1- // Ivan Carvalho
2- // Solution to https://dmoj.ca/problem/apio10p1
3- #include < cstdio>
4- #include < algorithm>
5- #include < vector>
6- using namespace std ;
7- typedef long long ll;
8- vector<ll> M,B;
9- int pointer;
10- bool bad (int l1,int l2,int l3)
11- {
12- /*
13- intersection(l1,l2) has x-coordinate (b1-b2)/(m2-m1)
14- intersection(l1,l3) has x-coordinate (b1-b3)/(m3-m1)
15- set the former greater than the latter, and cross-multiply to
16- eliminate division
17- */
18- return (B[l3]-B[l1])*(M[l1]-M[l2])<(B[l2]-B[l1])*(M[l1]-M[l3]);
19- }
20- // Adds a new line (with lowest slope) to the structure
21- void add (long long m,long long b)
22- {
23- // First, let's add it to the end
24- m *= -1 ;
25- b *= -1 ;
26- M.push_back (m);
27- B.push_back (b);
28- // If the penultimate is now made irrelevant between the antepenultimate
29- // and the ultimate, remove it. Repeat as many times as necessary
30- while (M.size ()>=3 &&bad (M.size ()-3 ,M.size ()-2 ,M.size ()-1 ))
31- {
32- M.erase (M.end ()-2 );
33- B.erase (B.end ()-2 );
34- }
35- }
36- // Returns the minimum y-coordinate of any intersection between a given vertical
37- // line and the lower envelope
38- long long query (long long x)
39- {
40- // If we removed what was the best line for the previous query, then the
41- // newly inserted line is now the best for that query
42- if (pointer>=M.size ())
43- pointer=M.size ()-1 ;
44- // Any better line must be to the right, since query values are
45- // non-decreasing
46- while (pointer<M.size ()-1 &&
47- M[pointer+1 ]*x+B[pointer+1 ]<M[pointer]*x+B[pointer])
48- pointer++;
49- return -(M[pointer]*x+B[pointer]);
50- }
51- int main (){
52- int TC = 1 ;
53- while (TC--){
54- M.clear ();
55- B.clear ();
56- pointer = 0 ;
57- int n;
58- scanf (" %d"
59- ll a,b,c;
60- scanf (" %lld %lld %lld"
61- add (0LL ,0LL );
62- ll davez, soma = 0 ;
63- for (int i=1 ;i<n;i++){
64- scanf (" %lld"
65- soma += davez;
66- ll best = query (soma) + a*soma*soma + b*soma + c;
67- add (-2 *a*soma,best + a*soma*soma - b*soma);
68- }
69- scanf (" %lld"
70- soma += davez;
71- ll best = query (soma) + a*soma*soma + b*soma + c;
72- printf (" %lld\n "
73- }
74- return 0 ;
75- }n 0 ;
76- }
1+ // Ivan Carvalho
2+ // Solution to https://dmoj.ca/problem/apio10p1
3+ #include < cstdio>
4+ #include < algorithm>
5+ #include < vector>
6+ using namespace std ;
7+ typedef long long ll;
8+ vector<ll> M,B;
9+ int pointer;
10+ bool bad (int l1,int l2,int l3)
11+ {
12+ /*
13+ intersection(l1,l2) has x-coordinate (b1-b2)/(m2-m1)
14+ intersection(l1,l3) has x-coordinate (b1-b3)/(m3-m1)
15+ set the former greater than the latter, and cross-multiply to
16+ eliminate division
17+ */
18+ return (B[l3]-B[l1])*(M[l1]-M[l2])<(B[l2]-B[l1])*(M[l1]-M[l3]);
19+ }
20+ // Adds a new line (with lowest slope) to the structure
21+ void add (long long m,long long b)
22+ {
23+ // First, let's add it to the end
24+ m *= -1 ;
25+ b *= -1 ;
26+ M.push_back (m);
27+ B.push_back (b);
28+ // If the penultimate is now made irrelevant between the antepenultimate
29+ // and the ultimate, remove it. Repeat as many times as necessary
30+ while (M.size ()>=3 &&bad (M.size ()-3 ,M.size ()-2 ,M.size ()-1 ))
31+ {
32+ M.erase (M.end ()-2 );
33+ B.erase (B.end ()-2 );
34+ }
35+ }
36+ // Returns the minimum y-coordinate of any intersection between a given vertical
37+ // line and the lower envelope
38+ long long query (long long x)
39+ {
40+ // If we removed what was the best line for the previous query, then the
41+ // newly inserted line is now the best for that query
42+ if (pointer>=M.size ())
43+ pointer=M.size ()-1 ;
44+ // Any better line must be to the right, since query values are
45+ // non-decreasing
46+ while (pointer<M.size ()-1 &&
47+ M[pointer+1 ]*x+B[pointer+1 ]<M[pointer]*x+B[pointer])
48+ pointer++;
49+ return -(M[pointer]*x+B[pointer]);
50+ }
51+ int main (){
52+ int TC = 1 ;
53+ while (TC--){
54+ M.clear ();
55+ B.clear ();
56+ pointer = 0 ;
57+ int n;
58+ scanf (" %d"
59+ ll a,b,c;
60+ scanf (" %lld %lld %lld"
61+ add (0LL ,0LL );
62+ ll davez, soma = 0 ;
63+ for (int i=1 ;i<n;i++){
64+ scanf (" %lld"
65+ soma += davez;
66+ ll best = query (soma) + a*soma*soma + b*soma + c;
67+ add (-2 *a*soma,best + a*soma*soma - b*soma);
68+ }
69+ scanf (" %lld"
70+ soma += davez;
71+ ll best = query (soma) + a*soma*soma + b*soma + c;
72+ printf (" %lld\n "
73+ }
74+ return 0 ;
75+ }
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