```cpp
#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
long long t;
long long n, m;
long long ans = 0;
long long a[100005];
long long logs[100005];
long long dp[100005][100];
inline long long read(long long &a){
int w=0,f=1;
char ch=getchar();
for(;ch<'0'||ch>'9';ch=getchar()) if(ch=='-') f=-1;
for(;ch>='0'&&ch<='9';ch=getchar()) w=(w<<3)+(w<<1)+(ch^48);
a=w*f;
}
void write(long long x){
if(x<0) putchar('-'),x=-x;
if(x>9) write(x/10);
putchar(x%10+'0');
}
extern inline long long fgcd(long long x, long long y);
long long ask(long long x, long long y){
long long lg = logs[y-x+1];
return fgcd(dp[x][lg], dp[y-(1<<lg)+1][lg]);
}
inline long long fgcd(long long x, long long y) {
if (x % y == 0) {
return y;
}
return fgcd(y, x%y);
}
int main() {
read(t);
while (t--) {
read(n);
logs = -1;
ans = 0;
for (long long i = 1; i <= n; i++) {
read(a[i]);
}
}
return 0;
}
```
应该是这样子
by EARS_TURE @ 2024-04-17 20:03:21
啊啊啊对不起写错了！！！
by EARS_TURE @ 2024-04-17 20:04:07
@[__0__](/user/794507)
```cpp
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
#define LL long long
queue<int>q,r;
int n;LL a[N],res;
LL gcd(LL a,LL b){return b==0?a:gcd(b,a%b);}
int main()
{
int T;scanf("%d",&T);
while(T--)
{
scanf("%d",&n);res=0;
for(int i=1;i<=n;i++)
{
scanf("%lld",&a[i]);LL last=0;res=max(res,a[i]);
while(!q.empty())
{
int x=q.front();q.pop();a[x]=gcd(a[x],a[i]);//x就是最小的满足区间[x,i]的gcd为a[x]的左端点
res=max(res,a[x]*(i-x+1));
if(a[x]==a[last]) continue;//一些左端点就可以合并掉
r.push(x);last=x;
}
while(!r.empty())
{
q.push(r.front());
r.pop();
}
if(a[last]!=a[i]) q.push(i);//i也可以作为一个新的左端点
}
printf("%lld\n",res);
while(!q.empty()) q.pop();
}
return 0;
}
```
by EARS_TURE @ 2024-04-17 20:06:53
@[EARS_TURE](/user/1080324) 有没有用ST表的做法？因为最近在学倍增。老师要求……
by __0__ @ 2024-04-18 17:32:15