HDU 4630 No Pain No Game（2013多校3 1010题 离线处理+树状数组求最值）

No Pain No Game

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17    Accepted Submission(s): 5

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

 

 

Source

2013 Multi-University Training Contest 3

 

 

Recommend

zhuyuanchen520

 

 

题目给出n个数，每个数的范围是1~n的。

n<=50000;

然后查询m次，m<=50000

每次查询[l,r]区间内，两个数的gcd的最大值.

 

n个数，如果把n个数的约数全部写出来。查询[l,r]之间的gcd的最大值，就相当于找一个最大的数，使得这个数是[l,r]之间至少是两个的约数。

 

对于一个数n,在sqrt(n)内可以找出所有约数。

 

我的做法是对查询进行离线处理。

将每个查询按照 l 从大到小排序。

然后 i 从 n~0 ,表示从后面不断扫这些数。

对于数a[i],找到a[i]的所有约数，对于约数x,在x上一次出现的位置加入值x.

这样的查询的时候，只要差值前 r 个数的最大值就可以了。

 

看代码吧，不解释了。

 

这么水竟然想了这么久，非常sad

  1 /*

  2  *  Author:kuangbin

  3  *  1010.cpp

  4  */

  5 

  6 #include <stdio.h>

  7 #include <algorithm>

  8 #include <string.h>

  9 #include <iostream>

 10 #include <map>

 11 #include <vector>

 12 #include <queue>

 13 #include <set>

 14 #include <string>

 15 #include <math.h>

 16 using namespace std;

 17 

 18 const int MAXN = 50010;

 19 int c[MAXN];

 20 int n;

 21 int lowbit(int x)

 22 {

 23     return x&(-x);

 24 }

 25 void add(int i,int val)

 26 {

 27     while(i <= n)

 28     {

 29         c[i] = max(c[i],val);

 30         i += lowbit(i);

 31     }

 32 }

 33 int Max(int i)

 34 {

 35     int s = 0;

 36     while(i > 0)

 37     {

 38         s = max(s,c[i]);

 39         i -= lowbit(i);

 40     }

 41     return s;

 42 }

 43 

 44 int a[MAXN];

 45 int b[MAXN];

 46 int ans[MAXN];

 47 

 48 struct Node

 49 {

 50     int l,r;

 51     int index;

 52 }node[MAXN];

 53 

 54 bool cmp(Node a,Node b)

 55 {

 56     return a.l > b.l;

 57 }

 58 

 59 int main()

 60 {

 61     //freopen("in.txt","r",stdin);

 62     //freopen("out.txt","w",stdout);

 63     int T;

 64     int m;

 65     int l,r;

 66     scanf("%d",&T);

 67     while(T--)

 68     {

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

 70         for(int i = 1;i <= n;i++)

 71             scanf("%d",&a[i]);

 72         scanf("%d",&m);

 73         for(int i = 0;i < m;i++)

 74         {

 75             scanf("%d%d",&node[i].l,&node[i].r);

 76             node[i].index = i;

 77         }

 78         sort(node,node+m,cmp);

 79         int i = n;

 80         int j = 0;

 81         memset(b,0,sizeof(a));

 82         memset(c,0,sizeof(c));

 83         while(j < m)

 84         {

 85             while(i > 0 && i>= node[j].l)

 86             {

 87                 for(int k =1;k*k <= a[i];k++)

 88                 {

 89                     if(a[i]%k == 0)

 90                     {

 91                         if(b[k]!=0)

 92                         {

 93                             add(b[k],k);

 94                         }

 95 

 96                         b[k] = i;

 97                         if(k != a[i]/k)

 98                         {

 99                             if(b[a[i]/k]!=0)

100                             {

101                                 add(b[a[i]/k],a[i]/k);

102                             }

103 

104                             b[a[i]/k]=i;

105                         }

106                     }

107                 }

108                 i--;

109             }

110             while(j < m && node[j].l > i)

111             {

112                 ans[node[j].index]=Max(node[j].r);

113                 j++;

114             }

115         }

116         for(int i = 0;i < m;i++)

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

118     }

119     return 0;

120 }