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  • Happy 2004(快速幂+乘法逆元)

    Happy 2004

    问题描述 :

    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.

    输入:

    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.

    输出:

    For each test case, in a separate line, please output the result of S modulo 29.

    样例输入:

    1
    10000
    0

    样例输出:

    6
    10

    设S(x)表示x的因子和。则题目求为:S(2004^X)mod 29
    因子和S是积性函数,即满足性质1。

    性质1 :如果 gcd(a,b)=1  则 S(a*b)= S(a)*S(b)
    2004^X=4^X * 3^X *167^X
    S(2004^X)=S(2^(2X)) * S(3^X) * S(167^X)

    性质2 :如果 p 是素数 则 S(p^X)=1+p+p^2+…+p^X = (p^(X+1)-1)/(p-1)
    因此:S(2004^X)=(2^(2X+1)-1) * (3^(X+1)-1)/2 * (167^(X+1)-1)/166
    167%29 == 22
    S(2004^X)=(2^(2X+1)-1) * (3^(X+1)-1)/2 * (22^(X+1)-1)/21

    性质3 :(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,这里M=29
    则inv(1)=1,inv(2)=15,inv(22)=15

    有上得:
    S(2004^X)=(2^(2X+1)-1) * (3^(X+1)-1)/2 * (22^(X+1)-1)/21
    =(2^(2X+1)-1) * (3^(X+1)-1)*15 * (22^(X+1)-1)*18

     1 #include <cstdio>
     2 #include <iostream>
     3 #include <algorithm>
     4 #include <cstring>
     5 #include <cmath>
     6 #include <queue>
     7 #include <map>
     8 #include <set>
     9 using namespace std;
    10 #define LL long long
    11 const int INF=0x3f3f3f3f;
    12 const double eps=1e-5;
    13 int p=29;
    14 LL pow_mod(LL x,LL n)
    15 {
    16     LL res=1;
    17     while(n>0)
    18     {
    19         if(n&1)    res=res*x%p;
    20         x=x*x%p;
    21         n>>=1;
    22     }
    23     return res;
    24 }
    25 int main()
    26 {
    27     LL x,i;
    28     while(cin>>x&&x)
    29     {
    30         int a=pow_mod(2,2*x+1);
    31         int b=pow_mod(3,x+1);
    32         int c=pow_mod(22,x+1);
    33         int s=((a-1)*(b-1)*(c-1)*15*18)%29;
    34         cout<<s<<endl;
    35      }
    36 
    37 }
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  • 原文地址:https://www.cnblogs.com/a1225234/p/4890051.html
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