P6826「EZEC-4」顶的和式

题意

传送门 P6826「EZEC-4」月下轻花舞

题解

先处理右侧的和式，依据个人习惯，将 $i$ 用 $x$ 表示 。设 $m=\lfloor {\log }_x^n\rfloor$，则有
$$\begin{array}{rl}\sum _{j=1}^n\lceil {\log }_x^j\rceil & =\sum _{j=1}^{n}k[k=\lceil {\log }_x^j\rceil ][{\log }_x^j\le k<{\log }_x^j+1]\\ & =\sum _{j,k}k[1\le k\le m][x^{k-1}<j\le x^k]+\sum _j(m+1)[x^m<j\le n]\\ & =\sum _{k=1}^{m}k(x^k-x^{k-1})+(m+1)(n-x^m)\\ & =\frac{1}{x-1}[1-(m+1)x^m+m{x}^{m+1}]+(m+1)(n-x^m)\end{array}$$
那么有
$$\sum _{x=l}^r(x-1)\sum _{j=1}^n\lceil {\log }_x^j\rceil =\sum _{x=l}^r[1+n(x-1)+mn(x-1)-x^{m+1}]$$ $m$ 的取值只有 $\log n$ 种，可以对满足 $m$ 相同的区间 $(\sqrt[m+1]{n},\sqrt[m]{n}]$ 依次求解。这样的区间为 $(n,+\infty ],(\sqrt{n},n],(\sqrt[3]{n},\sqrt{n}],\cdots$，取与 $(l,r]$ 的交集进行统计。对于满足 $x\le \sqrt[m]{n}$ 的最大 $x$，可以二分求解；依据 $\lfloor \lfloor n/x\rfloor /x\rfloor =\lfloor n/x^2\rfloor ,x>0,x\in \mathbb{Z}$ 的原理进行检验，基于与快速幂类似的倍增思想可以做到 $O(\log \log N)$。当 $m\ge 5$，有 $n<2^{60}\to \sqrt[m]{n}<2^{12}$，对这样值域较小的情况，可以暴力求前缀和；$m<5$ 的情况，应用闭合表达式。

总时间复杂度 $O(N{\log }^{2}N\log \log N)$。

#include <bits/stdc++.h>
using namespace std;
#define rep(i, l, r) for (int i = l, _ = r; i < _; ++i)
typedef long long ll;
const int maxlg = 1 << 6, maxn = 1 << 12;
const ll mod = 998244353, inf = 0x3f3f3f3f3f3f3f3f;
const ll coef[6][7] = {
    
    {
    
    0, 1, 0, 0, 0, 0, 0},
                       {
    
    0, 1, 1, 0, 0, 0, 0},
                       {
    
    0, 1, 3, 2, 0, 0, 0},
                       {
    
    0, 0, 1, 2, 1, 0, 0},
                       {
    
    0, -1, 0, 10, 15, 6, 0},
                       {
    
    0, 0, -1, 0, 5, 6, 2}};
const ll coef2[6] = {
    
    1, 2, 6, 4, 30, 12};
int T;
ll inv[50], sum[maxlg][maxn];

ll pow_mod(ll x, ll n)
{
    
    
    ll res = 1 % mod;
    x %= mod;
    while (n)
    {
    
    
        if (n & 1)
            res = res * x % mod;
        x = x * x % mod, n >>= 1;
    }
    return res;
}

ll F(int e, ll n)
{
    
    
    ll x = 1, res = 0;
    n %= mod;
    rep(i, 0, 7) res = (res + x * coef[e][i] % mod) % mod, x = x * n % mod;
    return res * inv[coef2[e]] % mod;
}

bool judge(ll x, int m, ll n)
{
    
    
    ll res = n;
    while (m)
    {
    
    
        if (m & 1)
            if ((res /= x) < 1)
                return 0;
        m >>= 1;
        if (x > (n / x) && m)
            return 0;
        x = x * x;
    }
    return 1;
}

ll find(int m, ll n, ll lim)
{
    
    
    ll lb = 0, ub = lim + 1;
    while (ub - lb > 1)
    {
    
    
        ll mid = (lb + ub) >> 1;
        if (judge(mid, m, n))
            lb = mid;
        else
            ub = mid;
    }
    return lb;
}

ll solve(ll l, ll r, ll n)
{
    
    
    --l;
    ll res = (r - l) % mod;
    res = (res + (n % mod) * (F(1, r - 1) - F(1, l - 1)) % mod) % mod;
    ll lb, ub = inf;
    rep(m, 0, maxlg)
    {
    
    
        if (ub < l)
            break;
        lb = find(m + 1, n, ub);
        if (!(r <= lb || ub <= l))
        {
    
    
            ll a = max(lb, l), b = min(ub, r);
            ll d = (F(1, b - 1) - F(1, a - 1)) * m % mod, d2;
            d = (n % mod) * d % mod;
            res = (res + d) % mod;
            if (m + 1 <= 5)
                d2 = (F(m + 1, b) - F(m + 1, a)) % mod;
            else
                d2 = (sum[m + 1][b] - sum[m + 1][a]) % mod;
            res = (res - d2) % mod;
        }
        ub = lb;
    }
    return (res + mod) % mod;
}

int main()
{
    
    
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    rep(j, 1, maxn)
    {
    
    
        ll x = 1;
        rep(i, 0, maxlg) sum[i][j] = (sum[i][j] + x) % mod, x = x * j % mod;
    }
    rep(i, 0, maxlg) rep(j, 1, maxn) sum[i][j] += sum[i][j - 1];
    rep(i, 1, 50) inv[i] = pow_mod(i, mod - 2);
    cin >> T;
    while (T--)
    {
    
    
        ll l, r, n;
        cin >> l >> r >> n;
        cout << solve(l, r, n) << '\n';
    }
}