HDU6025 Coprime Sequence —— 前缀和 & 后缀和

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6025

Coprime Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 666    Accepted Submission(s): 336

Problem Description

Do you know what is called ``Coprime Sequence''? That is a sequence consists of n positive integers, and the GCD (Greatest Common Divisor) of them is equal to 1.
``Coprime Sequence'' is easy to find because of its restriction. But we can try to maximize the GCD of these integers by removing exactly one integer. Now given a sequence, please maximize the GCD of its elements.

 

Input

The first line of the input contains an integer T(1≤T≤10), denoting the number of test cases.
In each test case, there is an integer n(3≤n≤100000) in the first line, denoting the number of integers in the sequence.
Then the following line consists of n integers a1,a2,...,an(1≤ai≤109), denoting the elements in the sequence.

 

Output

For each test case, print a single line containing a single integer, denoting the maximum GCD.

 

Sample Input

3
3
1 1 1
5
2 2 2 3 2
4
1 2 4 8

 

Sample Output

1
2
2

 

题解：

l[i]为前i个数的gcd， r[i]为后i个数的gcd。

假设被删除的数的下标为i， 则 删除该数后的gcd为： gcd(l[i-1], r[i+1])， 枚举i，取最大值。

学习之处：

当提到在序列里删除一段连续的数时，可以用前缀和+后缀和。

例如：http://blog.csdn.net/dolfamingo/article/details/71001021

代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const double eps = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int maxn = 1e5+10;

int n;
int a[maxn], l[maxn], r[maxn];

int gcd(int a, int b)
{
    return b==0?a:(gcd(b,a%b));
}

void solve()
{
    scanf("%d",&n);
    for(int i = 1; i<=n; i++)
        scanf("%d",&a[i]);

    l[1] = a[1]; r[n] = a[n];
    for(int i = 2; i<=n; i++)
        l[i] = gcd(l[i-1], a[i]);
    for(int i = n-1; i>=1; i--)
        r[i] = gcd(r[i+1], a[i]);

    int ans = 1;
    l[0] = a[2]; r[n+1] = a[n-1];
    for(int i = 1; i<=n; i++)
        ans = max(ans, gcd(l[i-1], r[i+1]) );
    cout<<ans<<endl;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        solve();
    }
    return 0;
}