## 题目描述

Bessie and the herd love chocolate so Farmer John is buying them some.

The Bovine Chocolate Store features N (1 <= N <= 100,000) kinds of chocolate in essentially unlimited quantities. Each type i of chocolate has price P_i (1 <= P_i <= 10^18) per piece and there are C_i (1 <= C_i <= 10^18) cows that want that type of chocolate.

Farmer John has a budget of B (1 <= B <= 10^18) that he can spend on chocolates for the cows. What is the maximum number of cows that he can satisfy? All cows only want one type of chocolate, and will be satisfied only by that type.

Consider an example where FJ has 50 to spend on 5 different types of chocolate. A total of eleven cows have various chocolate preferences:

Chocolate_Type Per_Chocolate_Cost Cows_preferring_this_type 1 5 3

2 1 1

3 10 4

4 7 2

5 60 1

Obviously, FJ can't purchase chocolate type 5, since he doesn't have enough money. Even if it cost only 50, it's a counterproductive purchase since only one cow would be satisfied.

Looking at the chocolates start at the less expensive ones, he can * purchase 1 chocolate of type #2 for 1 x 1 leaving 50- 1=49, then * purchase 3 chocolate of type #1 for 3 x 5 leaving 49-15=34, then * purchase 2 chocolate of type #4 for 2 x 7 leaving 34-14=20, then * purchase 2 chocolate of type #3 for 2 x 10 leaving 20-20= 0.

He would thus satisfy 1 + 3 + 2 + 2 = 8 cows.

## 输入格式

* Line 1: Two space separated integers: N and B

* Lines 2..N+1: Line i contains two space separated integers defining chocolate type i: P_i and C_i

## 输出格式

* Line 1: A single integer that is the maximum number of cows that Farmer John can satisfy

```5 50
5 3
1 1
10 4
7 2
60 1
```

`8 `

## 注意

1.unsigned long long！！！

2.读入最好用cin！！！

```#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int N=100010;

unsigned long long n,b,ans;

struct no {
unsigned long long p,c;
} a[N];

bool cmp(no a,no b) {
return a.p<b.p;
}

int main () {
//scanf("%llu%llu",&n,&b);
cin>>n>>b;
for(int i=1; i<=n; i++)
cin>>a[i].p>>a[i].c;
//scanf("%llu%llu",&a[i].p,&a[i].c);
sort(a+1,a+1+n,cmp);
for(int i=1; i<=n; i++)
if(b>a[i].p*a[i].c) {
ans+=a[i].c;
b-=a[i].p*a[i].c;
} else {
ans+=b/a[i].p;
break;
}
printf("%llu\n",ans);
return 0;
}
```