いっしょ / Be Together （暴力枚举）

题目链接：http://abc043.contest.atcoder.jp/tasks/arc059_a

Time limit : 2sec / Memory limit : 256MB

Score : 200 points

Problem Statement

Evi has N integers a1,a2,..,aN. His objective is to have N equal integers by transforming some of them.

He may transform each integer at most once. Transforming an integer x into another integer y costs him (x−y)2 dollars. Even if ai=aj(i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).

Find the minimum total cost to achieve his objective.

Constraints

• 1≦N≦100
• −100≦ai≦100

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Input

The input is given from Standard Input in the following format:

N
a1 a2 ... aN

Output

Print the minimum total cost to achieve Evi's objective.

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Sample Input 1

Copy

2
4 8

Sample Output 1

Copy

8

Transforming the both into 6s will cost (4−6)2+(8−6)2=8 dollars, which is the minimum.

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Sample Input 2

Copy

3
1 1 3

Sample Output 2

Copy

3

Transforming the all into 2s will cost (1−2)2+(1−2)2+(3−2)2=3 dollars. Note that Evi has to pay (1−2)2 dollar separately for transforming each of the two 1s.

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Sample Input 3

Copy

3
4 2 5

Sample Output 3

Copy

5

Leaving the 4 as it is and transforming the 2 and the 5 into 4s will achieve the total cost of (2−4)2+(5−4)2=5 dollars, which is the minimum.

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Sample Input 4

Copy

4
-100 -100 -100 -100

Sample Output 4

Copy

0

Without transforming anything, Evi's objective is already achieved. Thus, the necessary cost is 0.

题解：数据不大 枚举

 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <cstdio>
 #include <vector>
 #include <cstdlib>
 #include <iomanip>
 #include <cmath>
 #include <ctime>
 #include <map>
 #include <set>
 #include <queue>
 #include <stack>
 using namespace std;
 #define lowbit(x) (x&(-x))
 #define max(x,y) (x>y?x:y)
 #define min(x,y) (x<y?x:y)
 #define MAX 100000000000000000
 #define MOD 1000000007
 #define pi acos(-1.0)
 #define ei exp(1)
 #define PI 3.141592653589793238462
 #define INF 0x3f3f3f3f3f
 #define mem(a) (memset(a,0,sizeof(a)))
 typedef long long ll;
 ll gcd(ll a,ll b){
     return b?gcd(b,a%b):a;
 }
 bool cmp(int x,int y)
 {
     return x>y;
 }
 const int N=;
 const int mod=1e9+;
 int a[];
 int main()
 {
     std::ios::sync_with_stdio(false);
     int n;
     while(cin>>n){
         int s=;
         for(int i=;i<n;i++)
             cin>>a[i];
         sort(a,a+n);
         if(a[]==a[n-]) cout<<s<<endl;
         else{
             s=;
             int t=,j;
             for(int i=a[];i<a[n-];i++){
                 t=;
                 for(j=;j<n;j++){
                     t+=(i-a[j])*(i-a[j]);
                 }
                 if(t<s) s=t;
             }
             cout<<s<<endl;
         }
     }
     return ;
 }