HDU-1852-Beijing 2008-一个神奇的公式求逆元

Posted 2022-08-10 ofshk

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

 1 #include<stdio.h>
 2 typedef long long ll;
 3 
 4 ll ksm(ll x,ll n,ll mod)
 5 
 6     ll res=1;
 7     while(n>0)
 8     
 9         if(n&1)
10             res=res*x%mod;
11         x=x*x%mod;
12         n>>=1;
13     
14     return res%mod;
15 
16 
17 
18 //S=2008^x,所有因子和   get the sum S of all positive integer divisors of 2008 N.
19 //M=S%k(已知k)  the rest of the division of S by K. The result is kept as M
20 //求2008^M%k
21 
22 int main()
23 
24     ll n,k;
25     while(~scanf("%lld %lld",&n,&k))
26     
27         if(n==0&&k==0)
28             break;
29         ll k2=ksm(2,3*n+1,k*250);
30         if(k2-1<0)
31             k2=k2-1+k;
32         else
33             k2--;
34 
35         ll k251=ksm(251,n+1,250*k);
36         if(k251-1<0)
37             k251=k251-1+k;
38         else
39             k251--;
40 
41         ll M=k2*(k251)%(250*k)/250;
42      //   ll M=k2*(k251/k)%(250*k);
43   //  ll M=k2*(k251/250)%(250*k); WA，注意一下
44         ll ans=ksm(2008,M,k);
45         printf("%lld\n",ans);
46     
47     return 0;
48