Nyoj 928 small M factor and (number theory)

Small M factor and time limit: 1000 MS | memory limit: 65535 kb difficulty: 2

Description

When he was in class, Mr. M was a bit out of his mind, and the teacher was hard on his question. I heard that Mr. M is still very proud of finding the factor and the power of B. But after reading the question, I want to find the factor and the power of B. I am a little confused. Can you solve this problem?

Input

Multiple groups of test examples
Each row has two numbers a, B, (1 ≤ a, B ≤ 10 ^ 9)

Output

Output the factor and of B power of A, and take the remainder of 9901.

Sample Input

2 3

Sample output

15

Analysis: Perform prime factor decomposition for A. Assume that a = (P1 ^ A1) * (P2 ^ A2 )*...... * (Pk ^ AK), assuming the factor of A is sum,

Then sum = (1 + p1 + p1 ^ 2 + ...... + P1 ^ A1) * (1 + p2 + p2 ^ 2 + ...... + P2 ^ A2 )*...... * (1 + PK ^ 2 + ...... + PK ^ AK). After this formula is expanded, each item is a factor.

Therefore, for the factor and time of a ^ B, you only need to change AI to ai * B and then solve the problem. Because the power is relatively large, it needs to be obtained through sub-governance and quick power.

# Include <cstdio> # include <cmath> const int mod = 9901; int POW (int A, int N) {// fast power a ^ n int res = 1; A % = MOD; while (n) {If (N & 1) RES = res * A % MOD; A = A * A % MOD; n >>= 1 ;} return res;} int get_sum (int A, int N) {// evaluate (1 + A ^ 2 + ...... + A ^ N) if (n = 0) return 1; if (N & 1) return get_sum (A, n/2) * (POW (, (n/2 + 1) + 1) % MOD; else return (get_sum (A, n/2-1) * (POW (A, n/2) + 1) + POW (A, n) % MOD;} int main () {int A, B; while (~ Scanf ("% d", & A, & B) {int ans = 1; int M = (INT) SQRT (a + 0.5 ); for (INT I = 2; I <= m; ++ I) {if (a % I = 0) {int K = 0; while (a % I = 0) {k ++; A/= I;} ans = ans * get_sum (I, K * B) % mod ;}} if (A> 1) ans = ans * get_sum (a, B) % MOD; printf ("% d \ n", ANS);} return 0 ;}

Nyoj 928 small M factor and (number theory)