這題是區間DP的模板題，發佈題解的大佬用心良苦，故意把輸入改了，一般需要把整個程序邏輯理解透徹纔會知道哪裏要改。
具體邏輯大佬講的非常清楚，接下來是給代碼時間：
#include<algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <cstdio>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
#define SZ(x) ((int)x.size())
#define rep(i,a,n) for (int i = a; i < n; ++i)
#define per(i,a,n) for (int i = n-1; i >= a; --i)
#define devil_may_cry ios::sync_with_stdio(false)
using namespace std;
//head
constexpr int maxn = 300;
int f1[maxn][maxn], f2[maxn][maxn];
int s[maxn], num[maxn];
inline int d(int i, int j) { return s[j] - s[i - 1]; }
int main()//f[i][j] = max(f[i][k] + f[k + 1][j] + d(i, j), f[i][j])
{
int n;
cin >> n;
rep(i, 0, n)
{
scanf("%d",&num[i]);
num[i+n]=num[i];
}
rep(i, 1, n + n + 1)
{
s[i]=s[i-1]+num[i];
}
rep(q, 1, n)
{
for(int i = 1, j = i + q; (i < n + n) && (j < n + n); i++, j = i + q)
{
f2[i][j] = 1e9 + 7;
rep(k, i, j)
{
f1[i][j] = max(f1[i][k] + f1[k + 1][j] + d(i, j), f1[i][j]);
f2[i][j] = min(f2[i][k] + f2[k + 1][j] + d(i, j), f2[i][j]);
//rep(i, 1, n + n + 1) { rep(j, 1, n + n) printf("%4d", f1[i][j]); cout << endl; }
}
//cout << "--------------------------------------------------------------" << endl;
}
}
int ans1 = 0, ans2 = 1e9 + 7;
rep(i, 1, n)
{
ans1 = max(ans1, f1[i][i + n - 1]);
ans2 = min(ans2, f2[i][i + n - 1]);
}
cout << ans2 << endl;
cout << ans1 << endl;
return 0;
}