Beijing 2008
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 1284 Accepted Submission(s): 541
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
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define clr(a) memset(a,0,sizeof(a))
const int MAXN = 1e5+10;
const int INF = 0x3f3f3f3f;
const int N = 1010;
LL n,k,m;
LL qkm(LL a, LL b, LL mode)
{
LL sum = 1;
while (b) {
if (b & 1) {
sum = (sum * a) % mode;
b--;
}
b /= 2;
a = a * a % mode;
}
return sum;
}
int main(){
while(scanf("%d%d",&n,&k)!=EOF&&(n||k)){
LL a = qkm(2,3 * n + 1,250 * k) - 1;
LL b = qkm(251,n + 1,250 * k) - 1;
m = (a * b) % (250 * k);
m /= 250;
LL ans = qkm(2008,m,k);
printf("%lld\n",ans);
}
return 0;
}