解题报告：

维护前缀和以及后缀和，相等则记录为断点

dp[ i ]表示断点 i 中间全部合并，两侧最优合并的最小代价

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
#define CLR(x,y) memset(x,y,sizeof(x))
#define eps 1e-9
#define INF 0x3f3f3f3fusing namespace std;typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef complex<ld> point;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;template<class T>
inline bool read(T &n)
{T x = 0, tmp = 1; char c = getchar();while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();if(c == EOF) return false;if(c == '-') c = getchar(), tmp = -1;while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();n = x*tmp;return true;
}
template <class T>
inline void write(T n)
{if(n < 0){putchar('-');n = -n;}int len = 0,data[20];while(n){data[len++] = n%10;n /= 10;}if(!len) data[len++] = 0;while(len--) putchar(data[len]+48);
}
//-----------------------------------vector<pii> mat;
int a[5010],v[5010],n;
int dp[5010];int main()
{while(read(n)&&n){mat.clear();for(int i=1;i<=n;i++)read(a[i]);for(int i=1;i<=n;i++)read(v[i]);if(n==1){putchar('0');putchar('\n');continue;}int t=0,h=n+1,mid=v[n];ll pre=0,beh=0;while(t<h){if(pre>beh)beh+=a[--h];else if(pre<beh)pre+=a[++t];else{mat.push_back(make_pair(t,h));pre+=a[++t],beh+=a[--h];}}for(int i=0;i<mat.size();i++){dp[i]=v[mat[i].first-0]+v[n+1-mat[i].second];for(int j=1;j<i;j++)dp[i]=min(dp[i],dp[j]+v[mat[i].first-mat[j].first]+v[mat[j].second-mat[i].second]);mid=min(dp[i]+v[mat[i].second-mat[i].first-1],mid);}write(mid);putchar('\n');}return 0;
}