NOIP2015普及组 推销员

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唉，我连pj题都不会做，今年tg AFO稳了...

考虑枚举最远走到哪户，那么在能走到的范围内其他去的住户一定是 \(A_i\) 最大的那些。

这样就有一个\(O(n^2 \log n)\)的优秀做法了。

或者说，除了你要走到最远的住户以外，其它住户一定是能选择的\(A_i\)最大的住户。

那么对于选\(x\)个住户，就有两种情况：

1. 选的是\(A_i\)最大的（此基础上\(S_i\)也要最大）\(x\)个，

2. 选的是\(A_i\)最大的\(x-1\)个，另一个选的是剩下的里面\(A_i+S_i\)最大的。

正确性显然。

/*
Author: CNYALI_LK
LANG: C++
PROG: 2672.cpp
Mail: cnyalilk@vip.qq.com
*/
#include<bits/stdc++.h>
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
const double eps=1e-8;
const double pi=acos(-1.0);
typedef long long ll;
typedef pair<int,int> pii;
template<class T>int chkmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>int chkmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T>T mmin(T a,T b){return a<b?a:b;}
template<class T>T mmax(T a,T b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>int dcmp(T a,T b){return a>b;}
template<int *a>int cmp_a(int x,int y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
namespace io {
	const int SIZE = (1 << 21) + 1;
	char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
	// getchar
	#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
	// print the remaining part
	inline void flush () {
		fwrite (obuf, 1, oS - obuf, stdout);
		oS = obuf;
	}
	// putchar
	inline void putc (char x) {
		*oS ++ = x;
		if (oS == oT) flush ();
	}
	// input a signed integer
	inline void read (int &x) {
		for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
		for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
	}
	inline void read (char &x) {
		x=gc();
	}
	inline void read(char *x){
		while((*x=gc())=='\n' || *x==' '||*x=='\r');
		while(!(*x=='\n'||*x==' '||*x=='\r'))*(++x)=gc();
	}
	template<typename A,typename ...B>
	inline void read(A &x,B &...y){
		read(x);read(y...);
	}
	// print a signed integer
	inline void write (int x) {
		if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
		while (x) qu[++ qr] = x % 10 + '0',  x /= 10;
		while (qr) putc (qu[qr --]);
	}
	inline void write (char x) {
		putc(x);
	}
	inline void write(const char *x){
		while(*x){putc(*x);++x;}
	}
	inline void write(char *x){
		while(*x){putc(*x);++x;}
	}
	template<typename A,typename ...B>
	inline void write(A x,B ...y){
		write(x);write(y...);
	}
	//no need to call flush at the end manually!
	struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
pii a[100005];
int qs[100005],qmx[100005],hmx[100005];
int main(){
#ifdef cnyali_lk
	freopen("2672.in","r",stdin);
	freopen("2672.out","w",stdout);
#endif
	int n;
	read(n);
	for(int i=1;i<=n;++i){read(a[i].y);a[i].y*=2;}
	for(int i=1;i<=n;++i)read(a[i].x);
	sort(a+1,a+n+1,dcmp<pii>);
	for(int i=1;i<=n;++i){qmx[i]=max(qmx[i-1],a[i].y);qs[i]=qs[i-1]+a[i].x;}
	for(int i=n;i;--i)hmx[i]=max(hmx[i+1],a[i].x+a[i].y);
	for(int i=1;i<=n;++i)printf("%d\n",max(qmx[i]+qs[i],qs[i-1]+hmx[i]));
	return 0;
}