HDU1452 Happy 2004 （因子和）

Happy 2004

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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）

 

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首先2004 = 2*2*3*167

然后，利用因子和是积性函数的性质（蒟蒻准备专门写一篇持续更新的有关积性函数证明及学习的文章）：

σ(2004^x) =  σ(2^2x) * σ (3^x) * σ（167^x) Mod 29

∵167≡22（Mod29）

故σ(2004^x) =  σ(2^2x) * σ (3^x) * σ（22^x) Mod 29

                   =  [2^(2x+1)-1][3^(x+1)-1]/2*[22^(x+1)-1]/21

因为2的模29乘法逆元为15 ，22的模29乘法逆元为18

故σ(2004^x)  =  [2^(2x+1)-1][3^(x+1)-1]*15*[22^(x+1)-1]*18

即可用快速幂求解

#include<set>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int Mod = 29;
#define Rep(i,n) for(int i=1;i<=n;i++)
#define For(i,l,r) for(int i=l;i<=r;i++)

int ans,x;

int quickpow(int m,int n){
    int ans=1;
    while(n){
        if(n&1) ans=(ans*m)%Mod;
        m=(m*m)%Mod;
        n>>=1; 
    }
    return ans%Mod;
}

int main(){
    while(scanf("%d",&x),x){
        int ans=0;
        ans=(quickpow(2,2*x+1)-1)%Mod;
        ans=ans%Mod*(quickpow(3,x+1)-1)*15%Mod;
        ans=ans%Mod*(quickpow(22,x+1)-1)*18%Mod;
        printf("%d
",ans%Mod);
    }
    return 0;
}