[POJ1845] Sumdiv

This problem examines the formula theorems of three number theory:

Integer unique decomposition theorem: a= (P1^K1) * (P2^K2) **...* (PN^KN) (P (i) is prime)

factors and formulas (known as a= (P1^K1) * (P2^K2) **...* (PN^KN)), then the sum of all the factors of a is: sums = (1+P1+P1^2+...P1^K1) * (1+P2+P2^2+...P2^K2) *...* (1+PN+PN^2+...PN^KN)

The same comodule formula:

(a+b)%m= (a%m+b%m)%m

(a*b)%m= (a%m*b%m)%m

The code is as follows:

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <vector>5#include <algorithm>6#include <cmath>7 8 using namespacestd;9 TentypedefLong LongLL; One Const intMoD =9901; AVector<ll>p; -Vector<ll>K; - LL N, a, B; the  - //Fast Power - ll Quickmul (ll X, ll N) { -LL ans =1; +LL T =x; -      while(n) { +         if(N &1) { AAns = (ans * t)%MoD; at         } -t = t%MoD; -N >>=1; -     } -     returnans; - } in  - //Decomposition factorization to voidfactor (LL N) { + p.clear (); - k.clear (); the     intnn = (int) sqrt (n1.0); *      for(inti =2; I <= nn; i+=2) { $         if(n% i = =0) {//decompose N to turn N into p1^k1*p2^k2+...pn^knPanax Notoginseng P.push_back (i); -K.push_back (0); the              while(n% i = =0) { +K.back () + +; An = n/i; the             } +         } -         if(i = =2) { $i--; $         } -         if(n = =1){ -              Break; the         } -     }Wuyi     if(n! =1) {//Special Prime number the p.push_back (n); -K.push_back (1); Wu     } - } About  $ //The same comodule formula - //(a+b)%m= (a%m+b%m)%m - //(a*b)%m= (a%m*b%m)%m - //the 1+pi+pi^2+...+pi^n and the two-part seeking A ll Modu (ll X,ll y) { +     if(Y = =0) { the         return 1;  -     } $     if(Y%2==0) { the         return((Modu (x,y/2-1) * ((%mod) * (1+quickmul (x,y/2+1)) (%mod))%mod+quickmul (x,y/2)%mod)%MoD;  the     } the     if(Y%2!=0) { the         return(Modu (x,y/2) * ((%mod) * (1+quickmul (x,y/2+1))%mod)%MoD;  -     } in } the  the voidsolve () { AboutFactor (a);//decompose A to make a = P1^k1*p2^k2+...pn^kn the      for(inti =0; I < k.size (); i++) { theK[i] = k[i] * b;//so a^b = p1^ (k1*b) *p2^ (k2*b) ... pn^ (kn*b) the     } +LL sum =1; -      for(inti =0; I < k.size (); i++) { thesum = sum * MODU (P[i], k[i])%MoD;Bayi     } theprintf"%i64d\n", sum); the } -  - intMain () { the      while(~SCANF ("%d%d", &a, &b)) { the solve (); the     } the     return 0; -}

[POJ1845] Sumdiv