CodeForces：#448 div2 a Pizza Separation

传送门：http://codeforces.com/contest/895/problem/A

A. Pizza Separation

time limit per test1 second

memory limit per test256 megabytes

Problem Description

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.

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解题心得：

1. 这场比赛真的是打蒙了，A题啊，旁边一个小子念叨了一句dp，然后就被带歪了，按着0-1背包问题来写，结果人家说的很清楚啊，必须是连续的披萨块，一个小小的坑点就是披萨是圆的，首尾相连，一个模拟很轻松的解决。

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#include<bits/stdc++.h>
using namespace std;
const int maxn = 380;
int num[maxn];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
        scanf("%d",&num[i]);
    int Min = 360;
    int sum = 0;
    for(int i=0;i<n;i++)
    {
        int t = 0;
        int cnt = i;//枚举连续披萨的起点位置
        sum = 0;
        while(t < n)//收尾相连但是不能超过n块
        {
            sum += num[cnt];
            if(abs(360-sum*2) < Min)
                Min = abs(360-sum*2);
            cnt++;
            t++;
            if(cnt >= n)
                cnt = 0;
        }
    }
    printf("%d",Min);
    return 0;
}