Orac and LCM（数论）

题目链接： Orac and LCM

大致题意:

给你一个长度为 n 的数组，求 gcd {lcm({ai , aj}) | i < j}

解题思路:

gcd1=gcd[lcm(a1,a2),lcm(a1,a3)…lcm(a1,an)]=gcd(a1,lcm(a2,a3…an))

gcd2=gcd[lcm(a2,a3),lcm(a2,a4)…lcm(a2,an)]=gcd(a2,lcm(a3,a3…an))

…

推出res=gcd(gcd1,gcd2,…,gcdn)

lcm用后缀数组维护

时间复杂度： O(nlogn)

AC代码:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a % b);
}
ll lcm(ll a, ll b) {return a * b / gcd(a, b);
}
int n;
ll a[N], b[N];
int main()
{cin >> n;for (int i = 1; i <= n; ++i)cin >> a[i];for (int i = n; i >= 1; --i)b[i] = gcd(a[i], b[i + 1]);ll res = 0;for (int i = 1; i <= n; ++i)res = gcd(res, lcm(b[i + 1], a[i]));cout << res << endl;return 0;
}

END

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