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 ai mod aj=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 ai(0<ai<=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：
只想到从左到右去找最近的不合法点，没想到从右往左找，那么答案就出来了。其实数据范围那么小，就已经是一种暗示了。可以用数组记录下其最后出现的位置。注意扫的操作，要和记录同时进行。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;
}
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