CODE FESTIVAL 2016 Grand Final J - 123 Pairs

問題ページ

解法

ペアとなっている2要素を区間として，区間が交差しない部分は独立に考えてよさそう．各区間でどのようなペアのとり方が存在するのか考えると以下のパターンしかない．

(1) 1
(2) 2 3 … 3 2
(3) 3 1
(4) 3 3 3

2は(2)以外で使用できないので(2)はB/2個となる．(1)をw個とすると(3)はA-w個となる．(4)をz個とすると，(2)の間に使用する3は合計でC-(A-w)-3z個となる．
まとめると(1)をw個，(2)をB/2個，(3)をA-w個，(4)をz個，(2)の間に使用する3がC-(A-w)-3z個となる．(2)の間に使用する3についてはC-(A-w)-3z個をB/2個に分割するので 通りとなる．(1)～(4)を配置する方法は (w+B/2+A-w+z)!/(w!(B/2)!(A-w)!z!) 通りとなる．あとはw,zをO(N^2)通り全探索すればよい．

#include <bits/stdc++.h>

using namespace std;
using ll = long long;
// #define int ll
using PII = pair<ll, ll>;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()

template<typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template<typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template<typename T> bool IN(T a, T b, T x) { return a<=x&&x<b; }
template<typename T> T ceil(T a, T b) { return a/b + !!(a%b); }

template<typename T> vector<T> make_v(size_t a) { return vector<T>(a); }
template<typename T,typename... Ts>
auto make_v(size_t a,Ts... ts) {
    return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));
}
template<typename T,typename V> typename enable_if<is_class<T>::value==0>::type
fill_v(T &t, const V &v) { t=v; }
template<typename T,typename V> typename enable_if<is_class<T>::value!=0>::type
fill_v(T &t, const V &v ) { for(auto &e:t) fill_v(e,v); }

template<class S,class T>
ostream &operator <<(ostream& out,const pair<S,T>& a) {
    out<<'('<<a.first<<','<<a.second<<')'; return out;
}
template<class T>
ostream &operator <<(ostream& out,const vector<T>& a){
    out<<'[';
    for(const T &i: a) out<<i<<',';
    out<<']';
    return out;
}
template<class T>
ostream &operator <<(ostream& out, const set<T>& a) {
    out<<'{';
    for(const T &i: a) out<<i<<',';
    out<<'}';
    return out;
}
template<class T, class S>
ostream &operator <<(ostream& out, const map<T,S>& a) {
    out<<'{';
    for(auto &i: a) out<<i<<',';
    out<<'}';
    return out;
}

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0}; // DRUL
const int INF = 1<<30;
const ll LLINF = 1LL<<60;
const ll MOD = 1000000007;

template<ll MOD>
struct modint {
    ll x;
    modint(): x(0) {}
    modint(ll y) : x(y>=0 ? y%MOD : y%MOD+MOD) {}
    static constexpr ll mod() { return MOD; }
    // e乗
    modint pow(ll e) {
        ll a = 1, p = x;
        while(e > 0) {
            if(e%2 == 0) {p = (p*p) % MOD; e /= 2;}
            else {a = (a*p) % MOD; e--;}
        }
        return modint(a);
    }
    modint inv() const {
        ll a=x, b=MOD, u=1, y=1, v=0, z=0;
        while(a) {
            ll q = b/a;
            swap(z -= q*u, u);
            swap(y -= q*v, v);
            swap(b -= q*a, a);
        }
        return z;
    }
    // Comparators
    bool operator <(modint b) { return x < b.x; }
    bool operator >(modint b) { return x > b.x; }
    bool operator<=(modint b) { return x <= b.x; }
    bool operator>=(modint b) { return x >= b.x; }
    bool operator!=(modint b) { return x != b.x; }
    bool operator==(modint b) { return x == b.x; }
    // Basic Operations
    modint operator+(modint r) const { return modint(*this) += r; }
    modint operator-(modint r) const { return modint(*this) -= r; }
    modint operator*(modint r) const { return modint(*this) *= r; }
    modint operator/(modint r) const { return modint(*this) /= r; }
    modint &operator+=(modint r) {
        if((x += r.x) >= MOD) x -= MOD;
        return *this;
    }
    modint &operator-=(modint r) {
        if((x -= r.x) < 0) x += MOD;
        return *this;
    }
    modint &operator*=(modint r) {
    #if !defined(_WIN32) || defined(_WIN64)
        x = x * r.x % MOD; return *this;
    #endif
        unsigned long long y = x * r.x;
        unsigned xh = (unsigned) (y >> 32), xl = (unsigned) y, d, m;
        asm(
            "divl %4; \n\t"
            : "=a" (d), "=d" (m)
            : "d" (xh), "a" (xl), "r" (MOD)
        );
        x = m;
        return *this;
    }
    modint &operator/=(modint r) { return *this *= r.inv(); }
    // increment, decrement
    modint operator++() { x++; return *this; }
    modint operator++(signed) { modint t = *this; x++; return t; }
    modint operator--() { x--; return *this; }
    modint operator--(signed) { modint t = *this; x--; return t; }

