style rst des 一個數 output sequence amp efi comm

前綴後綴gcd，其實自己中用的是種奇怪的方法A掉的，不過先把這個學上，自己的方法有時間再填。

題意

告訴你N個數，求刪除一個數可以求得最大GCD。

N可能是100000。

思路

這道題其實很簡單，但是想不到這點就很難。

簡單的說就是先預處理，得到每個數字左邊的GCD和右邊的GCD.

1. befor(i)代表前i個數字的GCD, 復雜度 O(n*log(n))
2. after(i)代表i之後的數字的GCD. 復雜度 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

 1 #include<iostream>
 2 #include<stdio.h>
 3 
 4 #define ff 1000000007
 5 
 6 using namespace std;
 7 
 8 int main()
 9 {
10     int t,n,k;
11     scanf("%d",&t);
12     while(t--)
13     {
14         scanf("%d%d",&n,&k);
15         long long ans=0,temp;
16         for(int i = 1; i <= n; i++)
17         {
18             temp=1;
19             for(int j = 1; j <= k; j++)
20             {
21                 temp=(temp*i)%ff;
22             }
23             ans=(ans+temp)%ff;
24         }
25         printf("%lld\n",ans%ff);
26     }
27     
28     return 0;
29 }

2017 ccpc女生專場 1003 Coprime Sequence