一种枚举指定序列的方法 ————CF A. Pizza Separation

A. Pizza Separation

time limit per test

 1 second

memory limit per test

 256 megabytes

input

 standard input

output

 standard 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

490 90 90 90

output

0

input

3100 100 160

output

40

input

1360

output

360

input

4170 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.

———————————————————————————————————————————————————————————

题目的意思是给你一个序列找出一个区间是的它的和与剩下的所有数的和差值最小，求

差值

思路：暴力枚举所有情况

[cpp] view plain copy

1. #include <bits/stdc++.h>  
2.   
3. using namespace std;  
4.   
5. int main()  
6. {  
7.    int a[400];  
8.    int n;  
9.    while(cin>>n&&n)  
10.    {  
11.       for(int i=0;i<n;i++)  
12.       {  
13.           cin>>a[i];  
14.       }  
15.       int mn=360;  
16.       for(int i=0;i<n;i++)  
17.       {  
18.            int sum=0;  
19.         for(int j=i;j<n;j++)  
20.         {  
21.             sum+=a[j];  
22.   
23.             mn=min(mn,abs(360-2*sum));  
24.         }  
25.       }  
26.       cout<<mn<<endl;  
27.    }  
28.     return 0;  
29. }