    template<class T>
    friend modint operator*(T l, modint r) { return modint(l) *= r; }
    template<class T>
    friend modint operator+(T l, modint r) { return modint(l) += r; }
    template<class T>
    friend modint operator-(T l, modint r) { return modint(l) -= r; }
    template<class T>
    friend modint operator/(T l, modint r) { return modint(l) /= r; }
    template<class T>
    friend bool operator==(T l, modint r) { return modint(l) == r; }
    template<class T>
    friend bool operator!=(T l, modint r) { return modint(l) != r; }
    // Input/Output
    friend ostream &operator<<(ostream& os, modint a) { return os << a.x; }
    friend istream &operator>>(istream& is, modint &a) { return is >> a.x; }
    friend string to_frac(modint v) {
        static map<ll, PII> mp;
        if(mp.empty()) {
            mp[0] = mp[MOD] = {0, 1};
            FOR(i, 2, 1001) FOR(j, 1, i) if(__gcd(i, j) == 1) {
                mp[(modint(i) / j).x] = {i, j};
            }
        }
        auto itr = mp.lower_bound(v.x);
        if(itr != mp.begin() && v.x - prev(itr)->first < itr->first - v.x) --itr;
        string ret = to_string(itr->second.first + itr->second.second * ((int)v.x - itr->first));
        if(itr->second.second > 1) {
            ret += '/';
            ret += to_string(itr->second.second);
        }
        return ret;
    }
};
using mint = modint<1000000007>;

// 前計算O(N) クエリO(1)
mint combi(ll N, ll K) {
    const int maxN=5e5; // !!!
    static mint fact[maxN+1]={},factr[maxN+1]={};
    if (fact[0]==0) {
        fact[0] = factr[0] = 1;
        FOR(i, 1, maxN+1) fact[i] = fact[i-1] * i;
        factr[maxN] = fact[maxN].inv();
        for(ll i=maxN-1; i>=0; --i) factr[i] = factr[i+1] * (i+1);
    }
    if(K<0 || K>N) return 0; // !!!
    return factr[K]*fact[N]*factr[N-K];
}

mint frac[20001], ifrac[20001];
signed main(void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    ll n, a, b, c;
    cin >> n >> a >> b >> c;

    if(b%2) {
        cout << 0 << endl;
        return 0;
    }

    frac[0] = 1;
    FOR(i, 1, 20001) frac[i] = frac[i-1] * i;
    ifrac[20000] = frac[20000].inv();
    for(ll i=19999; i>=0; --i) ifrac[i] = ifrac[i+1] * (i+1);

    if(b == 0) {
        mint ret = 0;
        REP(w, a+1) {
            ll x = a-w, z = (c-x)/3;
            if(c-x<0 || (c-x)%3 != 0) continue;
            mint add = frac[w+x+z] * ifrac[w] * ifrac[x] * ifrac[z];
            ret += add;
        }
        cout << ret << endl;
        return 0;
    }

    mint ret = 0;
    REP(w, a+1) REP(z, c/3+1) {
        ll x = a-w, y = b/2, rest = c-x-3*z;
        if(x+3*z > c) continue;
        mint add = frac[w+x+y+z] * ifrac[w] * ifrac[x] * ifrac[y] * ifrac[z] * combi(rest+y-1, y-1);
        ret += add;
    }
    cout << ret << endl;

    return 0;
}