#10191. 「一本通 5.6 练习 4」打印文章

这题™是多组数据。。。还搞了半天。。。
最后只好这样打才过了。。。

while (scanf("%d%d",&n,&m)==2)
{
}

斜率DP，式子很容易推：

f[i]=min{f[j]+(sum[i]-sum[j])²+a[i]²};

其中sum[]表示a[]的前缀和

上标：

#include<cstdio>
#include<algorithm>
#define N 500010
#define db double
#define ll long long
using namespace std;
int n,m,g[N],l,len;
ll a[N],sum[N],f[N];

inline int read()
{
	int x=0; char c=getchar();
	while (c<'0' || c>'9') c=getchar();
	while (c>='0' && c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x;
}

ll sqr(ll x) {return x*x;}

ll solve(int x,int y) {return (db)(f[x]+sqr(sum[x])-f[y]-sqr(sum[y]))/(sum[x]-sum[y]);}

int main()
{
	freopen("#10191.in","r",stdin);
	freopen("#10191.out","w",stdout);
//	f[i]=min{f[j]+(sum[i]-sum[j])^2+a[i]^2};
	while (scanf("%d%d",&n,&m)==2)
	{
		for (int i=1;i<=n;i++)
			a[i]=read(),sum[i]=sum[i-1]+a[i];
		l=len=0;
		for (int i=1;i<=n;i++)
		{
			while (l<len && solve(g[l+1],g[l])<2*sum[i]) l++;
			f[i]=f[g[l]]+sqr(sum[i]-sum[g[l]])+m;
			while (l<len && solve(g[len],g[len-1])>solve(i,g[len])) len--;
			g[++len]=i;
		}
		printf("%lld
",f[n]);
	}
	return 0;
}