Sumdiv POJ

题目传送门

题意：给你一个A，B，问你A^B的所有正因子的和模9901的结果。

思路：这一个题目学了好多东西啊，先学习了逆元，发现如果是出发的话要用逆元来取模
a / b % MOD = (a % (MOD * b)) / b
然后这个题目是可以用唯一分解定理来做的
N=P1^a1 * P2^a2 * P3a3 * …… * Pn^an
唯一分解定理有一个结论，一个数字的所有的正因子的和为
(1 + P1 + P1^2 + … + P1^a1) * (1 + P2 + P2^2 + … + P2^a2) * … *
(1 + Pn + Pn^2 + … + Pn^an)
然后就可以用等比数列来做了，但是又遇到了一个新的问题，好像只用快速幂过不了啊，用了一下二分乘法终于过了这个题目。

#include <algorithm>#include <cmath>#include <cstdio>#include <cstring>#include <fstream>#include <iostream>#include <list>#include <map>#include <queue>#include <set>#include <sstream>#include <stack>#include <string>#include <vector>#define MAXN 20100#define MAXE 5#define INF 1000000000#define MOD 9901#define LL long long#define pi acos(-1.0)using namespace std;LL p[MAXN];bool prime[MAXN];int cnt;struct Node {    LL x;    LL y;};vector<Node> vec;void is_prime() {    cnt = 0;    for (int i = 0; i < MAXN; ++i) {        prime[i] = true;    }    for (int i = 2; i < MAXN - 10; i++) {        if (prime[i]) {            p[cnt++] = i;            for (int j = i + i; j < MAXN - 10; j += i)                prime[j] = false;        }    }}LL multi(LL a, LL b, LL m) {    LL ans = 0;    a %= m;    while (b) {        if (b & 1) {            ans = (ans + a) % m;            b--;        }        b >>= 1;        a = (a + a) % m;    }    return ans;}LL quick_mod(LL a, LL b, LL m) {    LL ans = 1;    a %= m;    while (b) {        if (b & 1) {            ans = multi(ans, a, m);            b--;        }        b >>= 1;        a = multi(a, a, m);    }    return ans;}LL solve(LL a, LL b) {    for (int i = 0; i < cnt; ++i) {        LL ans = 0;        while (a % p[i] == 0) {            ans++;            a /= p[i];        }        if (ans > 0) {            vec.push_back((Node) {p[i], ans});        }    }    LL sum = 1;    for (int i = 0; i < vec.size(); ++i) {        LL M = (vec[i].x - 1) * MOD;        sum = sum * (quick_mod(vec[i].x, vec[i].y * b + 1, M) + M - 1) / (vec[i].x - 1);        sum %= MOD;    }    if (a > 1) {        LL M = MOD * (a - 1);        sum *= (quick_mod(a, b + 1, M) + M - 1) / (a - 1);        sum %= MOD;    }    return sum % MOD;}int main() {    std::ios::sync_with_stdio(false);    is_prime();    LL a, b;    cin >> a >> b;    cout << solve(a, b) << endl;    return 0;}