POJ 1845 --Sumdiv 约数基本定理+乘法逆元

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).

题目大意：给你两个整数A，B。然后求A^B的所有约数的和。

题目解析：根据约数基本定理如何一个自然数都可以分解为A=(a1^n1)*(a2^n2)*(a3^n3)...(am^nm) 的形式，其中a1,a2,a3......为质数。所有约数的个数为（a1+1）*(a2+1).....*(am+1).而所有约数的和为（a1^0+a1^1+a1^2……a1^n1）*（a2^0+a2^1+a2^2……a2^n2）……*（am^0+am^1+am^2……am^nm）

然后就可以根据等比数列求每一项的和然后相乘   即(a1^(n1+1)-1)/(a1-1) mod 9901  根据前面逆元的知识：a/b mod c = (a mod (b*c))/ b就可以得到  (a1^(n1+1)-1)mod (9901*(a1-1)) / (a1-1)。

AC代码：

#include<iostream>#include<cstring>#include<cstdio>using namespace std;typedef long long LL;const int N=10005;const int MOD=9901;bool prime[N];int p[N];int cnt;void isprime(){cnt=0;memset(prime,true,sizeof(prime));for(int i=2;i<N;i++){if(prime[i]){p[cnt++]=i;for(int j=i+i;j<N;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;}void Solve(LL A,LL B){LL ans=1;for(int i=0;p[i]*p[i]<=A;i++){if(A%p[i]==0){int num=0;while(A%p[i]==0){num++;A/=p[i];}LL M=(p[i]-1)*MOD;ans*=(quick_mod(p[i],num*B+1,M)-1)/(p[i]-1);ans%=MOD;}}if(A>1){LL M=MOD*(A-1);ans*=(quick_mod(A,B+1,M)-1)/(A-1);ans%=MOD;}cout<<ans<<endl;}int main(){LL A,B;isprime();while(cin>>A>>B){Solve(A,B);}return 0;}