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

 

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

 

Sample Output

 

Source

2015 Multi-University Training Contest 1

求在[a , b] 中有多少个i 满足找不到另一个数j 使,a[i] % a[j] = 0

用L[]，R[]分别记录左右离起最近的因子，ans+=(R[i]-i+1)*(i-L[i]+1)/2 %MOD

//自己并没有想到怎么做 (╯▽╰)

#include<cstdio>

#include<cstring>

#include<iostream>

using namespace std ;

const int maxn = 100010 ;

const int mod = 1e9+7 ;

int a[maxn];

long long  last[maxn] ;

long long pre[maxn] ;

long long  l[maxn] ;

long long r[maxn] ;

int  main()

{

    //freopen("1001.txt" ,"r" ,stdin) ;

    int n;

    while(~scanf("%d",&n))

    {

        memset(pre,0,sizeof(pre));

        memset(last,0,sizeof(last));

        for(int i = 1;i <= n;i++)

        {

            l[i] = 1;

            r[i] = n;

            scanf("%d",&a[i]);

            for(int j = a[i];j < 10001;j+=a[i])

            {

                if(pre[j] && r[pre[j]] == n)

                    r[pre[j]] = i-1;

                pre[a[i]] = i;

            }

        }

        for(int i = n;i >=1;i--)

        {

            for(int j = a[i];j < 10001;j+=a[i])

            {

                if(last[j] && l[last[j]]==1)

                    l[last[j]] = i+1;

                last[a[i]] = i;

            }

        }

        long long ans = 0;

        for(long long int i = 1;i <= n;i++)

            ans= (ans+(((i-l[i]+1)%mod)*((r[i] - i +1)%mod))%mod)%mod;

        printf("%I64d\n",ans);

    }

    return 0 ;

}