Happy 2004
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 920 Accepted Submission(s): 648
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
Sample Output
6
10
Source
/*
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)
(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)/166
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(21)=18
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
*/ #include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef __int64 LL; const LL p = ; LL pow_mod(LL a,LL n)
{
LL ans=;
while(n)
{
if(n&) ans=(ans*a)%p;
n=n>>;
a=(a*a)%p;
}
ans=ans-;
if(ans<) ans=ans+p;
return ans;
}
void solve(LL x)
{
LL sum=;
sum=(sum*pow_mod(,x+)*)%p;
sum=(sum*pow_mod(,*x+))%p;
sum=(sum*pow_mod(,x+)*)%p;
printf("%I64d\n",sum);
}
int main()
{
LL x;
while(scanf("%I64d",&x)>)
{
if(x==)break;
solve(x);
}
return ;
}