DMOJ · ampl2024sp4

Spenser's Big Announcement

This C++ solution uses binary search for DMOJ ampl2024sp4 Spenser's Big Announcement. Read the reasoning, inspect the code, or try your own test case below.

ampl2024sp4Sorting & searchingBinary searchC++51 lines
Solution196of 248
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Approach

Binary search

Spenser's Big Announcement matches word wrapping into nonempty lines and binary search for minimum banner width.

Sorting & searching

Problem and code

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Written by benbenyaojifen. Try the problem first, then compare your approach with the code.

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Implementation

Problem_4_Spenser_s_Big_Announcement.cpp

C++

    #include <bits/stdc++.h>
    using namespace std;
    using ll = long long;
    using i128 = __int128;
    const int inf = 1e9;
    const ll INF = 1e18; //❄️
    int main() {
        ios::sync_with_stdio(0); cin.tie(0); 
        int n, l;
        ll s; cin >> n >> l >> s;
        vector<ll> v(n);
        ll sum = 0;
        for (int i = 0; i < n; i++) {
            cin >> v[i];
            sum += v[i];
        }
        ll lo = 0, hi = sum + (n - 1) * s;
        auto check = [&] (ll w) {
            int indx = 0;
            for (int i = 0; i < l && indx < v.size(); i++) {
                ll cur = w;
                while (indx < v.size() && cur >= v[indx]) {
                    if (cur == w) {
                        cur -= v[indx];
                        indx++;
                    } else {
                        if (cur >= v[indx] + s) {
                            cur -= v[indx] + s;
                            indx++;
                        } else {
                            break;
                        }
                    }
                }
            }
            return indx == v.size();
        };
        ll ans = INF;
        while (lo <= hi) {
            ll mid = lo + (hi - lo) / 2;
            int res = check(mid);
            if (res) {
                ans = min(ans, mid);
                hi = mid - 1;
            } else {
                lo = mid + 1;
            }
        }
        cout << ans << '\n';
        return 0;
    }
        

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