[题目链接]
https://www.lydsy.com/JudgeOnline/problem.php?id=4488
[算法]
不妨首先枚举左端点
注意到对于任意一个正整数n , 其质因子个数是log(n)级别的 , 故最大公约数最多降log(n)次
用稀疏表维护区间gcd
枚举左端点L , 二分求出每一段区间 , 更新答案即可
时间复杂度 : O(NlogN ^ 2)
[代码]
#include<bits/stdc++.h> using namespace std; #ifndef LOCAL #define eprintf(...) fprintf(stderr, _VA_ARGS_) #else #define eprintf(...) 42 #endif typedef long long ll; typedef pair<int , int> pii; typedef pair<ll , int> pli; typedef pair<ll , ll> pll; typedef long double ld; typedef unsigned long long ull; #define mp make_pair const int N = 1e5 + 10; const int MAXLOG = 20; #define rint register int int n; ll sp[N][MAXLOG] , a[N]; int bit[N] , lg[N]; ll ans; template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); } template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); } template <typename T> inline void read(T &x) { T f = 1; x = 0; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0'; x *= f; } inline ll gcd(ll x , ll y) { if (y == 0) return x; else return gcd(y , x % y); } inline ll query(int l , int r) { int k = lg[r - l + 1]; return gcd(sp[l][k] , sp[r - (1 << k) + 1][k]); } int main() { read(n); for (rint i = 1; i <= n; ++i) read(sp[i][0]); for (rint i = 1; i <= n; ++i) lg[i] = (int)((double)log(i) / log(2.0)); bit[0] = 1; for (rint i = 1; i < MAXLOG; ++i) bit[i] = bit[i - 1] << 1; for (rint i = 1; i < MAXLOG; ++i) { for (rint j = 1; j + (1 << i) - 1 <= n; ++j) { sp[j][i] = gcd(sp[j][i - 1] , sp[j + (1 << (i - 1))][i - 1]); } } for (int i = 1; i <= n; ++i) { int P = i; while (P <= n) { int nxt = P , l = P , r = n; while (l <= r) { int mid = (l + r) >> 1; if (query(i , mid) == query(i , P)) { nxt = mid; l = mid + 1; } else r = mid - 1; } chkmax(ans , 1LL * query(i , nxt) * (nxt - i + 1)); P = nxt + 1; } } printf("%lld " , ans); return 0; }