题干：

                                      Happy 2004

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1863    Accepted Submission(s): 1361

 

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

 

 

Source

ACM暑期集训队练习赛（六）

 

题意：求2004^x的所有因子和对29取余的结果。

 

题解：因为所有数都可以被分解成a=p1^c1*p2^c2*...*pk^ck，并有约数和定理：sum=(p1^0+...+p1^c1)*(p2^0+...+p2^c1)*...*(pk^0+...+pk^ck)。每一项可以通过等比数列求和得到。2004=2^2*3*167，所以答案就是2^(2*n+1)*3^(n+1)*167^(n+1)/(2*166)。由于需要取余，所以不能直接除要求出2*166的逆元改为相乘。

 

AC代码：

#include<bits/stdc++.h>
#define debug cout<<"aaa"<<endl
#define mem(a,b) memset(a,b,sizeof(a))
#define LL long long
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define MIN_INT (-2147483647-1)
#define MAX_INT 2147483647
#define MAX_LL 9223372036854775807i64
#define MIN_LL (-9223372036854775807i64-1)
using namespace std;
 
const int N = 100000 + 5;
const int mod = 29;
 
int e_gcd(int a,int b,int &x,int &y) {
	if(b==0) {
		x=1,y=0;return a;
	}
	int q=e_gcd(b,a%b,y,x);
	y-=a/b*x;
	return q;
	
}
 
int quick(int a,int b){
	int ans=1;
	while(b){
		if(b&1){
			ans=(ans*a)%mod;
		}
		b>>=1;
		a=(a*a)%mod;
	}
	return ans;
}
 
int main(){
	int n,x,y,a,b,c,ans;
	exgcd(166*2,29,x,y);
	while(~scanf("%d",&n)&&n){
		a=(quick(2,2*n+1)-1)%mod;
		b=(quick(3,n+1)-1)%mod;
		c=(quick(167,n+1)-1)%mod;
		ans=((a*b*c*x)%mod+mod)%mod;
		printf("%d\n",ans);
	}
	return 0;
}