HDU--1852（数论基础 积性函数）
HDU–1852
Beijing 2008
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/65535 K (Java/Others)
Problem Description
As we all know, the next Olympic Games will be held in Beijing in 2008. So the year 2008 seems a little special somehow. You are looking forward to it, too, aren’t you? Unfortunately there still are months to go. Take it easy. Luckily you meet me. I have a problem for you to solve. Enjoy your time.
Now given a positive integer N, get the sum S of all positive integer divisors of 2008N. Oh no, the result may be much larger than you can think. But it is OK to determine the rest of the division of S by K. The result is kept as M.
Pay attention! M is not the answer we want. If you can get 2008M, that will be wonderful. If it is larger than K, leave it modulo K to the output. See the example for N = 1，K = 10000: The positive integer divisors of 20081 are 1、2、4、8、251、502、1004、2008，S = 3780, M = 3780, 2008M % K = 5776.
Input
The input consists of several test cases. Each test case contains a line with two integers N and K (1 ≤ N ≤ 10000000, 500 ≤ K ≤ 10000). N = K = 0 ends the input file and should not be processed.
Output
For each test case, in a separate line, please output the result.
Sample Input
1 10000 0 0
Sample Output
5776
题意
：给你n和k，2008的n次方对k取余为m，求2008的m次方对k取余
**
a^n，先把a拆成所有的素数乘积，如2004=2*2*3*167,2008=2*2*2*251.
1: s(2004^n)=s(2^(2*n))*s(3^n)*s(167^n)
2: 对于一个素数p，则s(p^n)=1+p+p^2+p^3+...+p^n=(p^(n+1)-1)/(p-1) 等比数列求和公式。
**
**
这里用到了一个公式
x/d%k==x%(d*k)/d
**
快速幂模板：
```
#define LL long long
LL pow_mod(LL a,LL n,int k)
{
LL ans=1;
while(n)
{
if(n&1) ans=ans*a%k;
a=a*a%k;
n>>=1;
}
return ans;
}
```
通过上面的知识我们将2008^n分解成如下
s(2008^n)=s(2^(3*n))*s(251^(1*n))
next
s(2008^n)=(2^(3n+1)-1)/(2-1)*(251^(n+1)-1)/(251-1)
next
s(2008^n)%k=((2^(3n+1)-1)%((251-1)*k))*((251^(n+1)-1)%((251-1)*k))/(251-1) //据x/d%k==x%(d*k)/d得来
## 完整代码：
```
#include
#include
#include
#include
#include
#include
using namespace std;
#define LL long long
LL pow_mod(LL a,LL n,int k)
{
LL ans=1;
while(n)
{
if(n&1) ans=ans*a%k;
a=a*a%k;
n>>=1;
}
return ans;
}
int main()
{
int n,k;
while(scanf("%d %d",&n,&k)&&(n+k))
{
LL ans=1;
LL temp=pow_mod(2,3*n+1,250*k)-1;
LL temp1=pow_mod(251,n+1,250*k)-1;
ans=temp*temp1%(250*k)/250;
ans=pow_mod(2008,ans,k);
printf("%lld\n",ans);
}
return 0;
}
```