• Codeforces 837D


     先算出每个数的pop1(twonum),pop(fivenum)然后DP ans[i][j]表示选i个数有j个2时最多有多少个5

    转移方程是

            for(int j=k;j>=1;j--)
            {
            for(int w=pop1;w<5000;w++)
            {
            ans[j][w]=max(ans[j][w],ans[j-1][w-pop1]+pop);
            }
            }

    AC程序:

    #include <bits/stdc++.h>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    #include<queue>
    #define EPS 1.0e-9
    #define PI acos(-1.0)
    #define INF 30000000
    #define MOD 1000000007
    #define mem(a,b) memset((a),b,sizeof(a))
    #define TS printf("!!!
    ")
    #define pb push_back
    #define pai pair<int,int>
    //using ll = long long;
    //using ull= unsigned long long;
    //std::ios::sync_with_stdio(false);
    using namespace std;
    //priority_queue<int,vector<int>,greater<int>> que;
    typedef long long ll;
    typedef unsigned long long ull;
    const int maxn=1005;
    struct num
    {
      int two;
      int five;
    }number[205];
    ll ans[205][5001];
    int main()
    {
            mem(ans,-1);
            ans[0][0]=0;
        int n;
        int k;
        cin >> n >> k;
        int now;
        int pop,pop1;
        num temp;
        for(int i=1;i<=n;i++)
            {
            pop=pop1=0;
            cin >> now;
            while(now&&now%5==0)
            {
            pop++;
            now/=5;
            }
            while(now&&now%2==0)
            {
            pop1++;
            now/=2;
            }
            for(int j=k;j>=1;j--)
            {
            for(int w=pop1;w<5000;w++)
            {
            ans[j][w]=max(ans[j][w],ans[j-1][w-pop1]+pop);
            }
            }
            }
            ll anser=0;
            for(ll i=1;i<=5000;i++)
            anser=max(anser,min(i,ans[k][i]));
            cout<<anser<<endl;
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Aragaki/p/7350380.html
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