C. Annoying Present SB题

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 $\mathrm{nn as\; a\; birthday\; present\; once\; again!\; This\; is\; the\; third\; year\; in\; a\; row!}$

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 $\mathrm{mm changes\; of\; the\; following\; form.\; For\; some\; integer\; numbers xx and dd,\; he\; chooses\; an\; arbitrary\; position ii (1\le i\le n1\le i\le n)\; and\; for\; every j\in [1,n]j\in [1,n] adds x+d\cdot dist(i,j)x+d\cdot dist(i,j) to\; the\; value\; of\; the jj-th\; cell. dist(i,j)dist(i,j) is\; the\; distance\; between\; positions ii and jj (i.e. dist(i,j)=|i-j|dist(i,j)=|i-j|,\; where |x||x| is\; an\; absolute\; value\; of xx).}$

For example, if Alice currently has an array $[2,1,2,2][2,1,2,2] and\; Bob\; chooses\; position 33 for x=-1x=-1 and d=2d=2 then\; the\; array\; will\; become [2-1+2\cdot 2, 1-1+2\cdot 1, 2-1+2\cdot 0, 2-1+2\cdot 1][2-1+2\cdot 2, 1-1+2\cdot 1, 2-1+2\cdot 0, 2-1+2\cdot 1] = [5,2,1,3][5,2,1,3].\; Note\; that\; Bob\; can\text{'}t\; choose\; position ii outside\; of\; the\; array\; (that\; is,\; smaller\; than 11 or\; greater\; than nn).$

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 $\mathrm{nn and mm (1\le n,m\le 1051\le n,m\le 105)\; —\; the\; number\; of\; elements\; of\; the\; array\; and\; the\; number\; of\; changes.}$

Each of the next $\mathrm{mm lines\; contains\; two\; integers xixi and didi (-103\le xi,di\le 103-103\le xi,di\le 103)\; —\; the\; parameters\; for\; the ii-th\; change.}$

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 ${\mathrm{10-610-6.}}^{}$

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

  这题分析一下题目  就是对d的正负进行分类讨论 
　　d>=0 则选择 i=1； 反之 i=（n+1)/2;
　　代码量也特别少

　　

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
const int INF = 0x3fffffff;
typedef long long LL;
using namespace std;
LL n, m, x, d, ans = 0;
int main() {
    scanf("%lld%lld", &n, &m);
    for (int i = 0 ; i < m ; i++) {
        scanf("%lld%lld", &x, &d);
        if (d >= 0)    ans += n * x + d * (n - 1) * n / 2;
        else    ans += n * x + d * (n - (n + 1) / 2) * ((n + 1) / 2);
    }
    printf("%.8f
", 1.0 * ans / n);
    return 0;
}