BestCoder Round #39 1002——降低复杂度——Mutiple

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

大意：求j > i并且 a[j] % a[i]的j的最小值，求所有情况的和

用分解因数的方法不断地维护，就降成n√n的复杂度（alt+41420(小键盘））

1.n√n复杂度

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[10005];
int b[10005];
int main()
{
    int n;
    while(~scanf("%d",&n)){
        int sum = 0;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(int i  = 1; i <= n ; i++){
            scanf("%d",&a[i]);
        }

        for(int i = n ; i >= 1;  i--){
            sum += b[a[i]];
            for(int j = 1; j*j <= a[i]; j++){
               if(a[i] % j == 0){
                   b[j] = b[a[i]/j] = i;
               }
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}

 2.nlogn复杂度

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int MAX = 10010;
int f[MAX],a[MAX];
vector<int> g[MAX];
void inti()
{
    for(int i = 1; i <= MAX ; i++){
        for(int j = i ; j <= MAX; j += i){
            g[j].push_back(i);
        }
    }
}//g保存j的所有的因数
int main()
{
    inti();
    int n;
    while(~scanf("%d",&n)){
        for(int i = 1; i <= n ; i++)
            scanf("%d",&a[i]);
         memset(f,0,sizeof(f));
         int sum = 0;
         for(int i = n ;  i >= 1;  i--){
             sum += f[a[i]];
             for(int j = 0 ; j < g[a[i]].size();j++)
                 f[g[a[i]][j]] = i;
         } 
    printf("%d\n",sum);
    }
    return 0;
}

用vector预处理得到所有的因数