hdu 5288||2015多校联合第一场1001题

http://acm.hdu.edu.cn/showproblem.php?pid=5288

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

/**
hdu 5288||2015多校联合第一场1001题
题目大意：给定一个区间，找出f(i,j)的和
解题思路：http://blog.sina.com.cn/s/blog_15139f1a10102vnx5.html 和官方题解想的一样
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
typedef __int64 LL;
const int MAXN = 1e5+10;
const LL MOD = 1e9+7;
int a[MAXN],l,r,n;
vector<int> f[10010];
vector<int> vec[10010];
typedef vector<int>::iterator it;

int main()
{
    for(int i=1; i<=10000; i++)
        for(int j=1; j*i<=10000; j++)
            f[i*j].push_back(i);
    while(~scanf("%d",&n))
    {
        for(int i=0; i<=10000; i++)
            vec[i].clear();
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            vec[a[i]].push_back(i);
        }
        LL ans=0;
        for(int i=1; i<=n; i++)
        {
            l=1,r=n;
            for(it j=f[a[i]].begin(); j!=f[a[i]].end(); j++)
            {
                if(vec[*j].empty())continue;
                it t=lower_bound(vec[*j].begin(),vec[*j].end(),i);
                if(t==vec[*j].end())
                {
                    t--;
                    l=max((*t)+1,l);
                    continue;
                }
                if(*t==i)
                {
                    if(t!=vec[*j].begin())
                    {
                        t--;
                        l=max((*t)+1,l);
                        t++;
                    }
                    if(*t==i)t++;
                    if(t!=vec[*j].end())
                        r=min((*t)-1,r);
                }
                else
                {
                    r=min((*t)-1,r);
                    if(t!=vec[*j].begin())
                    {
                        t--;
                        l=max((*t)+1,l);
                    }
                }
            }
            ans = (ans + ((long long)(i - l+1) * (r - i+1) % MOD) ) % MOD;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}

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