HDU-1852-Beijing 2008-一个神奇的公式求逆元

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.

InputThe 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：
好好理解看清楚：
S=2008^x,所有因子和 get the sum S of all positive integer divisors of 2008 N. 
M=S%k(已知k) the rest of the division of S by K. The result is kept as M 
求2008^M%k

 #include<stdio.h>
 typedef long long ll;

 ll ksm(ll x,ll n,ll mod)
 {
     ll res=;
     while(n>)
     {
         if(n&)
             res=res*x%mod;
         x=x*x%mod;
         n>>=;
     }
     return res%mod;
 }

 //S=2008^x,所有因子和   get the sum S of all positive integer divisors of 2008 N.
 //M=S%k(已知k)  the rest of the division of S by K. The result is kept as M
 //求2008^M%k

 int main()
 {
     ll n,k;
     while(~scanf("%lld %lld",&n,&k))
     {
         if(n==&&k==)
             break;
         ll k2=ksm(,*n+,k*);
         if(k2-<)
             k2=k2-+k;
         else
             k2--;

         ll k251=ksm(,n+,*k);
         if(k251-<)
             k251=k251-+k;
         else
             k251--;

         ll M=k2*(k251)%(*k)/;
      //   ll M=k2*(k251/k)%(250*k);
   //  ll M=k2*(k251/250)%(250*k); WA，注意一下
         ll ans=ksm(,M,k);
         printf("%lld\n",ans);
     }
     return ;
 }