# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
961749 | LucaLucaM | Horses (IOI15_horses) | C++17 | 1552 ms | 12172 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include "horses.h"
#include <cassert>
#include <iostream>
#include <vector>
#include <algorithm>
typedef long long ll;
typedef __int128_t i128;
#define int ll
const int NMAX = 5e5;
const int mod = 1e9 + 7;
int n;
int x[NMAX + 1], y[NMAX + 1];
int solve() {
/**
dp[i][h] =def= care e profitul maxim daca dupa i ani am h cai
dp[i][h * x[i]] max= dp[i - 1][h]
dp[i][h] max= dp[i][h + 1] + y[i]
daca am bucati consecutive cu x[i] = 1, le pot grupa intr o singura bucata cu y sa fie maximu tuturor y urilor
1 <= y[i] <= 1e9
daca pot obtine 1e9 cai => tot inmultesc pana ajung la un sufix care are doar 1 la final
pt sufix ala cu doar 1 iau y maxim
asta is destul de sigur ca e usor de optimizat pt O((n + m) log) sau cv
ok, acum ramane cazu in care produsu e < 1e9
imi compactez x = 1
am n <= 60
fie pref[i] =def= nr maxim de cai din primii i ani
pref[i] = x[1] * x[2] * ... * x[i]
oare e optim sa fac cv gen
daca vreau sa vand, vand tot mai putin 1
(si ma rog, la final de tot ii vand pe toti)
ok bun am dat copy paste la notepad
ideea e ca am facut prost cazu in care am produsu mare
totusi, pare sa fie bine cazu in care am produsu mic
ma uit la cel mai mic sufix cu produsu > 1e9 -> daca compactez 1 urile, o sa aiba lungimea cel mult 30 (2^30 > 1e9)
=>
trebuie sa iau unu dintre astea
fac suff[i] practic dinamica pt produsu mic
max{big * pref[i] * y[i] + suff[i]}
big * pref[i] * y[i] + suff[i] - big * pref[j] * y[j] - suff[j] =
big * (pref[i] * y[i] - pref[j] * y[i]) + (suff[i] - suff[j])
ok aparent e optim sa vand tot intr o zi
**/
ll prod = 1;
ll ret = 0;
for (int i = 1; i <= n; i++) {
prod *= x[i];
ret = std::max(ret, (ll) prod * y[i]);
}
return ret;
}
signed init(signed N, signed X[], signed Y[]) {
n = N;
for (int i = 1; i <= n; i++) {
x[i] = X[i - 1];
y[i] = Y[i - 1];
}
return solve();
}
signed updateX(signed pos, signed val) {
x[++pos] = val;
return solve();
}
signed updateY(signed pos, signed val) {
y[++pos] = val;
return solve();
}
Compilation message (stderr)
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