HDOJ 5288 OO’s Sequence 水

Posted 2020-10-02 yfceshi

预处理出每一个数字的左右两边能够整除它的近期的数的位置

OO’s Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1880    Accepted Submission(s): 672

Problem Description

OO has got a array A of size n ,defined a function f(l,r) represent the number of i (l<=i<=r) , that there‘s no j(l<=j<=r,j<>i) satisfy a_i mod a_j=0,now OO want to know

∑i=1n∑j=inf(i,j) mod （109+7）.

 

Input

There are multiple test cases. Please process till EOF.
In each test case: 
First line: an integer n(n<=10^5) indicating the size of array
Second line:contain n numbers a_i(0<a_i<=10000)

 

Output

For each tests: ouput a line contain a number ans.

 

Sample Input

5
1 2 3 4 5

 

Sample Output

23

 

Author

FZUACM

 

Source

2015 Multi-University Training Contest 1

 

/* ***********************************************
Author        :CKboss
Created Time  :2015年07月24日 星期五 08时12分15秒
File Name     :HDOJ5288.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef long long int LL;

const int maxn=100100;
const LL MOD=(LL)(1e9+7);

int n;
int a[maxn];
int Left[maxn],Right[maxn];
vector<int> pos[10010];

void init() 
{ 
	for(int i=0;i<=10010;i++) 
	{
		pos[i].clear(); 
	}
	for(int i=0;i<n;i++)
	{
		Left[i]=0; Right[i]=n-1;
	}
}

void pre()
{
	for(int i=0;i<n;i++)
	{
		int x=a[i];
		for(int j=x;j<=10000;j+=x)
		{
			for(int k=0,sz=pos[j].size();k<sz;k++)
			{
				int z=pos[j][k];
				if(z==i) continue;
				else if(z<i)
				{
					Right[z]=min(Right[z],i-1);
				}
				else if(z>i)
				{
					Left[z]=max(Left[z],i+1);
				}
			}
		}
	}
}

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	while(scanf("%d",&n)!=EOF)
	{
		init();
		for(int i=0;i<n;i++) 
		{
			scanf("%d",a+i);
			pos[a[i]].push_back(i);
		}
		pre();
		LL ans=0;
		for(int i=0;i<n;i++)
		{
			LL L=i-Left[i]+1LL;
			LL R=Right[i]-i+1LL;
			ans=(ans+(L*R)%MOD)%MOD;
		}
		cout<<ans<<endl;
	}
    
    return 0;
}