思路：
f[i] 表示前i个的最小花费
转移：f[i] = f[j] + (i-j)*(i-j-1)/2 + A[i];
需要注意的是过程中数据超范围
代码一：
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 #define mem(a) memset(a,0,sizeof(a)) 5 #define mp(x,y) make_pair(x,y) 6 const int INF = 0x3f3f3f3f; 7 const ll INFLL = 0x3f3f3f3f3f3f3f3fLL; 8 inline ll read(){ 9 ll x=0,f=1;char ch=getchar(); 10 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 11 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 12 return x*f; 13 } 14 ////////////////////////////////////////////////////////////////////////// 15 const int maxn = 1e6+10; 16 17 int n; 18 ll A[maxn],f[maxn],q[maxn]; 19 20 double slope(ll j,ll k){ 21 return (1.0*(f[j]-f[k]-k*(k+1)/2+j*(j+1)/2))/(j-k); 22 } 23 24 int main(){ 25 n = read(); 26 for(int i=1; i<=n; i++) 27 A[i] = read(); 28 29 int L=0,R=0; 30 for(int i=1; i<=n; i++){ 31 while(L<R && slope(q[L+1],q[L])<i) L++; 32 ll j = q[L]; 33 f[i] = f[j] + (i-j)*(i-j-1)/2 + A[i]; 34 while(L<R && slope(i,q[R])<slope(q[R],q[R-1])) R--; 35 q[++R] = i; 36 } 37 38 cout << f[n] << endl; 39 40 return 0; 41 }
代码二：
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 #define mem(a) memset(a,0,sizeof(a)) 5 #define mp(x,y) make_pair(x,y) 6 const int INF = 0x3f3f3f3f; 7 const ll INFLL = 0x3f3f3f3f3f3f3f3fLL; 8 inline ll read(){ 9 ll x=0,f=1;char ch=getchar(); 10 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 11 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 12 return x*f; 13 } 14 ////////////////////////////////////////////////////////////////////////// 15 const int maxn = 1e6+10; 16 17 struct node{ 18 ll x,y; 19 }now,q[maxn]; 20 21 int n; 22 ll A[maxn]; 23 24 ll cross(node a,node b,node c){ 25 return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y); 26 } 27 28 int main(){ 29 n = read(); 30 for(int i=1; i<=n; i++) 31 A[i] = read(); 32 33 int L=0,R=0; 34 for(int i=1; i<=n; i++){ 35 while(L<R && q[L+1].y-i*q[L+1].x <= q[L].y-i*q[L].x) L++; 36 now.x = i; 37 now.y = q[L].y-(ll)i*q[L].x+(ll)i*i+A[i]; 38 while(L<R && cross(q[R-1],q[R],now)<=0) R--; 39 q[++R] = now; 40 } 41 42 cout << q[R].y-(ll)n*(n+1)/2 << endl; 43 44 return 0; 45 }