忘情水二分模板题，最优解对划分段数的导数满足单调性（原函数凸性）即可使用此方法。
详细题解洛谷里面就有，不啰嗦了。
二分的临界点让人有点头大。。。
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef double db; 5 const ll N=5e5+10,inf=0x3f3f3f3f3f3f3f3f; 6 ll n,m,hd,tl,a[N],S[N],dp[N],cnt[N]; 7 struct P {ll x,y,c;} q[N]; 8 P operator-(P a,P b) {return {a.x-b.x,a.y-b.y,a.c};} 9 ll cross(P a,P b) {return a.x*b.y-a.y*b.x;} 10 ll solve(ll m) { 11 hd=tl=0; 12 q[tl++]= {0,0,0}; 13 for(ll i=1; i<=n; ++i) { 14 for(; hd+1<tl&&cross((P) {1,2*(S[i]+1)},q[hd+1]-q[hd])<=0; ++hd); 15 dp[i]=q[hd].y-2*(S[i]+1)*q[hd].x+(S[i]+1)*(S[i]+1)+m; 16 cnt[i]=q[hd].c+1; 17 P np= {S[i],dp[i]+S[i]*S[i],cnt[i]}; 18 for(; hd+1<tl&&cross(q[tl-1]-q[tl-2],np-q[tl-1])<=0; --tl); 19 q[tl++]=np; 20 } 21 return cnt[n]; 22 } 23 ll bi(ll l,ll r) { 24 ll ret; 25 while(l<=r) { 26 ll mid=(l+r)>>1; 27 if(solve(mid)>=m)ret=dp[n]-m*mid,l=mid+1; 28 else r=mid-1; 29 } 30 return ret; 31 } 32 int main() { 33 scanf("%lld%lld",&n,&m); 34 for(ll i=1; i<=n; ++i)scanf("%lld",&a[i]); 35 for(ll i=1; i<=n; ++i)S[i]=S[i-1]+a[i]; 36 printf("%lld ",bi(0,inf)); 37 return 0; 38 }