Codeforces Round #448 (Div. 2) A. Pizza Separation【前缀和/枚举/将圆（披萨）分为连续的两块使其差最小】

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 toai. 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.

 

【题意】：将圆分为连续的两块使其差最小。

【分析】：因为是连续区间并且区间可以为空，3个for枚举t或者前缀和或者尺取。去掉连续的话就是01背包变形：http://blog.csdn.net/pegasuswang_/article/details/25081783

【代码】：

#include <bits/stdc++.h>

using namespace std;
int a[400],sum[400];
int main()
{
    int n;
    int ans=360;

    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
        sum[i]=a[i]+sum[i-1];
    }
    for(int i=0;i<n;i++)
    {
        for(int j=i;j<n;j++)//t-(360-t)=2*t-360,t为区间所取值，因为是连续区间并且区间可以为空，3个for枚举t或者前缀和或者尺取
        {
            ans=min(ans,abs(2*(sum[j]-sum[i-1])-360));
        }
    }
    cout<<ans<<endl;
    return 0;
}

 

#include <bits/stdc++.h>

using namespace std;
int a[1000],sum;//环的话 数组起码2倍大
int main()
{
    int n;
    int ans=360;

    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
        a[i+n]=a[i];
    }
    for(int i=0;i<n;i++)
    {
        sum=0; //注意置0的位置
        for(int j=i;j<i+n;j++)
        {
            sum+=a[j];
            ans=min(ans,abs( 2*sum-360) );
        }
    }
    cout<<ans<<endl;
    return 0;
}

/*
输入数据直接复制了一遍放到后面,然后枚举拿的起点i，拿的终点j。然后计算拿了多少，差值是多少。
*/