HDOJ 4630 No Pain No Game

树状数组维护+离线

最大gcd一定是某个数的一个约数，对所有询问按右端点排序，用树状数组维护即可。。

具体做法：记录每个约数出现的上一个位置pre，对于左端点落在pre之前的询问这个约数就可能是答案，所以我们对pre向前进行维护，而询问的时候从左端点往上找最大值。。。。

No Pain No Game

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1384    Accepted Submission(s): 585

Problem Description

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 a₁, a₂, ..., 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 a₁, a₂, ..., 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

1108 2 4 9 5 7 10 6 1 352 102 46 91 47 10

 

Sample Output

52243

 

Source

2013 Multi-University Training Contest 3

 

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>using namespace std;const int maxn=60000;vector<int> fact[maxn];void get_factor(){    for(int i=1;i<maxn;i++)    {        for(int j=i;j<maxn;j+=i)        {            fact[j].push_back(i);        }    }}int n,Q,T,tree[maxn],pre[maxn],arr[maxn],ans[maxn];struct ASK{    int l,r,id;}ask[maxn];bool cmp(ASK a,ASK b){   if(a.r!=b.r) return a.r<b.r;   return a.l<b.l;}int lowbit(int x){    return x&(-x);}void update(int p,int v) ///向下更新最大约数{    for(int i=p;i;i-=lowbit(i)) tree[i]=max(tree[i],v);}int query(int p){    int ret=0;    for(int i=p;i<maxn;i+=lowbit(i)) ret=max(ret,tree[i]);    return ret;}int main(){    scanf("%d",&T);    get_factor();while(T--){    scanf("%d",&n);    for(int i=1;i<=n;i++) scanf("%d",arr+i);    scanf("%d",&Q);    for(int i=0;i<Q;i++)    {        scanf("%d%d",&ask[i].l,&ask[i].r); ask[i].id=i;    }    sort(ask,ask+Q,cmp);    memset(tree,0,sizeof(tree));    memset(pre,-1,sizeof(pre));    int cnt=0;    for(int i=1;i<=n;i++)    {        for(int sz=fact[arr[i]].size(),j=0;j<sz;j++)        {            int fc=fact[arr[i]][j];            if(pre[fc]!=-1)            {                update(pre[fc],fc);            }            pre[fc]=i;        }        while(cnt<Q&&ask[cnt].r==i)        {            ans[ask[cnt].id]=query(ask[cnt].l);            cnt++;        }    }    for(int i=0;i<Q;i++) printf("%d\n",ans[i]);}    return 0;}