5.16 周结

 

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

 题意就是给一个数x，求2004的x次方的约数之和。

2004有三个素约数，2，3，167.  167mod29=22，可以把167换成22.

因为有 S(p^n)=1+p+p^2+...p^n=(p^(n+1)-1)/(p-1) 所以本题 就可以S(2004^x)=(2^(2*x+1)-1)(3^(x+1)-1)/2*(167^(x+1)-1)/166

因为（1*2*166）与 29 互质，所以不会影响答案，直接用上面的公式做就好了。

#include <iostream>  
using namespace std;  
  
 
int main()  
{  
    int n;  
    while (cin>>n && n!=0)  
    {  
		n%=28;
	int r,f,t,d;  
    f=t=d=1;  
    for (r=0;r<=n;++r)  
    {  
        f=f*(r==n?2:4)%29;   
        t=t*3%29;                
        d=d*22%29;              
    }  

         r=(f-1)*(t-1)*(d-1)%29;    
         int ans= r*9%29;
        cout<<ans<<endl;  
    }  
 
               

    return 0;  
} 

As we all know, the next Olympic Games will be held in Beijing in 2008. So the year 2008 seems a little special somehow. You are looking forward to it, too, aren't you? Unfortunately there still are months to go. Take it easy. Luckily you meet me. I have a problem for you to solve. Enjoy your time.

Now given a positive integer N, get the sum S of all positive integer divisors of 2008N. Oh no, the result may be much larger than you can think. But it is OK to determine the rest of the division of S by K. The result is kept as M.

Pay attention! M is not the answer we want. If you can get 2008M, that will be wonderful. If it is larger than K, leave it modulo K to the output. See the example for N = 1，K = 10000: The positive integer divisors of 20081 are 1、2、4、8、251、502、1004、2008，S = 3780, M = 3780, 2008M % K = 5776. 

 

Input
The input consists of several test cases. Each test case contains a line with two integers N and K (1 ≤ N ≤ 10000000, 500 ≤ K ≤ 10000). N = K = 0 ends the input file and should not be processed.
 

Output
For each test case, in a separate line, please output the result.
 

Sample Input
1 10000 0 0
 

Sample Output
5776
这题是类似的一题，可以除法可以通过求逆元。因为上面题的mod是29，gcd(2*166,29)==1，存在逆元。但是这里的gcd(250,k)不一定满足等于1，也就是说不一定存在逆元。

那么观察我们要求的m，m肯定有250这个分母，所以可以表示成m=x/250。m%k=(x/250)%k转化为(x%(250*k))/250。求变成了先除法再取余了。

#include<iostream>
#define LL long long
using namespace std;
const int N = 100000 + 5;
LL n,k,a,b,ans;
LL chong(LL a,LL b,LL mod){
	LL ans=1;
	while(b){
		if(b%2==1)
		ans=(ans*a)%mod;

		b>>=1;
		a=(a*a)%mod;
	}
	return ans;
}
 
int main(){
	while(cin>>n>>k){
		if(n==0&&k==0){
			break;
		}
		a=chong(2,3*n+1,250*k);
		b=chong(251,n+1,250*k);
		a--,b--;
		ans=(a*b)%(250*k);
		ans/=250;
		ans=((ans%k)+k)%k;
		cout<<chong(2008,ans,k)<<endl;
	}
	return 0;
}

本周做的几道题全是和素数，约数求和有关的。我还是把约数求和的模型给自己再写一遍。

#include <cstdio>
#include <cstring>
#include <iostream>
# typedef long int ll
const int size= 10000;
const int mod = 29;
ll sum(ll p,ll n);
ll power(ll p,ll n);

using namespace std;
int main()
{
	int A,B;
	int p[size];
	int n[size];
	while(cin>>A>>B)
	{
		int i,k=0;
		for(i=2;i*i<A;)
		{
			if(A%i==0)
			 p[k]=i;
			 n[k]=0;
			 while(!(A%i))
			 {
				 n[k]++;
				 A/=i;
			 }
			 k++;
		}
		if(i!=2)
			i+=2;
		else
			i++;
	}
	if(A!=1)
	{
		p[k]=A;
		n[k++]=1;
	}
	int ans=1;
	for(i=0;i<k;i++)
	ans=(ans*(sum(p[i],n[i]*B)%mod))%mod;

	cout<<ans<<endl;

}
ll sum(ll p,ll n){
	if(n==0)
		return 1;
	if(n%2)
		return (sum(p,n/2)*(1+power(p,n/2+1)))%mod;
	else
		return (sum(p,n/2-1)*(1+power(p,n/2+1))+power(p,n/2))%mod;
}
ll power(ll p,ll n)
{
	ll sq=1;
	ll chongle;
	while(n>0)
	{
		if(n%2)
			sq=(sq*p)%mod;
		n/=2;
		p=p*p%mod;
	}
	return sq;
}