hihoCoder #1759 取数字

main.cpp

#include <bits/stdc++.h>

using namespace std;
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(), x.end()
#define debug(x) cerr << #x <<": " << (x) << endl
//在每个函数的入口处执行一次，出口处执行一次。然后就可以快速得知是哪个地方段错误了
#define DEBUG printf("Passing [%s] in LINE %d\n",__FUNCTION__,__LINE__)
#ifdef LOCAL
#define see(x) cout  << #x << ": " << (x) << endl
#endif
#ifndef LOCAL
#define see(x)
#endif

#define RED     "\033[31m"
#define RESET "\033[0m"
#define alert(x) cerr << RED << x << RESET << endl
#ifndef LOCAL
#define see(x)
#endif

#define Rep(n) for(int _ = 0; _ != (n); ++_)
#define rep(i, a, b) for(int i = (a); i <= (b); ++i)
#define Rng(i, n) for(int i = 0; i != (n); ++i)
#define rng(i, a, b) for(int i = (a); i != (b); ++i)
#define RNG(i, a) for(auto &i: (a))

namespace std {
    template<class T>
    T begin(std::pair<T, T> p)
    {
        return p.first;
    }
    template<class T>
    T end(std::pair<T, T> p)
    {
        return p.second;
    }
}


#if __cplusplus < 201402L
template<class Iterator>
std::reverse_iterator<Iterator> make_reverse_iterator(Iterator it)
{
    return std::reverse_iterator<Iterator>(it);
}
#endif

template<class Range>
std::pair<std::reverse_iterator<decltype(begin(std::declval<Range>()))>, std::reverse_iterator<decltype(begin(std::declval<Range>()))>> make_reverse_range(Range &&r)
{
    return std::make_pair(make_reverse_iterator(::begin(r)), make_reverse_iterator(::end(r)));
}

#define RRNG(x, cont) for (auto &x: make_reverse_range(cont))



template<class T> int sign(const T &a) { return a == 0 ? 0 : a > 0 ? 1 : -1; }
template<class T> inline T min(T a, T b, T c){return min(min(a, b), c);}
template<class T> inline T max(T a, T b, T c){return max(max(a, b), c);}
template<class T> void Min(T &a, const T &b){ a = min(a, b); }
template<class T> void Max(T &a, const T &b){ a = max(a, b); }

template<typename T> void println(const T &t) { cout << t << '\n'; }
template<typename T, typename ...Args> void println(const T &t, const Args &...rest) { cout << t << ' '; println(rest...); }

template<typename T> void print(const T &t) { cout << t << ' '; }

template<typename T, typename ...Args> void print(const T &t, const Args &...rest) { cout << t; print(rest...); }

// this overload is chosen when there's only one argument
template<class T> void scan(T &t) { cin >> t; }
template<class T, class ...Args> void scan(T &a, Args &...rest) { cin >> a; scan(rest...); }


int cas;
const double pi = acos(-1);
int mod = 1e9 + 7;

template<class T>
void add_mod(T &a, const T &b) {
    a += b;
    if (a >= mod) a -= mod;
}
template<class T>
void sub_mod(T &a, const T &b){
    a -= b;
    if (a < 0) a += mod;
}


using ll = long long;
using ull = unsigned long long;
using vec = vector<ll>;
using mat = vector<vec>;
using pii = pair<int, int>;
using pdd = pair<double, double>;
using pip = pair<int, pii>;
using szt = size_t;
using vi = vector<int>;
using vl = vector<ll>;
using vb = vector<bool>;
using vpii = vector<pii>;

mat operator*(const mat &a, const mat &b) {
    mat c(a.size(), vec(b[0].size()));
    for (int i = 0; i < a.size(); i++) {
        for (int j = 0; j < a[0].size(); j++) {
            if (a[i][j]) { // optimization for sparse matrix
                for (int k = 0; k < b[0].size(); k++) {
                    add_mod(c[i][k], a[i][j] * b[j][k] % mod);
                }
            }
        }
    }
    return c;
}

vec operator*(const mat &a, const vec &b) {
    vec c(a.size());
    for (int i = 0; i < a.size(); i++) {
        for (int j = 0; j < a[0].size(); j++) {
            add_mod(c[i], a[i][j] * b[j] % mod);
        }
    }
    return c;
}

mat pow(mat a, ull n) {
    mat res(a.size(), vec(a[0].size()));
    for (int i = 0; i < a.size(); i++) {
        res[i][i] = 1;
    }
    while (n) {
        if (n & 1) {
            res = res * a;
        }
        a = a * a;
        n >>= 1;
    }
    return res;
}


ll POW(ll x, ll n) {
    ll res = 1;
    for (; n; n /= 2, x *= x, x %= mod) {
        if (n & 1) {
            res *= x;
            res %= mod;
        }
    }
    return res;
}

ll inv(ll x) {
    return POW(x, mod - 2);
}



// 2D rotation
void rotate(double &x, double &y, double theta) {
    double tx = cos(theta) * x - sin(theta) * y;
    double ty = sin(theta) * x + cos(theta) * y;
    x = tx, y = ty;
}
namespace bit {
    const int BIT_N = 1e5 + 5;

    int bit[BIT_N];

    int sum(int x) {
        int res = 0;
        while (x) {
            res += bit[x];
            x -= x & -x;
        }
        return res;
    }

    int sum(int l, int r) {
        if (l > r) return 0;
        return sum(r) - sum(l - 1);
    }

    void add(int x, int v, int n) {
        while (x <= n) {
            bit[x] += v;
            x += x & -x;
        }
    }
}

namespace util{
    int len(ll x){return snprintf(nullptr, 0, "%lld", x);}

    vi get_d(ll x){
        vi res;
        while(x) {
            res.pb(x%10);
            x /= 10;
        }
        reverse(all(res));
        return res;
    }
}



