【数论】【poj1845】Sumdiv

Description

Consider two natural numbers A and B. Let S be the sum of all natural divisors of A^B. Determine S modulo 9901 (the rest of the division of S by 9901).

Input

The only line contains the two natural numbers A and B, (0 <= A,B <= 50000000)separated by blanks.

Output

The only line of the output will contain S modulo 9901.

Sample Input

2 3

Sample Output

15

Hint

2^3 = 8.

The natural divisors of 8 are: 1,2,4,8. Their sum is 15.

15 modulo 9901 is 15 (that should be output).

——————中文版题目大意——————

给定A,B，求A^B的所有因数的和，再MOD 9901

思路

这题我们首先看，求的是A^B的因子之和

我们首先分解A

因为任何一个数都可以分解为素数的幂次方相乘:

所以：A=(p1^k1)*(p2^k2)………;

其中p是素数,k是这个数最多能被除几次（幂的次数）;

因子个数就是:

(1+2+……..k1)(1+2+……….k2) ……个

那么因子之和就是

(1+p1^1+…..p1^k1)(1+p2^2+…..p2^k2) ………..;

a^b=(p1^k1* b) * (p2^k2*b)…..;

1+p1^1+…..p1^k1这就是一个等比数列

我们分奇和偶讨论 提出他们的通项以减少次方数求解;

代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#define ll long long
#define mo 9901
using namespace std;
inline int read(){
int ret=0,f=1;char c=getchar();
for(;!isdigit(c);c=getchar())if(c=='-')f=-1;
for(;isdigit(c);c=getchar())ret=ret*10+c-'0';
return ret*f;
}
inline ll ksm(ll x,ll y){
ll ans=1ll;
x%=mo;
while(y){
if(y&1)(ans*=x)%=mo;
(x*=x)%=mo;
y>>=1;
}
return ans%mo;
}
ll sum(ll x,ll y){
if(!y)return 1;
else if(y&1)return (sum(x,y/2)*(1+ksm(x,y/2+1))%mo);
else return ((sum(x,y/2-1)*(1+ksm(x,y/2+1)))%mo+ksm(x,y/2)%mo)%mo;
}
ll n,m,a;
int main(){
while(scanf("%lld%lld",&a,&m)==2){
if(m==0||a<=1){printf("1\n");continue;}
ll ans=1;
ll n=sqrt(a+0.5);
for(int i=2;i<=n;++i){
if(!(a%i)){
ll cnt=0ll;
while(!(a%i)){
++cnt;
a/=i;
}
(ans*=sum(i,cnt*m))%=mo;
}
}
if(a>1)(ans*=sum(a,m))%=mo;
printf("%lld\n",ans);
}
return 0;
}