Submission #58459

#TimeUsernameProblemLanguageResultExecution timeMemory
58459BenqBootfall (IZhO17_bootfall)C++14
100 / 100
604 ms3220 KiB
#include <bits/stdc++.h> #include <ext/pb_ds/tree_policy.hpp> #include <ext/pb_ds/assoc_container.hpp> using namespace std; using namespace __gnu_pbds; typedef long long ll; typedef long double ld; typedef complex<ld> cd; typedef pair<int, int> pi; typedef pair<ll,ll> pl; typedef pair<ld,ld> pd; typedef vector<int> vi; typedef vector<ld> vd; typedef vector<ll> vl; typedef vector<pi> vpi; typedef vector<pl> vpl; typedef vector<cd> vcd; template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>; #define FOR(i, a, b) for (int i=a; i<(b); i++) #define F0R(i, a) for (int i=0; i<(a); i++) #define FORd(i,a,b) for (int i = (b)-1; i >= a; i--) #define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--) #define sz(x) (int)(x).size() #define mp make_pair #define pb push_back #define f first #define s second #define lb lower_bound #define ub upper_bound #define all(x) x.begin(), x.end() const int MOD = 1000000007; const ll INF = 1e18; const int MX = 500*500+1; typedef array<int,MX> B; int N, a[500]; bitset<MX> all; B b; int sum = 0; void ad(B& b, int x) { sum += x; FORd(i,x,MX) { b[i] += b[i-x]; if (b[i] >= MOD) b[i] -= MOD; } } void del(B& b, int x) { // b[i]+b[i-x] = b2[i], b[i] = b2[i]-b[i-x] sum -= x; FOR(i,x,MX) { b[i] -= b[i-x]; if (b[i] < 0) b[i] += MOD; } } int main() { ios_base::sync_with_stdio(0); cin.tie(0); all.flip(); cin >> N; b[0] = 1; F0R(i,N) { cin >> a[i]; ad(b,a[i]); } if (sum % 2 != 0 || !b[sum/2]) { cout << 0; exit(0); } F0R(i,N) { del(b,a[i]); bitset<MX> cur; for (int j = 0; 2*j < sum; ++j) if (b[j]) cur[sum-2*j] = 1; ad(b,a[i]); all &= cur; } vi v; FOR(i,1,MX) if (all[i]) v.pb(i); cout << sz(v) << "\n"; for (int i: v) cout << i << " "; } /* Look for: * the exact constraints (multiple sets are too slow for n=10^6 :( ) * special cases (n=1?) * overflow (ll vs int?) * array bounds */
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