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+10;

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=0;
    for(int i=n;i>0;i-=lowbit(i)){
        ans+=tree[i];
    }
    return ans;
}

int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        pos[a[i]]=i;//记录下标
    }
    for(int i=1;i<=n;i++){
        for(int j=a[i]*2;j<=n;j+=a[i]){
            gcd[max(i,pos[j])].pb(min(i,pos[j]));//预处理
        }
    }
    for(int i=1;i<=m;i++){
        int l,r;
        scanf("%d%d",&l,&r);
        op[r].pb(mp(l,i));//保存查询
    }
    for(int i=1;i<=n;i++){
        for(auto it:gcd[i]){
            add(it,1);//先把插入
        }
        for(auto it:op[i]){
            ans[it.second]=query(i)-query(it.first-1);//查询
        }
    }
    for(int i=1;i<=m;i++){
        printf("%d
",ans[i]);
    }
    return 0;
}