Codeforces731F Video Cards

考虑每个数在最大值内的倍数都求出来大概只有max(ai)ln(max(ai))个。

先排个序，然后对于每个数ai，考虑哪些数字可以变成ai*k。

显然就是区间[ai*k,ai*(k+1))内的数，这个二分一下就好了。

#include <bits/stdc++.h>
using namespace std;

int a[200010];

int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        scanf("%d", a + i);
    sort(a, a + n);
    long long ans = 0;
    int last = -1;
    for (int i = 0; i < n; i++)
    {
        if (a[i] == last)
            continue;
        long long temp = 0;
        for (int j = 0; (j - 1) * a[i] <= a[n - 1]; j++)
        {
            int down = j * a[i];
            int up = (j + 1) * a[i];
            int *pup = lower_bound(a, a + n, up);
            int *pdown = lower_bound(a, a + n, down);
            temp += 1LL * j * a[i] * (pup - pdown);
        }
        ans = max(ans, temp);
        last = a[i];
    }
    printf("%I64d", ans);
    return 0;
}