Poj1845 (number theory + binary fast modulo)

Evaluate the sum of all the factors of a ^ B and the modulo 9901, for example, 2 ^ 3 = 8 => 1 + 2 + 4 + 8 = 15. The final answer is 15% 9901 = 15.

Analysis: A ^ B is first decomposed into prime factor form: A ^ B = (P1 ^ k1) + (P2 ^ K2) + (P3 ^ K3) +...

Then the sum of all factors of a ^ B is: S = (1 + p1 ^ 1 + p1 ^ 2 + p1 ^ 3 +... + p1 ^ K) * (1 + p2 ^ 1 + p2 ^ 2 + p2 ^ 3 +... + p2 ^ K) * (1 + P3 ^ 1 + P3 ^ 2 + P3 ^ 3 +... + P3 ^ K )*(...

Computing 1 + P ^ 2 + P ^ 3 +... + P ^ K can be solved in binary mode;

If K is an even number, for example, K = 4, 1 + P ^ 2 + P ^ 3 + P ^ 4 = (1 + p) + P ^ 2*(1 + p * (1 + p ));

If K is an odd number, for example, K = 5, 1 + P ^ 2 + P ^ 3 + P ^ 4 + P ^ 5 = (1 + P ^ 2) + P ^ 3*(1 + P ^ 2 );

# Include <iostream> <br/> # include <cmath> <br/> using namespace STD; <br/> const int maxn = 10000; <br/> int, b; <br/> int mod = 9901; <br/>__ int64 POW (_ int64 x ,__ int64 N) // calculate P ^ n <br/>{< br/>__ int64 ret = 1, s = x; <br/> while (1) <br/> {<br/> If (N & 1) <br/> ret = (RET % mod) * (S % mod) % MOD; <br/> If (N >>= 1) <br/> S = (S % mod) * (S % mod) % MOD; <br/> else break; <br/>}< br/> return ret; <br/>}< br/>__ I Nt64 sum (_ int64 P, _ int64 N) // calculate 1 + P ^ 2 +... + P ^ N; <br/>{< br/> If (n = 0) <br/> return 1; <br/> If (N & 1) // n % 2 = 0; <br/> return (1 + POW (p, n/2 + 1) % mod) * (sum (p, n/2) % mod) % MOD; <br/> else <br/> return (1 + POW (p, n/2 + 1) % mod) * (sum (p, (n-1)/2) % mod) + POW (p, n/2) % mod) % MOD; <br/>}< br/> int main () <br/>{< br/>__ int64 S, ans; <br/> int I, P [maxn], c [maxn]; // P [I] indicates the I-th prime factor, and C [I] indicates its number. <Br/> while (scanf ("% d", & A, & B )! = EOF) <br/>{< br/> memset (p, 0, sizeof (p); <br/> memset (C, 0, sizeof (c )); <br/> // a ^ B = p1 ^ (C1 * B )*.... * PI ^ (CI * B) <br/> // ans = (1 + p1 + p1 ^ 2 +... + p1 ^ (C1 * B ))*()... * (1 + PI ^ 2 +... + PI ^ (CI * B) <br/> // decompose a into prime numbers <br/> for (I = 2; I * I <= A; I ++) // find that all prime factors of a exist in P [], <br/>{< br/> if (a % I = 0) <br/> {<br/> P [++ P [0] = I; <br/> while (a % I = 0) <br/>{< br/> A/= I; <br/> ++ C [p [0]; // number of corresponding prime factors. <Br/>}< br/> if (! = 1) // process the last number. <Br/> {<br/> P [++ P [0] = A; <br/> C [p [0] = 1; <br/>}< br/> ans = 1; <br/> for (I = 1; I <= P [0]; I ++) // sum = (P1 ^ (N1 * B + 1)-1)/(p1-1) * (P2 ^ (N2 * B + 1)-1)/(p2-2) *... <Br/> {<br/> ans = (ANS % mod) * (sum (P [I], C [I] * B) % mod) % MOD; <br/>}< br/> printf ("% i64d/N", ANS); <br/>}< br/> return 0; <br/>}