hdu 5288 OO’s Sequence(2015 Multi-University Training Contest 1)

OO’s Sequence

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1080 Accepted Submission(s): 403

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

Author

FZUACM

Source

2015 Multi-University Training Contest 1

题目大意：

给出一段序列。它的全部非空且里面的元素也为连续的子集中。不存在它能够整除数的元素的个数的和。

解题思路：

对序列的每一个元素分开考虑，对于每一个元素。往左边找能够找到连续区间长度为a的序列(包含这个元素),往右边可

以找到长度为b的序列(包含这个元素),那么这个元素出现了a*b次,将每一个元素出现的次数相加就能够了。怎样找这个区

间？发现序列元素值就10000。于是能够开个10000的数组。数组记录这个值最后一次出现的位置。然后枚举约数就

就能够了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int inf=0x7fffffff;
const int maxn=100000+1000;
const int mod=1000000000+7;
int a[maxn];
long long l[maxn];
long long r[maxn];
int h[maxn];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(h,0,sizeof(h));
        for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
        int cur=inf,te;
        for(int i=1;i<=n;i++)
        {
            cur=inf;
            for(int j=1;j*j<=a[i];j++)
            {
                if(a[i]%j==0)
                {
                    cur=min(cur,i-h[j]);
                    te=a[i]/j;
                    cur=min(cur,i-h[te]);
                }

            }
            l[i]=cur;
            h[a[i]]=i;
        }
        for(int i=1;i<=10500;i++)
        h[i]=n+1;
        for(int i=n;i>0;i--)
        {
            cur=inf;
            for(int j=1;j*j<=a[i];j++)
            {
                if(a[i]%j==0)
                {
                    cur=min(cur,h[j]-i);
                    te=a[i]/j;
                    cur=min(cur,h[te]-i);
                }
            }
            h[a[i]]=i;
            r[i]=cur;
        }
        long long ans=0;
        for(int i=1;i<=n;i++)
        {
            ans=(ans+(l[i]*r[i]))%mod;
        }
        printf("%I64d
",ans);
    }
    return 0;
}