[Tsinsen A1300]JZPKIL
卡常的题见过,这么变态的卡常题……可能出题人过于文明,,,下面是卡常记录的一部分:
下面开始颓式子:
\[
\begin{aligned}
&\quad\sum_{i = 1}^n (i, n)^x[i, n]^y\\
&=n^y\sum_{i = 1}^n (i, n)^{x - y} i^y\\
&=n^y\sum_{k | n} k^{x - y}\sum_{i = 1}^{\frac{n}{k}} (ik)^y [(i, \frac{n}{k})=1]\\
&=n^y\sum_{k | n} k^x\sum_{i = 1}^{\frac{n}{k}} i^y\sum_{d | i, d | \frac{n}{k}} \mu(d)\\
&=n^y\sum_{k | n} k^x\sum_{d | \frac{n}{k}}\mu(d) \sum_{i = 1}^{\frac{n}{kd}} (id)^y\\
&=n^y\sum_{k | n} k^x\sum_{d | \frac{n}{k}}\mu(d)d^y \sum_{i = 1}^{\frac{n}{kd}} i^y\\
&=n^y\cdot(\mathrm{id}^x\ast (\mu\cdot\mathrm{id}^y)\ast h_y)(n)
\end{aligned}
\]
其中\(h_y\)表示指数为\(y\)的等幂求和函数。
接下来的感觉就和BZOJ 3601差不多了……还是设想如果后面是积性函数那么肥肠好做。可惜不是,但\(h_y\)是一个\(y + 1\)次多项式,所以拆开每一项考虑。然后我们还是对每一类质因子分开算,利用\(\mu\cdot\mathrm{id}^y\)这个函数的性质(当整个卷积的参数是质数的幂\(p^k\)时,这个函数只有输入为\(1\)和\(p\)时结果不为0),就肥肠好做了。
但是这个题非常的坑……\(n\)超级大,所以质因数分解要用Pollard's rho……这样一来我们又需要写Miller-Rabin,再一来我们发现当\(n\)很大的时候可能会有一些取膜乘法的模数太大,导致一般的乘法会发生溢出,所以又要写快速乘……\(O(\log n)\)快速乘又会被卡,所以你要写gyz的\(O(1)\)快速乘……
还有一些卡常的建议:
- Pollard's rho和Miller-Rabin一定要写的精细……复杂度首先不要像我这种沙茶一样上来多挂一个log,然后常数也要注意优化。
- \(h_y\)说是\(y + 1\)次,其实有大约一半的项的系数都为0。
- \(x = y\)时直接利用伯努利数算就行。
- 模数小的地方就别xjb用快速乘了……
代码:
#include <cstdio> #include <cstring> #include <cstdlib> #include <cctype> #include <ctime> #include <algorithm> #include <utility> #include <random> #include <vector> using ll = long long; using ld = long double; const ll ha = 1000000007LL; std::mt19937 gen; inline ll mul(ll x, ll y, ll z = ha){ // x %= z; y %= z; if(z <= ha) return (x * y) % z; ll ret = (x * y - (ll)(x / (ld)z * y + 1e-3) * z + z) % z; return ret; } inline ll pow_mod(ll a, ll b, ll p = ha) { ll ans = 1LL, res = a; while(b) { if(1LL & b) ans = mul(ans, res, p); res = mul(res, res, p); b >>= 1; } return ans; } void exgcd(ll a, ll b, ll &x, ll &y) { if(!b) { x = 1LL; y = 0LL; } else { ll nx, ny; exgcd(b, a % b, nx, ny); x = ny; y = nx - (a / b) * ny; } } inline ll inv(ll v, ll p = ha) { ll x, y; exgcd(v, p, x, y); return (x + p) % p; } inline bool witness(ll a, ll u, ll t, ll p) { ll x = pow_mod(a, u, p); if(x == 1LL) return true; while(t --) { if(x != 1LL && x != (p - 1LL) && mul(x, x, p) == 1LL) return false; x = mul(x, x, p); } return (x == 1LL); } inline bool miller_rabin(ll x) { static const int JP[9] = {2, 3, 5, 7, 11, 13, 17, 19, 23}; if(x <= 23LL) { if(x == 2LL || x == 3LL || x == 5LL || x == 7LL || x == 11LL || x == 13LL || x == 17LL || x == 19LL || x == 23LL) { return true; } else { return false; } } if(x % 2LL == 0LL) return false; ll u = x - 1LL, t = 0LL; while(!