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)) ;

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 ), 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 }