思路

1. 每次要讓pos最小，且val最小的貓吃魚
2. 用小根堆維護

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath>
#include<cstdlib>
#include<ctime>
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
inline int read(){
	char ch=' ';int f=1;int x=0;
	while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
struct node
{
	int val,pos;
};
priority_queue <node> q;
bool operator < (node a,node b)
{
	if(a.pos==b.pos)
	{
		return a.val>b.val;
	}
	else
	{
		return a.pos>b.pos;
	}
}
int main()
{
	int m,n,x;
	m=read();n=read();x=read();
	int i,j;
	int mm=m;int cnt1=0;
	for(i=1;i<=n;i++)
	{
		node a;
		a.val=read();
		a.pos=a.val;
		q.push(a);
		if(x>=0&&a.pos>x) cnt1++;
		mm--;
		if(mm==0) break;
	}
	if(mm==0) 
	{
		cout<<'0'<<' '<<cnt1<<endl;
		return 0;
	}
	while(true)
	{
		node a=q.top();q.pop();
		if(a.pos>=x) break;
		if(a.pos<=x&&a.pos+a.val>x) cnt1++;
		mm--;
		if(mm==0) break;
		a.pos+=a.val;
		q.push(a);
	}
	cout<<mm<<' '<<cnt1<<endl;
	return 0;
}