Vasya And The Mushrooms CodeForces

大意: 给定2*n的矩阵, 每个格子有权值, 走到一个格子的贡献为之前走的步数*权值, 每个格子只能走一次, 求走完所有格子最大贡献.

沙茶模拟打了一个小时总算打出来了

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
//head




#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif


int n;
int a[2][N];
ll d[2][N], suf[N], s[2][N], b[2][2][N];

int main() {
	scanf("%d", &n);
	REP(i,1,n) scanf("%d", a[0]+i);
	REP(i,1,n) scanf("%d", a[1]+i);
	ll now = 0, clk = 0, ans = 0, s1 = 0, s2 = 0;
	REP(i,1,n) {
		ans += clk*a[now][i], d[now][i] = ans;
		++clk, now ^= 1;
		ans += clk*a[now][i], d[now][i] = ans;
		++clk;
	}
	PER(i,1,n) suf[i]=suf[i+1]+a[0][i]+a[1][i];
	clk = 0;
	REP(j,1,n) b[0][0][j]=clk*a[0][j],b[1][1][j]=clk*a[1][j],++clk;
	PER(j,1,n) b[0][1][j]=clk*a[1][j],b[1][0][j]=clk*a[0][j],++clk;
	REP(i,0,1) PER(j,1,n) s[i][j]=s[i][j+1]+b[i][0][j]+b[i][1][j];
	REP(i,0,1) REP(j,1,n) if (j&1^i) {
		ans=max(ans, d[i][j-1]+s[i][j]+(j-1)*suf[j]);
	}
	printf("%lld
", ans);
}