Happy 2004 （因数和公式+逆元）

最新推荐文章于 2021-07-23 16:35:28 发布

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本文探讨了如何求解特定正整数次幂的所有正因数之和，并对其进行了模运算简化。通过数学公式和C++代码实现，解决了计算2004的X次幂的所有正因数之和并对29取模的问题。

Happy 2004

 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

这道题是Beijing 2008这道题的基础版，这道题直接套用因数和公式即可
$2004 = 2^2 \cdot 3 \cdot 167$

$2004^x = 2^{2x} \cdot 3^x \cdot 167^x$

$f(2004^x) = \frac{2^{2x+1}-1}{1}\cdot \frac{3^{x+1}-1}{2} \cdot \frac{167^{x+1}-1}{166}$

因为取模29,29是素数，并且并且和2,166互质，因此均可求逆元

code:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 29;
const int maxn = 1e5+10;
int p[] = {2,3,167};
int a[] = {2,1,1};
int q_pow(int a,int b){
    int ans = 1;
    while(b){
        if(b & 1)
            ans = ans * a % mod;
        b >>= 1;
        a = a * a % mod;
    }
    return ans % mod;
}
int inv(int n){
    return q_pow(n,mod-2) % mod;
}
int main(){
    int x;
    while(~scanf("%d",&x) && x){
        int ans = 1;
        ans = (((q_pow(2,2*x+1) - 1) * ((q_pow(3,x+1) - 1) * inv(2) % mod)) % mod) * ((q_pow(167,x+1) - 1) * inv(166) % mod) % mod;
        printf("%d\n",ans);
    }
    return 0;
}