计蒜客 41391.query-二维偏序+树状数组(预处理出来满足情况的gcd) (The Preliminary Contest for ICPC Asia Xuzhou 2019 I.) 2019年徐州网络赛)

query

Given a permutation pp of length nn, you are asked to answer mm queries, each query can be represented as a pair (l ,r )(l,r), you need to find the number of pair(i ,j)(i,j) such that l \le i < j \le rl≤i<j≤r and \min(p_i,p_j) = \gcd(p_i,p_j )min(pi​,pj​)=gcd(pi​,pj​).

Input

There is two integers n(1 \le n \le 10^5)n(1≤n≤105), m(1 \le m \le 10^5)m(1≤m≤105) in the first line, denoting the length of pp and the number of queries.

In the second line, there is a permutation of length nn, denoting the given permutation pp. It is guaranteed that pp is a permutation of length nn.

For the next mm lines, each line contains two integer l_ili​ and r_i(1 \le l_i \le r_i \le n)ri​(1≤li​≤ri​≤n), denoting each query.

Output

For each query, print a single line containing only one integer which denotes the number of pair(i,j)(i,j).

样例输入复制

3 2
1 2 3
1 3
2 3

样例输出复制

2
0

题意很好理解。就是二维偏序+树状数组。

比赛时的队友们的代码写的太丑，学了一下别人的代码，写的很好。

参考链接：query（2019徐州网络赛）（一点思维+树状数组）

代码:

 //I.二维偏序+树状数组
 //提前预处理出来所有的倍数关系，然后按顺序，先插入然后查询
 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 #define pb push_back
 #define mp make_pair
 #define pii pair<int,int>
 const int maxn=1e5+;

 int a[maxn],pos[maxn],tree[maxn],ans[maxn];
 int n,m;
 vector<int> gcd[maxn];
 vector<pii> op[maxn];

 int lowbit(int x)
 {
     return x&(-x);
 }

 void add(int x,int val)
 {
     for(int i=x;i<=n;i+=lowbit(i)){
         tree[i]+=val;
     }
 }

 int query(int n)
 {
     int ans=;
     for(int i=n;i>;i-=lowbit(i)){
         ans+=tree[i];
     }
     return ans;
 }

 int main()
 {
     scanf("%d%d",&n,&m);
     for(int i=;i<=n;i++){
         scanf("%d",&a[i]);
         pos[a[i]]=i;//记录下标
     }
     for(int i=;i<=n;i++){
         for(int j=a[i]*;j<=n;j+=a[i]){
             gcd[max(i,pos[j])].pb(min(i,pos[j]));//预处理
         }
     }
     for(int i=;i<=m;i++){
         int l,r;
         scanf("%d%d",&l,&r);
         op[r].pb(mp(l,i));//保存查询
     }
     for(int i=;i<=n;i++){
         for(auto it:gcd[i]){
             add(it,);//先把插入
         }
         for(auto it:op[i]){
             ans[it.second]=query(i)-query(it.first-);//查询
         }
     }
     for(int i=;i<=m;i++){
         printf("%d\n",ans[i]);
     }
     return ;
 }