版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/qq_39972971/article/details/89180455
【思路要点】
- 不难看出,计算式中 对答案的贡献是一个积性函数。
- 可以直接用动态规划预处理 的函数值,再用 筛 计算答案。
- 时间复杂度 。
【代码】
#include<bits/stdc++.h> using namespace std; const int MAXN = 2e5 + 5; const int MAXK = 1e3 + 5; const int MAXE = 45; const int P = 998244353; typedef long long ll; typedef long double ld; typedef unsigned long long ull; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } int tot, prime[MAXN], f[MAXN], primek[MAXN], ps[MAXN]; int power(int x, int y) { if (y == 0) return 1; int tmp = power(x, y / 2); if (y % 2 == 0) return 1ll * tmp * tmp % P; else return 1ll * tmp * tmp % P * x % P; } void update(int &x, int y) { x += y; if (x >= P) x -= P; } ll n, val[MAXN]; int limit, m, k, cnt; int sum[MAXN], home1[MAXN], home2[MAXN]; int inv[MAXK], s[MAXK][MAXK], func[MAXE]; void init(int n, int Log) { for (int i = 2; i <= n; i++) { if (f[i] == 0) { prime[++tot] = f[i] = i; primek[tot] = power(i, k); ps[tot] = (ps[tot - 1] + primek[tot]) % P; } for (int j = 1; j <= tot && prime[j] <= f[i]; j++) { int tmp = prime[j] * i; if (tmp > n) break; f[tmp] = prime[j]; } } s[0][0] = inv[0] = 1; for (int i = 1; i <= k + 1; i++) { inv[i] = power(i, P - 2); for (int j = 1; j <= i; j++) s[i][j] = (s[i - 1][j - 1] + 1ll * j * s[i - 1][j]) % P; } static int dp[MAXE][MAXE][MAXE]; static int fac[MAXE], inf[MAXE]; fac[0] = inf[0] = 1; for (int i = 1; i <= Log; i++) { fac[i] = 1ll * fac[i - 1] * i % P; inf[i] = power(fac[i], P - 2); } dp[0][0][0] = 1; for (int i = 1; i <= Log; i++) for (int j = 0; j <= Log; j++) for (int k = 0; k <= Log; k++) { int res = 0; for (int l = 0; l <= j && l * i <= k; l++) update(res, 1ll * inf[l] * power(i + 1, l) % P * dp[i - 1][j - l][k - l * i] % P); dp[i][j][k] = res; } func[0] = 1; for (int i = 1; i <= Log; i++) { int res = 0; for (int j = 1; j <= Log; j++) { int tmp = dp[Log][j][i]; for (int k = 1; k <= j; k++) tmp = 1ll * tmp * (m - k + 1) % P; update(res, tmp); } func[i] = res; } } int calc(ll n) { n %= P; int ans = 0, now = n + 1; for (int i = 1; i <= k && i <= n; i++) { now = 1ll * now * (n - i + 1) % P; update(ans, 1ll * now * inv[i + 1] % P * s[k][i] % P); } return ans; } int solve(ll x, int y) { if (x <= 1 || prime[y] > x) return 0; int ans = 0, pos = 0; if (x <= limit) pos = home1[x]; else pos = home2[n / x]; ans = 1ll * func[1] * (sum[pos] - ps[y - 1] + P) % P; for (int i = y; i <= tot && 1ll * prime[i] * prime[i] <= x; i++) { int coef = primek[i]; ll now = prime[i], nxt = 1ll * prime[i] * prime[i]; for (int j = 1; nxt <= x; j++, now *= prime[i], nxt *= prime[i], coef = 1ll * coef * primek[i] % P) update(ans, (1ll * coef * func[j] % P * solve(x / now, i + 1) + 1ll * coef * primek[i] % P * func[j + 1]) % P); } return ans; } int main() { freopen("count.in", "r", stdin); freopen("count.out", "w", stdout); read(n), read(m), read(k); init(limit = sqrt(n) + 1, 40); for (ll i = 1, nxt; i <= n; i = nxt + 1) { ll tmp = n / i; nxt = n / tmp; val[++cnt] = tmp; if (tmp <= limit) home1[tmp] = cnt; else home2[n / tmp] = cnt; sum[cnt] = (calc(tmp) - 1 + P) % P; } for (int i = 1; i <= tot; i++) { for (int j = 1; 1ll * prime[i] * prime[i] <= val[j]; j++) { ll tmp = val[j] / prime[i]; int pos; if (tmp <= limit) pos = home1[tmp]; else pos = home2[n / tmp]; update(sum[j], P - 1ll * primek[i] * (sum[pos] - ps[i - 1] + P) % P); } } writeln((solve(n, 1) + 1) % P); return 0; }