石子合并

/*
 * @author  Panoss
 */
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<vector>
#include<ctime>
#include<stack>
#include<queue>
#include<list>
using namespace std;
#define DBG 1
#define fori(i,a,b) for(int i = (a); i < (b); i++)
#define forie(i,a,b) for(int i = (a); i <= (b); i++)
#define ford(i,a,b) for(int i = (a); i > (b); i--)
#define forde(i,a,b) for(int i = (a); i >= (b); i--)
#define forls(i,a,b,n) for(int i = (a); i != (b); i = n[i])
#define mset(a,v) memset(a, v, sizeof(a))
#define mcpy(a,b) memcpy(a, b, sizeof(a))
#define dout  DBG && cerr << __LINE__ << " >>| "
#define checkv(x) dout << #x"=" << (x) << " | "<<endl
#define checka(array,a,b) if(DBG) { \
    dout<<#array"[] | " << endl;     forie(i,a,b) cerr <<"["<<i<<"]="<<array[i]<<" |"<<((i-(a)+1)%5?" ":"\n");     if(((b)-(a)+1)%5) cerr<<endl; }
#define redata(T, x) T x; cin >> x
#define MIN_LD -2147483648
#define MAX_LD  2147483647
#define MIN_LLD -9223372036854775808
#define MAX_LLD  9223372036854775807
#define MAX_INF 18446744073709551615
inline int  reint() { int d; scanf("%d",&d); return d; }
inline long relong() { long l; scanf("%ld",&l); return l; }
inline char rechar() { scanf(" "); return getchar(); }
inline double redouble() { double d; scanf("%lf", &d); return d; }
inline string restring() { string s; cin>>s; return s; }

int f[105][105];
/*
f[i][j] 表示从第i堆到第j堆石子合并的最小总得分
考虑最后一次合并，即可推得状态转移方程
f[i][j] =min( f[i][k] + f[k+1][j] + a[i] + a[i+1] + ... + a[j])
其中 i<k<j  a[i]表示第i堆石子
*/
int a[105],n;

void read()
{
    forie(i,1,n)
        cin>>a[i];
}
int sum(int l,int r)
{
    int sum = 0;
    forie(i,l,r)
        sum +=a[i];
    return sum;
}
int DP()
{
    mset(f,-1);
    forie(i,1,n)   //初始化单堆得分为0
        f[i][i] = 0;
    forie(i,1,n-1) //相邻两堆直接加
        f[i][i+1] = a[i] + a[i+1];

    forie(length,3,n)
    {
        forie(i,1,n-length+1)
        {
            int j = i + length - 1;
            f[i][j] = f[i+1][j] + sum(i,j);
            fori(k,i+1,j)
                f[i][j] = min(f[i][k]+f[k+1][j]+sum(i,j),f[i][j]);
        }
    }
    return f[1][n];
}
int main()
{
    //freopen("data.txt","r",stdin);
    while(cin>>n,n)
    {
        read();
        int min_ans = DP();
        fori(i,1,n)  //n-1次求解
        {
            int temp = a[n];
            forde(j,n,1)  //后移
                a[j] = a[j-1];
            a[1] = temp;
            min_ans = min(min_ans,DP());
        }
        cout << min_ans << endl;
    }
    return 0;
}