This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#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+1;
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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