Happy 2004
不知道怎么转载就复制过来了
原地址是：http://www.cnblogs.com/372465774y/archive/2012/10/22/2733977.html
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,
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
10000
Sample Output
10
Source
ACM暑期集训队练习赛（六）
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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
",a*b*c%29);
}
return 0;
}