POJ 1845：Sumdiv 快速幂+逆元

Sumdiv

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 16466		Accepted: 4101

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 S modulo 9901.

Sample Input

2 3

Sample Output

Hint

2^3 = 8.
The natural divisors of 8 are: 1,2,4,8. Their sum is 15.
15 modulo 9901 is 15 (that should be output).

要求的是A^B的所有因子的和之后再mod 9901的值。

因为一个数A能够表示成多个素数的幂相乘的形式。即A=(a1^n1)*(a2^n2)*(a3^n3)...(am^nm)。所以这个题就是要求

(1+a1+a1^2+...a1^n1)*(1+a2+a2^2+...a2^n2)*(1+a3+a3^2+...a3^n2)*...(1+am+am^2+...am^nm) mod 9901。

对于每一个(1+a1+a1^2+...a1^n1) mod 9901

等于 (a1^(n1+1)-1)/(a1-1) mod 9901，这里用到逆元的知识：a/b mod c = (a mod (b*c))/ b

所以就等于(a1^(n1+1)-1)mod (9901*(a1-1)) / (a1-1)。

至于前面的a1^(n1+1)，快速幂。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;
#define M 9901

long long p[100008];
int prime[100008];

void isprime()
{
	int cnt = 0, i, j;
	memset(prime, 0, sizeof(prime));
	
	for (i = 2; i < 100008; i++)
	{
		if (prime[i] == 0)
		{
			p[++cnt] = i;
			for (j = 2 * i; j <100008;j=j+i)
			{
				prime[j] = 1;
			}
		}
	}
}
long long getresult(long long A,long long n,long long k)
{
	long long b = 1;
	while (n > 0)
	{
		if (n & 1)
		{
			b = (b*A)%k;
		}
		n = n >> 1;
		A = (A*A)%k;
	}
	return b;
}
void solve(long long A, long long B)
{
	int i;
	long long ans = 1;
	for (i = 1; p[i] * p[i] <= A; i++)
	{
		if (A%p[i] == 0)
		{
			int num = 0;
			while (A%p[i] == 0)
			{
				num++;
				A = A / p[i];
			}
			long long m = (p[i] - 1) * 9901;
			ans *= (getresult(p[i], num*B + 1, m) + m - 1) / (p[i] - 1);
			ans %= 9901;
		}
	}
	if (A > 1)
	{
		long long m = 9901 * (A - 1);
		ans *= (getresult(A, B + 1, m) + m - 1) / (A - 1);
		ans %= 9901;
	}
	cout << ans << endl;
}

int main()
{
	long long A, B;
	
	isprime();
	
	while (scanf("%lld%lld", &A, &B) != EOF)
	{
		  solve(A, B);
	}
	return 0;
}