結果
問題 | No.195 フィボナッチ数列の理解(2) |
ユーザー | rantd |
提出日時 | 2015-04-30 18:27:49 |
言語 | C++11 (gcc 11.4.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 3,680 bytes |
コンパイル時間 | 1,601 ms |
コンパイル使用メモリ | 170,760 KB |
実行使用メモリ | 5,376 KB |
最終ジャッジ日時 | 2024-07-05 17:12:30 |
合計ジャッジ時間 | 2,400 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge1 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 2 ms
5,248 KB |
testcase_01 | AC | 2 ms
5,376 KB |
testcase_02 | AC | 2 ms
5,376 KB |
testcase_03 | AC | 1 ms
5,376 KB |
testcase_04 | AC | 2 ms
5,376 KB |
testcase_05 | AC | 2 ms
5,376 KB |
testcase_06 | AC | 2 ms
5,376 KB |
testcase_07 | AC | 2 ms
5,376 KB |
testcase_08 | AC | 2 ms
5,376 KB |
testcase_09 | AC | 2 ms
5,376 KB |
testcase_10 | AC | 2 ms
5,376 KB |
testcase_11 | AC | 2 ms
5,376 KB |
testcase_12 | AC | 2 ms
5,376 KB |
testcase_13 | AC | 1 ms
5,376 KB |
testcase_14 | AC | 2 ms
5,376 KB |
testcase_15 | WA | - |
testcase_16 | AC | 2 ms
5,376 KB |
testcase_17 | AC | 1 ms
5,376 KB |
testcase_18 | AC | 2 ms
5,376 KB |
testcase_19 | AC | 2 ms
5,376 KB |
testcase_20 | AC | 2 ms
5,376 KB |
testcase_21 | WA | - |
testcase_22 | AC | 2 ms
5,376 KB |
testcase_23 | AC | 2 ms
5,376 KB |
testcase_24 | AC | 1 ms
5,376 KB |
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 = x * fib[j + 1] - y * fib[i + 1]; ll detb = y * fib[i] - x * fib[j]; 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; }