hdu1852

Beijing 2008

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 502 Accepted Submission(s): 172

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 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 分析：

// 这题主要求S
// 结论: S = (251^(n+1)-1) * (2^(3n+1)-1) / 250 
// 是两个等比数列和相乘 
// 
// 推理:
// 2008 = 2^3 * 251 
// 所以 2008^N 有 3N 个 2 和 N 个251 
// 所有仅由2组成的因子有
// 2^0 2^1 2^2 ... 2^(3N)
// 设集合 C = {2^0, 2^1, 2^2 ...,2^(3N)};
// SUM(C) =  2^(3n+1)-1

// 跟251组合产生的因子有
// 251^0 * C
// 251^1 * C
// ...
// 251^N * C

// 所有因子和为:
// S = (251^(n+1)-1))/250 * (2^(3n+1)-1)

// 计算S%K:

// S 很大, 不能保存在普通的数据类型中, 需要直接计算S%K
// 因为S有个分母250, 设 S = X/250
// 则S%K = (X/250)%K = (X%(250*K))/250
// 变成先求余数再除法的形式

程序：

#include"stdio.h"
#include"string.h"
__int64 pow(__int64 a,__int64 k,__int64 m)
{
     __int64 b=1;
     while(k>=1)
     {
          if(k&1)
               b=b*a%m;
          a=a*a%m;
          k=k/2;
     }
     return b;
}
int main()
{
     __int64 n,k,ans,m,p1,p2,s;
     while(scanf("%I64d%I64d",&n,&k),k||n)
     {
          p1=pow(2,3*n+1,k*250)-1;
          p2=(pow(251,n+1,k*250)-1);
          s=(p1*p2)%(250*k);
          m=s/250;
          //printf("%I64d %I64d %I64d\n",p1,p2,m);
          ans=pow(2008,m,k);
          printf("%I64d\n",ans);
     }
}

转载于:https://www.cnblogs.com/mypsq/p/4348253.html

原文链接：https://blog.csdn.net/weixin_30342827/article/details/97441378

本站声明:网站内容来源于网络,如有侵权,请联系我们,我们将及时处理。