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 | 2020-09-05 23:58:45 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 22 ms / 2,000 ms |
コード長 | 5,631 bytes |
コンパイル時間 | 2,506 ms |
コンパイル使用メモリ | 220,172 KB |
最終ジャッジ日時 | 2025-01-14 07:19:50 |
ジャッジサーバーID (参考情報) |
judge5 / judge3 |
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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
#pragma GCC optimize("O3")#include<bits/stdc++.h>using namespace std;using ll=long long;//typedef pair<ll,ll> P;template<class T> using V=vector<T>;#define fi first#define se second#define all(v) (v).begin(),(v).end()const ll inf=(1e18);//const ll mod=998244353;const ll mod=1000000007;ll gcd(ll a,ll b) {return b ? gcd(b,a%b):a;}ll lcm(ll c,ll d){return c/gcd(c,d)*d;}struct __INIT{__INIT(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(15);}} __init;template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }struct mint{using ull=unsigned long long int;ull v;mint(ll vv=0){s(vv%mod+mod);}mint& s(ull vv){v=vv<mod?vv:vv-mod;return *this;}//オーバーロードmint operator-()const{return mint()-*this;}//mint型にキャストmint&operator+=(const mint&val){return s(v+val.v);}mint&operator-=(const mint&val){return s(v+mod-val.v);}mint&operator*=(const mint&val){v=ull(v)*val.v%mod;return *this;}mint&operator/=(const mint&val){return *this*=val.inv();}mint operator+(const mint&val){return mint(*this)+=val;}mint operator-(const mint&val){return mint(*this)-=val;}mint operator*(const mint&val){return mint(*this)*=val;}mint operator/(const mint&val){return mint(*this)/=val;}mint pow(ll n)const{mint res(1),x(*this);while(n){if(n&1)res*=x;x*=x;n>>=1ll;}return res;}mint inv()const{return pow(mod-2);}//拡張ユークリッドの互除法/* mint inv()const{int x,y;int g=extgcd(v,mod,x,y);assert(g==1);if(x<0)x+=mod;return mint(x);}*/friend ostream& operator<<(ostream&os,const mint&val){return os<<val.v;}//出力bool operator<(const mint&val)const{return v<val.v;}bool operator==(const mint&val)const{return v==val.v;}bool operator>(const mint&val)const{return v>val.v;}};template<class T> struct Matrix{vector<vector<T>> A;Matrix() {}Matrix(size_t n,size_t m):A(n,vector<T>(m,0)) {}Matrix(size_t n):A(n,vector<T>(n,0)) {}size_t height() const{return (A.size());}size_t width() const{return (A[0].size());}inline const vector<T> &operator[](int k)const{return (A.at(k));}inline vector<T> &operator[](int k){return (A.at(k));}//単位行列static Matrix I(size_t n){Matrix mat(n);for(int i=0;i<n;i++)mat[i][i]=(T)(1);return mat;}Matrix &operator+=(const Matrix &B){size_t n=height(),m=width();assert(n==B.height()&&m==B.width());for(int i=0;i<n;i++){for(int j=0;j<m;j++){(*this)[i][j]+=B[i][j];}}return (*this);}Matrix &operator-=(const Matrix &B){size_t n=height(),m=width();assert(n==B.height()&&m==B.width());for(int i=0;i<n;i++){for(int j=0;j<m;j++){(*this)[i][j]-=B[i][j];}}return (*this);}Matrix &operator*=(const Matrix &B){size_t n=height(),m=B.width(),w=height();assert(w==B.height());vector<vector<T>> C(n,vector<T>(m,0));for(int i=0;i<n;i++){for(int j=0;j<m;j++){for(int k=0;k<w;k++){C[i][j]+=(*this)[i][k]*B[k][j];}}}A.swap(C);return (*this);}//累乗計算Matrix &operator^=(long long k){Matrix B=Matrix::I(height());while(k>0){if(k&1){B*=(*this);}(*this)*=(*this);k>>=1ll;}A.swap(B.A);return (*this);}Matrix operator+(const Matrix &B)const{return (Matrix(*this)+=B);}Matrix operator-(const Matrix &B)const{return (Matrix(*this)-=B);}Matrix operator*(const Matrix &B)const{return (Matrix(*this)*=B);}Matrix operator^(const long long k)const{return (Matrix(*this)^=k);}friend ostream &operator<<(ostream &os,Matrix &mat){size_t n=mat.height(),m=mat.width();for(int i=0;i<n;i++){os<<"[";for(int j=0;j<m;j++){os<<mat[i][j]<<(j+1==m?"]\n":",");}}return (os);}//行列式T det(){Matrix B(*this);assert(height()==width());T ret=(T)(1);for(int i=0;i<width();i++){int idx=-1;for(int j=i;j<width();j++){if(B[j][i]!=0)idx=j;}if(idx==-1)return (T)(0);if(i!=idx){ret*=(T)(-1);swap(B[i],B[idx]);}ret*=B[i][i];T v=B[i][i];for(int j=0;j<width();j++){B[i][j]/=v;}for(int j=i+1;j<width();j++){T a=B[j][i];for(int k=0;k<width();k++){B[j][k]-=B[i][k]*a;}}}return (ret);}};int main(){Matrix<mint> A(2),x(2,1);ll a,b,n;cin>>a>>b>>n;A.A[0][0]=mint(a);A.A[0][1]=mint(b);A.A[1][0]=mint(1);x.A[0][0]=mint(1);if(n==0){cout<<0<<"\n";return 0;}if(n==1){cout<<1<<"\n";return 0;}A^=(n-1);A*=x;cout<<A.A[0][0]<<"\n";}