結果
問題 | No.891 隣接3項間の漸化式 |
ユーザー |
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提出日時 | 2020-03-30 23:46:47 |
言語 | C++11 (gcc 13.3.0) |
結果 |
AC
|
実行時間 | 2 ms / 2,000 ms |
コード長 | 5,162 bytes |
コンパイル時間 | 1,459 ms |
コンパイル使用メモリ | 167,676 KB |
実行使用メモリ | 6,944 KB |
最終ジャッジ日時 | 2024-06-23 07:13:08 |
合計ジャッジ時間 | 2,568 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 39 |
ソースコード
#include <bits/stdc++.h>//typedef//-------------------------#include <bits/stdc++.h>const double pi = 3.141592653589793238462643383279;using namespace std;//conversion//------------------------------------------inline int toInt(string s){int v;istringstream sin(s);sin >> v;return v;}template <class T>inline string toString(T x){ostringstream sout;sout << x;return sout.str();}inline int readInt(){int x;scanf("%d", &x);return x;}//typedef//------------------------------------------typedef vector<int> VI;typedef vector<VI> VVI;typedef vector<string> VS;typedef pair<int, int> PII;typedef pair<long long, long long> PLL;typedef pair<int, PII> TIII;typedef long long LL;typedef unsigned long long ULL;typedef vector<LL> VLL;typedef vector<VLL> VVLL;//container util//------------------------------------------#define ALL(a) (a).begin(), (a).end()#define RALL(a) (a).rbegin(), (a).rend()#define PB push_back#define MP make_pair#define SZ(a) int((a).size())#define SQ(a) ((a) * (a))#define EACH(i, c) for (typeof((c).begin()) i = (c).begin(); i != (c).end(); ++i)#define EXIST(s, e) ((s).find(e) != (s).end())#define SORT(c) sort((c).begin(), (c).end())//repetition//------------------------------------------#define FOR(i, s, n) for (int i = s; i < (int)n; ++i)#define REP(i, n) FOR(i, 0, n)#define MOD 1000000007#define rep(i, a, b) for (int i = a; i < (b); ++i)#define trav(a, x) for (auto &a : x)#define all(x) x.begin(), x.end()#define sz(x) (int)(x).size()typedef long long ll;typedef pair<int, int> pii;typedef vector<int> vi;const double EPS = 1E-8;#define chmin(x, y) x = min(x, y)#define chmax(x, y) x = max(x, y)class UnionFind{public:vector<int> par;vector<int> siz;UnionFind(int sz_) : par(sz_), siz(sz_, 1){for (ll i = 0; i < sz_; ++i)par[i] = i;}void init(int sz_){par.resize(sz_);siz.assign(sz_, 1LL);for (ll i = 0; i < sz_; ++i)par[i] = i;}int root(int x){while (par[x] != x){x = par[x] = par[par[x]];}return x;}bool merge(int x, int y){x = root(x);y = root(y);if (x == y)return false;if (siz[x] < siz[y])swap(x, y);siz[x] += siz[y];par[y] = x;return true;}bool issame(int x, int y){return root(x) == root(y);}int size(int x){return siz[root(x)];}};ll modPow(ll x, ll n, ll mod = MOD){ll res = 1;while (n){if (n & 1)res = (res * x) % mod;res %= mod;x = x * x % mod;n >>= 1;}return res;}#define SIEVE_SIZE 5000000 + 10bool sieve[SIEVE_SIZE];void makeSieve(){for (int i = 0; i < SIEVE_SIZE; ++i)sieve[i] = true;sieve[0] = sieve[1] = false;for (int i = 2; i * i < SIEVE_SIZE; ++i)if (sieve[i])for (int j = 2; i * j < SIEVE_SIZE; ++j)sieve[i * j] = false;}bool isprime(ll n){if (n == 0 || n == 1)return false;for (ll i = 2; i * i <= n; ++i)if (n % i == 0)return false;return true;}const int MAX = 2000010;long long fac[MAX], finv[MAX], inv[MAX];// テーブルを作る前処理void COMinit(){fac[0] = fac[1] = 1;finv[0] = finv[1] = 1;inv[1] = 1;for (int i = 2; i < MAX; i++){fac[i] = fac[i - 1] * i % MOD;inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;finv[i] = finv[i - 1] * inv[i] % MOD;}}// 二項係数計算long long COM(int n, int k){if (n < k)return 0;if (n < 0 || k < 0)return 0;return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;}long long extGCD(long long a, long long b, long long &x, long long &y){if (b == 0){x = 1;y = 0;return a;}long long d = extGCD(b, a % b, y, x);y -= a / b * x;return d;}// 負の数にも対応した mod (a = -11 とかでも OK)inline long long mod(long long a, long long m){return (a % m + m) % m;}// 逆元計算 (ここでは a と m が互いに素であることが必要)long long modinv(long long a, long long m){long long x, y;extGCD(a, m, x, y);return mod(x, m); // 気持ち的には x % m だが、x が負かもしれないので}ll GCD(ll a, ll b){if (b == 0)return a;return GCD(b, a % b);}typedef vector<ll> vec;typedef vector<vec> mat;mat mul(mat &A, mat &B){mat C(A.size(), vec((int)B[0].size()));for (int i = 0; i < A.size(); ++i){for (int k = 0; k < B.size(); ++k){for (int j = 0; j < B[0].size(); ++j){C[i][j] = (C[i][j] + A[i][k] * B[k][j] % MOD) % MOD;}}}return C;}mat matPow(mat A, ll n){mat B(A.size(), vec((int)A.size()));for (int i = 0; i < A.size(); ++i){B[i][i] = 1;}while (n > 0){if (n & 1)B = mul(B, A);A = mul(A, A);n >>= 1;}return B;}int main(){cin.tie(0);ios::sync_with_stdio(false);cout << fixed << setprecision(20);ll a, b, n;cin >> a >> b >> n;mat v(2, vec(2));v[0][0] = a, v[0][1] = b;v[1][0] = 1, v[1][1] = 0;auto ans = matPow(v, n);cout << ans[1][0] << endl;return 0;}