Submission #575094

#TimeUsernameProblemLanguageResultExecution timeMemory
575094MohammadAghilBootfall (IZhO17_bootfall)C++14
65 / 100
1020 ms13388 KiB
      #include <bits/stdc++.h>
  #pragma GCC optimize ("Ofast,unroll-loops")
#pragma GCC target ("avx2")
    using namespace std;
  typedef long long ll;
   typedef pair<int, int> pp;
    #define er(args ...) cerr << __LINE__ << ": ", err(new istringstream(string(#args)), args), cerr << endl
      #define per(i,r,l) for(int i = (r); i >= (l); i--)
        #define rep(i,l,r) for(int i = (l); i < (r); i++)
           #define all(x) x.begin(), x.end()
              #define sz(x) (int)(x).size()
                  #define pb push_back
                      #define ss second
                           #define ff first
                                   void err(istringstream *iss){}template<typename T,typename ...Args> void err(istringstream *iss,const T &_val, const Args&...args){string _name;*iss>>_name;if(_name.back()==',')_name.pop_back();cerr<<_name<<" = "<<_val<<", ",err(iss,args...);}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const ll mod = 1e9 + 7, maxn = 5e2+5, lg = 20, inf = ll(1e9) + 5;
ll pw(ll a,ll b,ll md=mod){if(!b)return 1;ll k=pw(a,b>>1ll);return k*k%md*(b&1ll?a:1)%md;}

bitset<maxn*maxn> dp[maxn];

int main(){
     cin.tie(0) -> sync_with_stdio(0);
     int n; cin >> n;
     vector<int> a(n);
     int sum = 0;
     rep(i,0,n) cin >> a[i];
     int k = -1, f = 1;
     rep(i,0,n){
          int t = 0;
          while(a[i]%2 == 0) t++, a[i] >>= 1, f *= 1 + (k == -1);
          if(k == -1) k = t;
          if(t - k) return cout << "0\n", 0;
          sum += a[i];
     }
     rep(k,0,n+1){
          dp[k][0] = true;
          rep(i,0,n){
               if(i == k) continue;
               dp[k] |= dp[k]<<a[i];
          }
     }
     if((sum&1) || !dp[n][sum>>1]) return cout << "0\n", 0;
     vector<int> ans;
     rep(i,0,maxn*maxn) dp[n][i] = 1;
     rep(i,0,n) dp[i] >>= (sum - a[i] + 1)>>1, dp[n] &= dp[i];
     rep(i,0,sum+1) if(dp[n][i]) ans.pb((i<<1)+1);
     cout << sz(ans) << '\n';
     for(int c: ans) cout << c*f << ' ';
     if(sz(ans)) cout << '\n';
     return 0; 
}
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