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n數，從中選k個，使得它們乘積的後綴零最多

dp[i][j][k]表示（掃描前i個，從中選出j個，得到5的冪次為k）的2的冪次的值，類似背包問題

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <algorithm>
 5 #define INF 0x3f3f3f3f
 6 using namespace std;
 7 typedef long long LL;
 8 const int maxn = 2e2 + 10;
 
 9 const int maxm = 2e2 * 30;
10 int dp[maxn][maxm];
11 int N, K;
12 
13 int main() {
14     scanf("%d%d", &N, &K);
15     LL A;
16     for (int i = 0; i < maxn; i++) for (int j = 0; j < maxm; j++) dp[i][j] = -INF;
17     dp[0][0] = 0;
18     for (int i = 1; i <= N; i++) {
19         scanf("
 
%lld", &A);
20         LL B = A;
21         int p2 = 0;
22         int p5 = 0;
23         while (!(B % 2)) {
24             p2++;
25             B /= 2;
26         }
27         while (!(A % 5)) {
28             p5++;
29             A /= 5;
30         }
31         for (int j = K; j >= 1; j--) {
32             for
 
 (int k = p5; k < maxm; k++) {
33                 dp[j][k] = max(dp[j][k], dp[j - 1][k - p5] + p2);
34             }
35         }
36     }
37     int ans = 0;
38     for (int i = 1; i < maxm; i++) {
39         ans = max(ans, min(i, dp[K][i]));
40     }
41     printf("%d\n", ans);
42     return 0;
43 }

codeforces-837D. Round Subset