2257: [Jsoi2009]瓶子和燃料

最新推荐文章于 2020-05-21 21:28:29 发布

原创最新推荐文章于 2020-05-21 21:28:29 发布 · 130 阅读

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博客给出一个链接https://www.lydsy.com/JudgeOnline/problem.php?id=2257，并阐述解题思路，运用裴蜀定理（辗转相减求gcd），记录数的因子个数，因子个数大于等于k的即为所求。

https://www.lydsy.com/JudgeOnline/problem.php?id=2257

思路：裴蜀定理(就是辗转相减求gcd)；将数的因子都记录个数，大于等于k的就可以

#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-8
#define PI acos(-1.0)
#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 FAST_IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)

//----------------------------------------------------------
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;
}
//----------------------------------------------------------


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

map<int,int>num;
vector<int>v;
int main()
{
    //FAST_IO;

    //freopen("input.txt","r",stdin);

    int n,m;

    cin>>n>>m;


    for(int i=1;i<=n;i++)
    {
        int x;
        cin>>x;
        for(int j=1;j<=sqrt(x);j++)
        {
            if(x%j==0)
            {
                if(!num[j]) v.pb(j);
                num[j]++;
                if(j*j!=x) v.pb(x/j),num[x/j]++;
            }
        }
    }

    sort(v.begin(),v.end());
    int ans=0;
    for(int i=v.size()-1;i>=0;i--)
    {
        if(num[v[i]]>=m)
        {
            ans=v[i];
            break;
        }
    }
    cout<<ans<<endl;

    return 0;
}