hdu 1452 Happy 2004

Happy 2004

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 673    Accepted Submission(s): 481

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）

 

Recommend

lcy

求约数的和
首先 约束和函数是积性函数(就是如果m,n互质，则 f(mn)=f(m)f(n))
  还有约数个数函数也是积性函数 这2个比较好证明 直接带入就可以
S(x)代表x的约数和
S(20)=S(4)*S(5)
如果p是素数 S(p^n)=1+p+p^2+...p^n=(p^(n+1)-1)/(p-1);
所以本题 S(2004^x)=(2^(2*x+1)-1)(3^(x+1)-1)/2*(167^(x+1)-1)/166

而又同余性质  ： 若 a=b(mod m) 则 a^k=b^k (mod m)：
所以 167可以用 22代替，（对29 同余）
对于  a^k*b^h*...%m的题目 直接二进制快速运算
这里还有个 a^k/d % m
这就相当于 a^k*d-1%m
d-1 是 d的模m逆  就是  dd-1=1 mod m  ...1
这样的话               a^k/b=x mod m ...2
                  由1,2根据同余性质 a^k*d-1=x mod m
所以本题就Ok了
#include <iostream>

#include <map>

#include <stdio.h>

#include <math.h>

#include <string.h>

#include <stdlib.h>

#include <algorithm>

using namespace std;

int Mod(int a,int b)

{

    int t;

    for(t=1;b>0;b>>=1,a=(a*a)%29)

     if(b&1) t=(t*a)%29;

    return t;

}

int main()

{

    int n;

    int a,b,c;

    while(scanf("%d",&n),n)

    {

         a=(Mod(2,2*n+1)-1);

         b=(Mod(3,n+1)-1)*15; 15是2的mod 29逆

         c=(Mod(22,n+1)-1)*18;18是 21的mod 29逆

         printf("%d\n",a*b*c%29);

    }



    return 0;

}