codeforces-837D. Round Subset

传送门

n数，从中选k个，使得它们乘积的后缀零最多

dp[i][j][k]表示（扫描前i个，从中选出j个，得到5的幂次为k）的2的幂次的值，类似背包问题

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long LL;
const int maxn = 2e2 + 10;
const int maxm = 2e2 * 30;
int dp[maxn][maxm];
int N, K;

int main() {
    scanf("%d%d", &N, &K);
    LL A;
    for (int i = 0; i < maxn; i++) for (int j = 0; j < maxm; j++) dp[i][j] = -INF;
    dp[0][0] = 0;
    for (int i = 1; i <= N; i++) {
        scanf("%lld", &A);
        LL B = A;
        int p2 = 0;
        int p5 = 0;
        while (!(B % 2)) {
            p2++;
            B /= 2;
        }
        while (!(A % 5)) {
            p5++;
            A /= 5;
        }
        for (int j = K; j >= 1; j--) {
            for (int k = p5; k < maxm; k++) {
                dp[j][k] = max(dp[j][k], dp[j - 1][k - p5] + p2);
            }
        }
    }
    int ans = 0;
    for (int i = 1; i < maxm; i++) {
        ans = max(ans, min(i, dp[K][i]));
    }
    printf("%d
", ans);
    return 0;
}