Educational Codeforces Round 47 E. Intercity Travelling
解法
部分集合全通りを求めるのは無理なので見方を変えてi-1km~ikmの区間を難易度a[j]で通過する確率を求めるとしてみる。n=4で試してみると以下の図のようになる。
図を辺ごとに縦に区切るのではなく、難易度a[i]ごとに横で区切ってみる。すると答えはsum(a[i]*(1/2^(i-1))+(n-1)/2^i)*2^(n-1))=sum(a[i]*(n-i+2)/2^i*2^(n-1))=sum(a[i]*(n-i+2)*2^(n-i-1))となる。この式はO(N)で求めることができるので答えを求めることができた。
#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() #define endl '\n' 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<typename T> istream& operator >> (istream& is, vector<T>& vec){ for(T& x: vec) {is >> x;} return is; } template<class T> ostream &operator <<(ostream& out,const vector<T>& a){ out<<'['; for(T 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<<40; const ll MOD = 998244353; struct mint { ll x; mint(): x(0) { } mint(ll y) : x(y>=0 ? y%MOD : y%MOD+MOD) {} ll get() const { return x; } // e乗 mint 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 mint(a); } // Comparators bool operator <(mint b) { return x < b.x; } bool operator >(mint b) { return x > b.x; } bool operator<=(mint b) { return x <= b.x; } bool operator>=(mint b) { return x >= b.x; } bool operator!=(mint b) { return x != b.x; } bool operator==(mint b) { return x == b.x; } // increment, decrement mint operator++() { x++; return *this; } mint operator++(signed) { mint t = *this; x++; return t; } mint operator--() { x--; return *this; } mint operator--(signed) { mint t = *this; x--; return t; } // Basic Operations mint &operator+=(mint that) { x += that.x; if(x >= MOD) x -= MOD; return *this; } mint &operator-=(mint that) { x -= that.x; if(x < 0) x += MOD; return *this; } mint &operator*=(mint that) { x = (ll)x * that.x % MOD; return *this; } mint &operator/=(mint that) { x = (ll)x * that.pow(MOD-2).x % MOD; return *this; } mint &operator%=(mint that) { x = (ll)x % that.x; return *this; } mint operator+(mint that) const { return mint(*this) += that; } mint operator-(mint that) const { return mint(*this) -= that; } mint operator*(mint that) const { return mint(*this) *= that; } mint operator/(mint that) const { return mint(*this) /= that; } mint operator%(mint that) const { return mint(*this) %= that; } }; // Input/Output ostream &operator<<(ostream& os, mint a) { return os << a.x; } istream &operator>>(istream& is, mint &a) { return is >> a.x; } signed main(void) { cin.tie(0); ios::sync_with_stdio(false); ll n; cin >> n; vector<mint> a(n); REP(i, n) cin >> a[i]; mint ans = 0; FOR(i, 1, n+1) { if(i == n) ans += a[i-1] * (n-i+2) * mint(499122177); else ans += a[i-1] * (n-i+2) * mint(2).pow(n-i-1); } cout << ans << endl; return 0; }