Submission #836918

#TimeUsernameProblemLanguageResultExecution timeMemory
836918azberjibiouCookies (JOI23_cookies)C++17
7 / 100
327 ms52728 KiB
#include <bits/stdc++.h> #define all(v) v.begin(), v.end() #define gibon ios::sync_with_stdio(false); cin.tie(0); #define fi first #define se second #define pdd pair<long double, long double> #define pii pair<int, int> #define pll pair<ll, ll> #define ppi pair<pii, pii> #pragma GCC optimize("O3") #pragma GCC optimize("Ofast") #pragma GCC optimize("unroll-loops") typedef long long ll; using namespace std; const int mxN=15003; const int mxM=10004; const int MOD=1e9+7; const ll INF=8e18; int dx[4]={1, 0, -1, 0}, dy[4]={0, 1, 0, -1}; int N, M, K; int A[mxN], B[mxN], c[mxN], ci[mxN]; int cnt[mxN], S[mxN]; bitset <15003> X[147000]; bitset <15003> tX, nX; int ans; void solv(vector <int> v) { cout << v.size() << '\n'; priority_queue <pii> pq; for(int i=1;i<=N;i++) pq.emplace(A[i], i); for(int x : v) { cout << B[x] << " "; vector <pii> tmp; for(int i=0;i<B[x];i++) { tmp.push_back(pq.top()); pq.pop(); } for(auto [a, b] : tmp) cout << b << " "; cout << '\n'; for(auto [a, b] : tmp) if(a!=1) pq.emplace(a-1, b); } } int main() { gibon cin >> N; for(int i=1;i<=N;i++) cin >> A[i], K+=A[i]; for(int i=1;i<=N;i++) for(int j=1;j<=A[i];j++) cnt[j]++; for(int i=1;i<=K;i++) S[i]=S[i-1]+cnt[i]; for(int i=1;i<=K;i++) c[i]=c[i-1]+K/i; cin >> M; for(int i=1;i<=M;i++) cin >> B[i]; B[M+1]=K; for(int i=1;i<=M;i++) if(B[i]<=S[1]) X[i][B[i]]=1; if(B[M]==K) ans=1; for(int i=1;i<K;i++) if(ans==0) { tX.reset(); nX.reset(); for(int j=0;j<=S[i+1]-B[M];j++) tX[j]=1; for(int j=M;j>=1;j--) { for(int k=max(0, S[i+1]-B[j+1]+1);k<=S[i+1]-B[j];k++) tX[k]=1; if(B[j]>K/(i+1)) continue; nX|=X[c[i-1]+j]; X[c[i]+j]=((nX&tX)<<B[j]); } //for(int j=0;j<=K;j++) printf("dp[%d][%d]=(%d, %d)\n", i, j, dp[i][j].fi, dp[i][j].se); for(int j=1;j<=M;j++) { if(B[j]<=K/(i+1) && X[c[i]+j][K]==1) ans=i+1; } } if(ans==0) { cout << -1; return 0; } vector <int> v; int cur=K; for(int i=ans;i>=1;i--) { int nv=0; for(int j=1;j<=M;j++) if(B[j]<=K/i && X[c[i-1]+j][cur]) nv=j; v.push_back(nv); cur-=B[nv]; } solv(v); }
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