poj1845 Sumdiv

Sumdiv

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 23172		Accepted: 5769

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

Romania OI 2002

分析：这道题有点复杂，涉及到约数嘛，肯定是先要分解一下，将原数变成唯一分解定律的表达形式，a = p1^x1*p2^x2*...*pn^xn,a^b = p1^x1b*p2^x2b*...*pn^xnb,那么约数和怎么求呢？如果把a^b的所有约数用唯一分解定律的表达形式写出来，就会发现p1和p2,p3,...,pn的所有的乘积的组合形式乘了一次，因式分解一下，可以得到sum = (1 + p1 + p1^2 + ... + p1^x1b) * (1 + p2 + ...+p2^x2b)*......

括号里的式子是一个等比数列，我们可以套用公式来计算，涉及到取模，就要求逆元，不过这个地方有点小技巧，如果求-1的逆元，答案是-1,但是程序给的是1,也就是说我们最好是求正数的逆元，等比数列求和公式上下同取负号就好了.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>

using namespace std;

const int mod = 9901;
long long cnt[100010],wh[100010],tot,ans = 1;
int a,b;

void exgcd(long long a,long long b,long long &x,long long &y)
{
    if (!b)
    {
        x = 1;
        y = 0;
        return;
    }
    exgcd(b,a % b,x,y);
    long long t = x;
    x = y;
    y = t - (a / b) * y;
    return;
}

long long qpow(long long a,long long b)
{
    long long ans = 1;
    while (b)
    {
        if (b & 1)
        ans = (ans * a) % mod;
        a = (a * a) % mod;
        b >>= 1;
    }
    return ans;
}

int main()
{    
    scanf("%d%d",&a,&b);
    for (int i = 2; i <= sqrt(a); i++)
    {
        if (a % i == 0)
        {
            wh[++tot] = i;
            while (a % i == 0)
            {
            a /= i;
            cnt[i]++;
            }
        }
    }
    if (a)
    {
        wh[++tot] = a;
        cnt[a]++;
    }
    
    for (int i = 1; i <= tot; i++)
    {
        long long x,y;
        exgcd(wh[i] - 1,mod,x,y);
        ans = (ans * (( qpow(wh[i],cnt[wh[i]] * b + 1) - 1)* x) % mod) % mod;
    }
    printf("%lld
",ans);
    
    
    return 0;
}