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这个题的式子比较好推，

然而还是要吐槽，要把除法换成乘法，而且这个题可能出现斜率相等的情况，而且，不能直接用double判断，精度有问题，本蒟蒻WAWAWAWAWA,,,,得出的惨痛教训。。

#include <bits/stdc++.h>
#define LL long long
#define lowbit(x) x&(-x)
#define inf 0x3f3f3f3f
#define eps 1e-5
#define N 500005
using namespace std;
inline int ra()
{
    int x=0,f=1; char ch=getchar();
    while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();}
    while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
int n,M,sum[N],f[N],q[N];
int squ(int a)
{
    return a*a;
}
int up(int a, int b)
{
    return (f[b]+squ(sum[b])-f[a]-squ(sum[a]));
}
int down(int a, int b)
{
    return (2*(sum[b]-sum[a]));
}
int main(int argc, char const *argv[])
{
    while (scanf("%d%d",&n,&M)!=EOF)
    {
        for (int i=1; i<=n; i++) sum[i]=sum[i-1]+ra();
        int l=0,r=0;
        for (int i=1; i<=n; i++)
        {
            while (l<r && up(q[l],q[l+1])<=sum[i]*down(q[l],q[l+1])) l++;
            f[i]=f[q[l]]+squ(sum[i]-sum[q[l]])+M;
            while (l<r && up(q[r-1],q[r])*down(q[r],i)>=up(q[r],i)*down(q[r-1],q[r])) r--;
            q[++r]=i;
        }
        cout<<f[n]<<endl;
    }
    return 0;
}