Poj 1845 sumdiv (number theory, evaluate all the divisor sums of a ^ B)

Sumdiv

Time limit:1000 ms		Memory limit:30000 K
Total submissions:10071		Accepted:2357

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 in 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 shoshould be output ).

Source

Romania oi 2002 question: Calculate the result after modulo 9901 for the sum of all the approx. s of a ^ B.
According to the unique decomposition theorem, A can be obtained by means of Factorization: A = p1 ^ A1 * P2 ^ A2 * P3 ^ A3 * PN ^.
A ^ B = p1 ^ (A1 * B) * P2 ^ (A2 * B) *... * PN ^ (an * B );
Sum = [1 + p1 + p1 ^ 2 +... + p1 ^ (A1 * B)] * [1 + p2 + p2 ^ 2 +... + p2 ^ (A2 * B)] * [1 + Pn + PN ^ 2 +... + PN ^ (an * B)]. proportional Series 1 + PI ^ 2 + PI ^ 3 +... + PI ^ N can be obtained by binary (that is, the formula to be solved is divided into parts to solve)
If n is an odd number and a total number of even items exist and P is set to 3, (1 + p) + (P ^ 2 + P ^ 3) = (1 + p) + P ^ 2 (1 + p) = (1 + P ^ 2) * (1 + p) 1 + P ^ 2 + P ^ 3 + ........ + P ^ n = (1 + P ^ 2 + .... + P ^ (n/2) * (1 + P ^ (n/2 + 1 ));
If n is an even number, there are a total of odd values. If P is set to 4, (1 + p) + P ^ 2 + (P ^ 3 + P ^ 4) = (1 + p) + P ^ 2 + P ^ 3 (1 + p) = (1 + P ^ 3) * (1 + p) + P ^ 2 1 + P ^ 2 + P ^ 3 + ........ + P ^ n = (1 + P ^ 2 + .... + P ^ (n/2-1) * (1 + P ^ (n/2 + 1 ));

 /*  Poj 1845 sumdiv calculates the sum of all the approx. % 9901 of a ^ B  */  # Include <Stdio. h> # Include <Math. h># Include <Iostream> # Include <Algorithm> # Include < String . H> Using   Namespace  STD;  # Define Mod4 9901 //  **************************************** **  //  Prime Number filtering and number Decomposition  Const   Int Maxn =10000  ;  Int Prime [maxn + 1  ];  Void  Getprime () {memset (prime,  0 , Sizeof  (PRIME ));  For ( Int I = 2 ; I <= maxn; I ++ ){  If (! Prime [I]) prime [++ prime [0 ] = I;  For ( Int J = 1 ; J <= prime [ 0 ] & Prime [J] <= maxn/I; j ++ ) {Prime [prime [J] * I] = 1  ;  If (I % prime [J] = 0 ) Break  ;}}} Long   Long Factor [ 100 ] [ 2  ];  Int  Fatcnt;  Int Getfactors ( Long   Long  X) {fatcnt = 0  ;  Long   Long TMP = X; For ( Int I = 1 ; Prime [I] <= tmp/prime [I]; I ++ ) {Factor [fatcnt] [  1 ] = 0  ;  If (TMP % prime [I] = 0  ) {Factor [fatcnt] [  0 ] = Prime [I];  While (TMP % prime [I] =0  ) {Factor [fatcnt] [  1 ] ++ ; TMP /= Prime [I];} fatcnt ++ ;}}  If (TMP! = 1  ) {Factor [fatcnt] [  0 ] = TMP; factor [fatcnt ++] [ 1 ] = 1 ;}  Return  Fatcnt ;}  //  **************************************** **  Long   Long Pow_m ( Long   Long A, Long   Long N) //  Fast modulo operation  {  Long   Long Res = 1  ;  Long   Long TMP = A % MOD;  While  (N ){  If (N & 1 ) {Res * = TMP; Res % = MOD;} n >>= 1  ; TMP * = TMP; TMP % = MOD ;} Return  Res ;}  Long   Long Sum ( Long   Long P, Long   Long N) //  Calculate 1 + P ^ 2 + ''' + P ^ n  {  If (P = 0 ) Return   0  ; If (N = 0 ) Return   1  ;  If (N & 1 ) //  Odd  {  Return (( 1 + Pow_m (p, N/ 2 + 1 ) % Mod * sum (p, N/ 2 ) % Mod) %MOD ;}  Else   Return (( 1 + Pow_m (p, N/ 2 + 1 ) % Mod * sum (p, N/ 2 - 1 ) + Pow_m (p, N/ 2 ) % Mod) % MOD ;}  Int  Main (){  //  Freopen ("in.txt", "r", stdin ); //  Freopen ("out.txt", "W", stdout );      Int  A, B; getprime ();  While (Scanf ( "  % D  " , & A, & B )! = EOF) {getfactors ();  Long   Long Ans = 1  ;  For ( Int I = 0 ; I <fatcnt; I ++ ) {Ans * = (Sum (factor [I] [ 0 ], B * factor [I] [ 1 ]) % MoD); ans % = MOD;} printf (  "  % I64d \ n  "  , ANS );}  Return   0  ;}