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HDU4630:No Pain No Game(线段树)
2015-11-21 00:58:13 来源: 作者: 【 】 浏览:3
Tags:HDU4630:No Pain Game 线段
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 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

Author WJMZBMR
Source 2013 Multi-University Training Contest 3


题意: 求区间内的两两gcd里最大的
思路: 我们可以对于每个数求出因子,而在一个区间内出现超过两次的那么必然是一个gcd,这样我们只需要把所有出现超过两次的因子加入线段树中,并更新即可 对于查询,我们可以先按r从小到大排序,离线处理所有答案
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                 using namespace std; #define lson 2*i #define rson 2*i+1 #define ls l,mid,lson #define rs mid+1,r,rson #define UP(i,x,y) for(i=x;i<=y;i++) #define DOWN(i,x,y) for(i=x;i>=y;i--) #define MEM(a,x) memset(a,x,sizeof(a)) #define W(a) while(a) #define gcd(a,b) __gcd(a,b) #define LL long long #define N 50005 #define INF 0x3f3f3f3f #define EXP 1e-8 #define mpa make_pair #define lowbit(x) (x&-x) const int mod = 10007; struct node { int l,r,id; } op[N]; int tree[N<<2],a[N],ans[N]; int cmp(node a,node b) { return a.r
                
                 mid) ans2=query(L,R,rs); return max(ans1,ans2); } } int n,m; int pre[N]; int main() { int t,i,j,k,x; scanf(%d,&t); while(t--) { scanf(%d,&n); build(); for(i = 1; i<=n; i++) scanf(%d,&a[i]); scanf(%d,&m); for(i = 0; i
                 
                  

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