Codeforces731F Video Cards

Posted 2020-08-15

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。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 int a[200010];
 5 
 6 int main()
 7 {
 8     int n;
 9     scanf("%d", &n);
10     for (int i = 0; i < n; i++)
11         scanf("%d", a + i);
12     sort(a, a + n);
13     long long ans = 0;
14     int last = -1;
15     for (int i = 0; i < n; i++)
16     {
17         if (a[i] == last)
18             continue;
19         long long temp = 0;
20         for (int j = 0; (j - 1) * a[i] <= a[n - 1]; j++)
21         {
22             int down = j * a[i];
23             int up = (j + 1) * a[i];
24             int *pup = lower_bound(a, a + n, up);
25             int *pdown = lower_bound(a, a + n, down);
26             temp += 1LL * j * a[i] * (pup - pdown);
27         }
28         ans = max(ans, temp);
29         last = a[i];
30     }
31     printf("%I64d", ans);
32     return 0;
33 }