BestCoder Round #39 1002

Problem Description

WLD likes playing with a sequence a[1..N]. One day he is playing with a sequence of N integers. For every index i, WLD wants to find the smallest index F(i) ( if exists ), that i<F(i)≤n, and aF(i) mod ai = 0. If there is no such an index F(i), we set F(i) as 0.

Input

There are Multiple Cases.(At MOST 10)

For each case:

The first line contains one integers N(1≤N≤10000).

The second line contains N integers a1,a2,...,aN(1≤ai≤10000),denoting the sequence WLD plays with. You can assume that all ai is distinct.

Output

For each case:

Print one integer.It denotes the sum of all F(i) for all 1≤i<n

Sample Input

4
1 3 2 4

Sample Output

6

Hint

F(1)=2 F(2)=0 F(3)=4 F(4)=0

---------------------------------------------------------------------华丽的分割线-------------------------------------------------------------------

维护一个vis数组 vis[i]记录i上一次出现的位置，从右往左遍历给出的数字，每个数字num都从vis[num * 2]从左往右暴力vis数组

时间复杂度为O(n/1+n/2+…+n/n)=O(nlogn)

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = 10050;
int num[maxn]; int vis[maxn];
int main()
{
    int n;
    while(scanf("%d", &n) == 1) {
        int maxnum = 0;
        for(int i = 0 ; i < n  ; i++) scanf("%d", &num[i]), maxnum = max(maxnum, num[i]);
        memset(vis, 0, sizeof(vis));
        int ans = 0;
        for(int i = n - 1 ; i >= 0 ;i--) {
            vis[num[i]] = i + 1; int id = 0x3f3f3f3f;
            for(int j = num[i] * 2 ; j  <= maxnum ; j += num[i]) {
                if(vis[j])
                    id = min(id, vis[j]);
            }
            if(id != 0x3f3f3f3f) ans += id;
            ///printf("%d
", ans);
        }
        printf("%d
", ans);
    }
    return 0;
}