Happy 2004

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1102 Accepted Submission(s): 791


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

簡單的約數求和。2014^x,由素數基本定理，將2014分解，2014=2^2 *  3 * 167，則2014^x=2^(2*x) * 3^x  *   167^x,由約數和公式（後附）得2014^x的約數和為：[2^(2*x+1)-1]/(2-1)  * [3^(x+1)-1]/(3-1)  *  [167^(x+1)-1]/(167-1) = r/332，其中r=[2^(2*x+1)-1] * [3^(x+1)-1] *  [167^(x+1)-1]，答案就是（r/332）%29，在取模運算裡，除以一個數等於乘以這個數的逆元。而332關於模29的逆元是9，故答案就是（r*9）% 29.
在這裡還學到的知識：費馬小定理，對任意的素數p，任意整數a，有a^(p-1) mod p =1。 在本題裡29是素數，那麼對於a=2014就符合費馬小定理，由費馬小定理，a^28 mod 29 =1,則a^x mod 29 = a^(x mod 28) mod 29.   所以在代碼中處理時，先把x對28取余。
補：約數和公式，有素數基本定理，n=p1^a1 * p2^a2 ....... * pk^ak，約數和為：(p1^(a1+1)-1)/(p1-1) * (p2^(a2+1)-1)/(p2-1) .... * (pk^(ak+1)-1)/(pk-1)

#include  #include  #include  #include  using namespace std; typedef __int64 ll; const int MOD=29; quick_mod(int a,int b){ //a^b%MOD int res=1; a%=MOD; while(b){ if(b&1) res=(res*a)%MOD; b/=2; a=(a*a)%MOD; } return res; } int main() { int i,j,n,res,k; while(scanf(%d,&n),n){ int x1,x2,x3; x1=(2*n+1)%(MOD-1); x1=quick_mod(2,x1)-1; x2=x3=(n+1)%(MOD-1); x2=quick_mod(3,x2)-1; x3=quick_mod(167,x3)-1; res=(x1*x2*x3*9)%MOD; printf(%d ,res); } return 0; } 

補充逆元的求法：a*b mod m =1,則稱a為b的逆元，a，b互為逆元，a模m存在逆元的充要條件是：a和m互素。（這有線性同余方程的知識可證）。將a*b mod m =1變形：a*b - m*y =1,要求a的逆元的話就是求b，即線性同余方程的解，對a和m構造同余方程，然後求解即可。（用到了擴展歐幾裡得算法Exgcd）

void Exgcd(int a,int b,int &d,int &x,int &y){
	if(b==0){
		x=1;y=0;d=a;return ;
	}
	Exgcd(b,a%b,d,y,x);
	y-=(a/b)*x;
}

//求模n意義下a的逆元，如果不存在則返回-1
int inv(int a,int n){
	int d,x,y;
	Exgcd(a,n,d,x,y);
	return d==1?(x+n)%n:-1;
}

所以本題中332在模29意義下的逆元可以通過 inv(332,29)求得。