結果

問題 No.673 カブトムシ
ユーザー koprickykopricky
提出日時 2018-04-16 19:18:14
言語 C++11
(gcc 11.4.0)
結果
WA  
実行時間 -
コード長 6,053 bytes
コンパイル時間 1,502 ms
コンパイル使用メモリ 169,540 KB
実行使用メモリ 6,948 KB
最終ジャッジ日時 2024-06-27 04:02:52
合計ジャッジ時間 2,471 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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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 2 ms
5,376 KB
testcase_04 AC 2 ms
5,376 KB
testcase_05 WA -
testcase_06 AC 2 ms
5,376 KB
testcase_07 AC 1 ms
5,376 KB
testcase_08 AC 2 ms
5,376 KB
testcase_09 AC 1 ms
5,376 KB
testcase_10 AC 1 ms
5,376 KB
testcase_11 AC 2 ms
5,376 KB
testcase_12 WA -
testcase_13 AC 2 ms
5,376 KB
testcase_14 WA -
testcase_15 WA -
testcase_16 WA -
testcase_17 WA -
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ソースコード

diff #

#include <bits/stdc++.h>
#define ll long long
#define INF 1000000005
#define MOD 1000000007
#define EPS 1e-10
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;--i)
#define srep(i,s,t) for(int i=(int)(s);i<(int)(t);++i)
#define each(a,b) for(auto& (a): (b))
#define all(v) (v).begin(),(v).end()
#define len(v) (int)(v).size()
#define zip(v) sort(all(v)),v.erase(unique(all(v)),v.end())
#define cmx(x,y) x=max(x,y)
#define cmn(x,y) x=min(x,y)
#define fi first
#define se second
#define pb push_back
#define show(x) cout<<#x<<" = "<<(x)<<endl
#define spair(p) cout<<#p<<": "<<p.fi<<" "<<p.se<<endl
#define sar(a,n) cout<<#a<<":";rep(pachico,n)cout<<" "<<a[pachico];cout<<endl
#define svec(v) cout<<#v<<":";rep(pachico,v.size())cout<<" "<<v[pachico];cout<<endl
#define svecp(v) cout<<#v<<":";each(pachico,v)cout<<" {"<<pachico.first<<":"<<pachico.second<<"}";cout<<endl
#define sset(s) cout<<#s<<":";each(pachico,s)cout<<" "<<pachico;cout<<endl
#define smap(m) cout<<#m<<":";each(pachico,m)cout<<" {"<<pachico.first<<":"<<pachico.second<<"}";cout<<endl

using namespace std;

typedef pair<int,int> P;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<double> vd;
typedef vector<P> vp;
typedef vector<string> vs;

const int MAX_N = 100005;

