虚无的后缀

https://ac.nowcoder.com/acm/contest/911/H

思路：两个数相乘后缀是0的个数就是两个数2，5总个数的最小值；所以取n-k个使得剩下的最小值最大；

#include<algorithm>
#include<set>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstdio>
#include<map>
#include<stack>
#include<string>
#include<bits/stdc++.h>
using namespace std;

#define sfi(i) scanf("%d",&i)
#define sfs(i) scanf("%s",(i))
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define ull unsigned long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-16
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
#define MOD(x) ((x%mod)+mod)%mod
#define endl '\n'
#define pb push_back
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)

const int maxn=1e5+9;
const int mod=1e9+7;

inline ll read()
{
    ll f=1,x=0;
    char ss=getchar();
    while(ss<'0'||ss>'9')
    {
        if(ss=='-')f=-1;ss=getchar();
    }
    while(ss>='0'&&ss<='9')
    {
        x=x*10+ss-'0';ss=getchar();
    }    return f*x;
}

int a[maxn][3];
bool vis[maxn];
int main()
{
    FAST_IO;
    //freopen("input.txt","r",stdin);

    int n,k;
    int sum2=0;
    int sum5=0;
    cin>>n>>k;
    for(int i=1;i<=n;i++)
    {
        ll x;
        cin>>x;
        while(x%2==0) x/=2,sum2++,a[i][0]++;
        while(x%5==0) x/=5,sum5++,a[i][1]++;
    }
    for(int i=1;i<=n-k;i++)
    {
        int pos=0;
        int Max=0;
        for(int j=1;j<=n;j++)
        {
            if(vis[j]) continue;
            int k=min(sum2-a[j][0],sum5-a[j][1]);
            if(k>Max)
            {
                Max=k;
                pos=j;
            }
        }
        sum2-=a[pos][0];
        sum5-=a[pos][1];
        vis[pos]=1;
    }
    cout<<min(sum2,sum5)<<endl;
}