Deciphering Password

Description

Xiaoming has just come up with a new way for encryption, by calculating the key from a publicly viewable number in the following way: 
Let the public key N = A ^B, where 1 <= A, B <= 1000000, and a ₀, a ₁, a ₂, …, a _k-1 be the factors of N, then the private key M is calculated by summing the cube of number of factors of all ais. For example, if A is 2 and B is 3, then N = A ^B = 8, a ₀ = 1, a ₁ = 2, a ₂ = 4, a ₃ = 8, so the value of M is 1 + 8 + 27 + 64 = 100. 
However, contrary to what Xiaoming believes, this encryption scheme is extremely vulnerable. Can you write a program to prove it? 

 

Input

There are multiple test cases in the input file. Each test case starts with two integers A, and B. (1 <= A, B <= 1000000). Input ends with End-of-File. 
Note: There are about 50000 test cases in the input file. Please optimize your algorithm to ensure that it can finish within the given time limit. 

 

Output

For each test case, output the value of M (mod 10007) in the format as indicated in the sample output. 

 

Sample Input

 2 2
1 1
4 7 

 

Sample Output

 Case 1: 36
Case 2: 1
Case 3: 4393 

 

#include<cctype>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<map>
#define REP(i,a,b) for (i=(a);i<=(b);i++)
#define CLEAR(a) memset((a),0,sizeof((a)))
#pragma comment(linker, "/STACK:1024000000 , 1024000000")

using namespace std;

typedef long long LL;
const double pi = acos(-1.0);
const int maxn = 1010;
const int mod = 10007;

vector<int> vec;
map<int,bool> prime;

bool vis[maxn + 1];
int n, m, kase = 0;
LL ans;

void init();
void solve();
void outp();

void get_prime()
{
    vec.clear();
    prime.clear();
    CLEAR(vis);
    int k = 0;
    for (int i = 2; i < maxn; i++)
    {
        if (!vis[i])
        {
            vec.push_back(i);
            prime[i]=1;
        }
        for (int j = 1; j <= vec.size() && i * vec[j - 1] <= maxn; j++)
        {
            vis[i * vec[j - 1]] = 1;
        }
    }
}


int main()
{
    get_prime();
    while (~scanf("%d%d", &n, &m))
    {
        init();
        solve();
        outp();
    }
    return 0;
}

void solve()
{
    ans = 1;
    for(int k=0;vec[k]*vec[k]<=n&&k<vec.size();k++)
    {
        int e=0;
        while (!(n % vec[k]))
        {
            n /= vec[k];
            e++;
        }
        if (e)
            //ans*=((e*m+1)*(e*m+2)/2)*((e*m+1)*(e*m+2)/2);
        {
            LL tmp = ((LL)(e * m + 1) * (LL)(e * m + 2) / (LL)2) % mod;
            tmp = (tmp * tmp) % mod;
            ans = (ans * tmp) % mod;
        }
    }
    if (n>1)
    {
        LL tmp = ((LL)(m + 1) * (LL)(m + 2) / (LL)2) % mod;
        tmp = (tmp * tmp) % mod;
        ans = (ans * tmp) % mod;
    }
}

void init()
{
}

void outp()
{
    printf("Case %d: %I64d\n", ++kase, ans);
}