bzoj4052

gcd

跟那道cf题是一个原理。。。

每一时刻我们最多有log个gcd，那么我们用map存储每种gcd最左端，每次和新的数gcd就更新新的gcd的最左端，然后更新答案

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n, T;
map<ll, int> tmp_l, Left;
ll gcd(ll a, ll b) { return !b ? a : gcd(b, a % b); }
int main()
{
    int T;
    cin >> T;
    while(T --)
    {
        scanf("%d", &n);
        Left.clear();
        ll ans = 0;
        for(int i = 1; i <= n; ++i) 
        {
            ll x;
            scanf("%lld", &x);
            tmp_l.clear();
            for(map<ll, int> :: iterator it = Left.begin(); it != Left.end(); ++it)
            {
                ll tmp = gcd(it -> first, x);
                int le = it -> second;
                if(tmp_l.find(tmp) == tmp_l.end()) tmp_l[tmp] = le;
                else tmp_l[tmp] = min(tmp_l[tmp], le);
            }
            if(tmp_l.find(x) == tmp_l.end()) tmp_l[x] = i;
            for(map<ll, int> :: iterator it = tmp_l.begin(); it != tmp_l.end(); ++it)
            {
//              printf("gcd = %lld len = %d\n", it -> first, (i - it -> second + 1));
                ans = max(ans, it -> first * (long long)(i - it -> second + 1));
            }
            swap(tmp_l, Left);
        }
        printf("%lld\n", ans);
    }
    return 0;
}