2015 Multi-University Training Contest 1 OO’s Sequence

OO’s Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 312    Accepted Submission(s): 107

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

\[ \sum_{i=1}^{n}\sum_{j=i}^{n}f(i,j)mod(10^{9}+7) \]

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 

 

 解题：

 

需要先使用$\sqrt{n}$的时间复杂度把每个数的倍数所在的下标找到

然后我们对每个数，如果她的倍数所在的下标比它大，那么当前这个数是它的倍数左边的约数，求最大的左下标，

如果它的倍数所在的下标比它小，那么当前这个数是它的倍数右边的约数，求最小的右下标

 

$(i-L[i])\times (R[i] - i)$就是i所在的区间的个数，这个区间没有它的约数

 

 1 #include <bits/stdc++.h>

 2 using namespace std;

 3 typedef long long LL;

 4 const int maxn = 10010;

 5 const LL mod = 1000000007;

 6 vector<int>g[maxn],factor[maxn];

 7 int d[100010];

 8 LL L[100010],R[100010],n;

 9 int main() {

10     while(~scanf("%d",&n)) {

11         for(int i = 0; i < maxn; ++i) {

12             g[i].clear();

13             factor[i].clear();

14         }

15         for(int i = 1; i <= n; ++i) {

16             scanf("%d",d+i);

17             g[d[i]].push_back(i);

18             L[i] = 0;

19             R[i] = n+1;

20         }

21         for(int i = 1; i <= n; ++i) {

22             for(int j = 1,m = sqrt(d[i]); j <= m; ++j) {

23                 if(d[i]%j) continue;

24                 int tmp = d[i]/j;

25                 if(g[tmp].size()) factor[tmp].push_back(i);

26                 if(tmp == j) continue;

27                 if(g[j].size()) factor[j].push_back(i);

28             }

29         }

30         for(int i = 1; i <= n; ++i) {

31             for(int j = factor[d[i]].size()-1; j >= 0; --j) {

32                 int now = factor[d[i]][j];

33                 if(now < i) R[now] = min(R[now],(LL)i);

34                 if(now > i) L[now] = max(L[now],(LL)i);

35             }

36         }

37         LL ret = 0;

38         for(int i = 1; i <= n; ++i)

39             ret = (ret + (i - L[i])*(R[i] - i)%mod)%mod;

40         cout<<ret<<endl;

41     }

42     return 0;

43 }

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