[题解] [JSOI2015] 最大公约数

题面

题解

考虑到固定左端点, 那么所有的区间的 \(gcd\) 最多只有 \(log\) 种取值

证明: 从左至右. 若 \(gcd\) 变化, 每次至少都会 \(/ 2\)

对于每一个左端点二分找这个位置, ST 表查询区间 \(gcd\)

Code

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
const int N = 1e5 + 5;
typedef long long ll; 
using namespace std;

int n, lg[N];
ll st[20][N], ans; 

template < typename T >
inline T read()
{
	T x = 0, w = 1; char c = getchar();
	while(c < '0' || c > '9') { if(c == '-') w = -1; c = getchar(); }
	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return x * w; 
}

ll gcd(ll n, ll m) { return m ? gcd(m, n % m) : n; }

void getst()
{
	for(int i = 1; i <= 17; i++)
		for(int j = 1; j + (1 << i) - 1 <= n; j++)
			st[i][j] = gcd(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]); 
}

ll query(int l, int r)
{
	int x = lg[r - l + 1]; 
	return gcd(st[x][l], st[x][r - (1 << x) + 1]); 
}

int main()
{
	n = read <int> ();
	for(int i = 2; i <= n; i++) lg[i] = lg[i >> 1] + 1; 
	for(int i = 1; i <= n; i++)
		st[0][i] = read <ll> ();
	getst();
	for(int l, r, mid, i = 1; i <= n; i++)
		for(int res, j = i; j <= n; j = res + 1)
		{
			l = j, r = n;
			while(l <= r)
			{
				mid = (l + r) >> 1;
				if(query(i, j) == query(i, mid)) res = mid, l = mid + 1;
				else r = mid - 1; 
			}
			ans = max(ans, 1ll * (res - i + 1) * query(i, res)); 
		}
	printf("%lld\n", ans); 
	return 0; 
}