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.
讀不懂題意就是傻逼啊！！！！！！！
這個(gè)題目是要求每個(gè)因子的
因子的個(gè)數(shù)然后再立方和啊啊啊啊
8的因子有1 2 4 8，它們的因子數(shù)有1 2 3 4啊，立方和為1+8+27+64=100啊。
轉(zhuǎn)化為算術(shù)基本定理：
N=A^B
求N的每個(gè)因子的因子數(shù)：
任何一個(gè)大于1的數(shù)可以分解成 N=a1^p1*a2^p2*a3^p3*...*an^pn, N的約數(shù)總數(shù)為（p1+1)*(p2+1)*...*(pn+1),
（0,1,...,p1)(0,1,...,p2)...(0,1,...,pn)
不難發(fā)現(xiàn)（1^3+2^3+...+(p1+1)^3) (1^3+2^3+...+(p2+1)^3)...(1^3+2^3+...+(pn+1)^3)即為所求。
#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;
}