Programming contest

// Aalto wk6 D - Programming contest
// Task: n tasks in fixed order, task i takes m[i] minutes, solving it at time t
// gives p[i] - t points. Skip any tasks. Maximize total points.
// Key observations:
//   1. If you solve k tasks i1 < i2 < ... < ik, task ij's m is paid by itself
//      and every chosen task after it -> total = sum p - sum (#chosen at/after j) * m[j].
//   2. In an optimal answer the LAST task gives >= 0 points (else drop it), so
//      total time <= max p = 1e6. m is strictly increasing (m[i] >= i), so
//      k tasks take >= 1+2+..+k = k(k+1)/2 -> k <= ~1414.
// DP from the back: f[c] = best score of the suffix choosing exactly c tasks.
//   Taking task i as the c-th from the end costs c * m[i]:
//   f[c] = max(f[c], f[c-1] + p[i] - c * m[i])
// O(n * K) = 1e5 * 1415 ~ 1.4e8, fine.
// #ifdef ONLINE_JUDGE
// #pragma GCC optimize("O3,unroll-loops")
// #pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
// #endif
#include "bits/stdc++.h"
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define fill(arr,val) memset(arr,val,sizeof(arr))
#define FOR(i, a, b) for(__typeof(b) i = a, _b = b; i <= _b; ++i)
#define FORD(i, a, b) for(__typeof(a) i = a, _b = b; i >= _b; --i)
#define ALL(a) (a).begin(), (a).end()
#define YES cout << "YES\n"
#define NO cout << "NO\n"
#define ll long long
#define fi first
#define se second
#define pb push_back
#define pf push_front
#define ii pair<int,int>
#define iii pair<int,pair<int,int>>
#define dq deque<int>
#define nend '\n'
using namespace std;

const ll inf = 1e18;
const int dx[] = {1,0,-1,0};
const int dy[] = {0,1,0,-1};
const int MOD = 1e9 + 7;
const ll hashi = 2e9 + 11;

void init_code() {
    fast;
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
}

const int K = 1500;
ll f[K+1];
ll ans=-1e9;
void solve() {
    int n;
        cin >> n;
        vector<ll> m(n), p(n);
        for (auto& x : m) cin >> x;
        for (auto& x : p) cin >> x;
        FOR(i,0,n+1) f[i]=-inf;
        f[0] = 0;
        FORD(i,n-1,0) {
            int lim = min(K, n-i);
            FORD(c,lim,1) // backwards so f[c-1] is still the old value
                if (f[c-1] != -inf)
                    f[c] = max(f[c], f[c-1] + p[i] - c * m[i]);
        }

        FOR(i,0,n+1) ans=max(ans,f[i]);
        cout << ans;

}
int main() {
    init_code();
    int t=1;
    // cin >> t;
    while (t--) {
        solve();
    }
    
}

Test 1

Verdict: ACCEPTED

Test 2

Verdict: ACCEPTED

Test 3

Verdict: ACCEPTED

Test 4

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Test 5

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Test 6

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Test 7

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Test 8

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Test 9

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Test 10

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Test 11

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Test 12

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Test 13

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Test 14

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Test 15

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Test 16

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Test 17

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Test 18

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Test 19

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Test 20

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Test 21

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Test 22

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Test 23

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Test 24

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Test 25

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Test 26

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Test 27

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Test 28

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Test 29

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Test 30

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Test 31

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Test 32

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Test 33

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Test 34

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Test 35

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Test 36

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Test 37

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Test 38

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Test 39

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Test 40

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Test 41

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Test 42

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Test 43

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Test 44

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Test 45

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Test 46

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Test 47

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Test 48

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Test 49

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Test 50

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Test 51

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Test 52

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Test 53

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Test 54

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Test 55

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Test 56

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Test 57

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Test 58

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Test 59

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Test 60

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Test 61

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Test 62

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Test 63

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Test 64

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Test 65

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Test 66

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Test 67

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Test 68

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Test 69

Verdict: ACCEPTED

Test 70

Verdict: ACCEPTED

Test 71

Verdict: ACCEPTED

Test 72

Verdict: ACCEPTED

Test 73

Verdict: RUNTIME ERROR

Test 74

Verdict: RUNTIME ERROR

Test 75

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Test 76

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Test 77

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Test 78

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Test 79

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Test 80

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Test 81

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Test 82

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Test 83

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Test 84

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Test 85

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Test 86

Verdict: RUNTIME ERROR