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Life is a game,and you lose it,so you suicide.

But you can not kill yourself before you solve this problem: Given you a sequence of number a 1, a 2, …, a n.They are also a permutation of 1…n. You need to answer some queries,each with the following format: If we chose two number a,b (shouldn’t be the same) from interval [l, r],what is the maximum gcd(a, b)? If there’s no way to choose two distinct number(l=r) then the answer is zero. Input First line contains a number T(T <= 5),denote the number of test cases. Then follow T test cases. For each test cases,the first line contains a number n(1 <= n <= 50000). The second line contains n number a 1, a 2, …, a n. The third line contains a number Q(1 <= Q <= 50000) denoting the number of queries. Then Q lines follows,each lines contains two integer l, r(1 <= l <= r <= n),denote a query. Output For each test cases,for each query print the answer in one line. Sample Input 1 10 8 2 4 9 5 7 10 6 1 3 5 2 10 2 4 6 9 1 4 7 10 Sample Output 5 2 2 4 3 求区间任意两个数的最大gcd。做了几个gcd的题目，貌似都是离线处理的。 对于两个数的gcd，是因为这两个数都有相同的约数x，这样这两个数的gcd才有可能是x。这样我们用O（nsqrt n）的时间复杂度处理处所有的数约数。假如a[k]的一个约数是x，x上一次出现的位置是i。那么在i~k这一段的gcd有可能就会变成x。按着这个思路去更新，线段树去求gcd最大值。询问离线处理，当现在处理的位置恰好是某一次询问的右端点的话，就可以直接通过线段树求出最大值gcd来了。 代码如下：

#include
   
    #define ll long longusing namespace std;const int maxx=5e4+100;struct node{
   	int l;	int r;	int sum;}p[maxx<<2];struct Node{
   	int l;	int r;	int id;	bool operator<(const Node &a)const{
   		return a.r>r;	}}b[maxx];vector
    
      g[maxx];int a[maxx],ans[maxx];int n,m;/*------------预处理出所有数的约数-------------*/ inline void init(){
   	for(int i=1;i<=maxx;i++)		for(int j=i;j<=maxx;j+=i)			g[j].push_back(i);}/*---------------线段树----------------*/inline void pushup(int cur){
   	p[cur].sum=max(p[cur<<1].sum,p[cur<<1|1].sum);}inline void build(int l,int r,int cur){
   	p[cur].l=l;	p[cur].r=r;	p[cur].sum=0;	if(l==r) return ;	int mid=l+r>>1;	build(l,mid,cur<<1);	build(mid+1,r,cur<<1|1);}inline void update(int pos,int v,int cur){
   	int L=p[cur].l;	int R=p[cur].r;	if(L==R)	{
   		p[cur].sum=max(p[cur].sum,v);		return ;	}	int mid=L+R>>1;	if(pos<=mid) update(pos,v,cur<<1);	else update(pos,v,cur<<1|1);	pushup(cur);}inline int query(int l,int r,int cur){
   	int L=p[cur].l;	int R=p[cur].r;	if(l<=L&&R<=r) return p[cur].sum;	int mid=L+R>>1;	if(r<=mid) return query(l,r,cur<<1);	else if(l>mid) return query(l,r,cur<<1|1);	else return max(query(l,mid,cur<<1),query(mid+1,r,cur<<1|1));}int main(){
   	init();	int t,x;	scanf("%d",&t);	while(t--)	{
   		map
     
       mpp;		scanf("%d",&n);		for(int i=1;i<=n;i++) scanf("%d",&a[i]);		scanf("%d",&m);		for(int i=1;i<=m;i++) scanf("%d%d",&b[i].l,&b[i].r),b[i].id=i;		sort(b+1,b+1+m);		build(1,n,1);		for(int i=1,j=1;i<=m&&j<=n;j++)		{
   			for(int k=0;k
     
    
   

努力加油a啊，(^o)/~

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