Poj 1845 sumdiv quick power + cool + multiplication inverse element

Given a, B, calculate the sum of all the factors of a ^ B, and the modulus is 9901.

Question:

1: factorization prime factor

     A = p1 ^ A1 * P2 ^ A2 * P3 ^ A3 *...*Pn ^.
 HenceA ^ B = p1 ^ (A1 * B) * P2 ^ (A2 * B) *... * PN ^ (an * B );

2: sum of all the approx. values of a ^ B:

     Sum = [1 + p1 + p1 ^ 2 +... + p1 ^ (A1 * B)] * [1 + p2 + p2 ^ 2 +... + p2 ^ (A2 * B)] *... * [1 + Pn + PN ^ 2 +... + PN ^ (an * B)].

 For example200 = 2 ^ 3*5 ^ 2:Sum (200) = [1 + 2 + 4 + 8] * [1 + 5 + 25].

3) about mod inverse element (If AX = 1 mod m, it is called a multiplication inverse element of Mod f is X. It can also be expressed as ax limit 1 (mod m ).) However, gcd (a, m) = 1 is required for the reverse element. Therefore, we can directly calculate the power by binary + fast.

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;#define lint __int64#define MAXN 100000#define M 9901struct Factor { lint base, exp; };Factor f[MAXN]; lint fn;lint p[MAXN], a[MAXN], pn;void prime (){    int i, j; pn = 0;    memset(a,0,sizeof(a));    for ( i = 2; i < MAXN; i++ )    {        if ( !a[i] ) p[pn++] = i;        for ( j = 0; j < pn && i * p[j] < MAXN && (p[j] <= a[i] || a[i] == 0); j++ )            a[i*p[j]] = p[j];    }}void Factorization ( int num ){    fn = 0;    for ( int i = 0; i < pn && p[i] <= num; i++ )    {        if ( num % p[i] ) continue;        f[++fn].base = p[i];        f[fn].exp = 0;        while ( num % p[i] == 0 )        {            f[fn].exp++;            num /= p[i];        }    }    if ( num != 1 )    {        f[++fn].base = num;        f[fn].exp = 1;    }}int mod_exp ( int a, lint b ){    int ret = 1;    a = a % M;    while ( b >= 1 )    {        if ( b & 1 ) ret = ret * a % M;        a = a * a % M;        b >>= 1;    }    return ret;}int sum_exp ( int p, lint exp ){    if ( exp == 0 ) return 1;    lint tmp1, tmp2, mid = exp / 2;    if ( exp & 1 )    {        tmp1 = sum_exp (p, mid);        tmp2 = mod_exp (p, mid + 1);        return (tmp1+tmp2*tmp1) % M;    }    else    {        tmp1 = sum_exp (p, mid-1);        tmp2 = mod_exp (p, mid);        return (tmp1 + tmp2 + p*tmp2*tmp1) % M;    }}int main(){    prime();    int A, B, ret = 1;    scanf("%d%d",&A,&B);    if ( A == 0 ) {printf("0\n");return 0;}    if ( B == 0 || A == 1 ) {printf("1\n"); return 0;}    Factorization ( A );    for ( int i = 1; i <= fn; i++ )        ret = ret * sum_exp(f[i].base, f[i].exp*B) % M;    printf("%d\n",ret);}