DMOJ · coci16c4p3

COCI '16 Contest 4 #3 Kas

This C++ solution uses dynamic programming for DMOJ coci16c4p3 COCI '16 Contest 4 #3 Kas. Read the reasoning, inspect the code, or try your own test case below.

coci16c4p3Dynamic programmingC++32 lines
Solution118of 248
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Approach

Dynamic programming

Kas banknote balancing objective matches the difference-state dynamic program.

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

3_Kas.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);
        int mx = 0;
        for (int i = 0; i < n; i++) {
            cin >> v[i];
            mx += v[i];
        }
        vector<vector<int>> dp(n + 1, vector<int>(mx + 1, -inf));
        dp[0][0] = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j <= mx; j++) {
                dp[i][j] = dp[i - 1][j];
                if (dp[i - 1][abs(j - v[i - 1])] != -inf) {
                    dp[i][j] = max(dp[i][j], dp[i - 1][abs(j - v[i - 1])] + v[i - 1]);
                }
                if (j + v[i - 1] <= mx && dp[i - 1][j + v[i - 1]] != inf) {
                    dp[i][j] = max(dp[i][j], dp[i - 1][j + v[i - 1]] + v[i - 1]);
                }
            }
        }
        int ans = dp[n][0] / 2 + mx - dp[n][0];
        cout << ans << '\n';
        return 0;
    }
        

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