队列维护下凸包裸题
式子不太好推，但其实不用把式子全展开的.....
k单调递增，x单调递增，队列维护一下就行了
因为f[i]期望最小值，所以维护下凸包
好像记录一下凸包的坐标能减少常数
#include <cstdio>
#include <algorithm>
#include <cstring>
#define il inline
#define ll long long
#define N 51000
using namespace std;
int n,que[N];
ll L,f[N],c[N],s[N];
int gc()
{
int rett=0,fh=1;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')fh=-1;c=getchar();}
while(c>='0'&&c<='9'){rett=(rett<<3)+(rett<<1)+c-'0';c=getchar();}
return rett*fh;
}
il ll yy(ll j){return f[j]+2ll*(s[j]+j)*(1ll+L)+(s[j]+j)*(s[j]+j)+(1ll+L)*(1ll+L);}
il ll xx(ll j){return s[j]+j;}
il ll kk(ll i){return 2ll*(s[i]+i);}
int main()
{
n=gc(),L=gc();
for(int i=1;i<=n;i++) c[i]=gc();
for(int i=1;i<=n;i++) s[i]=s[i-1]+c[i];
int hd=1,tl=1;
for(ll i=1;i<=n;i++)
{
while(hd+1<=tl&&yy(que[hd])-kk(i)*xx(que[hd])>=yy(que[hd+1])-kk(i)*xx(que[hd+1]))
hd++;
f[i]=yy(que[hd])-kk(i)*xx(que[hd])+(s[i]+i)*(s[i]+i)-2ll*(s[i]+i)*(1ll+L);
while(hd+1<=tl&&(yy(i)-yy(que[tl]))*(xx(que[tl])-xx(que[tl-1])) <= (yy(que[tl])-yy(que[tl-1]))*(xx(i)-xx(que[tl])))
tl--;
que[++tl]=i;
}
printf("%lld\n",f[n]);
return 0;
}