Submission #575090

#	Time	Username	Problem	Language	Result	Execution time	Memory
575090		MohammadAghil	Bootfall (IZhO17_bootfall)	C++14	44 / 100	1008 ms	11936 KiB

bootfall

      #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], 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;
     rep(i,0,n) if(a[i]%2 != a[0]%2) return cout << "0\n", 0;
     vector<int> ans;
     rep(i,1,sum+1) if(i%2 == a[0]%2){
          bool ok = true;
          rep(j,0,n){
               int t = (sum - a[j] + i)>>1;
               if(t > sum || !dp[j][t]) {
                    ok = false;
                    break;
               }
          }
          if(ok) ans.pb(i);
     }
     cout << sz(ans) << '\n';
     for(int c: ans) cout << c << ' ';
     if(sz(ans)) cout << '\n';
     return 0; 
}

• Subtask 1: 6 / 6
• Subtask 2: 7 / 7
• Subtask 3: 15 / 15
• Subtask 4: 16 / 16
• Subtask 5: 0 / 21
• Subtask 6: 0 / 35