[HDOJ1492]Happy 2004（数论，快速幂，逆元，积性函数）

题目链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=1452

题意：求2004^n的所有因子和。

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
using namespace std;
#define fr first
#define sc second
#define cl clear
#define BUG puts("here!!!")
#define W(a) while(a--)
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rll(a) scanf("%I64d", &a)
#define Rs(a) scanf("%s", a)
#define Cin(a) cin >> a
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Clr(a, x) memset((a), (x), sizeof(a))
#define Full(a) memset((a), 0x7f7f7f, sizeof(a))
#define lrt rt << 1
#define rrt rt << 1 | 1
#define pi 3.14159265359
#define RT return
#define lowbit(x) x & (-x)
#define onecnt(x) __builtin_popcount(x)
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef pair<string, int> psi;
typedef pair<int, int> pll;
typedef map<string, int> msi;
typedef vector<int> vi;
typedef vector<int> vl;
typedef vector<vl> vvl;
typedef vector<bool> vb;

const LL mod = 29;
LL n;

LL mul(LL x, LL q) {
    LL ret = 1;
    while(q) {
        if(q & 1) ret = (ret * x) % mod;
        q >>= 1;
        x = (x * x) % mod;
    }
    return ret;
}

LL exgcd(LL a, LL b, LL &x, LL &y) {
    if(b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    else {
        LL ret = exgcd(b, a%b, x, y);
        LL tmp = x;
        x = y;
        y = tmp - a / b * y;
        return ret;
    }
}

LL iv(LL a) {
    LL x, y;
    exgcd(a, mod, x, y);
    return (x % mod + mod) % mod;
}

signed main() {
    // FRead();
    while(cin >> n && n) {
        cout << ((mul(2, 2*n+1)-1)%mod)*((mul(3, n+1)-1)*iv(2)%mod)*((mul(167,n+1)-1)*iv(166)%mod)%mod << endl;
    }
    RT 0;
}