HDU6025 Coprime Sequence （前缀后缀处理）

题意：

分析：

代码：

const int N = 1e5 + 7;
long long a
, pre
, beh
;
long long gcd(long long a, long long b) {
return b == 0 ? a : gcd(b, a % b);
}
int main()
{
int T;
scanf("%d", &T);
while(T--) {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; ++i)
scanf("%lld", a +i);
pre[1] = a[1];
for (int i= 2; i <= n; ++i)
pre[i] = gcd(a[i], pre[i - 1]);
beh
= a
;
for (int i = n - 1; i >= 1; --i)
beh[i] = gcd(a[i], beh[i + 1]);
int ans = max(pre[n-1], beh[2]);
for (int i = 2; i < n; ++i) {
ans = max(ans, (int)gcd(pre[i -1], beh[i + 1]));
}
printf("%d\n", ans);
}
return 0;
}