// powかmod_powかはその都度変更する
// 係数の順番に注意する!!!
// A[n] = 3*A[n-1]+2*A[n-2] とかなら a は {2,3} となる
template<typename T> class mat : public vector<vector<T> > {
private:
    int r,c;    //行,列
public:
    int row() const {
        return r;
    }
    int column() const {
        return c;
    }
    mat(int n,int m,T val = 0){
        this->r = n,this->c = m;
        rep(i,n){
            this->push_back(vector<T>(m,val));
        }
    }
    mat operator+(const mat& another){
        if(this->r != another.r && this->c != another.c){
            cout << "足し算失敗(サイズ不一致)" << endl;
            exit(1);
        }
        mat<T> X(this->r,this->c);
        rep(i,this->r){
            rep(j,this->c){
                X[i][j] = (*this)[i][j] + another[i][j];
            }
        }
        return X;
    }
    mat operator+(const T val){
        mat<T> X(this->r,this->c);
        rep(i,this->r){
            rep(j,this->c){
                X[i][j] = (*this)[i][j] + val;
            }
        }
        return X;
    }
    mat operator-(const mat& another){
        if(this->r != another.r && this->c != another.c){
            cout << "引き算失敗(サイズ不一致)" << endl;
            exit(1);
        }
        mat<T> X(this->r,this->c);
        rep(i,this->r){
            rep(j,this->c){
                X[i][j] = (*this)[i][j] - another[i][j];
            }
        }
        return X;
    }
    mat operator-(const T val){
        mat<T> X(this->r,this->c);
        rep(i,this->r){
            rep(j,this->c){
                X[i][j] = (*this)[i][j] - val;
            }
        }
        return X;
    }
    vector<T> operator*(const vector<T>& another){
        if(this->c != (int)another.size()){
            cout << "掛け算失敗(サイズ不一致)" << endl;
            exit(1);
        }
        vector<T> vec(this->r,0);
        rep(i,this->r){
            rep(j,this->c){
                vec[i] += (*this)[i][j] * another[j];
            }
        }
        return vec;
    }
    mat operator*(const mat& another){
        if(this->c != another.r){
            cout << "掛け算失敗(サイズ不一致)" << endl;
            exit(1);
        }
        mat<T> X(this->r,another.c);
        rep(i,this->r){
            rep(k,this->c){
                rep(j,another.c){
                    X[i][j] += (*this)[i][k]*another[k][j];
                }
            }
        }
        return X;
    }
    mat operator-(){
        mat<T> X(this->r,this->c);
        rep(i,this->r){
            rep(j,this->c){
                X[i][j] = -(*this)[i][j];
            }
        }
        return X;
    }
    void print(){
        rep(i,this->r){
            rep(j,(this->c)-1){
                cout << (*this)[i][j] << ",";
            }
            cout << (*this)[i][(this->c)-1] << endl;
        }
    }
};

template<typename T> mat<T> pow(mat<T> A,ll cnt)
{
    if(A.row() != A.column()){
        cout << "累乗不可" << endl;
    }
    int n = A.row();
	mat<T> B(n,n);
	rep(i,n){
		B[i][i] = 1;
	}
	while(cnt>0){
		if(cnt & 1){
			B = B*A;
		}
		A = A*A;
		cnt >>= 1;
	}
	return B;
}

template<typename T> mat<T> mod_mul(mat<T>& A,mat<T>& B)
{
    if(A.column() != B.row()){
        cout << "掛け算失敗(サイズ不一致)" << endl;
        exit(1);
    }
    mat<T> X(A.row(),B.column());
    rep(i,A.row()){
        rep(k,A.column()){
            rep(j,B.column()){
                X[i][j] = (X[i][j] + A[i][k]*B[k][j]) % MOD;
            }
        }
    }
    return X;
}

template<typename T> mat<T> mod_pow(mat<T> A,ll cnt)
{
    if(A.row() != A.column()){
        cout << "累乗不可" << endl;
    }
    int n = A.row();
	mat<T> B(n,n);
	rep(i,n){
		B[i][i] = 1;
	}
	while(cnt>0){
		if(cnt & 1){
			B = mod_mul(B,A);
		}
		A = mod_mul(A,A);
		cnt >>= 1;
	}
	return B;
}

// 第1引数 : A[n] = a[0] + a[1] * A[n-k] + a[2] * A[n-k+1] + ... + a[k] * A[n-1]
// 第2引数 : A[0] = x[0], A[1] = x[1], ... , A[k-1] = x[k-1]
template<typename T>
T solve(vector<T>& a, vector<T>& x, ll n)
{
    int k = (int)a.size() - 1;
    if(n < k){
        return x[n];
    }
    mat<T> A(k+1,k+1);
    rep(i,k+1){
        A[0][i] = a[k-i];
    }
    rep(i,k-1){
        A[i+1][i] = 1;
    }
    A[k][k] = 1;
    mat<T> B = mod_pow(A,n-k+1);
    T res = 0;
    rep(i,k){
        res = (res + B[0][i]*x[k-1-i]%MOD) % MOD;
    }
    res = (res + B[0][k]) % MOD;
    return res;
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll b,c,d;
    cin >> b >> c >> d;
    b %= MOD, c %= MOD, d %= MOD;
    vl a = {b*c%MOD,c}, x = {0};
    cout << solve<ll>(a,x,d) << "\n";
    return 0;
}
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