hdu 5288 （小技巧）

OO’s Sequence

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)

 

Output

For each tests: ouput a line contain a number ans.

 

Sample Input

5
1 2 3 4 5

 

Sample Output

23

 

题目大意： 给出n个数，求在任意区间内，i<>j 并且i%j！=0 的i的个数 

思        路： 利用pre数组存储每个数出现的序号
                    借用此从左到右，便利出每个数据的因子，找到在数据范围最右面的那个因子所在的序号
                   同理可知道左面的。
                  
 具体看代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int mod = 1e9+7;
#define ll long long
#define N 100005
ll left
,right
;
int n,num
,pre
;

int main()
{
while(scanf("%d",&n)!=EOF){
for(int i=1;i<=n;i++){
scanf("%d",&num[i]);
left[i]=1;
right[i]=n;
}

memset(pre,0,sizeof(pre));
for(int i=1;i<=n;i++){
for(int j=num[i];j<=10000;j+=num[i]){
if(pre[j]&&right[pre[j]]==n)
right[pre[j]]=i-1;
}
pre[num[i]]=i;
}
memset(pre,0,sizeof(pre));
for(int i=n;i>=1;i--){
for(int j=num[i];j<=10000;j+=num[i]){
if(pre[j]&&left[pre[j]]==1)
left[pre[j]]=i+1;
}
pre[num[i]]=i;
}

long long ans=0;
for(int i=1;i<=n;i++)
ans = (ans%mod+(long long)(i-left[i]+1)*(right[i]-i+1)%mod)%mod;
printf("%lld\n",ans);
}
return 0;
}