Poj 1845-sumdiv (number theory + unique decomposition theorem + Rapid power modulo)

Poj 1845-sumdiv (number theory + unique decomposition theorem + Rapid power modulo)

This is a good question of number theory and requires a good mathematical foundation.

Given a and B, calculate the sum of all the factors of a ^ B, and then mod 9901

Analysis:

A considerable part of the number theory knowledge is used here.

A. Unique Decomposition Theorem

Any integer can be decomposed into the product of several prime numbers.

A = (P1 ^ Q1 + p2 ^ Q2 +... + PN ^ qn) P is a prime number

A ^ B = (P1 ^ (Q1 * B) +P2 ^ (Q2 * B) +... + PN ^ (qN * B))

B. Appointment and Formula

Sum (A) = (1 + p1 + p1 ^ 2 + .... + p1 ^ Q1) * (1 + p2 + p2 ^ 2 + .... + p2 ^ q2 )*...... * (1 + Pn + PN ^ 2 + ..... + PN ^ qn)

Sum (a ^ B) = (1 + p1 + p1 ^ 2 + .... + p1 ^ (Q1 * B) * (1 + p2 + p2 ^ 2 + .... + p2 ^ (Q2 * B ))*...... * (1 + Pn + PN ^ 2 + ..... + PN ^ (qN * B ))

C. Fast Power modulo (omitted)

D. Simple Model Calculation Formula

(A * B) % m = (a % m) * (B % m) % m

E. recursively solve the first N values of the proportional series (or use the multiplication inverse element to obtain them. If not, add them immediately)

(1 + p1 + p1 ^ 2 +... + p1 ^ N) ParityFor discussion:

1 .)N is an odd number (N & 1) and has an even number in total. Assume n = 5.

(1 + p1 + p1 ^ 2 + p1 ^ 3 + p1 ^ 4 + p1 ^ 5) = (1 + p1 + p1 ^ 2 + p1 ^ 3 (1 + p1 + p1 ^ 2 ))

= (1 + p1 + p1 ^ 2 + P ^ (5/2 + 1) (1 + p1 + p1 ^ 2 ))

= (1 + P ^ (5/2 + 1 ))*(1 + p1 + p1 ^ 2)

The conversion to the general formula should be

(1 + p1 + p1 ^ 2 +... + p1 ^ N)

=(1 + p1 + p1 ^ 2 + .... + p1 ^ (n/2) + P ^ (n/2 + 1) * (1 + p1 + p1 ^ 2 + ..... p1 ^ (n/2 )))

= (1 + P ^ (n/2 + 1) * (1 + p1 + p1 ^ 2 +... + p1 ^ (n/2 ))

= (1 + fast_mod (P1, n/2 + 1) * sum (P1, n/2)

2 .)N is an even number with an odd number in total. Assume n = 6.

(1 + p1 + p1 ^ 2 + p1 ^ 3 + p1 ^ 4 + p1 ^ 5 + p1 ^ 6) = (1 + p1 + p1 ^ 2 + p1 ^ 3 + p1 ^ 4 (1 + p1 + p1 ^ 2 ))

= (1 + p1 + p2 + p1 ^ 4 (1 + p1 + p2) + p1 ^ 3)

= (1 + P ^ 4)(1 + p1 + p2) + p1 ^ 3)

Convert to the normal expression

(1 + p1 + p1 ^ 2 +... + p1 ^ N)

= (1 +P1 ^ (n/2 + 1 ))*(1 + p1 + p2 +... + P ^ (n/2-1) + p1 ^ (n/2)

= (1 + fast_mod (P1, n/2 + 1) * sum (P1, n/2-1) + fast_mod (P1, n/2)

F. The multiplication is against the RMB. The weak dishes are too weak. If not, add them immediately.

First, we need to know the first n terms and

Divide by (1-Q) and convert it into a multiplication inverse element.

Finally, the Code is provided.

/************************************ * *********** ID: whiteblock63lang: G ++ prog: poj-1845 sumdivdate: 15:56:09 ************************************** * **********/# include # include # include # include # include # define CLR (, b) memset (a, B, sizeof (A) using namespace STD; typedef long ll; typedef unsigned long ull; # define maxn 10000 # define mod 9901ll a, B; // fast_pow_modll fast_mod (ll a, LL B) {ll ans = 1; A = A % MOD; while (B) {If (B & 1) ans = ans * A % MOD; A = A * A % MOD; B >>= 1 ;} return ans ;} ll factors [maxn] [2]; int FN; // decomposition prime factor void gaoji (ll x) {fn = 0; For (ll I = 2; I * I <= x; ++ I) {If (X % I) continue; factors [++ FN] [0] = I; factors [FN] [1] = 0; while (X % I = 0) {factors [FN] [1] ++; X/= I ;}} if (X! = 1) {factors [++ FN] [0] = x; factors [FN] [1] = 1 ;}// recursively solves the proportional sequence ll sum (ll p, ll N) {If (n = 0) return 1; if (P = 0) return 0; If (N & 1) Return (sum (p, n/2) * (1 + fast_mod (p, n/2 + 1) % MOD; else return (sum (p, n/2-1) * (1 + fast_mod (p, n/2 + 1) + fast_mod (p, n/2) % MOD;} void orz () {scanf ("% LLD", &, & B); gaoji (a); LL ans = 1; for (INT I = 1; I <= FN; ++ I) {ans = ans * (sum (factors [I] [0], B * factors [I] [1]) % MOD;} printf ("% LLD \ n ", ans) ;}int main () {orz (); Return 0 ;}

Code Jun

Poj 1845-sumdiv (number theory + unique decomposition theorem + Rapid power modulo)