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=1;
    while(n>0)
    {
        if(n&1)
            res=res*x%mod;
        x=x*x%mod;
        n>>=1;
    }
    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==0&&k==0)
            break;
        ll k2=ksm(2,3*n+1,k*250);
        if(k2-1<0)
            k2=k2-1+k;
        else
            k2--;

        ll k251=ksm(251,n+1,250*k);
        if(k251-1<0)
            k251=k251-1+k;
        else
            k251--;

        ll M=k2*(k251)%(250*k)/250;
     //   ll M=k2*(k251/k)%(250*k);
  //  ll M=k2*(k251/250)%(250*k); WA，注意一下
        ll ans=ksm(2008,M,k);
        printf("%lld\n",ans);
    }
    return 0;
}