标签：name rac 个数 oid 奇数 ons mdi 一个 power
题意：求$A^{B}$的所有约数之和$mod\ 9901$
思路：由结论有，一个数$n$进行质因数分解得到$n={p_{1}}^{c_{1}} * {p_{2}}^{c_{2}} *...* {p_{k}}^{c_{k}}$，那么$n$的约数之和为
$$sum=(1+{p_{1}}^{1}+\cdots+{p_{1}}^{c_{1}})*(1+{p_{2}}^{1}+\cdots +{p_{2}}^{c_{2}})*\cdots*(1+{p_{k}}^{1}+\cdots+{p_{k}}^{c_{k}})$$
所以对$A$质因数分解后，那么$A^{B}$的约数之和
$$sum=(1+{p_{1}}^{1}+\cdots+{p_{1}}^{B*c_{1}})*(1+{p_{2}}^{1}+\cdots +{p_{2}}^{B*c_{2}})*\cdots*(1+{p_{k}}^{1}+\cdots+{p_{k}}^{B*c_{k}})$$
上式中每个括号内都是等比数列，利用分治法对等比数列求和，设$sum(p，c)=1+p+p^2+\cdots+p^{c}$
当$c$为奇数时
$$sum(p,c)=(1+p+\cdots+p^{\frac{c-1}{2}})+(p^{\frac{c+1}{2}}+\cdots+p^c)=(1+p^{\frac{c+1}{2}})*sum(p,\frac{c-1}{2})$$
当$c$为偶数时
$$sum(p,c)=(1+p+\cdots+p^{\frac{c}{2}-1})+(p^{\frac{c}{2}}+p^{\frac{c}{2}+1}\cdots+p^{c-1})+p^c=(1+p^{\frac{c}{2}})*sum(p,\frac{c}{2}-1)+p^c$$
当$c$等于$0$，结束递归， 返回$1$即可
#include <iostream> #include <algorithm> #include <cstdio> #include <cmath> using namespace std; typedef long long ll; const int N = 110; const ll mod = 9901; ll a, b; ll p[N], c[N], m; void divide(ll n) { m = 0; for (ll i = 2; i <= sqrt(n); i++) { if (0 == n % i) { p[++m] = i, c[m] = 0; while (0 == n % i) n /= i, c[m]++; } } if (n > 1) p[++m] = n, c[m] = 1; return; } ll power(ll a, ll b, ll p) { ll res = 1; while (b) { if (b & 1) res = (res * a) % p; a = (a * a) % p, b >>= 1; } return res % p; } ll sum(ll p, ll c) { if (0 == c) return 1; if (1 == c % 2) { ll tp1 = (1 + power(p, (c + 1) / 2, mod)) % mod; ll tp2 = sum(p, (c - 1) / 2) % mod; return tp1 * tp2 % mod; } else { ll tp1 = (1 + power(p, c / 2, mod)) % mod; ll tp2 = sum(p, c / 2 - 1) % mod; return (tp1 * tp2 % mod + power(p, c, mod)) % mod; } } int main() { scanf("%lld%lld", &a, &b); divide(a); if (0 == a) printf("0\n"); else { ll res = 1; for (int i = 1; i <= m; i++) res = res * sum(p[i], b * c[i]) % mod; printf("%lld\n", res); } return 0; }
标签：name rac 个数 oid 奇数 ons mdi 一个 power
原文地址：https://www.cnblogs.com/zzzzzzy/p/12221260.html
