HDU 1452 FZU 1053 Happy 2004(逆元函数+因子和函数+大指数取模化简公式)

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


一个数的因子和是一个积性函数

关于积性函数,即F(ab)=F(a)*F(b),在数论里有很多积性函数

来证明一下:

S(x)表示x的因子和。

如果x可以分成a,b(一定为素数),那么S(x)=S(a)*S(b)。

为什么一定要分成素数呢,因为一个素数的因子之后1和它本身,对于a,b 来说,就是1,a,1,b,那么x=a*b,x的因子只有1,a,,b,x这四个数,

(1+a)*(1+b)=1+a+b+a*b。看明白了么,这就是所谓的一个数的因子和是一个积性函数。

 

(a*b)/c %M= a%M * b%M * inv(c)
其中inv(c)即满足 (c*inv(c))%M=1的最小整数

 

a^(b)%c----->a(b%k(c)+k(c))%c  (b>=k(c)) 其中k(c)为c的欧拉值。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>

using namespace std;

typedef long long LL;

LL qpow(int x,int y){
    LL b = 1;
    for(;y;y>>=1){
        if(y&1) b = (b*x)%29;
        x=(x*x)%29;
    }b--;
    return b<0?b+29:b;
}

int main()
{
    int x;
    while(~scanf("%d",&x)&&x){
        LL a = 2*x+1;if(a>28) a%=28,a+=28;
        LL b = x+1;if(b>28) b%=28,b+=28;
        printf("%d\n",qpow(2,a)*qpow(3,b)*qpow(22,b)*9%29);
    }
    return 0;
}

  

posted @ 2015-08-05 17:11  Bug_Clearlove  阅读(403)  评论(0编辑  收藏  举报