#448 div2 a Pizza Separation

A. Pizza Separation 
time limit per test1 second 
memory limit per test256 megabytes 
inputstandard input 
outputstandard output 
Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut into n pieces. The i-th piece is a sector of angle equal to ai. Vasya and Petya want to divide all pieces of pizza into two continuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty.

Input 
The first line contains one integer n (1 ≤ n ≤ 360) — the number of pieces into which the delivered pizza was cut.

The second line contains n integers ai (1 ≤ ai ≤ 360) — the angles of the sectors into which the pizza was cut. The sum of all ai is 360.

Output 
Print one integer — the minimal difference between angles of sectors that will go to Vasya and Petya.

Examples 
input 
4 
90 90 90 90 
output 
0 
input 
3 
100 100 160 
output 
40 
input 
1 
360 
output 
360 
input 
4 
170 30 150 10 
output 
0 
Note 
In first sample Vasya can take 1 and 2 pieces, Petya can take 3 and 4 pieces. Then the answer is |(90 + 90) - (90 + 90)| = 0.

In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is |360 - 0| = 360.

In fourth sample Vasya can take 1 and 4 pieces, then Petya will take 2 and 3 pieces. So the answer is |(170 + 10) - (30 + 150)| = 0.

Picture explaning fourth sample:

Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector.

题解：写的时候没注意到是取连续的区间，喵的老了。暴力枚举区间就可以了 复杂度360*360

代码：

#include <iostream>
#include <math.h>
using namespace std;

int main(){
    int ans=;
    int sector[];//扇形数组
    int numberOfSector;//扇形个数
    cin>>numberOfSector;
    for(int i=;i<numberOfSector;i++){
        cin>>sector[i];
    }
    for(int i=;i<numberOfSector;i++){
        int temp=;//临时存放当前扇形度数和
        for(int j=;j<numberOfSector;j++){
            temp+=sector[(i+j)%numberOfSector];//关键步骤
            ans=min(ans,abs(-temp));
        }
    }
    cout<<ans*<<endl;

    return ;
}