Submission #942753

#	Time	Username	Problem	Language	Result	Execution time	Memory
942753		vjudge1	Fibonacci representations (CEOI18_fib)	C++17	0 / 100	1 ms	604 KiB

fib

#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,fma")
#include <bits/stdc++.h>
 
#define ff first
#define ss second
#define pb push_back
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define int long long
#define rnd(l, r) uniform_int_distribution<int>(l, r)(rng)
 
using namespace std;
 
void fp(string name){freopen((name+".in").c_str(),"r",stdin); freopen((name+".out").c_str(),"w",stdout);} 
int pow(int a,int b,int m){int ans=1;while(b){if(b&1){ans=(ans*a)%m;}b>>=1;a=(a*a)%m;}return ans;}
int binpow(int a,int b){int ans=1;while(b){if(b&1){ans=(ans*a);}b>>=1;a=(a*a);}return ans;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
 
const int N = 5005, mxc = 105, inf = 1e18, mod = 1e9 + 7;
 
vector <int> fib = {1, 2}, pref = {1, 3};

int ways(int x, int pos = 50){
	//cout << x <<" "<< pos << endl;
	//system("pause");
	if(pos == -1){
		return (x == 0);
	}
	if(pref[pos] < x)return 0;
	if(x < fib[pos])return ways(x, pos - 1);
	
	if(pos >= 2 && (pos == 2 && x - fib[pos] == 0 || x - fib[pos] <= fib[pos - 3]))
		return (ways(x - fib[pos], pos - 3)  * 2) % mod;
	else
		return ways(x - fib[pos], pos - 1) % mod;

}
main(){
	iostream::sync_with_stdio(false);	
    cin.tie(nullptr);
    cout.tie(nullptr);
    for(int i = 3; i <= 120; i++){
		fib.pb(fib[i - 2] + fib[i - 3]);
		pref.pb(fib.back() + pref.back());
	}
	int n;
	cin >> n;
	vector <int> v(n + 1);
	int cur = 0;
	for(int i = 1; i <= n; i++){
		cin >> v[i];
		cur += fib[v[i] - 1];
		//cout << fib[v[i] - 1] <<"-";
		int res = ways(cur);
		cout << res << endl;
	}
}	
/*
 * Before implementing the problem:
	
	Do I understand the problem correctly?
	Which places are tricky? What do I need to remember to implement them correctly?
	Which place is the heaviest by implementation? Can I do it simpler?
	Which place is the slowest? Where do I need to be careful about constant factors and where I can choose slower but simpler implementation?
	----------------------------------
	If not AC:
 
	Did you remember to do everything to handle the tricky places you thought about before?
	Is the solution correct?
	Do I understand the problem correctly?
	----------------------------------
	If you have a test on which the solution gives wrong answer:
 
	Is the solution doing what it was supposed to do?
	Is the problem in the code or in the idea?
*/

Compilation message (stderr)

fib.cpp: In function 'long long int ways(long long int, long long int)':
fib.cpp:33:27: warning: suggest parentheses around '&&' within '||' [-Wparentheses]
   33 |  if(pos >= 2 && (pos == 2 && x - fib[pos] == 0 || x - fib[pos] <= fib[pos - 3]))
      |                  ~~~~~~~~~^~~~~~~~~~~~~~~~~~~~
fib.cpp: At global scope:
fib.cpp:39:1: warning: ISO C++ forbids declaration of 'main' with no type [-Wreturn-type]
   39 | main(){
      | ^~~~
fib.cpp: In function 'void fp(std::string)':
fib.cpp:15:29: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   15 | void fp(string name){freopen((name+".in").c_str(),"r",stdin); freopen((name+".out").c_str(),"w",stdout);}
      |                      ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
fib.cpp:15:70: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
   15 | void fp(string name){freopen((name+".in").c_str(),"r",stdin); freopen((name+".out").c_str(),"w",stdout);}
      |                                                               ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 0 / 5
• Subtask 2: 0 / 20
• Subtask 3: 0 / 15
• Subtask 4: 0 / 10
• Subtask 5: 0 / 15
• Subtask 6: 0 / 35