1 solutions
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0
C++ :
#include<cmath> #include<cstdio> #include<iostream> #include<algorithm> using namespace std; int n,m,ans;//总需求量,提供的个数,总价; struct node { int a,b;//牛奶单价和产量 }a[5005];//定义结构体 bool cmp(node a,node b) { if(a.a!=b.a)return a.a<b.a; else return a.b>b.b; }//好我们定义一个判断函数,条件见思路1; int main() { cin>>n>>m; for(int i=1;i<=m;i++) cin>>a[i].a>>a[i].b; sort(a+1,a+1+m,cmp);//用刚定义的函数给它排序 int i=1; while(n)//从n开始减起,直达n为零停止 { if(a[i].b!=0)//当这头牛还没购买完 { a[i].b--;//这头牛可购买数减一 ans+=a[i].a;//总花费加上这头牛的单价(或者说当前最小单价) n--;//需求量减一 } else i++;//购买完了换头牛 } cout<<ans; return 0; }
Java :
import java.util.*; public class Main { public static void main(String[] args) { Scanner sc=new Scanner(System.in); int n=sc.nextInt(),m=sc.nextInt(); List<Map<String ,Integer>> l1=new ArrayList<>(); for (int i=1;i<=m;i++){ Map<String ,Integer> milk=new HashMap() ; milk.put("单价",sc.nextInt()); milk.put("数量",sc.nextInt()); l1.add(milk); } Collections.sort(l1, new Comparator<Map<String, Integer>>() { @Override public int compare(Map<String, Integer> o1, Map<String, Integer> o2) { return o1.get("单价")-o2.get("单价"); } }); int m1=0; // for (int i=0;i<l1.size();i++) { // System.out.println(l1.get(i)); // } for (int i=0;i<m;i++){ Map<String, Integer> shop = l1.get(i); if(shop.get("数量")>=n){ m1=m1+n*shop.get("单价"); n=0; break; }else { m1=m1+shop.get("数量")*shop.get("单价"); n=n-shop.get("数量");; } } System.out.println(m1); } }
Python :
class Cow: def __init__(self,price,num): self.price=price self.num=num def __lt__(self,other): return self.price<other.price or self.price==other.price and self.num>other.num def __str__(self): return "%d %d"%(self.price,self.num) total,n=list(map(int,input().split())) lst=list() for i in range(n): price,num=list(map(int,input().split())) lst.append(Cow(price,num)) lst.sort() num_sum=0 tqsum=0 i=0 for item in lst: if(total>=item.num): total = total-item.num tqsum=tqsum+item.num*item.price else: tqsum=tqsum+(total)*item.price total=0 break print(tqsum)
- 1
Information
- ID
- 9982
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
- Uploaded By