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.
贝西和其他奶牛们都喜欢巧克力，所以约翰准备买一些送给她们。奶牛巧克力专卖店里
有N种巧克力，每种巧克力的数量都是无限多的。每头奶牛只喜欢一种巧克力，调查显示，
有Ci头奶牛喜欢第i种巧克力，这种巧克力的售价是P。
约翰手上有B元预算，怎样用这些钱让尽量多的奶牛高兴呢？
输入输出格式
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
输入输出样例
输入样例#1：
5 50 5 3 1 1 10 4 7 2 60 1
输出样例#1：
8
题目大意：n头奶牛，每个奶牛有喜欢的巧克力，巧克力有价钱。问B元钱最多满足多少牛。
题解：简单贪心，每次购买价格最低的巧克力。
代码：
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#define LL long long
using namespace std;
LL n,b,ans;
struct W{
LL p,c;
bool operator < (const W &a)const{return p<a.p;}
}q[100005];
inline LL read(){
char ch=getchar();LL x=0,f=1;
for(;!isdigit(ch);ch=getchar())if(ch==‘-‘)f=-1;
for(;isdigit(ch);ch=getchar())x=x*10+ch-‘0‘;
return x*f;
}
int main(){
n=read();b=read();
for(int i=1;i<=n;i++)q[i].p=read(),q[i].c=read();
sort(q+1,q+n+1);
for(int i=1;i<=n;i++){
LL k=b/q[i].p;
if(k==0)break;
LL g=min(k,q[i].c);
b-=g*q[i].p;
ans+=g;
}
cout<<ans<<endl;
return 0;
}