poj 1845 Sumdiv

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Sumdiv

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 19529 Accepted: 4916

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

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：求A^B所有约数之和

思路：
由于乘性函数的性质
题目等价于求(1+a1+a1^2+a1^3+a1^4+.....+a1^(b1k)) *(1+a2+a2^2+a2^3+a2^4+......+a2^(b2k))+......
=s1*s2*s3......*sn
根据等比数列求和易知
(1-an^(bnk+1))/1-an
an为约数，bn为该约数的个数

由于出现除法取模，需要求逆元，记得讨论逆元不存在，逆元存在当且仅当gcd(x,MOD)==1

代码如下：

#include<cmath>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;typedef long long LL;const LL maxn=100005;const LL MOD=9901;LL a[maxn],b[maxn];LL quickpow(LL n,LL m){n%=MOD;if(n==0)return MOD-1;if(n==1)return 0;LL ans=1;while(m){if(m&1)ans*=n,ans%=MOD;m>>=1;n*=n,n%=MOD;}ans=(ans%MOD+MOD)%MOD;return ans-1;}LL getfac(LL n){LL index=0;for(int i=2;i*i<=n;i++){while(n%i==0&&n)b[index]++,n/=i;if(b[index])a[index++]=i;}if(n>1)b[index]++,a[index]=n,index++;return index;}LL x,y;long long exgcd(long long a, long long b, long long &x, long long &y){if (a == 0){x = 0;y = 1;return b;}else{long long tx, ty;long long d = exgcd(b%a, a, tx, ty);x = ty - (b / a)*tx;y = tx;return d;}}int main(){LL A,B;LL index=0;while(cin>>A>>B){memset(b,0,sizeof(b));LL cnt=getfac(A),ans=1;for(int i=0;i<cnt;i++){//if(a[i]%MOD==0)continue;if(a[i]%MOD==1){ans=ans*(b[i]*B+1),ans%=MOD;continue;}LL L=quickpow(a[i],b[i]*B+1);exgcd(a[i]-1,MOD,x,y);LL R=x;L=((L%MOD)+MOD)%MOD;R=((R%MOD)+MOD)%MOD;ans*=((L*R)%MOD+MOD)%MOD,ans%=MOD;}ans=((ans%MOD)+MOD)%MOD;cout<<ans<<endl;}}

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