結果

問題 No.195 フィボナッチ数列の理解(2)
ユーザー rantdrantd
提出日時 2015-04-30 18:39:24
言語 C++11
(gcc 11.4.0)
結果
AC  
実行時間 2 ms / 5,000 ms
コード長 3,753 bytes
コンパイル時間 1,434 ms
コンパイル使用メモリ 157,748 KB
実行使用メモリ 4,380 KB
最終ジャッジ日時 2023-09-19 04:41:19
合計ジャッジ時間 2,554 ms
ジャッジサーバーID
(参考情報)
judge13 / judge11
このコードへのチャレンジ
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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
4,380 KB
testcase_01 AC 1 ms
4,380 KB
testcase_02 AC 2 ms
4,376 KB
testcase_03 AC 2 ms
4,380 KB
testcase_04 AC 2 ms
4,376 KB
testcase_05 AC 1 ms
4,380 KB
testcase_06 AC 1 ms
4,380 KB
testcase_07 AC 2 ms
4,380 KB
testcase_08 AC 1 ms
4,380 KB
testcase_09 AC 1 ms
4,376 KB
testcase_10 AC 2 ms
4,376 KB
testcase_11 AC 2 ms
4,380 KB
testcase_12 AC 2 ms
4,380 KB
testcase_13 AC 2 ms
4,376 KB
testcase_14 AC 2 ms
4,380 KB
testcase_15 AC 1 ms
4,380 KB
testcase_16 AC 2 ms
4,380 KB
testcase_17 AC 2 ms
4,380 KB
testcase_18 AC 2 ms
4,380 KB
testcase_19 AC 1 ms
4,380 KB
testcase_20 AC 1 ms
4,380 KB
testcase_21 AC 2 ms
4,380 KB
testcase_22 AC 2 ms
4,376 KB
testcase_23 AC 2 ms
4,380 KB
testcase_24 AC 2 ms
4,380 KB
権限があれば一括ダウンロードができます

ソースコード

diff #

#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) {
                P tmp = solve(i, j, z[0], z[2]);
                bool good = 0;
                repu(i, 0, cnt) {
                    if (tmp.first * fib[i] + tmp.second * fib[i + 1] == z[1]) good = 1;
                }
                if (good) amin(ans, tmp);
            }
        }
        if (ans != inf) {
            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;
}
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