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

 

 

Recommend

We have carefully selected several similar problems for you:   5299  5298  5297  5296  5295 

哈哈！

 1 #include <math.h>

 2 #include <stdio.h>

 3 #include <string.h>

 4 #include <algorithm>

 5 using namespace std;

 6 

 7 const int mod=1e9+7;

 8 

 9 int main()

10 {

11     int n,i,j,k;

12     int str[100005],spfa[10005],l[100005],r[100005];

13     int s;

14     while(scanf("%d",&n)!=EOF)

15     {

16         for(i=1;i<=n;i++)

17             scanf("%d",&str[i]);

18         for(i=1;i<=10000;i++)

19             spfa[i]=0;

20         for(i=1;i<=n;i++)

21             l[i]=0,r[i]=n+1;

22         for(i=1;i<=n;i++)

23         {

24             for(j=1;j<=sqrt(str[i]);j++)

25             {

26                 if(str[i]%j==0)

27                 {

28                     if(spfa[j]>l[i])

29                         l[i]=spfa[j];

30                     if(spfa[str[i]/j]>l[i])

31                         l[i]=spfa[str[i]/j];

32                 }

33             }

34             spfa[str[i]]=i;

35         }

36         for(i=10000;i>=1;i--)

37             spfa[i]=n+1;

38         for(i=n;i>=1;i--)

39         {

40             for(j=1;j<=sqrt(str[i]);j++)

41             {

42                 if(str[i]%j==0)

43                 {

44                     if(spfa[j]<r[i])

45                         r[i]=spfa[j];

46                     if(spfa[str[i]/j]<r[i])

47                         r[i]=spfa[str[i]/j];

48                 }

49             }

50             spfa[str[i]]=i;

51         }

52         s=0;

53         for(i=1;i<=n;i++)

54         {

55             s=(s+(i-l[i])*(r[i]-i)%mod)%mod;

56         }

57         printf("%d\n",s);

58     }

59     return 0;

60 }

View Code