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)
1 #include<cstdio>
2 #include<algorithm>
3 #include<cstring>
4 using namespace std;
5 const int maxn = 10050;
6 int num[maxn]; int vis[maxn];
7 int main()
8 {
9 int n;
10 while(scanf("%d", &n) == 1) {
11 int maxnum = 0;
12 for(int i = 0 ; i < n ; i++) scanf("%d", &num[i]), maxnum = max(maxnum, num[i]);
13 memset(vis, 0, sizeof(vis));
14 int ans = 0;
15 for(int i = n - 1 ; i >= 0 ;i--) {
16 vis[num[i]] = i + 1; int id = 0x3f3f3f3f;
17 for(int j = num[i] * 2 ; j <= maxnum ; j += num[i]) {
18 if(vis[j])
19 id = min(id, vis[j]);
20 }
21 if(id != 0x3f3f3f3f) ans += id;
22 ///printf("%d\n", ans);
23 }
24 printf("%d\n", ans);
25 }
26 return 0;
27 }