HDU - 1852 - Beijing 2008（因子和）

最新推荐文章于 2019-12-20 11:11:31 发布

CNG Steve·Curcy

最新推荐文章于 2019-12-20 11:11:31 发布

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分类专栏：数论文章标签：因子和

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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.

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
题目链接
 参考题解
 求因子和1
求因子和2
题意：求2008ⁿ的所有因子和s对k取余的值作为m，然后求2008^m对k取余。
使用公式x/y%k=x%(ky)/y，求因子和的公式参考以上两个题解。2008=2^3*251，则：2008ⁿ=2³ⁿ 251ⁿ所以因子和为：s=(2^ (3ⁿ+1)-1)(251^(n+1)-1)/250
那么m=s%k=(2^ (3ⁿ+1)-1)(251^(n+1)-1)%(250*k)/250,其实就是一个等比数列求和的公式，套用就可以了。
AC代码：

#include <cstdio>
#include <cstring>
#define ll long long
using namespace std;

ll quick_pow(ll base, ll index, ll mod)
{
    ll ans = 1;
    while(index)
    {
        if(index & 1)   ans = (ans * base) % mod;
        base = (base * base) % mod;
        index >>= 1;
    }
    return ans % mod;
}

int main()
{
    ll n, k;
    while(scanf("%lld%lld", &n, &k), n + k)
    {
        ll factor_com = (quick_pow(2, 3 * n + 1, 250 * k) - 1) * (quick_pow(251, n + 1, 250 * k) - 1) % (250 * k);
        factor_com = (factor_com % (250 * k) + 250 * k) % (250 * k);
        factor_com /= 250;
        printf("%lld\n", (quick_pow(2008, factor_com, k) + k) % k);
    }
    return 0;
}