Codeforces 837D 动态规划

Codeforces 837D 动态规划

传送门：https://codeforces.com/contest/837/problem/D

题意：

给你n个数，问你从这n个数中取出k个数，这k个数的乘积的末尾最多有多少个0

题解：

要想让乘积的末尾有0，实际上就是2的倍数和5的倍数相乘才能得到贡献，所以每个数对答案的贡献实际上就是这个数中包含的2的个数还有这个数中包含的5的数对答案的贡献

设定dp状态为

(dp[i][j]表示从n个数中选出i个数，其中有j个5的个数，最多有多少个2)

边界 dp[0][0]=0, else dp[i][j]=-INF

转移：dp[i][j]=max(dp[i][j],dp[i-1][j-cnt5[i]]+cnt2[i])

代码：

#include <set>
#include <map>
#include <cmath>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define ls rt<<1
#define rs rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********
")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]
"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]
"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]
"
const int maxn = 3e5 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
LL quick_pow(LL x, LL y) {
    LL ans = 1;
    while(y) {
        if(y & 1) {
            ans = ans * x % mod;
        } x = x * x % mod;
        y >>= 1;
    } return ans;
}
struct node {
    int cnt2;
    int cnt5;
} a[maxn];
int dp[205][205 * 64];
int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    int n, k;
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; i++) {
        LL x;
        scanf("%lld", &x);
        while(x % 2 == 0) {
            a[i].cnt2++;
            x /= 2;
        }
        while(x % 5 == 0) {
            a[i].cnt5++;
            x /= 5;
        }
    }
    LL sum = 0;
    memset(dp, -INF, sizeof(dp));
    dp[0][0]=0;
    for(int i = 1; i <= n; i++) {
        sum += a[i].cnt5;
                // debug1(a[i].cnt2);
            // debug1(sum);
        for(int j = min(k, i); j >= 1; j--) {
            for(int k = sum; k >= a[i].cnt5; k--) {
                // debug2(k,a[i].cnt5);
                dp[j][k] = max(dp[j][k], dp[j - 1][k - a[i].cnt5] + a[i].cnt2);
                // debug3(j,k,dp[j][k]);
            }
        }
    }
    LL ans = 0;
    for(int i = 1; i <= sum; i++) {
        ans = max(ans, 1LL * min(i, dp[k][i]));
    }
    printf("%lld
", ans);
    return 0;
}