HDU 5288 OO‘s sequence （技巧）

题目链接：http://acm.hdu.edu.cn/showproblem.php?

pid=5288

题面：

OO’s Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 985    Accepted Submission(s): 375

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

 

解题：

仅仅想到从左到右去找近期的不合法点。没想到从右往左找，那么答案就出来了。事实上数据范围那么小，就已经是一种暗示了。能够用数组记录下其最后出现的位置。注意扫的操作，要和记录同一时候进行。注意小心处理1的情况就好。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <map>
#include <vector>
#include <cmath>
#include <algorithm>
#define mod 1000000007
#define maxn 100010
#define LL long long
using namespace std;
int t,le[100010],ri[100010],store[100010],pre[100010],tmp,root;
LL ans;
int main()
{
	while(~scanf("%d",&t))
	{
		ans=0;
		for(int i=1;i<=t;i++)
		  scanf("%d",&store[i]);
		for(int i=1;i<=t;i++)
			pre[i]=0;
		for(int i=1;i<=t;i++)
		{
           tmp=1;
		   root=sqrt((double)store[i]);
		     for(int j=1;j<=root;j++)
		     {
				 if(store[i]%j==0)
				 {
					 tmp=max(tmp,pre[j]+1);
					 tmp=max(tmp,pre[store[i]/j]+1);
				 }
		     }
		     le[i]=tmp;
		   pre[store[i]]=i;
		}
		for(int i=1;i<=t;i++)
			pre[i]=t+1;
		for(int i=t;i>=1;i--)
		{
			tmp=t;
            root=sqrt((double)store[i]);
			  for(int j=1;j<=root;j++)
			  {
				  if(store[i]%j==0)
				  {
					  tmp=min(pre[j]-1,tmp);
					  tmp=min(tmp,pre[store[i]/j]-1);
				  }
			  }
			  ri[i]=tmp;
			pre[store[i]]=i;
		}
		/*for(int i=1;i<=t;i++)
			cout<<i<<" "<<le[i]<<" "<<ri[i]<<endl;*/
		for(int i=1;i<=t;i++)
		{
				ans=(ans+1LL*(i-le[i]+1)*(ri[i]-i+1))%mod;
		}
		printf("%lld\n",ans);
	}
	return 0;
}