hdu5288(2015多校1)OO’s Sequence

OO’s Sequence

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 353 Accepted Submission(s): 117

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) m o d （ 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

f(l,r)求[l,r]区间内存在多少i（区间内没有i的约数）。公式中给出的区间，也就是所有存在的区间。

记录l[i]，r[i]，i的约数存在的位置，要求最接近i的。可以用数组记录下出现的最接近当前位置的数。得到l[i],r[i]后，那么i可以被计数的区间也就有了，左边界为l[i]到i，右边界为i到r[i]，那么i一共会被记录(i-l[i])*(r[i]-i)次，累加所有的计数。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
#define LL __int64
const int MOD = 1e9+7 ;
int a[100010] , l[100010] , r[100010] ;
int Map[10010] ;
int main() {
    int i , j , n ;
    LL ans ;
    while( scanf("%d", &n) != EOF ) {
        for(i = 1 ; i <= n ; i++)
            scanf("%d", &a[i]) ;
        memset(Map,0,sizeof(Map)) ;
        for(i = 1 ; i <= n ; i++) {
            for(j = 1 , l[i] = 0 ; j*j <= a[i] ; j++) {
                if( a[i]%j ) continue ;
                if( Map[j] ) l[i] = max(l[i],Map[j]) ;
                if( Map[a[i]/j] ) l[i] = max(l[i],Map[a[i]/j]) ;
            }
            Map[a[i]] = i ;
        }
        memset(Map,0,sizeof(Map)) ;
        for(i = n ; i > 0 ; i--) {
            for(j = 1 , r[i] = n+1 ; j*j <= a[i] ; j++) {
                if( a[i]%j ) continue ;
                if( Map[j] ) r[i] = min(r[i],Map[j]) ;
                if( Map[a[i]/j] ) r[i] = min(r[i],Map[a[i]/j]) ;
            }
            Map[a[i]] = i ;
        }
        for(i = 1 , ans = 0 ; i <= n ; i++) {
            ans = (ans + (i-l[i])*(r[i]-i) ) % MOD ;
        }
        printf("%I64d\n", ans) ;
    }
    return 0 ;
}

hdu5288(2015多校一)OO’s Sequence

OO’s Sequence

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