HDU - 6025 Coprime Sequence（前缀，后缀，gcd）

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本文介绍了一道算法题目，要求在给定的整数序列中删除一个元素以最大化剩余元素的最大公因数（GCD）。文章提供了问题描述、输入输出格式及样例，并附带了完整的C++代码实现。

Coprime Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1098 Accepted Submission(s): 547

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

Sample Output

Source

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题意：给出一段序列，删去其中一个数使剩余数字的gcd最大；

#include <bits/stdc++.h>
using namespace std;

const int N = 100000 + 10;
int t, n;
int pre[N], suf[N], a[N];

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

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