Happy 2004
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uDescription
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.
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.
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
/*/ 题意： 求2004^x的所有的因子和然后 mod 29。 一开始没有头绪，后面百度了下，N的因子和是积性函数，所以就开始用积性函数的方法解题。 积性函数 ： 当gcd(a,b)=1时 s(a*b)=s(a)*s(b);
如果p是素数[或素数取X模后的数] s(p^n)=1+p+p^2+...+p^n= (p^(n+1)-1) /(p-1) /*/ /*/ 2004=2*2*3*167 质因子 /*/ /*/ σ(2004^x)=σ(2^2x)*σ(3^x)*σ(167^x) mod 29 积性函数。 /*/ /*/ 167%29 = 22 /*/ /*/ σ(2004^x)=σ(2^2x)*σ(3^x)*σ(22^x) mod 29 /*/ /*/ σ(2^2x)= (2^(2x+1)-1)/(2-1)=(2^(2x+1)-1)/1 /*/ /*/ σ(3^x) = (3^(x+1)-1)/(3-1) =(3^(x+1)-1) /2 /*/ /*/ σ(22^x)=(22^(x+1)-1)/(22-1)=(22^(x+1)-1)/21 /*/ /*/ σ(2004^x)=(2^(2x+1)-1)*(3^(x+1)-1)/2*(22^(x+1)-1)/21 /*/ /*/ 1/2 对于29的逆元x=15 1=2*15 -29*1 /*/ /*/ 1/21对于29的逆元x=18 1=21*18-29*13 /*/ /*/ σ(2004^x)=((2^(2x+1)-1)*(3^(x+1)-1)*15*(22^(x+1)-1)*18)%29 /*/ /*/ 然后开始用快速幂求解。 /*/ /*/ 这个题目的重点在在于这个数学推导过程。 /*/ // AC代码：
#include"cmath"
#include"string"
#include"cstdio"
#include"cstring"
#include"iostream"
#include"algorithm"
using namespace std;
typedef long long LL;
LL mod_pow(LL x,LL y,LL p) {
LL rt=1;
LL t=x;
while(y) {
if(y&1)rt=(rt*t)%p;
t=t*t%p;
y>>=1;
}
return rt;
}
int main() {
LL x,ans;
while(~scanf("%I64d",&x)) {
if(x==0)break;
/*/ σ(2004^x)=((2^(2x+1)-1)*(3^(x+1)-1)*15*(22^(x+1)-1)*18)%29 /*/
ans=1;
ans*=(mod_pow(2,2*x+1,29)-1)%29;
ans*=(mod_pow(3,x+1,29)-1)*15%29;
ans*=(mod_pow(22,x+1,29)-1)*18%29;
printf("%I64d
",ans%29);
}
return 0;
}