Codeforces 1009C： Annoying Present

C. Annoying Present

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Alice got an array of length $n$ as a birthday present once again! This is the third year in a row!

And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.

Bob has chosen $m$ changes of the following form. For some integer numbers $x$ and $d$, he chooses an arbitrary position $i$ ($1\le i\le n$) and for every $j\in [1,n]$ adds $x+d\cdot dist(i,j)$ to the value of the $j$-th cell. $dist(i,j)$ is the distance between positions $i$ and $j$ (i.e. $dist(i,j)=|i-j|$, where $|x|$ is an absolute value of $x$).

For example, if Alice currently has an array $[2,1,2,2]$ and Bob chooses position $3$ for $x=-1$ and $d=2$ then the array will become $[2-1+2\cdot 2,1-1+2\cdot 1,2-1+2\cdot 0,2-1+2\cdot 1]$ = $[5,2,1,3]$. Note that Bob can't choose position $i$ outside of the array (that is, smaller than $1$ or greater than $n$).

Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.

What is the maximum arithmetic mean value Bob can achieve?

Input

The first line contains two integers $n$ and $m$ ($1\le n,m\le {10}^5$) — the number of elements of the array and the number of changes.

Each of the next $m$ lines contains two integers $x_i$ and $d_i$ ($-{10}^3\le x_i,d_i\le {10}^3$) — the parameters for the $i$-th change.

Output

Print the maximal average arithmetic mean of the elements Bob can achieve.

Your answer is considered correct if its absolute or relative error doesn't exceed ${10}^{-6}$.

Examples

input

Copy

2 3
-1 3
0 0
-1 -4

output

Copy

-2.500000000000000

input

Copy

3 2
0 2
5 0

output

Copy

7.000000000000000

题意：一个数组有n个数，每个数初始值为0。对该数组进行m次操作，每一次操作将数组中的a[i]=a[i]+x+d*abs(i-j)，求m次操作后该数组所有元素和的最大的平均值是多少（看了好久的题，最后看了猛神的博客才把题意弄懂(；´д｀)ゞ）。

实现：

          因为数组的输出值全是0，所以用ans代表当前数组的元素和（ans初始值为0）。因为每次操作，每个元素都要先加上x，

          所以ans=ans+x*n

          接下来求该操作中的d*abs(i-j)，因为i是一直变化的，每次操作是j是在随意的位置，但是一旦确定就不改变，所以可以看做是求数轴上1~n每个数到某一定点的距离。

          因为d的正负不确定，还要使d*∑abs(i-j)的值最大，所以要对j的位置进行讨论：

          如果d是正数，距离要尽可能大，所以j应该放在端点处；如果是负数，要使距离尽可能小，即需要放在中间位置

PS：最后注意精度问题，好像double会卡39组数据，int第五组都过不去，要用long long，最后再将long long转换成浮点型

AC代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#define ll long long
#define ms(a) memset(a,0,sizeof(a))
#define pi acos(-1.0)
#define INF 0x3f3f3f3f
const double E=exp(1);
const int maxn=1e6+10;
using namespace std;
int main(int argc, char const *argv[])
{
	ios::sync_with_stdio(false);
	ll n,m;
	ll x,d;
	cin>>n>>m;
	ll ans=0;
	while(m--)
	{
		cin>>x>>d;
		ans+=n*x;//求数组里元素的总和
		if(d>0)//距离应该最大，j取到两端点处,距离和为：(0+(n-1)*n)/2=n*(n-1)/2
		{
			ans+=d*(n*(n-1)/2);
		}
		else//此时j应在中间，分元素个数为奇数，偶数两种情况
		{
			if(n%2)//如果是奇数，放在正中间
			{
				 ll mid=(n+1)/2;
				 ll x=mid*(mid-1);
				 ans+=d*x;
			}
			else//偶数的话中间两个位置都可以
			{
				ll mid=n/2;
				ll x=(1+mid)*mid/2;
				ans+=d*(2*x-mid);
			}
		}
	}
	printf("%.15lf
",ans*1.0/n);
	return 0;
}