Hdu 5288 OO ' s Sequence 2015 + Small League a problem

OO ' s SequenceTime limit:4000/2000 MS (java/others) Memory limit:131072/131072 K (java/others)
Total submission (s): 1751 Accepted Submission (s): 632


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
Recommendwe has carefully selected several similar problems for you:5299 5298 5297 5296 5295 Test instructions is simple: Ask how many AI have satisfied the condition in any interval if According to the test instructions program should write: four-layer loop--(do not timeout to eat the keyboard) although the final optimization to N^2/2 also time-out data is too large

 #include <stdio.h>int main () {int n,i,j,k,kk,a[100050];        while (scanf ("%d", &n)!=eof) {for (i=1; i<=n; i++) scanf ("%d", &a[i]);        Long Long suma=0;                 For (I=1, i<=n; i++) {for (j=i; j<=n; J + +) {for (k=i; k<=j; k++)//equivalent to I                    {int flag=0;                        for (kk=i; KK <= J; kk++)//equivalent to J {if (a[k]%a[kk] = = 0 && k!=kk)                            {flag=1;                        Break                        }} if (flag!=1) {suma++;                    suma%=100000007;            }} printf ("%d%d=%d%d=%d\n", I,j,a[i],a[j],suma);    }} printf ("%i64d\n", Suma); }}

Brain Hole big Open: Change a thought is not test instructions beg is to find those interval can satisfy the AI value exists? Which means that AI can provide several answers.
Defines two arrays L R represents the left and right of the number of I
The position of the number closest to his value and the value is the a[i] factor
So the first number I can provide is the answer (r[i]-i) * (l[i]-i)
In fact, this method has a loophole if I give you 10w 1 program on my knees ╮(╯▽╰)╭ no such data so be assured to do it boldly

#include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include < math.h>using namespace Std;const int M = 10e5 + 5;const Long mod = 1e9+7;int vis[m],a[m],l[m],r[m];int main () {int n    ;        while (~SCANF ("%d", &n)) {memset (l,0,sizeof (l));        Memset (r,0,sizeof (R));        memset (vis,0,sizeof (VIS));            for (int i = 1;i <= n; ++i) {scanf ("%d", &a[i]);            R[i] = n+1;                    for (int j = a[i];j <= 10000; j+=a[i])//Find the nearest factor {if (Vis[j]) {                    R[VIS[J]] = i;                VIS[J] = 0;        }} Vis[a[i]] = i;        } memset (Vis,0,sizeof (VIS)); for (int i = n;i >= 1;-i) {for (int j = a[i];j <= 10000; J+=a[i]) {if (                    Vis[j]) {L[vis[j]] = i;                VIS[J] = 0;            }} Vis[a[i]] = i;        } long long ans = 0;        for (int i = 1;i <= n; ++i) {ans = ((ans + (long Long) (r[i]-i) * (Long Long) (I-l[i])%mod);    } printf ("%i64d\n", ans); }}

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Hdu 5288 OO ' s Sequence 2015 + Small League a problem