hdu 2421

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Deciphering Password

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1550    Accepted Submission(s): 374

Problem 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 a₀, a₁, a₂, …, 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, a₀ = 1, a₁ = 2, a₂ = 4, a₃ = 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 21 14 7

 

Sample Output

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

 

Source

2008 Asia Hangzhou Regional Contest Online

代码：

#include<stdio.h>#include<iostream>#include<cstring>#define MAXN 1010using namespace std;int prim[MAXN],p[200];void init() {    int i, j, k = 0;    memset(prim,1,sizeof(prim));    for(i=2;i<35;++i)    {        if(prim[i])            for(j=i;j*i<MAXN;++j)                prim[i*j]=0;    }    for(i=2;i<MAXN;++i)        if(prim[i])            p[k++] = i; } int main() {     int m,n,sum=1,ans,t=1,tmp,i,j;     init();    while(~scanf("%d%d",&m,&n))     {        sum=1;      for(i=0;i<169;i++)       {            tmp=0;         while(m%p[i]==0&&m!=1)           {               m/=p[i];               tmp++;           }         tmp=(tmp*n%10007+1)%10007;         tmp=tmp*(tmp+1)/2%10007;         sum=sum*(tmp*tmp%10007)%10007;        if(m==1) break;       }       if(m!=1)         {             tmp=(n+1)%10007;             tmp=tmp*(tmp+1)/2%10007;             sum=sum*(tmp*tmp%10007)%10007;         }      printf("Case %d: %d\n",t++,sum);     }     return 0; }