OO’s Sequence

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3612 Accepted Submission(s): 1330

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

\sum i = 1 n \sum j = i n f (i, j) m o d （ 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

【题意】给一个序列，求共有多少个找不到任意两个不同数是整除关系的连续子序列，结果mod 1e9+7

【解题方法】维护一下每一个a[i]向左和向右可以到达的最大距离l[i],r[i]，那么答案就是对于每一个i，对(i-l[i])*(r[i]-i)求和就可以了。

【AC 代码】

//
//Created by just_sort 2016/9/12 16:50
//Copyright (c) 2016 just_sort.All Rights Reserved
//

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 1e5+7;
const LL mod = 1e9+7;
int a[maxn],vis[maxn],l[maxn],r[maxn];

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        memset(vis,0,sizeof(vis));
        int maxx = 0;
        for(int i=1; i<=n; i++) l[i]=0,r[i]=n+1;
        for(int i=1; i<=n; i++) scanf("%d",&a[i]);
        for(int i=1; i<=n; i++){
            maxx = max(maxx,a[i]);
            for(int j=a[i]; j<=maxx; j+=a[i]){
                if(vis[j] && r[vis[j]]>i) r[vis[j]] = i;
            }
            vis[a[i]] = i;
        }
        maxx = 0;
        memset(vis,0,sizeof(vis));
        for(int i=n; i>=1; i--){
            maxx = max(maxx,a[i]);
            for(int j=a[i]; j<=maxx; j+=a[i]){
                if(vis[j] && l[vis[j]]<i) l[vis[j]] = i;
            }
            vis[a[i]] = i;
        }
//        for(int i=1; i<=n; i++){
//            cout<<l[i]<<" "<<r[i]<<endl;
//        }
        LL ans = 0;
        for(int i=1; i<=n; i++){
            ans = (ans+1LL*(i-l[i])*(r[i]-i)%mod)%mod;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}