POJ 1845Sumdiv（数论）

Sumdiv

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 20041		Accepted: 5060

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

15

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).

Source

膜拜大神 http://blog.csdn.net/rowanhaoa/article/details/8591077

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>

using namespace std;

long long p[100000];
long long k[100000];

long long pows(long long n, long long m)//快速幂
{
long long t = 1;
while(m)
{
if(m%2 != 0)
{
t = (t*n)%9901;
m--;
}
n = (n*n)%9901;
m /= 2;
}
//cout<<t<<endl;
return t;
}

long long mou(long long x,long long y)
{
if(y==0)
return 1;
if(y%2==0)
return (((mou(x,y/2-1)%9901)*((1+pows(x,y/2+1))%9901))%9901+pows(x,y/2)%9901)%9901;
if(y%2!=0)
return (mou(x,y/2)%9901)*((1+pows(x,y/2+1))%9901)%9901;
}
int main()
{
long long m, n, i;
while(~scanf("%lld %lld", &n, &m))
{
int j = 0;
for(i = 2; i <= (int)sqrt(n*1.0); i += 2)
{
if(n%i==0)
{
p[j] = i;
k[j] = 0;
while(n%i==0)
{
k[j]++;
n /= i;
}
}
j++;
if(i == 2)
{
i--;
}
if(n==1)
{
break;
}
}
if(n != 1)
{
p[j] = n;
k[j++] = 1;
}
//以上是分解质因数。
for(i = 0; i < j; i++)
{
//printf("%d %d\n", p[i], k[i]);
k[i] *= m;
}
long long sum = 1;
for(i = 0; i < j; i++)
{
sum = sum * mou(p[i], k[i])%9901;
}
printf("%lld\n", sum);
}
return 0;
}