Submission #244810

#	Time	Username	Problem	Language	Result	Execution time	Memory
244810		Vimmer	Kas (COCI17_kas)	C++14	40 / 100	177 ms	44416 KiB

kas

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

//#pragma GCC optimize("unroll-loops")
//#pragma GCC optimize("-O3")
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("fast-math")
//#pragma GCC optimize("no-stack-protector")

#define F first
#define S second
#define sz(x) int(x.size())
#define pb push_back
#define N 100001
#define M ll(998244353)
#define inf 1e9 + 1e9


using namespace std;
//using namespace __gnu_pbds;

typedef long double ld;
typedef long long ll;
typedef short int si;
typedef array <int, 3> a3;

//typedef tree <int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;



int main()
{
    //freopen("input.txt", "r", stdin); //freopen("output4.txt", "w", stdout);

    ios_base::sync_with_stdio(0); istream::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    int n;

    cin >> n;

    int c[n], sumr = 0;;

    for (int i = 0; i < n; i++) {cin >> c[i]; sumr += c[i];}

    sort(c, c + n);

    reverse(c, c + n);

    bool dp[n + 1][N];

    memset(dp, 0, sizeof(dp));

    dp[0][0] = 1;

    for (int i = 0; i < n; i++)
      for (int j = 0; j < N; j++)
        if (dp[i][j])
            {
                dp[i + 1][j] = 1;

                dp[i + 1][j + c[i]] = 1;
            }

    int sm = 0, sum = 0, ans = 1e9;

    for (int i = n; i >= 0; i--)
    {
        if (i != n) sm += c[i];

        int val = sumr - sm;

        if (val % 2 || !dp[i][val / 2]) continue;

        sum = val / 2;

        ans = min(ans, sum + sm);
    }

    reverse(c, c + n);

    memset(dp, 0, sizeof(dp));

    dp[0][0] = 1;

    for (int i = 0; i < n; i++)
      for (int j = 0; j < N; j++)
        if (dp[i][j])
            {

                dp[i + 1][j] = 1;


                dp[i + 1][j + c[i]] = 1;
            }

    sm = 0; sum = 0;

    for (int i = n; i >= 0; i--)
    {
        if (i != n) sm += c[i];

        int val = sumr - sm;

        if (val % 2 || !dp[i][val / 2]) continue;

        sum = val / 2;

        ans = min(ans, sum + sm);
    }

    cout << ans << endl;

}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 0 / 10
• Subtask 5: 0 / 10
• Subtask 6: 0 / 10
• Subtask 7: 0 / 10
• Subtask 8: 10 / 10
• Subtask 9: 0 / 10
• Subtask 10: 0 / 10