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[];

 int main()
 {
     int n;
     scanf("%d", &n);
     for (int i = ; i < n; i++)
         scanf("%d", a + i);
     sort(a, a + n);
     long long ans = ;
     int last = -;
     for (int i = ; i < n; i++)
     {
         if (a[i] == last)
             continue;
         long long temp = ;
         for (int j = ; (j - ) * a[i] <= a[n - ]; j++)
         {
             int down = j * a[i];
             int up = (j + ) * 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 ;
 }