2017 ccpc女生专场 1003 Coprime Sequence

前缀后缀gcd，其实自己比赛中用的是种奇怪的方法A掉的，不过先把这个学上，自己的方法有时间再填。

题意

告诉你N个数，求删除一个数可以求得最大GCD。

N可能是100000。

思路

这道题其实很简单，但是想不到这点就很难。

简单的说就是先预处理，得到每个数字左边的GCD和右边的GCD.

1. befor(i)代表前i个数字的GCD, 复杂度 O(n*log(n))
2. after(i)代表i之后的数字的GCD. 复杂度 O(n*log(n))
3. ans = max(after(2), befor(1)+after(3), ..., befor(n-2)+after(n), befor(n-1)) ;

Coprime Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 250    Accepted Submission(s): 145

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

 

#include<iostream>
#include<stdio.h>

#define ff 1000000007

using namespace std;

int main()
{
    int t,n,k;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&k);
        long long ans=0,temp;
        for(int i = 1; i <= n; i++)
        {
            temp=1;
            for(int j = 1; j <= k; j++)
            {
                temp=(temp*i)%ff;
            }
            ans=(ans+temp)%ff;
        }
        printf("%lld
",ans%ff);
    }
    
    return 0;
}