Submission #1001402

#	Time	Username	Problem	Language	Result	Execution time	Memory
1001402		GrindMachine	Bigger segments (IZhO19_segments)	C++17	100 / 100	171 ms	94340 KiB

segments

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42
#endif

/*

refs:
https://codeforces.com/blog/entry/64479?#comment-484350

dp[i][j] = max k s.t [k..i] is the last segment and #of jumps = j
only considering the best 2 j values works because:
dp[i][j] >= dp[i][j-2] (for all valid j)
can be proved by induction

*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

void solve(int test_case)
{
    ll n; cin >> n;
    vector<ll> a(n+5);
    rep1(i,n) cin >> a[i];
    vector<ll> p(n+5);
    rep1(i,n) p[i] = p[i-1]+a[i];

    vector<map<ll,ll>> dp(n+5);
    dp[1][0] = 0;

    rep1(i,n){
        while(sz(dp[i]) > 2){
            dp[i].erase(dp[i].begin());
        }

        for(auto [j,k] : dp[i]){
            // append
            amax(dp[i+1][j],dp[i][j]);

            // jump
            ll sum1 = p[i]-p[k];
            ll pos = lower_bound(p.begin()+1,p.begin()+n+1,sum1+p[i])-p.begin();
            amax(dp[pos][j+1],(ll)i);
        }
    }

    ll ans = dp[n].rbegin()->first+1;
    cout << ans << endl;
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

• Subtask 1: 13 / 13
• Subtask 2: 14 / 14
• Subtask 3: 10 / 10
• Subtask 4: 36 / 36
• Subtask 5: 27 / 27