DMOJ · ucc20p4

Bubble Tea

This C++ solution uses dynamic programming for DMOJ ucc20p4 Bubble Tea. Read the reasoning, inspect the code, or try your own test case below.

ucc20p4Dynamic programmingC++26 lines
Solution038of 248
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Approach

Dynamic programming

Bubble Tea's consecutive grouping and 25/50-percent discount rules match the code's dynamic programming transitions.

Dynamic programming

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

P_4_Bubble_Tea.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; cin >> n;
        vector<int> v(n);
        for (int i = 0; i < n; i++) cin >> v[i];
        vector<int> dp(n + 1);
        for (int i = 1; i <= n; i++) {
            dp[i] = dp[i - 1] + v[i - 1];
            if (i >= 2) {
                dp[i] = min(dp[i], dp[i - 2] + max(v[i - 1], v[i - 2]) + min(v[i - 1], v[i - 2]) / 4 * 3);
                if (i >= 3) {
                    vector<int> a = {v[i - 1], v[i - 2], v[i - 3]};
                    sort(a.begin(), a.end());
                    dp[i] = min(dp[i], dp[i - 3] + a[0] / 2 + a[1] + a[2]);
                }
            }
        }
        cout << dp[n] << '\n';
        return 0;
    }
        

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