結果

問題 No.336 門松列列
ユーザー anta
提出日時 2016-01-15 22:51:42
言語 C++11（廃止可能性あり）
(gcc 13.3.0)
結果
AC  
実行時間 56 ms / 2,000 ms
コード長 3,523 bytes
コンパイル時間 943 ms
コンパイル使用メモリ 84,876 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-06-12 01:09:36
合計ジャッジ時間 1,806 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 * 4
other AC * 8
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

#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) { return *this *= that.inverse(); }
    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; }
    ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }
    ModInt inverse() const {
        signed a = x, b = MOD, u = 1, v = 0;
        while(b) {
            signed t = a / b;
            a -= t * b; std::swap(a, b);
            u -= t * v; std::swap(u, v);
        }
        if(u < 0) u += Mod;
        ModInt res; res.x = (unsigned)u;
        return res;
    }
};
template<int MOD> ModInt<MOD> operator^(ModInt<MOD> a, long long k) {
    if(k < 0) return (a ^ (-k)).inverse();
    ModInt<MOD> r = 1;
    while(k) {
        if(k & 1) r *= a;
        a *= a;
        k >>= 1;
    }
    return r;
}
typedef ModInt<1000000007> mint;
vector<mint> fact, factinv;
void nCr_computeFactinv(int N) {
    N = min(N, mint::Mod - 1);
    fact.resize(N + 1); factinv.resize(N + 1);
    fact[0] = 1;
    rer(i, 1, N) fact[i] = fact[i - 1] * i;
    factinv[N] = fact[N].inverse();
    for(int i = N; i >= 1; i --) factinv[i - 1] = factinv[i] * i;
}
mint nCr(int n, int r) {
    if(n >= mint::Mod)
        return nCr(n % mint::Mod, r % mint::Mod) * nCr(n / mint::Mod, r / mint::Mod);
    return r > n ? 0 : fact[n] * factinv[n - r] * factinv[r];
}
mint solve(int n) {
    mint sum;
    rer(m, 0, n / 2) {
        int k = n + 1 - 2 * m;
        mint t;
        rer(j, 0, k)
            t += nCr(k, j) * (j % 2 == 0 ? 1 : -1) * (mint(k - 2 * j) ^ (n + 1));
        t /= k;
        t *= (m % 2 == 0 ? 1 : -1);
        t *= mint(2) ^ (1 - k);
        sum += t;
    }
    return sum;
}
int main() {
    nCr_computeFactinv(100000);
    int N;
    while(~scanf("%d", &N)) {
        mint ans;
        if(N <= 2) {
            ans = 0;
        } else {
            ans = solve(N);
        }
        printf("%d\n", ans.get());
    }
    return 0;
}
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