이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#define rf(x) (x)=0;while(*p<48)p++;while(47<*p)(x)=((x)<<3)+((x)<<1)+(*p++&15);
//#define rf(x) (x)=0;while(*p<48)im=*p=='-';while(47<*p)(x)=((x)<<3)+((x)<<1)+(*p++&15);if(im)(x)=-(x);
#define pb push_back
#define eb emplace_back
#define sz(V) ((int)(V).size())
#define allv(V) ((V).begin()),((V).end())
#define befv(V) ((V)[(sz(V)-2)])
#define sorv(V) sort(allv(V))
#define revv(V) reverse(allv(V))
#define univ(V) (V).erase(unique(allv(V)),(V).end())
#define clv(V) (V).clear()
#define upmin(a,b) (a)=min((a),(b))
#define upmax(a,b) (a)=max((a),(b))
#define rb(x) ((x)&(-(x)))
#define cb(x) (x)=(!(x))
#define INF (0x3f3f3f3f)
#define INFLL (0x3f3f3f3f3f3f3f3fll)
#define INFST (0x7f7f7f7f)
#define INFLLST (0x7f7f7f7f7f7f7f7fll)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<int, ll> pil;
typedef pair<ll, int> pli;
typedef pair<ld, ld> pdd;
typedef complex<ld> base;
const ld EPS = (ld)1e-7;
const ld PI = acos(0) * 2;
bool isZero(const ld& x) { return abs(x) <= EPS; }
int sign(const ld& x) { return isZero(x) ? 0 : (0 < x ? 1 : -1); }
ll gcd(ll a, ll b) { for(;b;a%=b,swap(a,b)){} return abs(a); }
pll operator + (const pll& a, const pll& b) { return pll(a.first+b.first, a.second+b.second); }
pll operator - (const pll& a, const pll& b) { return pll(a.first-b.first, a.second-b.second); }
pll operator * (const pll& a, const ll& b) { return pll(a.first*b, a.second*b); }
ll operator * (const pll& a, const pll& b) { return a.first*b.second - b.first*a.second; }
ll ccw(const pll& a, const pll& b, const pll& c) { return a*b + b*c + c*a; }
void fg(vector<int> G[], int a, int b) { G[a].pb(b); G[b].pb(a); }
void fg(vector<pii> G[], int a, int b, int c) { G[a].pb({b, c}); G[b].pb({a, c}); }
const int MAXN = 1005;
ll dp[MAXN][MAXN];
bitset<MAXN> chk[MAXN];
int A[MAXN], B[MAXN];
int N;
ll f(int s, int e) {
if(1 == s && e == N) return 0;
ll &ret = dp[s][e];
if(chk[s][e]) return ret;
chk[s][e] = true;
int k = e - s + 2;
if(e < N) upmax(ret, f(s, e+1) + (A[e+1] == k ? B[e+1] : 0));
if(1 < s) upmax(ret, f(s-1, e) + (A[s-1] == k ? B[s-1] : 0));
return ret;
}
int main() {
ios::sync_with_stdio(false);
cin >> N;
for(int i = 1; i <= N; i++) cin >> A[i];
for(int i = 1; i <= N; i++) cin >> B[i];
for(int i = 1; i <= N; i++)
printf("%lld ", f(i, i) + (1 == A[i] ? B[i] : 0));
puts("");
return 0;
}
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