第6回 ドワンゴからの挑戦状 予選 B - Fusing Slimes

問題ページ
答え = (N-1)! * 期待値
期待値 スライム が を通る確率

スライム j が [i,i+1] を通る確率
→ j+1~i が j より前に選ばれる確率で 1/(i-j)
→ jだけの式にしたほうが楽なので変形すると 1/j

～ の後に が来る確率が なのとか立式するパート実質 AGC028 B - Removing Blocks - ferinの競プロ帳 なのに解けないのマジで…

(N-1)! * 期待値 を総和に言い換えたら立式複雑になって間違えた
実験してエスパーしようとしたときに差分見たけど明らかに積で見るべきだろ
こういうの大体立式ガチャになるんだけどどうすればまともな式が高速につくれるの…

#include <bits/stdc++.h>  
using namespace std;  
using ll = long long;  
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> void chmin(T &a, const T &b) { a = min(a, b); }  
template<typename T> void chmax(T &a, const T &b) { a = max(a, b); }  
struct FastIO {FastIO() { cin.tie(0); ios::sync_with_stdio(0); }}fastiofastio;  
#ifdef DEBUG_   
#include "../program_contest_library/memo/dump.hpp"  
#else  
#define dump(...)  
#endif  
const ll INF = 1LL<<60;  
  
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; }  
    // 平方剰余のうち一つを返す なければ-1  
    friend modint sqrt(modint a) {  
        if(a == 0) return 0;  
        ll q = MOD-1, s = 0;  
        while((q&1)==0) q>>=1, s++;  
        modint z=2;  
        while(1) {  
            if(z.pow((MOD-1)/2) == MOD-1) break;  
            z++;  
        }  
        modint c = z.pow(q), r = a.pow((q+1)/2), t = a.pow(q);  
        ll m = s;  
        while(t.x>1) {  
            modint tp=t;  
            ll k=-1;  
            FOR(i, 1, m) {  
                tp *= tp;  
                if(tp == 1) { k=i; break; }  
            }  
            if(k==-1) return -1;  
            modint cp=c;  
            REP(i, m-k-1) cp *= cp;  
            c = cp*cp, t = c*t, r = cp*r, m = k;  
        }  
        return r.x;  
    }  
  
    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) {   
        is >> a.x;  
        a.x = ((a.x%MOD)+MOD)%MOD;  
        return is;  
    }  
    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 fact[N]*factr[N-K];  
}  
  
int main(void) {  
    ll n;  
    cin >> n;  
    vector<ll> x(n);  
    REP(i, n) cin >> x[i];  
  
    vector<mint> inv(n+1);  
    FOR(i, 1, n+1) inv[i] = mint(i).inv();  
    FOR(i, 1, n+1) inv[i] += inv[i-1];  
  
    mint ret = 0;  
    REP(i, n-1) ret += inv[i+1] * (x[i+1]-x[i]);  
    REP(i, n-1) ret *= i+1;  
    cout << ret << endl;  
  
    return 0;  
}