結果

問題 No.142 単なる配列の操作に関する実装問題
ユーザー heno239
提出日時 2020-08-07 15:01:22
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,628 ms / 5,000 ms
コード長 4,392 bytes
コンパイル時間 1,047 ms
コンパイル使用メモリ 114,268 KB
実行使用メモリ 9,472 KB
最終ジャッジ日時 2024-09-23 06:54:19
合計ジャッジ時間 7,267 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge2
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ファイルパターン 結果
other AC * 5
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<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
#include<complex>
#include<numeric>
using namespace std;
//#define int long long
typedef long long ll;
typedef unsigned long long ul;
typedef unsigned int ui;
constexpr ll mod = 1000000007;
const ll INF = mod * mod;
typedef pair<int, int>P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-12;
const ld pi = acos(-1.0);
ll mod_pow(ll x, ll n, ll m) {
    ll res = 1;
    while (n) {
        if (n & 1)res = res * x % m;
        x = x * x % m; n >>= 1;
    }
    return res;
}
struct modint {
    ll n;
    modint() :n(0) { ; }
    modint(ll m) :n(m) {
        if (n >= mod)n %= mod;
        else if (n < 0)n = (n % mod + mod) % mod;
    }
    operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint& a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint& a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint& a, modint b) { a.n = ((ll)a.n * b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
    if (n == 0)return modint(1);
    modint res = (a * a) ^ (n / 2);
    if (n % 2)res = res * a;
    return res;
}
ll inv(ll a, ll p) {
    return (a == 1 ? 1 : (1 - p * inv(p % a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }
const int max_n = 1 << 17;
modint fact[max_n], factinv[max_n];
void init_f() {
    fact[0] = modint(1);
    for (int i = 0; i < max_n - 1; i++) {
        fact[i + 1] = fact[i] * modint(i + 1);
    }
    factinv[max_n - 1] = modint(1) / fact[max_n - 1];
    for (int i = max_n - 2; i >= 0; i--) {
        factinv[i] = factinv[i + 1] * modint(i + 1);
    }
}
modint comb(int a, int b) {
    if (a < 0 || b < 0 || a < b)return 0;
    return fact[a] * factinv[b] * factinv[a - b];
}
ul rec[64][64];
ul z;
const int m = 64;
bool b[2000000];
void solve() {
    rep(i, 64) {
        rec[i][i] = (ul)1 << i;
        for (int j = i + 1; j < 64; j++) {
            rec[i][j] = rec[i][j - 1] ^ (ul)1 << j;
        }
    }
    z = (ul)-1;
    int n; ll s, x, y, z;
    cin >> n >> s >> x >> y >> z;
    ll cur = s;
    rep(i, n) {
        b[i] = cur % 2;
        cur = (cur * x + y) % z;
    }
    
    ul a[2000000 / m] = {};
    ul memo[100000 / 64 + 5];
    rep(i, 2000000 / m) {
        rep(j, m) {
            if (b[i*64+j]){
                a[i] |= (ul)1 << j;
            }
        }
    }
    //rep(i, n)cout << b[i]; cout << "\n";
    //cout << a[0] << "\n";
    auto calc = [&](int le, int ri)->ul {
        int dl = le >>6,dr= ri>>6;
        int rl = le & 63, rr = ri & 63;
        if (dl == dr) {
            return ((rec[rl][rr] & a[dl]) >> rl);
        }
        else {
            return ((rec[rl][63] & a[dl]) >> rl) ^ ((rec[0][rr] & a[dr]) << (63 ^ rl)+1);
        }
    };
    int q; cin >> q;
    rep(i, q) {
        int l1, r1, l2, r2;
        cin >> l1 >> r1 >> l2 >> r2; l1--; r1--; l2--; r2--;
        int dl2 = l2>>6, dr2 = r2>>6;
        int rl2 = l2 & 63, rr2 = r2 & 63;
        if (dl2 == dr2) {
            a[dl2] ^= calc(l1, r1) << rl2;
        }
        else {
            int tmp = 0;
            memo[0] = calc(l1, l1+(64-rl2)-1) << rl2;
            tmp++;
            int tle = l1 + (64 - rl2);
            int tri =tle + 63;
            for (int i = dl2 + 1; i < dr2; i++) {
                memo[tmp] = calc(tle,tri); tmp++;
                tle += 64; tri += 64;
            }
            memo[tmp] = calc(tle, r1);
            
            int cur = dl2;
            rep(j, tmp + 1) {
                a[cur] ^= memo[j]; cur++;
            }
            
        }
        //cout << a[0] << "\n";
    }
    rep(i, 2000000 / 64) {
        rep(j, 64) {
            if (a[i] & ((ul)1 << j)) {
                b[i * 64 + j] = true;
            }
            else b[i * 64 + j] = false;
        }
    }
    string ans; ans.resize(n);
    rep(i, n) {
        if (b[i])ans[i] = 'O';
        else ans[i] = 'E';
    }
    cout << ans << "\n";
}
signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    //cout << fixed << setprecision(10);
    //init_f();
    //int t; cin >> t; rep(i, t)
    solve();
    return 0;
}
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