题目大意:
给出n,k,s为2008的n次幂的所有因子和,m为s%k,求2008的m次幂%k
分析:
2008 = 2^3 * 251;
故 2008 ^ n = 2 ^ 3n * 251 ^ n;
设集合C= {2^0 , 2^1 , …… , 2^3n};
sum(C) = 2^(3n+1) - 1;
集合W = {251^0 , 251^1 , …… ,251^n};
sum(W) = (251^(n+1) - 1 )/250;
则所有因子和为: S = sum(C) * sum (W);
因为S太,故直接取模;
S = sum(C) * sum(W) % K;
因为 sum(W)存在除法 , 所以需要对K* 250 取模;
故 S = sum(C) * sum(W) %( K*250) /250;
N^M % mod 的求法:
long long Pow(long long n ,long long m, int mod)
{
long long res = 1;
while(m >= 1)
{
if(m & 1)
{
res = (res * n ) % mod;
}
n = n * n % mod ;
m >>= 1;
}
return res;
}
#include <iostream> #include <string> #include <cstring> #include <queue> #include <stack> #include <cmath> #include <algorithm> #include <cstdio> using namespace std; #define MOD 3780 long long Pow(long long n ,long long m, int mod) { long long res = 1; while(m >= 1) { if(m & 1) { res = (res * n ) % mod; } n = n * n % mod ; m >>= 1; } return res; } int main() { int N , K; while(cin >> N >> K && ( N || K)) { long long temp1 , temp2; temp1 = Pow(2 , 3 * N + 1 , K * 250) - 1; temp2 = Pow(251 , N + 1 , K * 250) - 1; long long res = (temp1 * temp2) % (250 * K); res /= 250 ; res = Pow(2008 , res , K); cout << res <<endl; } }