一眼望去不會。
考慮問題中的\(f(i,j)=|\sum_{p=i}^{j}?a_p? |\)的實際意義。
其實就是前綴和相減的絕對值。
\(f(i,j)=|\ sum[j]-sum[i-1]\ |\)
\(f(i,j)=max(sum[j]-sum[i-1],sum[i-1]-sum[j])\)
那加上x呢。
\(f(i,j)=max[(sum[j]+xj)-(sum[i-1]+x(i-1)),(sum[i-1]+x(i-1))-(sum[j]+xj)]\)
\(sum[i]+xi\)聯想到直線。
我們把這些直線放到坐標系裏。
(這個圖是從別人那裏蒯過來的)

這是把樣例轉為坐標系中直線的效果圖。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<map>
using namespace std;
const int N=201000;
const double eps=1e-8;
double sum[N];
struct node{
    double x,y;
    node(double xx=0,double yy=0){
        x=xx;y=yy;
    }
};
int top1,top2;
struct Line{
    node u,v;
    double w;
}a[N],stack1[N],stack2[N];
node operator +(node a,node b){
    return node(a.x+b.x,a.y+b.y);
}
node operator -(node a,node b){
    return node(a.x-b.x,a.y-b.y);
}
node operator *(node a,double b){
    return node(a.x*b,a.y*b);
}
double chaji(node a,node b){
    return a.x*b.y-a.y*b.x;
}
node jiao(Line a,Line b){
    double A=chaji(b.u-a.u,b.v-a.u);
    double B=chaji(b.v-a.v,b.u-a.v);
    return a.u+(a.v-a.u)*(A/(A+B));
}
bool judge1(Line a,Line b,Line c){
    node x=jiao(b,c);
    if(chaji(x-a.u,a.v-a.u)+eps>=0)return true;
    else return false;
}
bool judge2(Line a,Line b,Line c){
    node x=jiao(b,c);
    if(chaji(x-a.u,a.v-a.u)+eps<0)return true;
    else return false;
}
int read(){
    int sum=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){sum=sum*10+ch-'0';ch=getchar();}
    return sum*f;
}
int n,m;
double w(Line a,double x){
//  cout<<a.w<<endl;
    return a.u.y+a.w*(x-(-2.0*n));
}
map<int,double>ans;
long long lastans;
int main(){
    n=read();m=read();
    for(int i=1;i<=n;i++)sum[i]=sum[i-1]+read();
    for(int i=1;i<=n;i++){
        a[i].u=node(-2.0*n,sum[i]+i*-2.0*n);
        a[i].v=node(2.0*n,sum[i]+i*2.0*n);
        a[i].w=i;
    }
    a[0].u=node(-2.0*n,0);a[0].v=node(2.0*n,0);
    a[0].w=0;
    for(int i=0;i<=n;i++){
        while(top1>1&&judge1(a[i],stack1[top1-1],stack1[top1]))top1--;
        stack1[++top1]=a[i];
    }
    for(int i=n;i>=0;i--){
        while(top2>1&&judge2(a[i],stack2[top2-1],stack2[top2]))top2--;
        stack2[++top2]=a[i];
    }
    int now1=1;int now2=1;
    for(int i=-2*n;i<=n*2;i++){
        while(now1<top1&&w(stack1[now1+1],i)>=w(stack1[now1],i))now1++;
        while(now2<top2&&w(stack2[now2+1],i)<=w(stack2[now2],i))now2++;
        double A=w(stack1[now1],i);
        double B=w(stack2[now2],i);
        ans[i]=A-B;
    }
    lastans=0;
    while(m--){
        int x=read();
        x=(x+lastans)%(4*n+1)-2*n;
        lastans=ans[x];
        printf("%.0lf\n",ans[x]);
    }
    return 0;
}
 

luogu P4756 Added Sequence（凸包+思維）