洛谷P1880 [NOI1995]石子合并

这题是区间DP的模板题，发布题解的大佬用心良苦，故意把输入改了，一般需要把整个程序逻辑理解透彻才会知道哪里要改。

具体逻辑大佬讲的非常清楚，接下来是给代码时间：

#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;
}