Submission #1112935

#	Time	Username	Problem	Language	Result	Execution time	Memory
1112935		adiyer	Bootfall (IZhO17_bootfall)	C++17	100 / 100	136 ms	3060 KiB

bootfall

#pragma optimize ("g",on)
#pragma GCC optimize("inline")
#pragma GCC optimize ("Ofast")
#pragma GCC optimize ("unroll-loops")
#pragma GCC optimize ("03")
 
#include <bits/stdc++.h>
 
#define ios ios_base::sync_with_stdio(0); cin.tie(0);
#define pb push_back
 
using namespace std;
 
typedef int ll;
 
const int N = 5e2;
const int MAX = N * N;
 
short int n;
short int a[N];
 
int s;

int dp[MAX + 1];

vector < int > ans;

void sol(){
	cin >> n, dp[0] = 1;
	for(int i = 1; i <= MAX; i++) ans.pb(i);
	for(short int i = 0; i < n; i++){ 
		cin >> a[i], s += a[i];
		for(ll j = MAX; j >= a[i]; j--) dp[j] += dp[j - a[i]];
	}	
	if(s % 2 || dp[s / 2] == 0){
		cout << "0\n";
		return;
	}
	for(short int i = 0; i < n; i++){
		s -= a[i];
		for(int j = a[i]; j <= MAX; j++) dp[j] -= dp[j - a[i]];
		int j = 0;
		while(j < ans.size()){
			if(s < ans[j] || (s - ans[j]) % 2 || !dp[(s - ans[j]) / 2]){
				if(j < ans.size()) swap(ans[j], ans[ans.size() - 1]);
				ans.pop_back();
			}
			else{
				j++;
			}
		}
		s += a[i];
		for(int j = MAX; j >= a[i]; j--) dp[j] += dp[j - a[i]];
	}
	sort(ans.begin(), ans.end());
	cout << ans.size() << '\n';
	for(int x : ans) cout << x << ' ';
}
 
signed main(){
	ios                     
	sol();
// 	slow();
//  stress();
}

Compilation message (stderr)

bootfall.cpp:1: warning: ignoring '#pragma optimize ' [-Wunknown-pragmas]
    1 | #pragma optimize ("g",on)
      | 
bootfall.cpp: In function 'void sol()':
bootfall.cpp:43:11: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   43 |   while(j < ans.size()){
      |         ~~^~~~~~~~~~~~
bootfall.cpp:45:10: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   45 |     if(j < ans.size()) swap(ans[j], ans[ans.size() - 1]);
      |        ~~^~~~~~~~~~~~

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