HDU 1452 Happy 2004（积性函数 逆元）

数学题就是99%的时间在纸上1%的时间写代码。

Orz...

设sum(x)为x的因子和，由于sum(x)为积性函数，从而有sum(a*b) = sum(a) * sum(b)，其中a和b互素。

故：

2004 = 2^2 x 3 x 167 => 2004^x = 2^(2x) * 3^x * 167 ^x => sum = (2^(2x) - 1) * (3^x - 1) / 2 * (167^x - 1) / 166 =(2^(2x)
- 1) * (3^x - 1) / 2 * ((167%29)^x - 1) / (166%29) = (2^(2x) - 1) * (3^x - 1) / 2 * (22^x - 1) / 21.

(a * b) / c % MOD = (a % MOD) * (b % MOD) * inv(c).

inv(c)为 c * t = 1 % MOD c模MOD的逆元。

只有两个数，很容易算出inv(2) = 15, inv(21) = 22.

关于逆元的求解，详细的可以看：/article/2371887.html

所以sum(2004^x) = (2^(2x) - 1) * (3^x - 1) * 15 * (167^x - 1) * 18.

Orz...

于是主要的代码就几行了。

代码：

// Header.
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <set>
#include <map>
using namespace std;

// Macro
typedef long long LL;
#define TIME cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s." << endl;
#define IN freopen("/Users/apple/input.txt", "r", stdin);
#define OUT freopen("/Users/apple/out.txt", "w", stdout);
#define mem(a, n) memset(a, n, sizeof(a))
#define rep(i, n) for(int i = 0; i < (n); i ++)
#define REP(i, t, n) for(int i = (t); i < (n); i ++)
#define FOR(i, t, n) for(int i = (t); i <= (n); i ++)
#define ALL(v) v.begin(), v.end()
#define Min(a, b) a = min(a, b)
#define Max(a, b) a = max(a, b)
#define put(a) printf("%d\n", a)
#define ss(a) scanf("%s", a)
#define si(a) scanf("%d", &a)
#define sii(a, b) scanf("%d%d", &a, &b)
#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define VI vector<int>
#define pb push_back
const int inf = 0x3f3f3f3f, N = 1e2 + 5, MOD = 29;
// Macro end

int T, cas = 0;
int n, m;
// Imp
int Pow(int a, int b) {
int ret = 1;
while(b) {
if(b & 1) ret = (ret * a) % MOD;
a = (a * 2) % MOD;
b >>= 1;
}
return ret % MOD;
}

int main(){
#ifdef LOCAL
IN // OUT
#endif

while(si(n) != EOF, n) {
int a = Pow(2, 2 * n + 1) - 1,
b = Pow(3, n + 1) - 1,
c = Pow(22, n + 1) - 1;
int t = a * b * c * 15 * 18;
cout << a * b * c * 15 * 18 % MOD << endl;
}

return 0;
}