結果

問題 No.194 フィボナッチ数列の理解(1)
ユーザー snteasntea
提出日時 2016-10-10 12:00:16
言語 C++11
(gcc 13.3.0)
結果
AC  
実行時間 23 ms / 5,000 ms
コード長 4,284 bytes
コンパイル時間 2,606 ms
コンパイル使用メモリ 176,360 KB
実行使用メモリ 11,008 KB
最終ジャッジ日時 2024-11-22 00:49:46
合計ジャッジ時間 3,550 ms
ジャッジサーバーID
(参考情報)
judge2 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 37
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 #
プレゼンテーションモードにする

#ifdef LOCAL111
#else
#define NDEBUG
#endif
#include <bits/stdc++.h>
const int INF = 1e9;
using namespace std;
template<typename T, typename U> ostream& operator<< (ostream& os, const pair<T,U>& p) { cout << '(' << p.first << ' ' << p.second << ')'; return os;
    }
#define endl '\n'
#define ALL(a) (a).begin(),(a).end()
#define SZ(a) int((a).size())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) FOR(i,0,n)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)
#define RBP(i,a) for(auto& i : a)
#ifdef LOCAL111
#define DEBUG(x) cout<<#x<<": "<<(x)<<endl
template<typename T> void dpite(T a, T b){ for(T ite = a; ite != b; ite++) cout << (ite == a ? "" : " ") << *ite; cout << endl;}
#else
#define DEBUG(x) true
template<typename T> void dpite(T a, T b){ return; }
#endif
#define F first
#define S second
#define SNP string::npos
#define WRC(hoge) cout << "Case #" << (hoge)+1 << ": "
#define rangej(a,b,c) ((a) <= (c) and (c) < (b))
#define rrangej(b,c) rangej(0,b,c)
template<typename T> void pite(T a, T b){ for(T ite = a; ite != b; ite++) cout << (ite == a ? "" : " ") << *ite; cout << endl;}
template<typename T> bool chmax(T& a, T b){if(a < b){a = b; return true;} return false;}
template<typename T> bool chmin(T& a, T b){if(a > b){a = b; return true;} return false;}
typedef long long int LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
typedef pair<LL,LL> LP;
void ios_init(){
//cout.setf(ios::fixed);
//cout.precision(12);
#ifdef LOCAL111
return;
#endif
ios::sync_with_stdio(false); cin.tie(0);
}
const LL mod = 1e9+7;
//library
template<typename T>
class Matrix {
public:
vector<vector<T> > v;
Matrix(int n){
v.resize(n,vector<T>(n,0));
}
Matrix operator+ (Matrix x){
Matrix res(v.size());
for(int i = 0; i < (int)v.size(); i++){
for(int j = 0; j < (int)v.size(); j++){
res.v[i][j] = (x.v[i][j]+v[i][j])%mod;
}
}
return res;
}
Matrix operator-(Matrix x){
Matrix res(v.size());
for(int i = 0; i < (int)v.size(); i++){
for(int j = 0; j < (int)v.size(); j++){
res.v[i][j] = (v[i][j]-x.v[i][j])%mod;
}
}
}
Matrix operator*(Matrix x){
Matrix res(v.size());
for(int i = 0;i < (int)v.size(); i++){
for(int j = 0; j < (int)v.size(); j++){
for(int k = 0; k < (int)v.size(); k++){
res.v[i][j] += v[i][k]*x.v[k][j]%mod;
res.v[i][j] %= mod;
}
}
}
return res;
}
vector<T> operator*(vector<T> x){
assert(x.size() == v.size());
vector<T> res(v.size());
for(int i = 0; i < (int)v.size(); i++){
T tmp = 0;
for(int j = 0; j < (int)v.size(); j++){
tmp += v[i][j]*x[j]%mod;
tmp %= mod;
}
res[i] = tmp;
}
return res;
}
vector<T>& operator[](int x){
return v[x];
}
};
template<typename T>
Matrix<T> pow(Matrix<T> x,long long n){
Matrix<T> tmp = x, res(SZ(x.v));
for(int i = 0; i < SZ(x.v); i++) res.v[i][i] = 1;
while(n != 0){
if(n&1){
res = res*tmp;
}
tmp = tmp*tmp;
n >>= 1;
}
return res;
}
//libary
int main()
{
ios_init();
int n;
LL k;
while(cin >> n >> k) {
DEBUG(n);
vector<LL> a(n);
REP(i,n) cin >> a[i];
function<void()> solvemat = [&](){
Matrix<LL> A(n);
REP(i,n) A[0][i] = 1;
REP(i,n-1) A[i+1][i] = 1;
Matrix<LL> B(n+1);
B[0][0] = B[0][1] = 1;
REP(i,n) B[1][i+1] = 1;
REP(i,n-1) B[i+2][i+1] = 1;
REP(i,n+1) dpite(ALL(B[i]));
if(k <= n){
k--;
cout << a[k] << ' ' << accumulate(a.begin(),a.begin()+k,0LL) << endl;
}else{
A = pow(A,k-n);
vector<LL> b = a;
b.push_back(accumulate(a.begin(),--a.end(),0LL));
reverse(ALL(a));
auto ans = A*a;
// reverse(ALL(a));
reverse(ALL(b));
dpite(ALL(b));
B = pow(B,k-n+1);
auto ans2 = B*b;
dpite(ALL(ans2));
cout << ans[0] << ' ' << ans2[0] << endl;
}
};
function<void()> solvedp = [&](){
if(k <= n){
cout << a[k-1] << endl;
}else{
vector<LL> dp(k,0);
LL sum = 0;
REP(i,SZ(a)){
sum = (sum+a[i])%mod;
dp[i]= a[i];
}
FOR(i,n,k){
dp[i] = sum;
sum = ((sum-dp[i-n]+mod)%mod+dp[i])%mod;
}
LL ans = 0;
REP(i,k){
ans = (ans+dp[i])%mod;
}
cout << dp[k-1] << ' ' << ans << endl;
}
};
if(n <= 31){
solvemat();
}else{
solvedp();
}
}
return 0;
}
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