NOIp模拟赛Day17 月亮模拟VIII A

题目描述

给出 A 和 B，求：

∑d|ABd

输入描述

两个非负整数A，B

输出描述

仅一个正整数，表示答案

输入样例

2 3

输出样例

15

样例解释

23=8，而8的因子有1，2，4，8，而1+2+4+8=15

数据范围

A，B≤1012

---

质因数分解 AB

AB=pk11pk22...pknn

则

Sum=(1+p1+p21+...+pk11)(1+p2+p22+...+pk22)...(1+pn+p2n+...+pknn)

等比数列求和公式或分治计算 Sum
O(log2n)

---

#include <cstdio>#include <algorithm>#include <cmath>typedef long long ll;const int N = (int)1e6+5, mo = 9901;ll A, B;int p[N], num[N];ll ans;inline ll pow(ll x, ll n) {    x %= mo;    ll tmp = x, ret = 1;    while(n) {        if(n&1) ret = (ret*tmp)%mo;        tmp = (tmp*tmp)%mo;        n >>= 1;    }    return ret;}inline ll Sum(ll x, ll n) {return (pow(x,n+1)-1)*pow(x-1,mo-2)%mo;}inline ll Re() {    ll x = 0; char ch=getchar(); bool f = 0;    for(; ch>'9'||ch<'0'; ch=getchar()) if(ch=='-') f = 1;    for(; ch>='0'&&ch<='9'; ch=getchar()) x = (x<<1)+(x<<3)+ch-48;    if(f) return -x;    return x;}inline void prework() {    int lim = (int)sqrt(A);    for(int i=2;i<=lim;++i)        if(A%i==0) {            p[++p[0]] = i;            while(A%i==0) {                num[p[0]]++;                A /= i;            }        }    if(A>1) {        p[++p[0]] = A;        num[p[0]] = 1;    }}int main() {    A = Re(); B = Re();    prework();    ans = 1;    for(int i=1;i<=p[0];++i) ans = ans*Sum(p[i],num[i]*B)%mo;    printf("%lld\n",ans);    return 0;}