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 a1, a2, ..., an.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.
But you can not kill yourself before you solve this problem:
Given you a sequence of number a1, a2, ..., an.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, ..., an.
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.
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, ..., an.
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
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 ",ans[i]); 118 } 119 return 0; 120 }