HDU6025-Coprime Sequence

Coprime Sequence

                                                                           Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

                                                                                                             Total Submission(s): 0    Accepted Submission(s): 0


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

 

题意：一共n个数，去掉其中一个，问剩余n-1个数gcd的最大值

解题思路：rmq或者求前缀和后缀

RMQ:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

int a[100006],n;
int dp[100006][20];

int gcd(int x,int y)
{
if(x>y) swap(x,y);
while(y%x)
{
int k=y%x;
y=x;
x=k;
}
return x;
}

void init()
{
for(int i=1;i<=n;i++) dp[i][0]=a[i];
for(int i=1;(1<<i)<=n;i++)
for(int j=1;j+(1<<i)-1<=n;j++)
dp[j][i]=gcd(dp[j][i-1],dp[j+(1<<(i-1))][i-1]);

}

int query(int l,int r)
{
int k=0;
while(1<<(k+1)<=r-l+1) k++;
return gcd(dp[l][k],dp[r-(1<<k)+1][k]);
}

int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
init();
int ma=-1;
for(int i=2;i<n;i++)
{
int k=query(1,i-1),kk=query(i+1,n);
ma=max(ma,gcd(k,kk));
}
ma=max(ma,query(2,n));
ma=max(ma,query(1,n-1));
printf("%d\n",ma);
}
return 0;
}

前缀+后缀

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

int a[100005],x[100005],y[100005];

int gcd(int a,int b)
{
return b==0?a:gcd(b,a%b);
}

int main()
{
int t,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=0; i<n; i++)
scanf("%d",&a[i]);
x[0]=a[0];
for(int i=1; i<n; i++)
x[i]=gcd(x[i-1],a[i]);
y[n-1]=a[n-1];
for(int i=n-2; i>=0; i--)
y[i]=gcd(y[i+1],a[i]);
int ans=max(y[1],x[n-2]);
for(int i=1; i<n-1; i++)
ans=max(ans,gcd(x[i-1],y[i+1]));
printf("%d\n",ans);
}
return 0;
}