传送门 首先得到转移方程： f[i]=max(f[j]+a*(sum[i]-sum[j-1])^2+b*(sum[i]-sum[j-1])+c) 然后我们斜率优化一发： 如果j>k且j比k更优 f[j]-f[k]+a*sum[j]^2-a*sum[k]^2+b*(sum[k]-sum[j])>2*a*(sum[j]-sum[k])*sum[i] 然后……….就没有了
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<cmath> #define ll long long #define N 1000005 using namespace std; ll sum[N],f[N],n,a,b,c,x; int q[N],l,r; inline int read(){ int k=0; char ch=getchar(); for (;ch<'0'||ch>'9';ch=getchar()); for (;ch>='0'&&ch<='9';ch=getchar()) k=k*10+ch-48; return k; } ll sqr(ll x){return x*x;} double slop(int k,int j){ return (double)(f[j]-f[k]+a*(sqr(sum[j])-sqr(sum[k]))+b*(sum[k]-sum[j]))/(double)(2*a*(sum[j]-sum[k])); } int main(){ n=read(); scanf("%lld%lld%lld",&a,&b,&c); for (int i=1;i<=n;i++){ x=read(); sum[i]=sum[i-1]+x; } for (int i=1;i<=n;i++){ while (l<r&&slop(q[l],q[l+1])<sum[i]) l++; int t=q[l]; f[i]=f[t]+a*sqr(sum[i]-sum[t])+b*(sum[i]-sum[t])+c; while (l<r&&slop(q[r-1],q[r])>slop(q[r],i)) r--; q[++r]=i; } printf("%lld",f[n]); }