Hdoj 5288 OO ' s Sequence water

Preprocessing the position of the left and right sides of each number to divide its nearest number.

OO ' s SequenceTime limit:4000/2000 MS (java/others) Memory limit:131072/131072 K (java/others)
Total submission (s): 1880 Accepted Submission (s): 672


Problem Descriptionoo has got a array a of size n, defined a function f (l,r) represent the number of I (L&LT;=I&LT;=R), t Hat there ' s no J (l<=j<=r,j<>i) satisfy AI mod aj=0,now OO want to know ∑ i = 1 n ∑j=in F (I,J) mod  ( ten 9 +7).

Inputthere is 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)

Outputfor each tests:ouput a line contain a number ans.
Sample Input

51 2 3) 4 5

Sample Output

23

Authorfzuacm
Source2015 multi-university Training Contest 1

/* ***********************************************author:ckbosscreated time:2015 July 24 Friday 08:12 15 seconds file Name : hdoj5288.cpp************************************************ * * #include <iostream> #include <cstdio># Include <cstring> #include <algorithm> #include <string> #include <cmath> #include <cstdlib > #include <vector> #include <queue> #include <set> #include <map>using namespace Std;typedef Long long int ll;const int Maxn=100100;const ll mod= (LL) (1e9+7); int N;int A[maxn];int left[maxn],right[maxn];vector< int> pos[10010];void Init () {for (int i=0;i<=10010;i++) {pos[i].clear ();} for (int i=0;i<n;i++) {left[i]=0; right[i]=n-1;}} void Pre () {for (int i=0;i<n;i++) {int x=a[i];for (int. j=x;j<=10000;j+=x) {for (int k=0,sz=pos[j].size (); k<sz;k+ +) {int z=pos[j][k];if (z==i) Continue;else if (z<i) {right[z]=min (right[z],i-1);} else if (z>i) {Left[z]=max (left[z],i+1);}}}} int main () {//freopen ("In.txt", "R", stdin);//freopen ("OUT.txt", "w", stdout), while (scanf ("%d", &n)!=eof) {init (); for (int i=0;i<n;i++) {scanf ("%d", a+i); Pos[a[i]].push_back (i);} Pre (); ll ans=0;for (int i=0;i<n;i++) {ll l=i-left[i]+1ll; LL r=right[i]-i+1ll;ans= (ans+ (l*r)%mod)%mod;}        Cout<<ans<<endl;} return 0;}

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Hdoj 5288 OO ' s Sequence water