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 2008
N. 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 2008
M, 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, 2008
M % 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
思路：不要用库函数的pow很慢，用快速幂，只要log（n）的时间。
另外，这道题不能用乘法逆元去处理k，因为乘法逆元要求必须互素。
所以我们可以运用这个公式
x/d%k==x%(d*k)/d这样就可以直接得到答案了。注意取模就要全部都对(d*k)取了~
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
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;
}
