OO’s Sequence

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

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

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

using namespace std;

int L[100100],R[100010];
int n;
int a[100010];
int visit[100100];

void updatalr()
{
    memset(visit,0,sizeof(visit));
    for(int i=1;i<=n;i++)
    {
        int maxx = 0;
        int t = sqrt(a[i]);
        for(int j=t;j>=1;j--)
        {
            if(a[i]%j == 0)
            {
                maxx = max(maxx,visit[j]);
                maxx = max(maxx,visit[a[i]/j]);
            }
        }
        L[i] = maxx;
        visit[a[i]] = i;
    }

   for(int i=0;i<100001;i++)
   {
       visit[i] = n + 1;
   }
    for(int i=n;i>=1;i--)
    {
        int minn = n+1;
        int t = sqrt(a[i]);
        for(int j=t;j>=1;j--)
        {
            if(a[i]%j == 0)
            {
                minn = min(minn,visit[j]);
                minn = min(minn,visit[a[i]/j]);
            }
        }
        R[i] = minn;
        visit[a[i]] = i;
    }
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        __int64 sum = 0;
        memset(visit,0,sizeof(visit));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        updatalr();
        for(int i=1;i<=n;i++)
        {
            __int64 LL = (__int64)(i - L[i]);
            __int64 RR = (__int64)(R[i] - i);
            sum = (sum + (LL*RR))%1000000007;
        }
        printf("%I64d\n",sum);
    }
    return 0;
}

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