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; }