POJ - 3910 - GCD Determinant（数论——GCD矩阵）

Problem  POJ - 3910 - GCD Determinant

Time Limit: 1000 mSec

Problem Description

We say that a set S = {x1, x2, ..., xn} is factor closed if for any xi ∈ S and any divisor d of xi we have d ∈ S. Let’s build a GCD matrix (S) = (sij), where sij = GCD(xi, xj) – the greatest common divisor of xi and xj. Given the factor closed set S, find the value of the determinant:

Input

The input file contains several test cases. Each test case starts with an integer n (0 < n < 1000), that stands for the cardinality of S. The next line contains the numbers of S: x1, x2, ..., xn. It is known that each xi is an integer, 0 < xi < 2*10 ⁹. The input data set is correct and ends with an end of file.

Output

For each test case find and print the value Dn mod 1000000007.

Sample Input

2
1 2
3
1 3 9
4
1 2 3 6

Sample Output

1
12
4

 

题解：这种结论题真是顶不住，记录下来当作积累吧，题目是在UVALive上交的，POJ的编译器实在有点老。。。

附上大佬的链接，有gcd矩阵的相关结论：

https://zhuanlan.zhihu.com/p/53447466

 

#include <bits/stdc++.h>

using namespace std;

#define REP(i, n) for (int i = 1; i <= (n); i++)
#define sqr(x) ((x) * (x))

const int maxn = 1000 + 10;
const int maxm = 200000 + 10;
const int maxs = 8;

typedef long long LL;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;

const LL unit = 1LL;
const int INF = 0x3f3f3f3f;
const LL Inf = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-14;
const double inf = 1e15;
const double pi = acos(-1.0);
const LL mod = 1000000007;

LL get_phi(LL n)
{
    LL ans = n;
    LL m = sqrt(n + 0.5);
    for(LL i = 2; i <= m; i++)
    {
        if(n % i == 0)
        {
            while(n % i == 0)
            {
                n  /= i;
            }
            ans = ans * (i - 1) / i;
        }
        if(n == 1)
            break;
    }
    if(n > 1)
        ans = ans * (n - 1) / n;
    return ans;
}

int n;
LL x;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    while(cin >> n)
    {
        LL ans = 1;
        for(int i = 0; i < n; i++)
        {
            cin >> x;
            ans *= get_phi(x);
            ans %= mod;
        }
        cout << ans << endl;
    }
    return 0;
}