codeforces Round 35 (Rated for Div. 2)
Educational Codeforces Round 35 (Rated for Div. 2) https://codeforces.com/contest/911 A 模拟 #include& ...
随机推荐
- 如何配置Notepad++的C_C++语言开发环境
相信很多人用notepad++,但把其配置成为C/C++还是需要小折腾一下的.本人在网上找了很长时间,也没有一个统一的答案,而且很多人说的方法根本不管用,而且也不够通用,所以还是自己摸索了一下,分享给 ...
- Linux 模块管理
查看模块信息 modinfo module-name 加载模块 insmod module-name 卸载模块 rmmod module-name 生成模块依赖 cd /lib/module/`una ...
- Java8 改进的匿名内部类:
1.匿名内部类适合创建那种只需要一次使用的类 2.匿名内部类定义格式: new 实现接口() | 父类构造器(实参列表){ //匿名内部类类体部分 } 3.从上面定义格式可以看出,匿名内部类必须实现一 ...
- [PHP] 转义字符 Escape character
\n is a symbol for new line \t is a symbol for tab and \r is for 'return'
- Xcode.Subproject.And.Framework
1. Easy Xcode Static Library Subprojects and Submodules http://www.blog.montgomerie.net/easy-xcode-s ...
- python使用wmi模块获取windows下的系统信息监控系统-乾颐堂
Python用WMI模块获取Windows系统的硬件信息:硬盘分区.使用情况,内存大小,CPU型号,当前运行的进程,自启动程序及位置,系统的版本等信息. 本文实例讲述了python使用wmi模块获取w ...
- Zookeeper 系列(三)Zookeeper API
Zookeeper 系列(三)Zookeeper API 本节首先介绍 Zookeeper 的 Shell 命令,再对 Java 操作 Zookeeper 的三种方式进行讲解,本节先介绍 Zookee ...
- div和span元素的区别
2个都是用来划分区间但是没有实际语义的标签,差别就在于div是块级元素,不会其他元素在同一行;span是内联元素,可以与其他元素位于同一行. DIV 和 SPAN 元素最大的特点是默认都没有对元素内的 ...
- 2018.09.23 bzoj3143: [Hnoi2013]游走(dp+高斯消元)
传送门 显然只需要求出所有边被经过的期望次数,然后贪心把边权小的边定城大的编号. 所以如何求出所有边被经过的期望次数? 显然这只跟边连接的两个点有关. 于是我们只需要求出两个点被经过的期望次数. 对于 ...
- 2018.08.21 NOIP模拟 unlock(模拟+找规律)
unlock 描述 经济危机席卷全球,L国也收到冲击,大量人员失业. 然而,作为L国的风云人物,X找到了自己的新工作.从下周开始,X将成为一个酒店的助理锁匠,当然,他得先向部门领导展示他的开锁能力. ...