POJ-1845 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).

Source

Romania OI 2002

分析： 用到了因数和公式，（1+p1+p1^2...p1^n）*........(1+q1+q1^2...q1^n) ，然后因为9901,比较小，所以存在几个比较坑的数据（因数膜9901为1时），求逆元的时候特判下就

过了。

#include <cmath> #include <cstdio>#include <iostream>using namespace std;int a,b,num;long long q[30],f[30];long long ksm(long long x,long long y){long long ans = 1;while(y){if(y & 1 == 1) ans = ans * x % 9901;x = x * x % 9901;y >>= 1; }return ans;}int main(){cin>>a>>b;if(a == 0) {cout<<0<<endl;return 0;} for(int i = 2;i <= sqrt(double(a));i++){if(a % i == 0) f[++num] = i;while(a % i == 0){q[num]++;a = a / i;}} if(a != 1) {q[++num] = 1;f[num] = a;}long long ans = 1;for(int i = 1;i <= num;i++){if(f[i] % 9901 == 1) ans = ans * (b*q[i] + 1) % 9901; else ans = (ans * (ksm(f[i],b*q[i]+1) + 9900) % 9901) * ksm(f[i]-1,9899) % 9901;}cout<<ans<<endl;}