hdu 4630 树状数组 ****

题意：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.

思路：这题的处理方式和hdu4358有点像。我们用一个pre[x]表示约数x的倍数上次出现的位置，将查询按区间的右节点升序排序。num[i]的约 数为j，如果pre[j]为0，就将pre[j]置为i;否则就update(pre[j],j)，表示的意思是约数j肯定不是第一次出现,将 pre[j]以前的区间更新最大约数。如果查询区间的右边界在i处，那么左边界在pre[j]以前就肯定就能取到j。因为num[pre[j]]和 num[i]有一个公共约数j，且pre[j]和i被该查询区间所覆盖。

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #define Maxn 50010

 6 #define lowbit(x) (x&(-x))

 7 using namespace std;

 8 int C[Maxn],n,num[Maxn],pre[Maxn],q,ans[Maxn];

 9 struct QT{

10     int l,r,i;

11     int operator <(const QT &temp) const

12     {

13         return r<temp.r;

14     }

15 }qt[Maxn];

16 int Sum(int pos)//往后找

17 {

18     int sum=0;

19     while(pos<=n)

20     {

21         sum=max(sum,C[pos]);

22         pos+=lowbit(pos);

23     }

24     return sum;

25 }

26 void update(int pos,int val)//更新pos以前的位置

27 {

28     while(pos)

29     {

30         C[pos]=max(C[pos],val);

31         pos-=lowbit(pos);

32     }

33 }

34 int main()

35 {

36     int t,i,j;

37     scanf("%d",&t);

38     while(t--)

39     {

40         memset(C,0,sizeof(C));

41         memset(pre,0,sizeof(pre));

42         scanf("%d",&n);

43         for(i=1;i<=n;i++)

44             scanf("%d",num+i);

45         scanf("%d",&q);

46         for(i=1;i<=q;i++)

47         {

48             scanf("%d%d",&qt[i].l,&qt[i].r);

49             qt[i].i=i;

50         }

51         sort(qt+1,qt+1+q);

52         int r=1;

53         for(i=1;i<=n;i++)

54         {

55             if(r>q) break;

56             for(j=1;j*j<=num[i];j++)

57             {

58                 if(num[i]%j) continue;

59                 if(pre[j]) update(pre[j],j);

60                 pre[j]=i;

61                 if(j*j==num[i]) break;

62                 int k=num[i]/j;

63                 if(pre[k]) update(pre[k],k);

64                 pre[k]=i;

65             }

66             while(qt[r].r==i&&r<=q)

67             {

68                 ans[qt[r].i]=Sum(qt[r].l);

69                 r++;

70             }

71         }

72         for(i=1;i<=q;i++)

73             printf("%d\n",ans[i]);

74     }

75     return 0;

76 }