(u & 1LL)) { u >>= 1; t ++; } for(int i = 0; i < 9; i ++) { if(!witness(JP[i], u, t, x)) return false; } return true; } inline ll gcd(ll a, ll b) { while(b != 0LL) { ll na = b, nb = a % b; a = na, b = nb; } return a; } inline ll abs(ll x) { if(x < 0LL) { return -x; } else { return x; } } inline ll pollard_rho(ll p) { if(p % 2LL == 0LL) return 2LL; while(true) { ll a = (gen() % (p - 1LL) + 1LL), c = (gen() % (p - 1LL) + 1LL); ll b = a; if(c == 2LL) c = 3LL; do { a = (mul(a, a, p) + c) % p; b = (mul(b, b, p) + c) % p; b = (mul(b, b, p) + c) % p; if(a == b) break; ll d = gcd(p, abs(b - a)); if(d > 1LL && d < p) return d; } while(true); } } void find_fac(ll x, std::vector<ll> &A) { if(miller_rabin(x)) { A.push_back(x); return; } ll p = pollard_rho(x); // while(p == x) p = pollard_rho(x); find_fac(p, A); find_fac(x / p, A); } using pii = std::pair<ll, ll>; inline void fact(ll x, std::vector<pii> &V) { std::vector<ll> A; find_fac(x, A); std::sort(A.begin(), A.end()); ll las = 1LL, t = 0LL; for(int i = 0; i < A.size(); i ++) { ll v = A[i]; if(v != las) { if(las > 1LL) { V.push_back(std::make_pair(las, t)); } las = v; t = 1LL; } else { t ++; } } if(las > 1LL) V.push_back(std::make_pair(las, t)); #ifdef LOCAL printf("Brk %lld...\n", x); for(auto p : V) { printf("Pair (%lld, %lld)\n", p.first, p.second); } #endif } const int maxn = 3505; ll C[maxn][maxn]; inline void process_C() { C[0][0] = 1LL; for(int i = 1; i <= 3005; i ++) { C[i][0] = C[i][i] = 1LL; for(int j = 1; j < i; j ++) { C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % ha; } } } ll B[maxn]; inline void process_B() { B[0] = 1LL; for(int i = 1; i <= 3001; i ++) { B[i] = 1LL; for(int j = 0; j < i; j ++) { B[i] = (B[i] - (C[i][j] * ((B[j] * inv(i - j + 1)) % ha)) % ha + ha) % ha; } } } std::vector<pii> V; ll n; int x, y; inline ll calc(ll k, int z) { if(k == 0LL) return 0LL; ll ans = 1LL; for(auto pr : V) { ll p = pr.first, t = pr.second; p %= ha; ll v1 = 0LL; for(ll i = 0; i <= t; i ++) { v1 += (pow_mod(p, i * (ll)x) * pow_mod(p, (t - i) * (ll)z)) % ha; v1 %= ha; } ll v2 = 0LL; for(ll i = 0; i <= t - 1LL; i ++) { v2 += (pow_mod(p, i * (ll)x) * pow_mod(p, (t - i - 1LL) * (ll)z)) % ha; v2 %= ha; } v2 = (v2 * (ha - pow_mod(p, y))) % ha; v1 = (v1 + v2) % ha; ans = (ans * v1) % ha; } return (ans * k) % ha; } inline ll calc2() { n %= ha; ll ans = 0LL; for(int i = 0; i <= y; i ++) { ll delta = (C[y + 1][i] * B[i]) % ha; delta = (delta * pow_mod(n, y + 1 - i)) % ha; ans = (ans + delta) % ha; } ans = (ans * inv(y + 1)) % ha; ans = (ans * pow_mod(n % ha, y)) % ha; return ans; } int main() { gen.seed(time(0)); double d = clock(); process_C(); process_B(); #ifdef LOCAL double dd = clock(); printf("Time used : %.3lf\n", (dd - d) / (double(CLOCKS_PER_SEC))); #endif int T; scanf("%d", &T); while(T --) { scanf("%I64d%d%d", &n, &x, &y); if(x == y) { printf("%I64d\n", calc2()); continue; } V.clear(); fact(n, V); ll ans = 0LL; for(int i = 0; i <= y; i ++) { ans = (ans + calc((((C[y + 1][i] * inv(y + 1)) % ha) * B[i]) % ha, y + 1 - i)) % ha; } ans = (ans * pow_mod(n % ha, y)) % ha; printf("%I64d\n", ans); } return 0; }
Sep 08, 2022 10:07:02 PM
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