895A - Pizza Separation - CodeForces Solution


brute force implementation *1200

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Python Code:

n=int(input())
a=[int(i) for i in input().split()]
min_sum=360
for i in range(n):
    for j in range(i+1,n):
        min_sum=min(min_sum,max(360-2*sum(a[i:j]),2*sum(a[i:j])-360))
print(min_sum)

C++ Code:

#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
#define hello() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define nl '\n'
#define elif else if
#define pb push_back
#define mod 1000000007
#define cntbit __builtin_popcount
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) (a*b)/__gcd(a,b)
int main()
{
    hello();

    int n;
    cin>>n;


    int arr[n+2];
    for(int i=1; i<=n; i++)
        cin>>arr[i];
    int ans=360;
    int sum=0;
    for(int i=1; i<=n; i++)
    {
        sum=0;
        for(int j=i; j<n; j++)
        {
            sum+=arr[j];
        
        ans=min(ans,abs(360-(2*sum)));
        }

    }

    cout<<ans<<endl;
}


 			 	  			 	  			 	 	 		  				


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