이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define in(x) freopen(x, "r", stdin)
#define out(x) freopen(x, "w", stdout)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-O3")
#define F first
#define S second
#define pb push_back
#define N +100500
#define M ll(1e9 + 7)
#define sz(x) (int)x.size()
#define re return
#define oo ll(1e18)
#define el '\n'
#define pii pair <int, int>
using namespace std;
//using namespace __gnu_pbds;
//typedef tree <int, null_type, less_equal <int> , rb_tree_tag, tree_order_statistics_node_update> ordered_set;
typedef long long ll;
typedef long double ld;
int a[N][20], cur, i, j, k, n, m, point[N], ans[N];
vector <pii> times;
int main()
{
    srand(time(0));
    ios_base::sync_with_stdio(0);
    iostream::sync_with_stdio(0);
    ios::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);
//    in("input.txt");
//    out("output.txt");
    cin >> n;
    for (i = 1; i <= n; i++)
    {
        cin >> a[i][1] >> a[i][2] >> a[i][3];
    }
    cin >> m;
    for (i = 1; i <= m; i++) cin >> point[i];
    for (j = 1; j <= m; j++)
    {
        for (i = 1; i <= n; i++)
        {
            int k = 1;
            cur = 0;
            if (point[j] > 100) cur = cur + a[i][1] * 100, k++;
            if (point[j] > 200) cur = cur + a[i][2] * 100, k++;
            cur += a[i][k] * (point[j] % 100);
            times.pb({cur, i});
        }
        sort(times.begin(), times.end());
        int kl = 0;
        for (i = 0; i < n; i++)
        {
            int k = times[i].S;
            int tm = times[i].F + kl % 20;
            int pos = point[j] + kl % 20;
            if (pos <= 100) tm += (100 - pos) * a[k][1], pos = 100;
            if (pos <= 200) tm += (200 - pos) * a[k][2], pos = 200;
            if (pos <= 300) tm += (300 - pos) * a[k][3];
            ans[k] = tm;
            if (i < n - 1 && times[i].F == times[i + 1].F) kl--;
            else kl = i;
            kl++;
        }
    }
    for (i = 1; i <= n; i++) cout << ans[i] << el;
}
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