이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define sz(a) (int)a.size()
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define ff(i,a,b) for(int i=a;i<=b;i++)
#define fb(i,b,a) for(int i=b;i>=a;i--)
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
const int maxn = 305;
const int inf = 1e9 + 5;
template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// os.order_of_key(k) the number of elements in the os less than k
// *os.find_by_order(k) print the k-th smallest number in os(0-based)
int n, m, K;
int A[maxn];
int B[maxn];
const int N = 300 * 300;
int dp[N + 5];
int main()
{
ios::sync_with_stdio(false);
cout.tie(nullptr);
cin.tie(nullptr);
cin >> n >> m >> K;
int zbirA = 0, zbirB = 0, mn = inf;
ff(i,1,n)cin >> A[i], zbirA += A[i], mn = min(mn, A[i]);
ff(i,1,m)cin >> B[i], zbirB += B[i];
if(zbirA > zbirB || mn < K)return cout << "Impossible", 0;
memset(dp, -inf, sizeof(dp));
dp[0] = 0;
ff(i,1,m){
int X = B[i];
fb(j,N,1){
if(j >= X)dp[j] = max(dp[j], dp[j - X] + min(X, n));
}
}
ff(j,zbirA,N){
if(dp[j] >= n * K)return cout << j - zbirA, 0;
}
cout << "Impossible" << endl;
return 0;
}
/**
// probati bojenje sahovski ili slicno
**/
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