hdu 1452 Happy 2004 膜拜这推导过程

Happy 2004

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 920    Accepted Submission(s): 648

Problem Description

Consider a positive integer X,and let S be the sum of all positive integer divisors of 2004^X. Your job is to determine S modulo 29 (the rest of the division of S by 29).

Take X = 1 for an example. The positive integer divisors of 2004^1 are 1, 2, 3, 4, 6, 12, 167, 334, 501, 668, 1002 and 2004. Therefore S = 4704 and S modulo 29 is equal to 6.

 

 

Input

The input consists of several test cases. Each test case contains a line with the integer X (1 <= X <= 10000000). 

A test case of X = 0 indicates the end of input, and should not be processed.

 

 

Output

For each test case, in a separate line, please output the result of S modulo 29.

 

 

Sample Input

1

10000

0

 

 

Sample Output

6

10

 

Source

ACM暑期集训队练习赛（六）

 

 1 /*

 2   性质1 ：

 3   如果 gcd(a,b)=1 则 S(a*b)= S(a)*S(b)

 4   2004^X=4^X * 3^X *167^X

 5   S(2004^X)=S(2^(2X)) * S(3^X) * S(167^X)

 6   性质2 ：如果 p 是素数 则 S(p^X)=1+p+p^2+...+p^X = (p^(X+1)-1)/(p-1)

 7   (2^(2X+1)-1) * (3^(X+1)-1)/2 * (167^(X+1)-1)/166

 8   167%29 = 22

 9   S(2004^X)=(2^(2X+1)-1) * (3^(X+1)-1)/2 * (22^(X+1)-1)/166

10   性质3 ：(a*b)/c %M= a%M * b%M * inv(c)

11   其中inv(c)即满足 (c*inv(c))%M=1的最小整数,这里M=29

12   则inv(1)=1，inv(2)=15，inv(21)=18

13   有上得：

14   S(2004^X)=(2^(2X+1)-1) * (3^(X+1)-1)/2 * (22^(X+1)-1)/21

15   =(2^(2X+1)-1) * (3^(X+1)-1)*15 * (22^(X+1)-1)*18

16 */

17 

18 #include<iostream>

19 #include<stdio.h>

20 #include<cstring>

21 #include<cstdlib>

22 using namespace std;

23 typedef __int64 LL;

24 

25 const LL p = 29;

26 

27 LL pow_mod(LL a,LL n)

28 {

29     LL ans=1;

30     while(n)

31     {

32         if(n&1) ans=(ans*a)%p;

33         n=n>>1;

34         a=(a*a)%p;

35     }

36     ans=ans-1;

37     if(ans<0) ans=ans+p;

38     return ans;

39 }

40 void solve(LL x)

41 {

42     LL sum=1;

43     sum=(sum*pow_mod(3,x+1)*15)%p;

44     sum=(sum*pow_mod(2,2*x+1))%p;

45     sum=(sum*pow_mod(22,x+1)*18)%p;

46     printf("%I64d\n",sum);

47 }

48 int main()

49 {

50     LL x;

51     while(scanf("%I64d",&x)>0)

52     {

53         if(x==0)break;

54         solve(x);

55     }

56     return 0;

57 }