Poj1845:sumdiv (factor and + inverse + mass factor decomposition) good problem

Title Link: http://poj.org/problem?id=1845

Definition: The K value satisfying a*k≡1 (mod p) is a multiplicative inverse of p. Why do we have to multiply the inverse element? When we ask for (A/b) mod P's value, and a is large, and cannot directly obtain a A/b value, we will use the multiplication inverse. We can use the B to multiply the inverse k of p, multiply a by the K-mode p, ie (a*k) mod p. The result is equivalent to (A/b) mod p. Problem analysis: Let a^b factor and MODK, because it is large number can not directly seek, because the factor and function is multiplicative function, so first of all the quality factor decomposition, intoN=p1^a1*p2^a2*p3^a3****ps^as, then

S (n) =[(p1^a1+1-1)/(P1-1)]*[(p2^a2+1-1)/(p2-1)]*[(p3^a3+1-1)/(p3-1)]***[(ps^as+1-1)/(Ps-1)]; (Factor and)

And because S (n)%mod equals each part of the modulo, so can be gradually solved, such as (p1^a1+1-1)/(p1-1)%mod, here to use the division to take the model, so to use the concept of multiplicative inverse,

i.e. ( A/b)%p= (a *b^ ( -1)%p) , and because (a^b)% P = ((a% p) ^b)% P ,

So (p1^a1+1-1)/(P1-1)%mod== (((p1%mod) ^a1+1-1)%mod* (p1-1) ^-1)%mod;

Of course there is the premise of inverse element is gcd (a,p) ==1;

#include <iostream>#include<stdio.h>#include<string.h>#include<algorithm>#include<math.h>#defineN 500010#defineMoD 9901typedef __int64 LL;using namespacestd;ll a,b,x,y;ll ans[n],num[n],top;ll Pow (ll x,ll k) {ll T=1;  while(k) {if(k&1) t= ((t%mod) * (x%mod))%MoD; K>>=1; X= ((x%mod) * (x%mod))%MoD; }    returnt;}voidExtend (__int64 a,__int64 b,__int64 &x1, __int64 &y1) {    if(b==0) {x1=1; Y1=0; return ; } Extend (B,a%b,x1,y1); ll T=X1; X1=Y1; Y1=t-(A/b) *Y1; return ;}voidsolve () {ll sum=1, a,xx;  for(intI=0; i<top; i++)    {        if(ans[i]%mod==0)Continue;//Key two judgments, related to seeking inverse element. If ans[i]%mod=0, then there is a hierarchical formula can be seen, the original is less than 0, so can only use the original, the result is 1if(ans[i]%mod==1)//i.e. mod|                                  (ans[i]-1), because ans[i]>=2, so ans[i] can not be equal to 1, this is gcd (ans[i]-1,mod) ==mod, there is no inverse, can not be used to expand Euclid to seek the inverse of the yuan {            At this time for (1+ans[i]^1+ans[i]^2+.....+ans[i]^num[i])%mod= (num[i]+1)%mod; Sum= (sum* (num[i]+1))%MoD; Continue; } A=pow (ans[i],num[i]+1); A= (A-1)%MoD; Extend (Ans[i]-1, mod,x,y);//because Ans[i] is a prime number, ans[i]-1 is even, so ans[i]-1 and 9901 coprimexx= (x%mod+mod)%MoD; A= ((a%mod) * (xx%mod))%MoD; Sum= (sum*a)%MoD; } printf ("%i64d\n", sum);}intMain () { while(SCANF ("%i64d%i64d", &a,&b)! =EOF) {        if(a==0) {printf ("0\n"); Continue; }        Else if(a==1|| b==0) {printf ("1\n"); Continue; } ll T=A; Top=0; memset (num,0,sizeof(num));  for(intI=2; i*i<=a; i++)        {            if(t%i==0) {Num[top]++; Ans[top]=i; T/=i;  while(t%i==0) {Num[top]++; T/=i; } Top++; }        }        if(t>1) {Num[top]++; Ans[top++]=T; }         for(intI=0; i<top; i++) {Num[i]*=b;    } solve (); }    return 0;}

Poj1845:sumdiv (factor and + inverse + mass factor decomposition) good question