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 ai mod aj=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 ai(0<ai<=10000)
In each test case:
First line: an integer n(n<=10^5) indicating the size of array
Second line:contain n numbers ai(0<ai<=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
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\n",ans);
}
return 0;
}