Happy 2004 hdu1452

Posted 2020-10-30 行远山

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2021    Accepted Submission(s): 1474

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

 

数论题，推理挺麻烦的：

题目的大意是说求2004^n的全部因子之和。

根据唯一分解定理2004=(2^2)*3*167,则2004^n=(2^2n)*(3^n)*(167^n);

又有结论：一个因数的因子和是一个积性函数。设f(x)为x的因子和，则f(ab)=f(a)*f(b);

则f(2004^n)=f(2^2n)*f(3^n)*f(167^n).

继续上结论：如果一个数是素数，那么f(a^n)=1+a+a^2+a^3+.......a^n=(a^(n+1)-1)/(a-1);

考虑到 167%29=22；

则f(2004^n)=(2^(2n+1)-1) * (3^(n+1)-1)/2 * (22^(n+1)-1)/21;

接着上逆元： (a*b/c)%mod=a%mod*b%mod*inv(c);

其中inv(c)表示(c*inv(c))%mod=1的最小整数. mod=29，则inv(1)=1;

inv(2)=15;inv(21)=18;

则原式=(2^(2n+1)-1)*(3^(n+1)-1)%mod*inv(2)* (22^(n+1)-1)*inv(21)

15*18%29=9---------------则原式(2^(2n+1)-1) * (3^(n+1)-1)* (22^(n+1)-1)*9%29;

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int ans=0;
const int mod=29;
ll quick_mod(ll k,ll n)
{
    ll res=1;
    while(n>0)
    {
        if(n&1)
        {
            res=(res*k)%mod;
        }
        k=k*k%mod;
        n>>=1;
    }
    res--;
    if(res<0) res+=29;
    return res;
}
int main()
{
    ll n;
    while(~scanf("%lld",&n)&&n)
    {
        ans=quick_mod(2,2*n+1)*quick_mod(3,n+1)*quick_mod(22,n+1)*9%mod;
        printf("%d\n",ans);
    }
    return 0;
}