// #include <ext/pb_ds/priority_queue.hpp>

// typedef __gnu_pbds :: priority_queue<pip, less<pip>, __gnu_pbds::thin_heap_tag > Heap;

// Heap h;

// Heap::point_iterator pos[N][N];
const ll LINF = LLONG_MAX/10;
const int INF = INT_MAX/10;
const int M = 3000 + 5;


const int N = 2e5+5;
int main() {
    // Single Cut of Failure taut me
    // std::fixed 使得所有实数（默认）输出6位小数，即使实际小数位数多于6位。
    cout << setprecision(10) << std::fixed; 
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
#ifdef LOCAL
    freopen("main.in", "r", stdin);
//    freopen("main.out", "w", stdout);
#endif

    int n;
    scan(n, mod);

    int m = sqrt(n);
    vi minp(m+1);
    vi p;
    p.pb(1);
    vl f(m+1);
    f[1] = 1;
    vi cnt(m+1); // 最小质因子的幂次
    vl s1(m+1), s2(m+1), t1(m+1), t2(m+1);
    vi ub(m+1); // ub[j] 大于j的最小质数在p中的下标
//    1152921504606846976
    vl inv(40);

    inv[1] = 1;   // 预处理逆元
    for (int i = 2; i < 40; i++) {
        inv[i]=(mod-mod/i)*inv[mod%i]%mod;
    }

    const ll inv2 = (mod+1)/2;

    rng(i, 2, m+1) {
        if(!minp[i]) {
            minp[i] = i;
            for (int j = i - 1; j >= p.back(); j--) {
                ub[j] = p.size();
            }
            p.pb(i);
            cnt[i] = 1;
            f[i] = inv2;
        }
        for(int j=1; j < p.size(); j++) {
            int x= p[j];
            if(x > minp[i]) break;
            ll t = 1LL * x * i;
            if(t > m) break;
            minp[t] = x;
            if(x < minp[i]) {
                cnt[t] = 1;
                f[t] = f[x] * f[i] % mod;
            }
            else{
                cnt[t] = cnt[i] + 1;
                ll _t = POW(x, cnt[t]);
                if(_t == t){
                    f[t] = inv[cnt[t]+1];
                }
                else{
                    f[t] = f[t/_t] * f[_t] % mod;
                }
            }
        }
    }

    for (int i = m; i >= p.back(); i--) {
        ub[i] = p.size();
    }

    int t = n / (m+1);

    for (int i = 1; i <= m; i++) {
        s1[i] = 1;
    }

    // 各种优化
    vi v(m+1);
    for (int i = 1; i <= t; i++) {
        v[i] = n/i;
    }

    for (int i = 1; i <= t; i++) {
        s2[i] =  1;
    }

    auto get_s = [inv2, n, m, &p, &s1, &s2, &ub](int i, ll j)->ll {
        if (i == p.size() || j < p[i]) return 1;
        // j >= p[i]
        if (j < p[i] * p[i]) {
            int pos = j <= m ? ub[j] : p.size();
            return ((pos - i) * inv2 + 1) % mod;
        }
        return j <= m ? s1[j] : s2[n / j];
    };


    for(int i=p.size()-1; i>=1; i--) {
        int p2 = p[i] * p[i];
        for (int k = 1; k <= t; ++k) {
            int j = v[k];
            if (j >= p2) {
                ll x = p[i];
                int c = 1;
                s2[k] = get_s(i + 1, j);
                while (j >= x) { //这一部分的复杂度怎么算？
                    s2[k] += get_s(i + 1, j / x) * inv[c + 1] % mod;
                    s2[k] %= mod;
                    x *= p[i];
                    ++c;
                }
            } else break;
        }
        for (int j = m; j >= p2; j--) {
            ll k = p[i];
            int c = 1;
            s1[j] = get_s(i + 1, j);
            while (j >= k) { //这一部分的复杂度怎么算？
                s1[j] += get_s(i + 1, j / k) * inv[c + 1] % mod;
                s1[j] %= mod;
                k *= p[i];
                ++c;
            }
        }
    }

    for (int i = 1; i <= t; i++) {
        t2[i] = v[i];
    }
    for (int i = 1; i <= m; i++) {
        t1[i] = i;
    }

    vi pos1(m+1), pos2(m+1);

    auto get_t = [inv2, n, m, &pos1, & pos2, &p, &t1, &t2](int i, ll j)->ll{
        if (i == 0) return j;
        if (j <= p[i]) return 1;
        if (j >= p[i]*p[i]) return j <= m ? t1[j] : t2[n/j];
        return j <= m ? t1[j] - (i - pos1[j]) : t2[n/j] - (i - pos2[n/j]);
    };

    for(int i = 1; i < p.size(); i++){
        int p2 = p[i] * p[i];
        rng(k, 1, t+1) {
            int j = v[k];
            if(j >= p2) {
                t2[k] = get_t(i-1, j) - get_t(i-1, j/p[i]);
                pos2[k] = i;
            }
            else break;
        }
        for (int j = m; j >= p2; j--) {
            t1[j] = get_t(i-1, j) - get_t(i-1, j/p[i]);
            pos1[j] = i;
        }
    }

    ll ans = get_s(1, n);
    for (int i = 1; i <= m; i++) {
        add_mod(ans, f[i] * (get_t(p.size()-1, n/i) - 1) % mod * inv2 % mod); // 注意：是 p.size()-1 不是 p.size()
    }
    println(ans);
    return 0;
}

实现洲阁筛的过程中，遇到几个问题：

1. reversed range-based for
2. recursive lambda expression
3. C++ name lookup 的原理？
4. 线性时间求逆元