暴力+降复杂度 BestCoder Round #39 1002 Mutiple

题目传送门

/*
    设一个b[]来保存每一个a[]的质因数的id，从后往前每一次更新质因数的id，
        若没有，默认加0，nlogn复杂度；
    我用暴力竟然水过去了:)
*/
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;

const int MAXN = 1e4 + 10;
const int INF = 0x3f3f3f3f;
int a[MAXN], b[MAXN];

int main(void)        //BestCoder Round #39 1002 Mutiple
{
    //freopen ("1002.in", "r", stdin);

    int n;

    while (scanf ("%d", &n) == 1)
    {
        memset (b, 0, sizeof (b));
        for (int i=1; i<=n; ++i)    scanf ("%d", &a[i]);

        long long ans = 0;
        for (int i=n; i>=1; --i)
        {
            ans += b[a[i]];
            for (int j=1; j*j<=a[i]; ++j)
            {
                if (a[i] % j == 0)
                {
                    b[j] = i;    b[a[i]/j] = i;
                }
            }
        }

        printf ("%lld
", ans);
    }

    return 0;
}

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;

const int MAXN = 1e4 + 10;
const int INF = 0x3f3f3f3f;
int a[MAXN];

int main(void)        //BestCoder Round #39 1002 Mutiple
{
    //freopen ("1002.in", "r", stdin);

    int n;
    long long sum = 0;

    while (scanf ("%d", &n) == 1)
    {
        for (int i=1; i<=n; ++i)    scanf ("%d", &a[i]);

        sum = 0;    int k = 0;
        for (int i=1; i<=n-1; ++i)
        {
            k = 0;
            for (int j=i+1; j<=n; ++j)
            {
                if (a[j] % a[i] == 0)
                {
                    k = j;    break;
                }
            }
            sum += k;
        }

        printf ("%lld
", sum);
    }

    return 0;
}