多校7.21A——维护技巧——OO’s Sequence

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

 

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/*
 题意：对于所有的子区间找a[i]在这个区间没有因子的所有的情况
 对于一个数a[i]它的价值是L[i]：从左往右最接近的因子的下标，R[[i]：从右往左最接近的因子下标,a[i]在这个区间内都是可以的那么总共的种类区间有(i - L[i]) * (R[i] - i)
总复杂度O(n根号n)
用index数组来记录该因子出现的位置
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

const int MAX = 100000 + 10;
const int mod = 1e9 + 7;
int a[MAX];
int L[MAX], R[MAX], index[MAX];

int main()
{
    int n;
    while(~scanf("%d", &n)){
        for(int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        memset(index, -1, sizeof(index));
        for(int i = 1; i <= n; i++){
            int temp = sqrt(1.0 * a[i]);
            int l = 0;
            for(int j = 1; j <= temp; j++){
                if(a[i] % j == 0){
                    if(index[j] != -1) 
                        l = max(index[j], l);
                    if(index[a[i] / j] != -1)
                        l = max(index[a[i] / j], l);
                }
            }
                index[a[i]] = i;
                L[i] = l;
        }
        memset(index, -1, sizeof(index));
            for(int i = n; i >= 1 ;i--){
                int temp = sqrt(1.0 * a[i]);
                int r = n + 1;
                for(int j = 1; j <= temp; j++){
                    if(a[i] % j == 0){
                        if(index[j] != -1)
                            r = min(index[j], r);
                        if(index[a[i] / j] != -1)
                            r = min(index[a[i] / j], r);
                    }
                }
                index[a[i]] = i;
                R[i] = r;
            }
         int l1, r1;
         int  sum = 0;
        for(int i = 1; i <= n; i++){
            l1 = i - L[i], r1 = R[i] - i ;
            sum = (l1 * r1 % mod + sum) % mod;
        }
        printf("%d
", sum);
    }
    return 0;
}