HDU - 6025 Coprime Sequence （gcd前缀后缀处理）

Do you know what is called ``Coprime Sequence''? That is a sequence consists of  nnpositive 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.

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

这个题意呢，给你一系列质序列（就是他们的最大公约数为1）。然后呢，让你选一个数然后去掉，让他们的最大公约数变的最大。

思路：一看数据那么大，肯定不能直接算。但是又要将所有的组合都要算一遍，怎么办呢？用数组存头到尾存一遍他们的最大公约数，用pre[]代替好了，然后呢，怎样去一个数呢，把他“隔”过去好了，问题是怎样实现呢。

1.用·s[]逆着存一遍。

2、如果求去掉下标为i的数字后整个数列的gcd，直接将该数字前的所有数字的gcd（prefix[i-1]）和该数字后所有数字的gcd(s[i+1])再求一下gcd就好了。

代码如下：

#include
#include
#include
using namespace std;
int a[100005],n;
int pre[100005],s[100005];

int gcd(int a,int b)
{
    if(b==0)
        return a;
     else return gcd(b,a%b);
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0; i=0; i--)
        {
            s[i]=gcd(s[i+1],a[i]);
        }
        int ans=max(s[1],pre[n-2]);

        for(int i=1; i

转载于:https://www.cnblogs.com/zitian246/p/9123610.html