CodeForces-1016C Vasya And The Mushrooms

传送门：CodeForces-1016C

思路分析

实际上路径只会是一段上下+顺时针或逆时针转一圈
预处理出每列开始顺时针逆时针的贡献，然后枚举上下走的部分

样例输入

3
1 2 3
6 5 4

样例输出

70

AC代码

#include <bits/stdc++.h>
#define ls x<<1
#define rs x<<1|1
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;
const int N = 3e5+10;

ll a[2][N],suf[2][N],s[N];

int main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int n;
    read(n);
    for(int i=0; i<2; i++) for(int j=1; j<=n; j++) read(a[i][j]);
    for(int i=n; i>=1; i--) s[i]=s[i+1]+a[0][i]+a[1][i];
    int ch=n;
    for(int i=n; i>=1; i--) {
        suf[0][i]=suf[0][i+1]+a[0][i]*(i-1)+a[1][i]*ch;
        ch++;
    }
    ch=n;
    for(int i=n; i>=2; i--) {
        suf[1][i]=suf[1][i+1]+a[1][i]*i+a[0][i]*(ch+1);
        ch++;
    }
    ll sum=0;
    ll ans=0;
    ll cnt=0;
    for(int i=1; i<=n; i++) {
        if(i&1) ans=max(ans,sum+suf[0][i]+s[i]*(i-1));
        else ans=max(ans,sum+suf[1][i]+s[i]*(i-2));
        sum+=a[(i+1)%2][i]*cnt;
        cnt++;
        sum+=a[i%2][i]*cnt;
        cnt++;
    }
    ans=max(ans,sum);
    cout<<ans<<endl;



    return 0;
}