POJ 1845 Sumdiv (factor decomposition + fast Power + two-point summation)

Test instructions: Give you A, B, let a^b all the factors and modulo 9901

Idea: A can be split into the product of the element factor: a = p1^x1 * p2^x2 *...* PN^XN

Then a^b = p1^ (b*x1) * p2^ (B*X2) *...* pn^ (B*XN)

So a^b all the vegetarian factors and is

(p1^0 + p1^1 + p1^2 + ... + p1^ (b*x1)) * (p2^0 + p2^1 + ... + p2^ (b*x2)) * ... * (pn^0 + pn^1 + ... + pn^ (B*XN))

It can be seen that each parenthesis is geometric series, but do not use the geometric series formula, because there is division (at first I use division, then the inverse of the mold, WA to cool dead), because not necessarily meet the multiplicative inverse of the conditions required, divisor and modulus may be non-reciprocity (divisor may be modulus of how many times). Since the formula cannot be used, it is necessary to draw on the two points. For example, the following formula sums: A1+a2+a3+a4 = A1+A2+A2 (A1+A2). Through this formula found, as long as the A2 on the line, and then as long as the calculation of a a1+a2, you can dispense with half of the computational capacity. Then the same a1+a2 can continue to score.

Now spread to the general formula. A1+a2+...+an

1) n is even: A1+a2+...+an = a1+a2+ ... +a (N/2) + A (N/2) (A1+a2+...+a (N/2))

2) n is odd: A1+a2+...+an = a1+a2+ ... +a (N/2) + A (N/2) (A1+a2+...+a (N/2)) + an

The introduction of these can be recursive solution.

Note : To find a factor, hit a prime table, only to hit the sqrt (n) on the line, because it is only possible in sqrt (n), if there is a greater than sqrt (n) two factor, the product is naturally greater than n, so only need to sqrt (n) can be. Because even if there is a large prime and a small prime number multiplied, then in the case of the small prime number, only a large number of prime, which will go directly to the bad judgment.

#include <cstdio>#include<iostream>#include<cstring>#include<algorithm>using namespaceStd;typedefLong Longll;Const intMAXN =10010;Constll mod =9901ll;ll A, B;structFactor {ll FAC; ll CNT;} FACTOR[MAXN];inttot;BOOLPRIME[MAXN + -];intPR[MAXN];//Prime Listintpr_cnt;voidInit_prime () {memset (prime,true,sizeof(prime)); prime[0] = prime[1] =false;  for(inti =2; I * I <= MAXN; i++)        if(Prime[i]) for(intj = i + i; J < Maxn; J + =i) prime[j]=false; Pr_cnt=0;  for(inti =2; I <= MAXN; i++)        if(Prime[i]) pr[pr_cnt++] =i;}voidInit ()//Find all the vegetarian factors{tot=0; Memset (Factor,0,sizeof(factor));  for(inti =0; I < pr_cnt && Pr[i] <= A; i++)    {        if(A% pr[i] = =0) {FACTOR[TOT].FAC=Pr[i];  while(A% pr[i] = =0) {factor[tot].cnt++; A/=Pr[i]; } factor[tot].cnt*=B; Tot++; }    }    if(A >1) {FACTOR[TOT].FAC=A; Factor[tot++].cnt =B; }}ll Quickpow (ll A, ll B, ll MoD) {ll ans=1LL;  while(b) {if(B &1) ans = ans * a%MoD; A= A * A%MoD; b>>=1; }    returnAns%MoD;} ll Binary_pow (ll A, ll B, ll MoD)//Calculate the geometric series and{    if(b = =0)return1LL; if(b = =1)returnA; ll ans=0; if(B &1) {ans=Quickpow (A, B, MoD); Ans= (ans + (Quickpow (A, B/2, MoD) + 1LL)% mod * BINARY_POW (A, B/2, MoD))%MoD; }    Elseans= (Quickpow (A, B/2, MoD) + 1LL)% mod * BINARY_POW (A, B/2, MoD)%MoD; returnans;}voidsolve () {if(B = =0) {puts ("1"); return; }    if(A = =0) {puts ("0"); return;    } init (); ll ans=1LL;  for(inti =0; i < tot; i++) {ans= ans * (Binary_pow (FACTOR[I].FAC, factor[i].cnt, MoD) + 1LL)%MoD; } cout<< ans <<Endl;}intMain () {init_prime ();  while(Cin >> A >>B) {solve (); }    return 0;}

POJ 1845 Sumdiv (factor decomposition + fast Power + two-point summation)