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 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.
summing the cube of number of factors of all ais.
summing the cube of number of factors of all ais.
summing the cube of number of factors of all ais.
summing the cube of number of factors of all ais.
summing the cube of number of factors of all ais.
summing the cube of number of factors of all ais.
summing the cube of number of factors of all ais.
summing the cube of number of factors of all ais.
summing the cube of number of factors of all ais.
读不懂题意就是傻逼啊！！！！！！！
这个题目是要求每个因子的
因子的个数然后再立方和啊啊啊啊
8的因子有1 2 4 8，它们的因子数有1 2 3 4啊，立方和为1+8+27+64=100啊。
转化为算术基本定理：
N=A^B
求N的每个因子的因子数：
任何一个大于1的数可以分解成 N=a1^p1*a2^p2*a3^p3*...*an^pn, N的约数总数为（p1+1)*(p2+1)*...*(pn+1),
（0,1,...,p1)(0,1,...,p2)...(0,1,...,pn)
不难发现（1^3+2^3+...+(p1+1)^3) (1^3+2^3+...+(p2+1)^3)...(1^3+2^3+...+(pn+1)^3)即为所求。
#include<stdio.h>
#include<string.h>
#include<math.h>
#define maxn 1000005
int p[1015];
int b[1015];
int tot;
int eular()
{
memset(b,0,sizeof(b));
int i=2;tot=0;
while (i<1010)
{
while (b[i]) i++;
p[tot++]=i;
int j=i;
while (j<1010)
{
b[j]=1;
j+=i;
}
}
tot--;
return 0;
}
int main()
{
long long A,B;
int t=0;
eular();
while (scanf("%I64d%I64d",&A,&B)==2)
{
printf("Case %d: ",++t);
B%=10007;
long long ans=1;
long long t,tt;
int i=0;
while (i<tot && A>1)
{
t=0;
while (A%p[i]==0)
t++,A/=p[i];
tt=(t*B+1)*(t*B+2)/2 % 10007;
tt=tt*tt % 10007;
ans=(ans*tt) % 10007;
i++;
}
if (A>1)
{
tt=(B+1)*(B+2)/2 % 10007;
tt=tt*tt % 10007;
ans=(ans*tt)%10007;
}
printf("%I64d\n",ans);
}
return 0;
}