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
求區(qū)間任意兩個(gè)數(shù)的最大gcd。做了幾個(gè)gcd的題目，貌似都是離線處理的。
對(duì)于兩個(gè)數(shù)的gcd，是因?yàn)檫@兩個(gè)數(shù)都有相同的約數(shù)x，這樣這兩個(gè)數(shù)的gcd才有可能是x。這樣我們用O（nsqrt n）的時(shí)間復(fù)雜度處理處所有的數(shù)約數(shù)。假如a[k]的一個(gè)約數(shù)是x，x上一次出現(xiàn)的位置是i。那么在i~k這一段的gcd有可能就會(huì)變成x。按著這個(gè)思路去更新，線段樹去求gcd最大值。詢問離線處理，當(dāng)現(xiàn)在處理的位置恰好是某一次詢問的右端點(diǎn)的話，就可以直接通過線段樹求出最大值gcd來了。
代碼如下：

#include<bits/stdc++.h> #define ll long long using 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<int> g[maxx]; int a[maxx],ans[maxx]; int n,m; /*------------預(yù)處理出所有數(shù)的約數(shù)-------------*/ inline void init() {for(int i=1;i<=maxx;i++)for(int j=i;j<=maxx;j+=i)g[j].push_back(i); } /*---------------線段樹----------------*/ 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<int,int> 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<g[a[j]].size();k++){x=g[a[j]][k];if(mpp[x]!=0) update(mpp[x],x,1);mpp[x]=j;}for(i;i<=m&&j==b[i].r;i++) ans[b[i].id]=query(b[i].l,b[i].r,1);}for(int i=1;i<=m;i++) printf("%d\n",ans[i]);}return 0; }

努力加油a啊，(^o)/~

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