#1181486 (C++23) No.3662 yuu Hates Sigma Problem

• 問題: No.3662 yuu Hates Sigma Problem
• ユーザー: AD010
• 提出日時: 2026-08-30 16:07:20
• 言語: C++23 (gcc 15.3.0 + boost 1.92.0)
• 結果: AC
• 実行時間: 235 ms / 2,000 ms
• コード長: 6,915 bytes
• コンパイル時間: 2,294 ms
• コンパイル使用メモリ: 338,440 KB
• 実行使用メモリ: 223,744 KB
• 最終ジャッジ日時: 2026-08-30 16:07:35
• 合計ジャッジ時間: 8,225 ms
• ジャッジサーバーID （参考情報）: judge1_0 / judge2_0

サブタスク	配点	結果
subtask1.	20 %	AC * 19
subtask2.	30 %	AC * 13
subtask3.	50 %	AC * 49
合計	2.5 * 100% = 250 点

ソースコード

#include<bits/stdc++.h>
using namespace std;

#define ll long long
#define ull unsigned long long
#define ld  long double

using LL = long long; using ULL = unsigned long long;
using VI = vector<int>; using VVI = vector<VI>; using VVVI = vector<VVI>;
using VL = vector<LL>; using VVL = vector<VL>; using VVVL = vector<VVL>;
using VB = vector<bool>; using VVB = vector<VB>; using VVVB = vector<VVB>;
using VD = vector<double>; using VVD = vector<VD>; using VVVD = vector<VVD>;
using VC = vector<char>; using VS = vector<string>; using VVC = vector<VC>;
using PII = pair<int,int>; using PLL = pair<LL,LL>; using PDD = pair<double,double>; using PIL = pair<int,LL>;
using MII = map<int,int>; using MLL = map<LL,LL>;
using SI = set<int>; using SL = set<LL>;
using MSI = multiset<int>; using MSL = multiset<LL>;
template<class T> using MAXPQ = priority_queue<T>;
template<class T> using MINPQ = priority_queue< T, vector<T>, greater<T> >;

const ll MOD = 1000000007;
const ll MOD2 = 998244353;
const ll INF = 1LL << 60;
#define PI  3.14159265358979323846

#define FOR(i, a, b) for(int i = (a); i < (b); ++i)
#define REP(i, n) FOR(i, 0, n)
#define EACH(e, v) for(auto &e : v)
#define RITR(it, v) for(auto it = (v).rbegin(); it != (v).rend(); ++it)
#define ALL(v) v.begin(),v.end()

vector<ll> x8={1,1,1,0,0,-1,-1,-1},y8={1,0,-1,1,-1,1,0,-1};
int dx4[4]={1,-1,0,0}, dy4[4]={0,0,1,-1};

/*
memo
-uf,RMQ(segtree),BIT,BIT2,SegTree,SegTreeLazy
-isprime,Eratosthenes,gcdlcm,factorize,divisors,modpow,moddiv
nCr(+modnCr,inverse,extend_euclid.powmod),tobaseB,tobase10
-dijkstra,Floyd,bellmanford,sccd,topological,treediamiter
-compress1,compress2,rotate90

-co,ci,fo1,fo2,fo3,fo4
-bitsearch,binaryserach
-bfs
-SegTreedec,SegTreeLazydec
*/

template <typename X>
struct SegTree{
    using FX = function<X(X,X)>; //Xを2つ受け取りXを返す関数の型
    long long n,N;
    FX fx; 
    X ex;
    vector<X> dat;

    //要素数N,二項演算fx,単位元ex;
    SegTree(long long _n, FX _fx, X _ex){
        init(_n, _fx, _ex);
    }

    void init(long long _n, FX _fx, X _ex){
        N = _n, fx = _fx; ex = _ex;
        long long x = 1;
        while(_n > x) x *= 2;
        n = x;
        dat.assign(n*2,_ex);
    }

    //i番目の要素にアクセス(0-indexed),O(1)
    X operator[](long long i){return dat[n+i];}
    X get(long long i){return dat[n+i];}

    //i番目の要素をxにアップデート(0-indexed),O(logN)
    void set(long long i, X x){
        i += n;
        dat[i] = x;
        while(i >>= 1){
            dat[i] = fx(dat[2*i],dat[2*i+1]);
        }
    }

