hdu 1452(因子和+逆元)

Happy 2004

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1393    Accepted Submission(s): 1018

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

n的因子和为s(n)

令g(p, e) = (p^(e+1) - 1) / (p-1)，则s(n) = g(p1, e1) * g(p2, e2) * ... * g(pk, ek)

 

这个题只不过是求n^x的因子和而已，假设 n = p1^e1p2^e2...pn^en 那么 n^x=p1^e1*xp2^e2*x...pn^en*x  由于这题x的范围比较大,所以要用快速取模，又因为这题有除法取模,而模的数又是一个质数，所以要用到逆元.

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <math.h>
#include <queue>
using namespace std;
typedef long long LL;
const int N = ;
LL p[]={,,};
LL e[]={,,};
LL extend_gcd(LL a,LL b,LL &x,LL &y){
    if( b ==  ) {
        x = ;
        y = ;
        return a;
    }
    else{
        LL x1,y1;
        LL d = extend_gcd(b,a%b,x1,y1);
        x = y1;
        y= x1-a/b*y1;
        return d;
    }
}
LL mod_reverse(LL a,LL n)
{
    LL x,y;
    LL d=extend_gcd(a,n,x,y);
    if(d==) return (x%n+n)%n;
    else return -;
}
LL pow_mod(LL a,LL n,LL mod){
    LL ans = ;
    while(n){
        if(n&) ans = ans*a%mod;
        a= a*a%mod;
        n=n>>;
    }
    return ans;
}
LL g(LL p,LL e){
    LL inv = mod_reverse(p-,);
    LL ans = (inv*(pow_mod(p,e+,)-))%;
    return ans;
}
int main()
{
    LL n;
    while(scanf("%lld",&n)!=EOF,n){
        LL m = ;
        LL sum = ;
        for(int i=;i<;i++){
            sum = (sum*g(p[i],e[i]*n))%;
        }
        printf("%lld\n",sum);
    }
    return ;
}