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
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.
Sample Input 1
2
4 8
Sample Output 1
8
Transforming the both into 6s will cost (4−6)2+(8−6)2=8 dollars, which is the minimum.
Sample Input 2
3
1 1 3
Sample Output 2
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.
Sample Input 3
3
4 2 5
Sample Output 3
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.
Sample Input 4
4
-100 -100 -100 -100
Sample Output 4
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 ;
}