    //[l,r)で二項演算を作用した結果(0-indexed),O(logN)
    X prod(long long l, long long r){
        X vleft = ex, vright = ex;
        for(long long left = l+n, right = r+n; left < right; left >>= 1, right >>= 1){
            if(left & 1) vleft = fx(vleft,dat[left++]);
            if(right & 1) vright = fx(dat[--right],vright);
        }
        return fx(vleft,vright);
    }
    //[0,N)まで二項演算を作用,O(1)
    X all_prod() {return dat[1];}

    //x=prod(l,r),f(x)=trueとなる最大のrを求める(f(ex)=true, 0-indexed),O(logN)
    long long max_right(const function<bool(X)> f, long long l = 0){
        if(l == N) return N;
        l += n;
        X sum = ex;
        do{
            while(l % 2 == 0) l >>= 1;
            if(!f(fx(sum,dat[l]))){
                while(l < n){
                    l = l * 2;
                    if(f(fx(sum,dat[l]))){
                        sum = fx(sum,dat[l]);
                        l++;
                    }
                }
                return l - n;
            }
            sum = fx(sum,dat[l]);
            l++;
        }while((l & -l) != l);
        return N;
    }
  
    //x=prod(l,r),f(x)=trueとなる最小のlを求める(f(ex)=true, 0-indexed),O(logN)
    long long min_left(const function<bool(X)> f, long long r = -1){
        if(r == 0) return 0;
        if(r == -1) return N;
        r += n;
        X sum = ex;
        do{
            r--;
            while(r > 1 && (r % 2)) r >>= 1;
            if(!f(fx(dat[r],sum))){
                while(r < n){
                    r = r * 2 + 1;
                    if(f(fx(dat[r],sum))){
                        sum = fx(dat[r],sum);
                        r--;
                    }
                }
                return r + 1 - n;
            }
            sum = fx(dat[r],sum);
        }while((r & -r) != r);
        return 0;
    } 
};


template <int mod>
class mint {
public:
    long long x;
    constexpr mint(long long x=0) : x((x%mod+mod)%mod) {}
    constexpr mint operator-() const { 
      return mint(-x);
    }
    constexpr mint& operator+=(const mint& a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    constexpr mint& operator-=(const mint& a) {
        if ((x += mod-a.x) >= mod) x -= mod;
        return *this;
    }
    constexpr mint& operator*=(const mint& a) {
        (x *= a.x) %= mod;
        return *this;
    }
    constexpr mint operator+(const mint& a) const {
        mint res(*this);
        return res+=a;
    }
    constexpr mint operator-(const mint& a) const {
        mint res(*this);
        return res-=a;
    }
    constexpr mint operator*(const mint& a) const {
        mint res(*this);
        return res*=a;
    }
    constexpr mint pow(long long t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }

    // for prime mod
    constexpr mint inv() const {
        return pow(mod-2);
    }
    constexpr mint& operator/=(const mint& a) {
        return (*this) *= a.inv();
    }
    constexpr mint operator/(const mint& a) const {
        mint res(*this);
        return res/=a;
    }
};
using mint1 = mint<1000000007>;
using mint2 = mint<998244353>;
using VM = vector<mint2>;


int main(){
    cin.tie(0);
    ios_base::sync_with_stdio(0);
    ll N; cin >> N;
    VL a(N); for(ll i = 0; i < N; i++) cin >> a[i]; 
    ll B = 30;
    VVL cl1(N+1,VL(B)),cr1(N+1,VL(B)),cl0(N+1,VL(B)),cr0(N+1,VL(B));
    for(ll i = 0; i < N; i++){
        for(ll j = 0; j < B; j++){
            cl1[i+1][j] += cl1[i][j];
            cl0[i+1][j] += cl0[i][j];
            if(i&(1<<j)) cl1[i+1][j]++;
            else cl0[i+1][j]++;
        }
    }

    for(ll i = N; i >= 1; i--){
        for(ll j = 0; j < B; j++){
            cr1[i-1][j] += cr1[i][j];
            cr0[i-1][j] += cr0[i][j];
            if((i-1)&(1<<j)) cr1[i-1][j]++;
            else cr0[i-1][j]++;
        }
    }

    mint2 ans = 0;
    for(ll i = 0; i < N; i++){
        mint2 c = 0;
        for(ll j = 0; j < B; j++){
            if(i&(1<<j)){
                c = cl0[i][j] + cr0[i+1][j];
            }
            else{
                c = cl1[i][j] + cr1[i+1][j];
            }
            mint2 p = 1<<j;
            ans += mint2(a[i])*c*p;
        }
    }
    
    cout << ans.x << '\n';
}