Do you know what is called ``Coprime Sequence''? That is a sequence consists of npositive
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.
InputThe first line of the input contains an integer T(1≤T≤10),
denoting the number of test cases. ``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.
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. OutputFor 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 8Sample Output
1 2 2
这个题意呢,给你一系列质序列(就是他们的最大公约数为1)。然后呢,让你选一个数然后去掉,让他们的最大公约数变的最大。
思路:一看数据那么大,肯定不能直接算。但是又要将所有的组合都要算一遍,怎么办呢?用数组存头到尾存一遍他们的最大公约数,用pre[]代替好了,然后呢,怎样去一个数呢,把他“隔”过去好了,问题是怎样实现呢。
1.用·s[]逆着存一遍。
2、如果求去掉下标为i的数字后整个数列的gcd,直接将该数字前的所有数字的gcd(prefix[i-1])和该数字后所有数字的gcd(s[i+1])再求一下gcd就好了。
代码如下:
#include<stdio.h> #include<algorithm> #include<string.h> using namespace std; int a[100005],n; int pre[100005],s[100005]; int gcd(int a,int b) { if(b==0) return a; else return gcd(b,a%b); } int main() { int t; scanf("%d",&t); while(t--) { scanf("%d",&n); for(int i=0; i<n; i++) scanf("%d",&a[i]); pre[0]=a[0]; for(int i=1; i<n; i++) { pre[i]=gcd(pre[i-1],a[i]); } s[n-1]=a[n-1]; for(int i=n-2; i>=0; i--) { s[i]=gcd(s[i+1],a[i]); } int ans=max(s[1],pre[n-2]); for(int i=1; i<n-1; i++) { ans=max(ans,gcd(pre[i-1],s[i+1])); } printf("%d\n",ans); } return 0; }