結果

問題 No.344 ある無理数の累乗
ユーザー anta
提出日時 2016-02-12 23:06:42
言語 C++11
(gcc 13.3.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 3,318 bytes
コンパイル時間 752 ms
コンパイル使用メモリ 86,472 KB
実行使用メモリ 6,948 KB
最終ジャッジ日時 2024-09-22 04:51:34
合計ジャッジ時間 1,713 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
other AC * 30
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ソースコード

diff #
プレゼンテーションモードにする

#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <set>
#include <map>
#include <queue>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <cctype>
#include <cassert>
#include <limits>
#include <functional>
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
#if defined(_MSC_VER) || __cplusplus > 199711L
#define aut(r,v) auto r = (v)
#else
#define aut(r,v) __typeof(v) r = (v)
#endif
#define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)
#define all(o) (o).begin(), (o).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset(m,v,sizeof(m))
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3fLL
using namespace std;
typedef vector<int> vi; typedef pair<int, int> pii; typedef vector<pair<int, int> > vpii; typedef long long ll;
template<typename T, typename U> inline void amin(T &x, U y) { if(y < x) x = y; }
template<typename T, typename U> inline void amax(T &x, U y) { if(x < y) x = y; }
template<int MOD>
struct ModInt {
    static const int Mod = MOD;
    unsigned x;
    ModInt() : x(0) {}
    ModInt(signed sig) { int sigt = sig % MOD; if(sigt < 0) sigt += MOD; x = sigt; }
    ModInt(signed long long sig) { int sigt = sig % MOD; if(sigt < 0) sigt += MOD; x = sigt; }
    int get() const { return (int)x; }
    ModInt &operator+=(ModInt that) { if((x += that.x) >= MOD) x -= MOD; return *this; }
    ModInt &operator-=(ModInt that) { if((x += MOD - that.x) >= MOD) x -= MOD; return *this; }
    ModInt &operator*=(ModInt that) { x = (unsigned long long)x * that.x % MOD; return *this; }
    ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
    ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
    ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
};
typedef ModInt<1000> mint;
mint linearlyRecurrentSequenceValue(long long K, const vector<mint> &initValues, const vector<mint> &annPoly) {
    assert(K >= 0);
    if(K < (int)initValues.size())
        return initValues[(int)K];
    int d = (int)annPoly.size() - 1;
    assert(d >= 0);
    assert(annPoly[d].get() == 1);
    assert(d <= (int)initValues.size());
    if(d == 0)
        return mint();
    vector<mint> coeffs(d), square;
    coeffs[0] = 1;
    int l = 0;
    while((K >> l) > 1) ++ l;
    for(; l >= 0; -- l) {
        square.assign(d * 2 - 1, mint());
        for(int i = 0; i < d; ++ i)
            for(int j = 0; j < d; ++ j)
                square[i + j] += coeffs[i] * coeffs[j];
        for(int i = d * 2 - 2; i >= d; -- i) {
            mint c = square[i];
            if(c.x == 0) continue;
            for(int j = 0; j < d; ++ j)
                square[i - d + j] -= c * annPoly[j];
        }
        for(int i = 0; i < d; ++ i)
            coeffs[i] = square[i];
        if(K >> l & 1) {
            mint lc = coeffs[d - 1];
            for(int i = d - 1; i >= 1; -- i)
                coeffs[i] = coeffs[i - 1] - lc * annPoly[i];
            coeffs[0] = mint() - lc * annPoly[0];
        }
    }
    mint res;
    for(int i = 0; i < d; ++ i)
        res += coeffs[i] * initValues[i];
    return res;
}
int main() {
    vector<mint> init = { 1, 2, 7, 20, 55 };
    vector<mint> phi = { 2, 2, -3, -2, 1 };
    int n;
    while(~scanf("%d", &n)) {
        mint ans = linearlyRecurrentSequenceValue(n, init, phi);
        printf("%d\n", ans.get());
    }
    return 0;
}
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