[JSOI 2015] 最大公约数

[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;
    
}