多校1 OO’s Sequence 1001

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 256    Accepted Submission(s): 94

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

 

Source

2015 Multi-University Training Contest 1

 

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#include <math.h>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

const int mod=1e9+7;

int main()
{
    int n,i,j,k;
    int str[100005],spfa[10005],l[100005],r[100005];
    int s;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
            scanf("%d",&str[i]);
        for(i=1;i<=10000;i++)
            spfa[i]=0;
        for(i=1;i<=n;i++)
            l[i]=0,r[i]=n+1;
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=sqrt(str[i]);j++)
            {
                if(str[i]%j==0)
                {
                    if(spfa[j]>l[i])
                        l[i]=spfa[j];
                    if(spfa[str[i]/j]>l[i])
                        l[i]=spfa[str[i]/j];
                }
            }
            spfa[str[i]]=i;
        }
        for(i=10000;i>=1;i--)
            spfa[i]=n+1;
        for(i=n;i>=1;i--)
        {
            for(j=1;j<=sqrt(str[i]);j++)
            {
                if(str[i]%j==0)
                {
                    if(spfa[j]<r[i])
                        r[i]=spfa[j];
                    if(spfa[str[i]/j]<r[i])
                        r[i]=spfa[str[i]/j];
                }
            }
            spfa[str[i]]=i;
        }
        s=0;
        for(i=1;i<=n;i++)
        {
            s=(s+(i-l[i])*(r[i]-i)%mod)%mod;
        }
        printf("%d
",s);
    }
    return 0;
}