开始的时候数据范围算错了~ 我以为整个序列 2 和 5 的个数都不超过 70 ~ 一个非常水的 dp code: #include <bits/stdc++.h> #define M 75 #define N 201 #define LL long long using namespace std; void setIO(string s) { string in=s+".in"; string out=s+".out"; freopen(in.c_st…

Round Subset 题目链接:http://codeforces.com/problemset/problem/837/D 数据范围:略. 题解: $dp$比较显然. 但是卡空间,有两种方法: 第一种是滚动数组,第二种是反向枚举. 这两种都可以达到:当前更新需要的数组位置仍然是上个版本的数组.…

Let's call the roundness of the number the number of zeros to which it ends. You have an array of n numbers. You need to choose a subset of exactly k numbers so that the roundness of the product of the selected numbers will be maximum possible. Input…

设dp[i][j][k]表示前i个数中选j个并且因子含有k个2的能获得的最多的5的个数 则dp[i][j][k]=max(dp[i-1][j][k],dp[i-1][j-1][k-cnt2]+cnt5) 滚掉一维 #include <iostream> #include <cstring> #include <cstdio> using namespace std; typedef long long ll; int n, m, allcnt, dp[205][1200…

CF837D. Round Subset Let's call the roundness of the number the number of zeros to which it ends. You have an array of n numbers. You need to choose a subset of exactly k numbers so that the roundness of the product of the selected numbers will be ma…

837D - Round Subset 思路:dp.0是由2*5产生的. ①dp[i][j]表示选i个数,因子2的个数为j时因子5的个数. 状态转移方程:dp[i][j]=max(dp[i][j],dp[i-1][j-c2]+c5). 初始化:dp[0][0]=0,dp[i][j]=-INF(i!=0||j!=0).因为所有状态都是由dp[0][0]转移过来的,所以除此之外的dp[i][j]都得初始化为-INF,防止对答案产生影响. 代码1: #include<bits/stdc++.h> u…

D - Round Subset 思路:背包: 代码: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define ll long long ll n,m,dp[][],tmp,p2,p5,pos,sum,ans; int main() { //freopen("data.txt",&q…

题目链接  Round Subset 题意  在n个数中选择k个数,求这k个数乘积末尾0个数的最大值. 首先我们预处理出每个数5的因子个数c[i]和2的因子个数d[i] 然后就可以背包了. 设f[i][j]为选i个数,5的因子总和为j时,2的因子总和的最大值. 则状态转移方程为 $f[i][j] = max(f[i - 1][j - c[k]] + d[k])$ 注意边界条件 时间复杂度$O(5200nk)$ #include <bits/stdc++.h> using namespace s…

D. Round Subset time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Let's call the roundness of the number the number of zeros to which it ends. You have an array of n numbers. You need to cho…

/* CodeForces 837D - Round Subset [ DP ] | Educational Codeforces Round 26 题意: 选k个数相乘让末尾0最多 分析: 第i个数字有a[i]个2, b[i] 个5 以其中一维作体积另一维作价值01背包即可 */ #include <bits/stdc++.h> using namespace std; int dp[205][20005]; int get2(long long x) { int s = 0; while…