結果
| 問題 |
No.195 フィボナッチ数列の理解(2)
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2015-04-30 18:36:07 |
| 言語 | C++11(廃止可能性あり) (gcc 13.3.0) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 3,680 bytes |
| コンパイル時間 | 1,478 ms |
| コンパイル使用メモリ | 170,748 KB |
| 実行使用メモリ | 5,376 KB |
| 最終ジャッジ日時 | 2024-07-05 17:12:39 |
| 合計ジャッジ時間 | 2,251 ms |
|
ジャッジサーバーID (参考情報) |
judge2 / judge1 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 12 WA * 10 |
ソースコード
#include <bits/stdc++.h>
using namespace std;
#define repu(i, begin, end) for (__typeof(begin) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define repe(i, begin, end) for (__typeof(begin) i = (begin); i != (end) + 1 - 2 * ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define mem(a, x) memset(a, x, sizeof(a))
#define all(a) a.begin(), a.end()
#define count_bits(x) __builtin_popcount(x)
#define count_bitsll(x) __builtin_popcountll(x)
#define least_bits(x) __builtin_ffs(x)
#define least_bitsll(x) __builtin_ffsll(x)
#define most_bits(x) 32 - __builtin_clz(x)
#define most_bitsll(x) 64 - __builtin_clz(x)
vector<string> split(const string &s, char c) {
vector<string> v;
stringstream ss(s);
string x;
while (getline(ss, x, c)) v.push_back(x);
return v;
}
#define error(args...) { vector<string> _v = split(#args, ','); err(_v.begin(), args); }
void err(vector<string>::iterator it) {}
template<typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
cerr << it -> substr((*it)[0] == ' ', it -> length()) << " = " << a << '\n';
err(++it, args...);
}
typedef long long ll;
const int MOD = 1000000007;
template<class T> inline T tmin(T a, T b) {return (a < b) ? a : b;}
template<class T> inline T tmax(T a, T b) {return (a > b) ? a : b;}
template<class T> inline void amax(T &a, T b) {if (b > a) a = b;}
template<class T> inline void amin(T &a, T b) {if (b < a) a = b;}
template<class T> inline T tabs(T a) {return (a > 0) ? a : -a;}
template<class T> T gcd(T a, T b) {while (b != 0) {T c = a; a = b; b = c % b;} return a;}
const int INF = 1000000000;
typedef pair<ll, ll> P;
const P inf = P(INF, INF);
vector<ll> fib;
P solve(int i, int j, ll x, ll y) {
ll det = fib[i] * fib[j + 1] - fib[i + 1] * fib[j];
ll deta = y * fib[i] - x * fib[j];
ll detb = x * fib[j + 1] - y * fib[i + 1];
if (tabs(deta) % tabs(det) != 0) return inf;
if (tabs(detb) % tabs(det) != 0) return inf;
ll ra = deta / det, rb = detb / det;
if (ra > 0 && rb > 0) return P(ra, rb);
return inf;
}
ll extgcd(ll a, ll b, ll &x, ll &y) {
ll d = a;
if (b != 0) {
d = extgcd(b, a % b, y, x);
y -= (a / b) * x;
}
else {
x = 1; y = 0;
}
return d;
}
ll mod_inverse(ll a, ll m) {
ll x, y;
extgcd(a, m, x, y);
return (m + x % m) % m;
}
int main(int argc, char *argv[]) {
ios_base::sync_with_stdio(false);
int z[3];
fib.push_back(1);
fib.push_back(0);
int cnt = 2;
while (true) {
ll val = fib[cnt - 1] + fib[cnt - 2];
if (val > INF) break;
fib.push_back(val);
cnt++;
}
repu(i, 0, 3) cin >> z[i];
sort(z, z + 3);
P ans = inf;
if (z[0] != z[2]) {
repu(i, 0, cnt - 1) repu(j, 0, cnt - 1) {
if (i != j) amin(ans, solve(i, j, z[0], z[2]));
}
if (ans != inf) {
bool good = 0;
repu(i, 0, cnt) {
if (ans.first * fib[i] + ans.second * fib[i + 1] == z[1]) good = 1;
}
if (good) printf("%lld %lld\n", ans.first, ans.second);
else printf("-1\n");
}
}
else {
P ans = P(z[0], 1);
amin(ans, P(1, z[0]));
if (z[0] > 1) amin(ans, P(1, z[0] - 1));
repu(i, 3, cnt - 1) {
ll x = mod_inverse(fib[i], fib[i + 1]);
ll ra = (x * z[0]) % fib[i + 1];
if (ra == 0) ra = fib[i + 1];
ll rb = (z[0] - ra * fib[i]) / fib[i + 1];
if (rb > 0) amin(ans, P(ra, rb));
}
printf("%lld %lld\n", ans.first, ans.second);
}
return 0;
}