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

331 1 152 2 2 3 241 2 4 8

 

Sample Output

122

 

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 longconst 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;  }