P14635 [NOIP2025] 糖果店 / candy

呃呃呃 10 min 切 t1 大战暴力场 /dk。

首先每个糖果拿一轮价值为 2，发现除了这个啥都不会贡献了，所以只有 \((x+y)_{\min}\) 会拿一整轮，剩下的就是按 \(x\) 排序后一个前缀只拿 \(x\)。枚举拿的哪个前缀看剩下最多能拿几轮 \((x+y)_{\min}\) 即可。

记得判这个前缀要 \(\le m\) 啊，不写也能过所有大样例 /ll。

#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
#define fin(x) freopen(#x".in","r",stdin)
#define fout(x) freopen(#x".out","w",stdout)
#define fr(x) fin(x),fout(x);
#define Fr(x,y) fin(x),fout(y)
#define INPUT(_1,_2,FILE,...) FILE
#define IO(...) INPUT(__VA_ARGS__,Fr,fr)(__VA_ARGS__)
using namespace std;
using namespace __gnu_pbds;
#define mp make_pair
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define cfast ios::sync_with_stdio(false);cin.tie(0),cout.tie(0)
#define ll long long
#define ull unsigned long long
#define intz(x,y) memset((x),(y),sizeof((x)))
char *p1,*p2,buf[100000];
#define nc() (p1==p2 && (p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++)
#define tup(x) array<int,(x)>
inline ll read(){
    ll x=0,f=1;char ch=nc();
    while(ch<48||ch>57){if(ch=='-')f=-1;ch=nc();}
    while(ch>=48&&ch<=57)x=x*10+ch-48,ch=nc();
   	return x*f;
}
//void write(int x){cout<<x<<' ';}
//void write(pii x){cout<<"P("<<x.fi<<','<<x.se<<")\n";}
//void write(vector<auto>x){for(auto i:x)write(i);cout<<'\n';}
//void write(auto *a,int l,int r){for(int i=l;i<=r;i++)write(a[i]);cout<<'\n';}
inline ll lowbit(ll x){return x&-x;}
inline int pcount(ll x){
	for(int i=0,res=0;;res+=(x>>i)&1,i++)
		if(i>60)return res;
}

//struct mt{
//	ll v;
//	mt(){v=0;}
//	mt(int x){this->v=x;}
//	inline mt operator+(mt x){return {(v+x.v)%mod};}
//	inline mt operator-(mt x){return {(v+mod-x.v)%mod};}
//	inline mt operator*(mt x){return {1ll*v*x.v%mod};}
//};
//inline void add(mt &x,mt y){x=x+y;}
//mt qp(mt x,int y){mt res(1);for(;y;x=x*x,y>>=1)if(y&1)res=res*x;return res;}
const int N=1e5+5;
struct cand{ll x,y;}a[N];ll s[N];
inline void UesugiErii(){
	int n;ll m,mn=1e15,ans=0;cin>>n>>m;
	for(int i=1;i<=n;i++)
		cin>>a[i].x>>a[i].y,mn=min(mn,a[i].x+a[i].y);
	sort(a+1,a+1+n,[](cand x,cand y){return x.x<y.x;});
	for(int i=1;i<=n;i++)s[i]=s[i-1]+a[i].x;
	for(int i=0;i<=n;i++){
		if(s[i]>m)break;
		ans=max(ans,i+(m-s[i])/mn*2);
	}
	cout<<ans;
}
signed main(){
	//IO(candy);
	cfast;
	int _=1;//cin>>_;
	for(;_;_--)UesugiErii();
	return 0;
}